Research Article Representations of Generalized Inverses...

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Research Article Representations of Generalized Inverses and Drazin Inverse of Partitioned Matrix with Banachiewicz-Schur Forms Xiaoji Liu, 1 Hongwei Jin, 1 and Jelena VišnjiT 2 1 Faculty of Science, Guangxi University for Nationalities, Nanning 530006, China 2 Faculty of Medicine, University of Niˇ s, 18000 Niˇ s, Serbia Correspondence should be addressed to Xiaoji Liu; [email protected] Received 15 July 2016; Accepted 6 September 2016 Academic Editor: Gen Qi Xu Copyright © 2016 Xiaoji Liu et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Representations of {1, 2, 3}-inverses, {1, 2, 4}-inverses, and Drazin inverse of a partitioned matrix = [ ] related to the generalized Schur complement are studied. First, we give the necessary and sufficient conditions under which {1, 2, 3}-inverses, {1, 2, 4}-inverses, and group inverse of a 2×2 block matrix can be represented in the Banachiewicz-Schur forms. Some results from the paper of Cvetkovi´ c-Ili´ c, 2009, are generalized. Also, we expressed the quotient property and the first Sylvester identity in terms of the generalized Schur complement. 1. Introduction Let C × denote the set of all × complex matrices. We use (), (), and () to denote the range, the null space, and the rank of a matrix , respectively. e smallest nonnegative integer such that ( +1 ) = ( ) is the index of which is denoted by ind(). If ind() = , the Drazin inverse of is the unique matrix ∈ C × satisfying = , = , +1 = , (1) for all ≥. e Drazin inverse of is denoted by . In particular, when ind() = 1, the matrix is called the group inverse of and it is denoted by # . Recall that the Moore-Penrose inverse of a matrix ∈ C × is a matrix ∈ C × which satisfies (1) = , (2) = , (3) () ∗ = , (4) () ∗ = . (2) e Moore-Penrose inverse of is unique and it is denoted by † . For any ∈ C × , let {, , . . . , } denote the set of all ∈ C × which satisfy equations (), (), . . . , () of (2). In this case is a {, , . . . , }-inverse of , which we denote by (,,...,) . Evidently, {1, 2, 3, 4} = { † }. For a complex matrix of the form =[ ]∈ C (+)×(+) , (3) in the case when is invertible, the Schur complement of in is defined by = − −1 . Similarly, if is invertible, then the Schur complement of in is defined by =− −1 . It is well-known that if ∈ C × is nonsingular then the invertibility of a matrix is equivalent to the invertibility of the Schur complement of in and in that case the inverse of is given by −1 =[ −1 + −1 −1 −1 − −1 −1 − −1 −1 −1 ]. (4) Expression (4) is called the Banachiewicz-Schur form of the matrix . e notation of Schur complement was Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2016, Article ID 9236281, 14 pages http://dx.doi.org/10.1155/2016/9236281

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Research ArticleRepresentations of Generalized Inverses and Drazin Inverse ofPartitioned Matrix with Banachiewicz-Schur Forms

Xiaoji Liu1 Hongwei Jin1 and Jelena VišnjiT2

1Faculty of Science Guangxi University for Nationalities Nanning 530006 China2Faculty of Medicine University of Nis 18000 Nis Serbia

Correspondence should be addressed to Xiaoji Liu xiaojiliu72126com

Received 15 July 2016 Accepted 6 September 2016

Academic Editor Gen Qi Xu

Copyright copy 2016 Xiaoji Liu et alThis is an open access article distributed under theCreativeCommonsAttributionLicense whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Representations of 1 2 3-inverses 1 2 4-inverses and Drazin inverse of a partitioned matrix 119872 = [ 119860 119861119862 119863 ] related to thegeneralized Schur complement are studied First we give the necessary and sufficient conditions under which 1 2 3-inverses1 2 4-inverses and group inverse of a 2times2 block matrix can be represented in the Banachiewicz-Schur forms Some results fromthe paper of Cvetkovic-Ilic 2009 are generalized Also we expressed the quotient property and the first Sylvester identity in termsof the generalized Schur complement

1 Introduction

LetC119898times119899 denote the set of all119898times119899 complex matricesWe use119877(119860)119873(119860) and 119903(119860) to denote the range the null space andthe rank of a matrix119860 respectivelyThe smallest nonnegativeinteger 119896 such that 119903(119860119896+1) = 119903(119860119896) is the index of 119860 which isdenoted by ind(119860) If ind(119860) = 119903 the Drazin inverse of 119860 isthe unique matrix119883 isin C119898times119898 satisfying

119883119860119883 = 119883119860119883 = 119883119860

119860119896+1119883 = 119860119896(1)

for all 119896 ge 119903 The Drazin inverse of 119860 is denoted by 119860119863 Inparticular when ind(119860) = 1 thematrix119860119863 is called the groupinverse of 119860 and it is denoted by 119860

Recall that the Moore-Penrose inverse of a matrix 119860 isinC119898times119899 is a matrix119883 isin C119899times119898 which satisfies

(1) 119860119883119860 = 119860(2) 119883119860119883 = 119883(3) (119860119883)lowast = 119860119883(4) (119883119860)lowast = 119883119860

(2)

The Moore-Penrose inverse of 119860 is unique and it is denotedby 119860dagger

For any 119860 isin C119898times119899 let 119860119894 119895 119896 denote the set of all119883 isin C119899times119898 which satisfy equations (119894) (119895) (119896) of (2) Inthis case 119883 is a 119894 119895 119896-inverse of 119860 which we denote by119860(119894119895119896) Evidently 1198601 2 3 4 = 119860dagger

For a complex matrix119872 of the form

119872 = [119860 119861119862 119863] isin C

(119898+119901)times(119899+119901) (3)

in the case when 119860 is invertible the Schur complement of 119860in119872 is defined by 119878 = 119863minus119862119860minus1119861 Similarly if119863 is invertiblethen the Schur complement of119863 in119872 is defined by 119866 = 119860 minus119861119863minus1119862

It is well-known that if119860 isin C119898times119898 is nonsingular then theinvertibility of a matrix119872 is equivalent to the invertibility ofthe Schur complement of 119860 in119872 and in that case the inverseof119872 is given by

119872minus1 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] (4)

Expression (4) is called the Banachiewicz-Schur formof the matrix 119872 The notation of Schur complement was

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 9236281 14 pageshttpdxdoiorg10115520169236281

2 Mathematical Problems in Engineering

first introduced by Schur [1] G E Trapp first defined thegeneralized Schur complement where the ordinary inversewas replaced by the generalized inverse For a matrix 119872 isinC(119898+119901)times(119899+119901) given by (3) and any fixed generalized inverse119860minus isin 1198601 the generalized Schur complement of 119860 in119872 isdefined by

119878 (119860minus) = 119863 minus 119862119860minus119861 (5)

Similarly for some fixed 119863minus isin 1198631 the generalizedSchur complement of119863 in119872 is defined by

119866 (119863minus) = 119860 minus 119861119863minus119862 (6)

The Schur complements and generalized Schur comple-ments were studied by a number of authors which haveapplications in statistics matrix theory electrical networktheory discrete-time regulator problem sophisticated tech-niques and some other fields It has been evident that theSchur complement plays an important role in many aspectsof a matrix theory (see [2ndash8])

Let

119883 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (7)

where 119860minus isin 1198601 119878 = 119863 minus 119862119860minus119861 and 119878minus isin 1198781 and let

119884 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (8)

where119863minus isin 1198631 119866 = 119860 minus 119861119863minus119862 and 119866minus isin 1198661For convenience we will firstly state the following nota-

tions which are helpful in the proofs We denote

119864120572 = 119868 minus 120572minus120572119865120572 = 119868 minus 120572120572minus120572120587 = 119868 minus 120572120572119863

120572 isin 119860 119878 119863 119866 (9)

In this paper we will establish necessary and sufficientconditions under which matrices 119883 and 119884 given by (7)and (8) respectively belong to a given class of generalizedinverses of the matrix 119872 given by (3) Our interest focuseson the cases when 1 2 3-inverses 1 2 4-inverses andgroup inverse of a 2 times 2 block matrix 119872 are given by theBanachiewicz-Schur form Also we give representations ofthe Drazin inverse of 2 times 2 block matrix 119872 under someconditions relating to the Schur complement of 119872 whichgeneralized the results studied by Cvetkovic-Ilic [9] Also thequotient property and the first Sylvester identity based on theMP-inverse group inverse and Drazin inverse are given

We need some auxiliary lemmas

Lemma 1 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198721 3 if and only if

119860minus isin 119860 1 3 119878minus isin 119878 1 3

119865119860119861 = 0119865119878119862 = 0

(10)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Lemma 2 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198721 4 if and only if

119860minus isin 119860 1 4 119878minus isin 119878 1 4

119861119864119878 = 0119862119864119860 = 0

(11)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Lemma 3 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198722 if and only if

119860minus isin 119860 1 2 119878minus isin 119878 2 (12)

Lemma 4 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119883 isin 1198721 2 3 if and only if

119860minus isin 119860 1 2 3 119878minus isin 119878 1 2 3

119865119860119861 = 0119865119878119862 = 0

(13)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Proof The conclusion follows by Lemmas 1 and 2

Lemma 5 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119883 isin 1198721 2 4 if and only if

119860minus isin 119860 1 2 4 119878minus isin 119878 1 2 4

119861119864119878 = 0119862119864119860 = 0

(14)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119864119860 and 119864119878Using the same method as in [10] we can prove the

following

Mathematical Problems in Engineering 3

Lemma 6 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 3 if and only if

119863minus isin 119863 1 2 3 119866minus isin 119866 1 2 3 119865119866119861 = 0119865119863119862 = 0

(15)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 7 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 4 if and only if

119863minus isin 119863 1 2 4 119866minus isin 119866 1 2 4

119861119864119863 = 0119862119864119866 = 0

(16)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 8 (see [11]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 = 119872 if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(17)

2 Representations of 1 2 3-Inverse 1 2 4-Inverse and Group Inverse in terms ofBanachiewicz-Schur Forms

Baksalary and Styan [10] presented the necessary and suf-ficient conditions under which 1 2 1 2 1 3 and1 4-inverses of block matrix 119872 can be represented by theBanachiewicz-Schur form

Sheng and Chen [12] considered sufficient conditionsunder which 119872dagger 119872119863 and 119872 can be represented by bothof the Banachiewicz-Schur forms at the same time

Cvetkovic-Ilic [9] gave necessary and sufficient condi-tions which ensure the representation of the MP-inverse ofa block matrix119872 by both of the Banachiewicz-Schur forms

In this section we will present necessary and sufficientconditions under which 1 2 3 1 2 4-inverses and thegroup inverse of a partitioned matrix 119872 can be representedby the Banachiewicz-Schur form

For convenience we first introduce the following nota-tions

1198731 119894 119895 119896 = 119883

= [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] 119860minus

isin 119860 119894 119895 119896 119878 = 119863 minus 119862119860minus119861 119878minus isin 119878 119894 119895 119896

1198732 119894 119895 119896 = 119884

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] 119863minus

isin 119863 119894 119895 119896 119866 = 119860 minus 119861119863minus119862 119866minus isin 119866 119894 119895 119896

(18)

Theorem 9 Let 119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 3 sube 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0

(19)

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(20)

(ii) 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119878minus isin 1198781 2 3

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

(21)

and for any 119863minus isin 1198631 2 3 and any 119866minus isin 1198661 2 3 thereexists 119860minus isin 1198601 2 3 such that

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(22)

(iii) 11987311 2 3 = 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3

119865119860119861 = 0119865119863119862 = 0 (23)

4 Mathematical Problems in Engineering

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that 119865119863 = 119865119878 119864119863119878minus = 0 and 119864119878119863minus = 0and for any 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 there exists119860minus isin 1198601 2 3 such that 119865119860 = 119865119866 119864119860119866minus = 0 and 119864119866119860minus = 0Proof (i) (lArr) By Lemma 6 we have that arbitrary 119884 isin11987321 2 3 belongs to1198721 2 3 Now we only need to provethat11987311 2 3 sube 11987321 2 3 Take arbitrary

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (24)

By assumption for 119878minus1 there exists 1 2 3-inverse of119863whichwe denote by 119863minus1 such that 119865119863 = 1198651198781 119864119863119878minus1 = 0 and1198641198781119863minus1 = 0 Let 119866 = 119860 minus 119861119863minus1119862 We will prove that 119866minus1 =119860minus1 + 119860minus1119861119878minus1119862119860minus1 isin 1198661 2 3 By computation we get

119866119866minus1 = (119860 minus 119861119863minus1119862) (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119860119860minus1 + 119860119860minus1119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1minus 119861119863minus1119862119860minus1119861119878minus1119862119860minus1

= 119860119860minus1 + 119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1 minus 119861119863minus1119863119878minus1119862119860minus1+ 119861119863minus1 1198781119878minus1119862119860minus1 = 119860119860minus1

(25)

which implies 119866119866minus1119866 = 119860119860minus1(119860 minus 119861119863minus1119862) = 119860 minus 119861119863minus1119862 = 119866Also

119866minus1119866119866minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119860119860minus1= 119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (26)

so119866minus1 isin 1198661 2 3 Since 119865119863 = 1198651198781 119864119863119878minus1 = 0 and 1198641198781119863minus1 = 0a simple computation shows that

119863minus1119862119866minus1 = 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119863minus1119862119860minus1 + 119863minus1119862119860minus1119861119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1 (119863 minus 1198781) 119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1119863119878minus1119862119860minus1minus 119863minus1 1198781119878minus1119862119860minus1 = 119878minus1119862119860minus1

119866minus1119861119863minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119862119860minus1119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1 (119863 minus 1198781)119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119863119863minus1minus 119860minus1119861119878minus1 1198781119863minus1 = 119860minus1119861119878minus1

119863minus1 + 119863minus1119862119866minus1119861119863minus1 = 119863minus1+ 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1

= 119863minus1 + 119863minus1119862119860minus1119861119863minus1+ 119863minus1119862119860minus1119861119878minus1119862119860minus1119861119863minus1

= 119863minus1 + 119863minus1 (119863 minus 1198781)119863minus1+ 119863minus1 (119863 minus 1198781) 119878minus1 (119863 minus 1198781)119863minus1

= 119878minus1 (27)

Thus

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ]

= [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732(28)

Therefore1198731 sube 1198732 sube 1198721 2 3(rArr) Since 11987311 2 3 sube 1198721 2 3 and 11987321 2 3 sube1198721 2 3 by Lemmas 4 and 6 we get that for arbitrary119860minus isin 1198601 2 3 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3the following hold 119865119860119861 = 0 119865119866119861 = 0 and 119865119863119862 = 0Moreover by11987311 2 3 sube 11987321 2 3 it follows that for any119883 isin 11987311 2 3 there exists some 119884 isin 11987321 2 3 such that119883 = 119884 Hence for any 119860minus isin 1198601 2 3 119878minus isin 1198781 2 3 thereexist119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 such that

[119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(29)

A simple computation shows that

119878119878minus = (119863 minus 119862119860minus119861) (119863minus + 119863minus119862119866minus119861119863minus)= 119863119863minus + 119863119863minus119862119866minus119861119863minus minus 119862119860minus119861119863minusminus 119862119860minus119860119866minus119861119863minus + 119862119860minus119866119866minus119861119863minus

= 119863119863minus + 119862 (119868 minus 119860minus119860)119866minus119861119863minus= 119863119863minus + 119862 (119868 minus 119860minus119860)119860minus119861119878minus = 119863119863minus

(30)

thus we get 119865119878 = 119865119863 Since119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119878minus119862119860minus119861119863minus

= 119863minus + 119878minus (119863 minus 119878)119863minus= 119863minus + 119878minus119863119863minus minus 119878minus119878119863minus = 119878minus

(31)

Mathematical Problems in Engineering 5

we get119863minus = 119878minus119878119863minus that is 119864119878119863minus = 0 Similarly

119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119863minus119862119860minus119861119878minus= 119863minus + 119863minus (119863 minus 119878) 119878minus= 119863minus + 119863minus119863119878minus minus 119863minus119878119878minus = 119878minus

(32)

which implies 119864119863119878minus = 0(ii) (lArr) Similar to (i) since 119865119860119861 = 0 and 119865119878119862 = 0we have 11987311 2 3 sube 1198721 2 3 Now we will prove that11987321 2 3 sube 11987311 2 3 Take arbitrary1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732 1 2 3 (33)

For 119866minus1 there exists 1 2 3-inverse of 119860 which we denote by119860minus1 such that 119865119860 = 1198651198661 119864119860119866minus1 = 0 and 1198641198661119860minus1 = 0 Let 119878 =119863 minus 119862119860minus1119861 A simple computation shows that

119878minus1 = 119863minus1 + 119863minus1119862119866minus1119861119863minus1 isin 119878 1 2 3 119878minus1119862119860minus1 = 119863minus1119862119866minus1 119860minus1119861119878minus1 = 119866minus1119861119863minus1

119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (34)

Hence

1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1]

= [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (35)

Therefore 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 The reversepart of the proof is similar to (i)(iii) Combining (i) and (ii) we get (iii)

The case for 1 2 4-inverses is treated completely analo-gously and the corresponding result follows by taking adjoint

Theorem 10 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 4 sube 11987321 2 4 sube 1198721 2 4 if and only iffor arbitrary 119860minus isin 1198601 2 4 119863minus isin 1198631 2 4 and 119866minus isin1198661 2 4

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(36)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(37)

(ii) 11987321 2 4 sube 11987311 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4 and 119878minus isin 1198781 2 4

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(38)

and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that

119864119860 = 119864119866119866minus119865119860 = 0119860minus119865119866 = 0

(39)

(iii) 11987311 2 4 = 11987321 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4

119862119864119860 = 0119861119864119863 = 0 (40)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that 119864119878 = 119864119863 119878minus119865119863 = 0 and 119863minus119865119878 = 0and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that 119864119860 = 119864119866 119866minus119865119860 = 0 and119860minus119865119866 = 0

Similarly we get the following results

Theorem 11 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 3 given by (7) there exists119884 isin 11987321 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119860minus isin 1198601 2 3 and 119878minus isin 1198781 2 3 and there exist 119863minus isin1198631 2 3 and 119866minus isin 1198661 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(41)

(ii) For arbitrary 119884 isin 11987321 2 3 given by (8) there exists119883 isin 11987311 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 and there exist 119860minus isin1198601 2 3 and 119878minus isin 1198781 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 2: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

2 Mathematical Problems in Engineering

first introduced by Schur [1] G E Trapp first defined thegeneralized Schur complement where the ordinary inversewas replaced by the generalized inverse For a matrix 119872 isinC(119898+119901)times(119899+119901) given by (3) and any fixed generalized inverse119860minus isin 1198601 the generalized Schur complement of 119860 in119872 isdefined by

119878 (119860minus) = 119863 minus 119862119860minus119861 (5)

Similarly for some fixed 119863minus isin 1198631 the generalizedSchur complement of119863 in119872 is defined by

119866 (119863minus) = 119860 minus 119861119863minus119862 (6)

The Schur complements and generalized Schur comple-ments were studied by a number of authors which haveapplications in statistics matrix theory electrical networktheory discrete-time regulator problem sophisticated tech-niques and some other fields It has been evident that theSchur complement plays an important role in many aspectsof a matrix theory (see [2ndash8])

Let

119883 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (7)

where 119860minus isin 1198601 119878 = 119863 minus 119862119860minus119861 and 119878minus isin 1198781 and let

119884 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (8)

where119863minus isin 1198631 119866 = 119860 minus 119861119863minus119862 and 119866minus isin 1198661For convenience we will firstly state the following nota-

tions which are helpful in the proofs We denote

119864120572 = 119868 minus 120572minus120572119865120572 = 119868 minus 120572120572minus120572120587 = 119868 minus 120572120572119863

120572 isin 119860 119878 119863 119866 (9)

In this paper we will establish necessary and sufficientconditions under which matrices 119883 and 119884 given by (7)and (8) respectively belong to a given class of generalizedinverses of the matrix 119872 given by (3) Our interest focuseson the cases when 1 2 3-inverses 1 2 4-inverses andgroup inverse of a 2 times 2 block matrix 119872 are given by theBanachiewicz-Schur form Also we give representations ofthe Drazin inverse of 2 times 2 block matrix 119872 under someconditions relating to the Schur complement of 119872 whichgeneralized the results studied by Cvetkovic-Ilic [9] Also thequotient property and the first Sylvester identity based on theMP-inverse group inverse and Drazin inverse are given

We need some auxiliary lemmas

Lemma 1 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198721 3 if and only if

119860minus isin 119860 1 3 119878minus isin 119878 1 3

119865119860119861 = 0119865119878119862 = 0

(10)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Lemma 2 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198721 4 if and only if

119860minus isin 119860 1 4 119878minus isin 119878 1 4

119861119864119878 = 0119862119864119860 = 0

(11)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Lemma 3 (see [10]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 isin 1198722 if and only if

119860minus isin 119860 1 2 119878minus isin 119878 2 (12)

Lemma 4 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119883 isin 1198721 2 3 if and only if

119860minus isin 119860 1 2 3 119878minus isin 119878 1 2 3

119865119860119861 = 0119865119878119862 = 0

(13)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119865119860 and 119865119878Proof The conclusion follows by Lemmas 1 and 2

Lemma 5 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119883 isin 1198721 2 4 if and only if

119860minus isin 119860 1 2 4 119878minus isin 119878 1 2 4

119861119864119878 = 0119862119864119860 = 0

(14)

the last two conditions being independent of the choice of 119860minus isin1198601 and 119878minus isin 1198781 involved in 119864119860 and 119864119878Using the same method as in [10] we can prove the

following

Mathematical Problems in Engineering 3

Lemma 6 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 3 if and only if

119863minus isin 119863 1 2 3 119866minus isin 119866 1 2 3 119865119866119861 = 0119865119863119862 = 0

(15)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 7 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 4 if and only if

119863minus isin 119863 1 2 4 119866minus isin 119866 1 2 4

119861119864119863 = 0119862119864119866 = 0

(16)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 8 (see [11]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 = 119872 if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(17)

2 Representations of 1 2 3-Inverse 1 2 4-Inverse and Group Inverse in terms ofBanachiewicz-Schur Forms

Baksalary and Styan [10] presented the necessary and suf-ficient conditions under which 1 2 1 2 1 3 and1 4-inverses of block matrix 119872 can be represented by theBanachiewicz-Schur form

Sheng and Chen [12] considered sufficient conditionsunder which 119872dagger 119872119863 and 119872 can be represented by bothof the Banachiewicz-Schur forms at the same time

Cvetkovic-Ilic [9] gave necessary and sufficient condi-tions which ensure the representation of the MP-inverse ofa block matrix119872 by both of the Banachiewicz-Schur forms

In this section we will present necessary and sufficientconditions under which 1 2 3 1 2 4-inverses and thegroup inverse of a partitioned matrix 119872 can be representedby the Banachiewicz-Schur form

For convenience we first introduce the following nota-tions

1198731 119894 119895 119896 = 119883

= [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] 119860minus

isin 119860 119894 119895 119896 119878 = 119863 minus 119862119860minus119861 119878minus isin 119878 119894 119895 119896

1198732 119894 119895 119896 = 119884

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] 119863minus

isin 119863 119894 119895 119896 119866 = 119860 minus 119861119863minus119862 119866minus isin 119866 119894 119895 119896

(18)

Theorem 9 Let 119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 3 sube 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0

(19)

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(20)

(ii) 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119878minus isin 1198781 2 3

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

(21)

and for any 119863minus isin 1198631 2 3 and any 119866minus isin 1198661 2 3 thereexists 119860minus isin 1198601 2 3 such that

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(22)

(iii) 11987311 2 3 = 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3

119865119860119861 = 0119865119863119862 = 0 (23)

4 Mathematical Problems in Engineering

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that 119865119863 = 119865119878 119864119863119878minus = 0 and 119864119878119863minus = 0and for any 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 there exists119860minus isin 1198601 2 3 such that 119865119860 = 119865119866 119864119860119866minus = 0 and 119864119866119860minus = 0Proof (i) (lArr) By Lemma 6 we have that arbitrary 119884 isin11987321 2 3 belongs to1198721 2 3 Now we only need to provethat11987311 2 3 sube 11987321 2 3 Take arbitrary

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (24)

By assumption for 119878minus1 there exists 1 2 3-inverse of119863whichwe denote by 119863minus1 such that 119865119863 = 1198651198781 119864119863119878minus1 = 0 and1198641198781119863minus1 = 0 Let 119866 = 119860 minus 119861119863minus1119862 We will prove that 119866minus1 =119860minus1 + 119860minus1119861119878minus1119862119860minus1 isin 1198661 2 3 By computation we get

119866119866minus1 = (119860 minus 119861119863minus1119862) (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119860119860minus1 + 119860119860minus1119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1minus 119861119863minus1119862119860minus1119861119878minus1119862119860minus1

= 119860119860minus1 + 119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1 minus 119861119863minus1119863119878minus1119862119860minus1+ 119861119863minus1 1198781119878minus1119862119860minus1 = 119860119860minus1

(25)

which implies 119866119866minus1119866 = 119860119860minus1(119860 minus 119861119863minus1119862) = 119860 minus 119861119863minus1119862 = 119866Also

119866minus1119866119866minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119860119860minus1= 119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (26)

so119866minus1 isin 1198661 2 3 Since 119865119863 = 1198651198781 119864119863119878minus1 = 0 and 1198641198781119863minus1 = 0a simple computation shows that

119863minus1119862119866minus1 = 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119863minus1119862119860minus1 + 119863minus1119862119860minus1119861119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1 (119863 minus 1198781) 119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1119863119878minus1119862119860minus1minus 119863minus1 1198781119878minus1119862119860minus1 = 119878minus1119862119860minus1

119866minus1119861119863minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119862119860minus1119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1 (119863 minus 1198781)119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119863119863minus1minus 119860minus1119861119878minus1 1198781119863minus1 = 119860minus1119861119878minus1

119863minus1 + 119863minus1119862119866minus1119861119863minus1 = 119863minus1+ 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1

= 119863minus1 + 119863minus1119862119860minus1119861119863minus1+ 119863minus1119862119860minus1119861119878minus1119862119860minus1119861119863minus1

= 119863minus1 + 119863minus1 (119863 minus 1198781)119863minus1+ 119863minus1 (119863 minus 1198781) 119878minus1 (119863 minus 1198781)119863minus1

= 119878minus1 (27)

Thus

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ]

= [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732(28)

Therefore1198731 sube 1198732 sube 1198721 2 3(rArr) Since 11987311 2 3 sube 1198721 2 3 and 11987321 2 3 sube1198721 2 3 by Lemmas 4 and 6 we get that for arbitrary119860minus isin 1198601 2 3 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3the following hold 119865119860119861 = 0 119865119866119861 = 0 and 119865119863119862 = 0Moreover by11987311 2 3 sube 11987321 2 3 it follows that for any119883 isin 11987311 2 3 there exists some 119884 isin 11987321 2 3 such that119883 = 119884 Hence for any 119860minus isin 1198601 2 3 119878minus isin 1198781 2 3 thereexist119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 such that

[119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(29)

A simple computation shows that

119878119878minus = (119863 minus 119862119860minus119861) (119863minus + 119863minus119862119866minus119861119863minus)= 119863119863minus + 119863119863minus119862119866minus119861119863minus minus 119862119860minus119861119863minusminus 119862119860minus119860119866minus119861119863minus + 119862119860minus119866119866minus119861119863minus

= 119863119863minus + 119862 (119868 minus 119860minus119860)119866minus119861119863minus= 119863119863minus + 119862 (119868 minus 119860minus119860)119860minus119861119878minus = 119863119863minus

(30)

thus we get 119865119878 = 119865119863 Since119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119878minus119862119860minus119861119863minus

= 119863minus + 119878minus (119863 minus 119878)119863minus= 119863minus + 119878minus119863119863minus minus 119878minus119878119863minus = 119878minus

(31)

Mathematical Problems in Engineering 5

we get119863minus = 119878minus119878119863minus that is 119864119878119863minus = 0 Similarly

119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119863minus119862119860minus119861119878minus= 119863minus + 119863minus (119863 minus 119878) 119878minus= 119863minus + 119863minus119863119878minus minus 119863minus119878119878minus = 119878minus

(32)

which implies 119864119863119878minus = 0(ii) (lArr) Similar to (i) since 119865119860119861 = 0 and 119865119878119862 = 0we have 11987311 2 3 sube 1198721 2 3 Now we will prove that11987321 2 3 sube 11987311 2 3 Take arbitrary1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732 1 2 3 (33)

For 119866minus1 there exists 1 2 3-inverse of 119860 which we denote by119860minus1 such that 119865119860 = 1198651198661 119864119860119866minus1 = 0 and 1198641198661119860minus1 = 0 Let 119878 =119863 minus 119862119860minus1119861 A simple computation shows that

119878minus1 = 119863minus1 + 119863minus1119862119866minus1119861119863minus1 isin 119878 1 2 3 119878minus1119862119860minus1 = 119863minus1119862119866minus1 119860minus1119861119878minus1 = 119866minus1119861119863minus1

119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (34)

Hence

1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1]

= [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (35)

Therefore 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 The reversepart of the proof is similar to (i)(iii) Combining (i) and (ii) we get (iii)

The case for 1 2 4-inverses is treated completely analo-gously and the corresponding result follows by taking adjoint

Theorem 10 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 4 sube 11987321 2 4 sube 1198721 2 4 if and only iffor arbitrary 119860minus isin 1198601 2 4 119863minus isin 1198631 2 4 and 119866minus isin1198661 2 4

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(36)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(37)

(ii) 11987321 2 4 sube 11987311 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4 and 119878minus isin 1198781 2 4

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(38)

and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that

119864119860 = 119864119866119866minus119865119860 = 0119860minus119865119866 = 0

(39)

(iii) 11987311 2 4 = 11987321 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4

119862119864119860 = 0119861119864119863 = 0 (40)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that 119864119878 = 119864119863 119878minus119865119863 = 0 and 119863minus119865119878 = 0and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that 119864119860 = 119864119866 119866minus119865119860 = 0 and119860minus119865119866 = 0

Similarly we get the following results

Theorem 11 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 3 given by (7) there exists119884 isin 11987321 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119860minus isin 1198601 2 3 and 119878minus isin 1198781 2 3 and there exist 119863minus isin1198631 2 3 and 119866minus isin 1198661 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(41)

(ii) For arbitrary 119884 isin 11987321 2 3 given by (8) there exists119883 isin 11987311 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 and there exist 119860minus isin1198601 2 3 and 119878minus isin 1198781 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 3: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 3

Lemma 6 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 3 if and only if

119863minus isin 119863 1 2 3 119866minus isin 119866 1 2 3 119865119866119861 = 0119865119863119862 = 0

(15)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 7 Let 119872 119878 and 119883 be defined by (3) (5) and (7)respectively Then 119884 isin 1198721 2 4 if and only if

119863minus isin 119863 1 2 4 119866minus isin 119866 1 2 4

119861119864119863 = 0119862119864119866 = 0

(16)

the last two conditions being independent of the choice of119863minus isin1198631 and 119866minus isin 1198661 involved in 119865119863 and 119865119866Lemma 8 (see [11]) Let 119872 119878 and 119883 be defined by (3) (5)and (7) respectively Then 119883 = 119872 if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(17)

2 Representations of 1 2 3-Inverse 1 2 4-Inverse and Group Inverse in terms ofBanachiewicz-Schur Forms

Baksalary and Styan [10] presented the necessary and suf-ficient conditions under which 1 2 1 2 1 3 and1 4-inverses of block matrix 119872 can be represented by theBanachiewicz-Schur form

Sheng and Chen [12] considered sufficient conditionsunder which 119872dagger 119872119863 and 119872 can be represented by bothof the Banachiewicz-Schur forms at the same time

Cvetkovic-Ilic [9] gave necessary and sufficient condi-tions which ensure the representation of the MP-inverse ofa block matrix119872 by both of the Banachiewicz-Schur forms

In this section we will present necessary and sufficientconditions under which 1 2 3 1 2 4-inverses and thegroup inverse of a partitioned matrix 119872 can be representedby the Banachiewicz-Schur form

For convenience we first introduce the following nota-tions

1198731 119894 119895 119896 = 119883

= [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] 119860minus

isin 119860 119894 119895 119896 119878 = 119863 minus 119862119860minus119861 119878minus isin 119878 119894 119895 119896

1198732 119894 119895 119896 = 119884

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] 119863minus

isin 119863 119894 119895 119896 119866 = 119860 minus 119861119863minus119862 119866minus isin 119866 119894 119895 119896

(18)

Theorem 9 Let 119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 3 sube 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0

(19)

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(20)

(ii) 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3 and 119878minus isin 1198781 2 3

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

(21)

and for any 119863minus isin 1198631 2 3 and any 119866minus isin 1198661 2 3 thereexists 119860minus isin 1198601 2 3 such that

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(22)

(iii) 11987311 2 3 = 11987321 2 3 sube 1198721 2 3 if and only if forarbitrary 119860minus isin 1198601 2 3119863minus isin 1198631 2 3

119865119860119861 = 0119865119863119862 = 0 (23)

4 Mathematical Problems in Engineering

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that 119865119863 = 119865119878 119864119863119878minus = 0 and 119864119878119863minus = 0and for any 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 there exists119860minus isin 1198601 2 3 such that 119865119860 = 119865119866 119864119860119866minus = 0 and 119864119866119860minus = 0Proof (i) (lArr) By Lemma 6 we have that arbitrary 119884 isin11987321 2 3 belongs to1198721 2 3 Now we only need to provethat11987311 2 3 sube 11987321 2 3 Take arbitrary

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (24)

By assumption for 119878minus1 there exists 1 2 3-inverse of119863whichwe denote by 119863minus1 such that 119865119863 = 1198651198781 119864119863119878minus1 = 0 and1198641198781119863minus1 = 0 Let 119866 = 119860 minus 119861119863minus1119862 We will prove that 119866minus1 =119860minus1 + 119860minus1119861119878minus1119862119860minus1 isin 1198661 2 3 By computation we get

119866119866minus1 = (119860 minus 119861119863minus1119862) (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119860119860minus1 + 119860119860minus1119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1minus 119861119863minus1119862119860minus1119861119878minus1119862119860minus1

= 119860119860minus1 + 119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1 minus 119861119863minus1119863119878minus1119862119860minus1+ 119861119863minus1 1198781119878minus1119862119860minus1 = 119860119860minus1

(25)

which implies 119866119866minus1119866 = 119860119860minus1(119860 minus 119861119863minus1119862) = 119860 minus 119861119863minus1119862 = 119866Also

119866minus1119866119866minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119860119860minus1= 119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (26)

so119866minus1 isin 1198661 2 3 Since 119865119863 = 1198651198781 119864119863119878minus1 = 0 and 1198641198781119863minus1 = 0a simple computation shows that

119863minus1119862119866minus1 = 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119863minus1119862119860minus1 + 119863minus1119862119860minus1119861119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1 (119863 minus 1198781) 119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1119863119878minus1119862119860minus1minus 119863minus1 1198781119878minus1119862119860minus1 = 119878minus1119862119860minus1

119866minus1119861119863minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119862119860minus1119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1 (119863 minus 1198781)119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119863119863minus1minus 119860minus1119861119878minus1 1198781119863minus1 = 119860minus1119861119878minus1

119863minus1 + 119863minus1119862119866minus1119861119863minus1 = 119863minus1+ 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1

= 119863minus1 + 119863minus1119862119860minus1119861119863minus1+ 119863minus1119862119860minus1119861119878minus1119862119860minus1119861119863minus1

= 119863minus1 + 119863minus1 (119863 minus 1198781)119863minus1+ 119863minus1 (119863 minus 1198781) 119878minus1 (119863 minus 1198781)119863minus1

= 119878minus1 (27)

Thus

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ]

= [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732(28)

Therefore1198731 sube 1198732 sube 1198721 2 3(rArr) Since 11987311 2 3 sube 1198721 2 3 and 11987321 2 3 sube1198721 2 3 by Lemmas 4 and 6 we get that for arbitrary119860minus isin 1198601 2 3 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3the following hold 119865119860119861 = 0 119865119866119861 = 0 and 119865119863119862 = 0Moreover by11987311 2 3 sube 11987321 2 3 it follows that for any119883 isin 11987311 2 3 there exists some 119884 isin 11987321 2 3 such that119883 = 119884 Hence for any 119860minus isin 1198601 2 3 119878minus isin 1198781 2 3 thereexist119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 such that

[119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(29)

A simple computation shows that

119878119878minus = (119863 minus 119862119860minus119861) (119863minus + 119863minus119862119866minus119861119863minus)= 119863119863minus + 119863119863minus119862119866minus119861119863minus minus 119862119860minus119861119863minusminus 119862119860minus119860119866minus119861119863minus + 119862119860minus119866119866minus119861119863minus

= 119863119863minus + 119862 (119868 minus 119860minus119860)119866minus119861119863minus= 119863119863minus + 119862 (119868 minus 119860minus119860)119860minus119861119878minus = 119863119863minus

(30)

thus we get 119865119878 = 119865119863 Since119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119878minus119862119860minus119861119863minus

= 119863minus + 119878minus (119863 minus 119878)119863minus= 119863minus + 119878minus119863119863minus minus 119878minus119878119863minus = 119878minus

(31)

Mathematical Problems in Engineering 5

we get119863minus = 119878minus119878119863minus that is 119864119878119863minus = 0 Similarly

119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119863minus119862119860minus119861119878minus= 119863minus + 119863minus (119863 minus 119878) 119878minus= 119863minus + 119863minus119863119878minus minus 119863minus119878119878minus = 119878minus

(32)

which implies 119864119863119878minus = 0(ii) (lArr) Similar to (i) since 119865119860119861 = 0 and 119865119878119862 = 0we have 11987311 2 3 sube 1198721 2 3 Now we will prove that11987321 2 3 sube 11987311 2 3 Take arbitrary1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732 1 2 3 (33)

For 119866minus1 there exists 1 2 3-inverse of 119860 which we denote by119860minus1 such that 119865119860 = 1198651198661 119864119860119866minus1 = 0 and 1198641198661119860minus1 = 0 Let 119878 =119863 minus 119862119860minus1119861 A simple computation shows that

119878minus1 = 119863minus1 + 119863minus1119862119866minus1119861119863minus1 isin 119878 1 2 3 119878minus1119862119860minus1 = 119863minus1119862119866minus1 119860minus1119861119878minus1 = 119866minus1119861119863minus1

119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (34)

Hence

1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1]

= [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (35)

Therefore 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 The reversepart of the proof is similar to (i)(iii) Combining (i) and (ii) we get (iii)

The case for 1 2 4-inverses is treated completely analo-gously and the corresponding result follows by taking adjoint

Theorem 10 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 4 sube 11987321 2 4 sube 1198721 2 4 if and only iffor arbitrary 119860minus isin 1198601 2 4 119863minus isin 1198631 2 4 and 119866minus isin1198661 2 4

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(36)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(37)

(ii) 11987321 2 4 sube 11987311 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4 and 119878minus isin 1198781 2 4

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(38)

and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that

119864119860 = 119864119866119866minus119865119860 = 0119860minus119865119866 = 0

(39)

(iii) 11987311 2 4 = 11987321 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4

119862119864119860 = 0119861119864119863 = 0 (40)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that 119864119878 = 119864119863 119878minus119865119863 = 0 and 119863minus119865119878 = 0and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that 119864119860 = 119864119866 119866minus119865119860 = 0 and119860minus119865119866 = 0

Similarly we get the following results

Theorem 11 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 3 given by (7) there exists119884 isin 11987321 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119860minus isin 1198601 2 3 and 119878minus isin 1198781 2 3 and there exist 119863minus isin1198631 2 3 and 119866minus isin 1198661 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(41)

(ii) For arbitrary 119884 isin 11987321 2 3 given by (8) there exists119883 isin 11987311 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 and there exist 119860minus isin1198601 2 3 and 119878minus isin 1198781 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 4: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

4 Mathematical Problems in Engineering

and for any119860minus isin 1198601 2 3 and any 119878minus isin 1198781 2 3 there exists119863minus isin 1198631 2 3 such that 119865119863 = 119865119878 119864119863119878minus = 0 and 119864119878119863minus = 0and for any 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 there exists119860minus isin 1198601 2 3 such that 119865119860 = 119865119866 119864119860119866minus = 0 and 119864119866119860minus = 0Proof (i) (lArr) By Lemma 6 we have that arbitrary 119884 isin11987321 2 3 belongs to1198721 2 3 Now we only need to provethat11987311 2 3 sube 11987321 2 3 Take arbitrary

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (24)

By assumption for 119878minus1 there exists 1 2 3-inverse of119863whichwe denote by 119863minus1 such that 119865119863 = 1198651198781 119864119863119878minus1 = 0 and1198641198781119863minus1 = 0 Let 119866 = 119860 minus 119861119863minus1119862 We will prove that 119866minus1 =119860minus1 + 119860minus1119861119878minus1119862119860minus1 isin 1198661 2 3 By computation we get

119866119866minus1 = (119860 minus 119861119863minus1119862) (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119860119860minus1 + 119860119860minus1119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1minus 119861119863minus1119862119860minus1119861119878minus1119862119860minus1

= 119860119860minus1 + 119861119878minus1119862119860minus1 minus 119861119863minus1119862119860minus1 minus 119861119863minus1119863119878minus1119862119860minus1+ 119861119863minus1 1198781119878minus1119862119860minus1 = 119860119860minus1

(25)

which implies 119866119866minus1119866 = 119860119860minus1(119860 minus 119861119863minus1119862) = 119860 minus 119861119863minus1119862 = 119866Also

119866minus1119866119866minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119860119860minus1= 119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (26)

so119866minus1 isin 1198661 2 3 Since 119865119863 = 1198651198781 119864119863119878minus1 = 0 and 1198641198781119863minus1 = 0a simple computation shows that

119863minus1119862119866minus1 = 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1)= 119863minus1119862119860minus1 + 119863minus1119862119860minus1119861119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1 (119863 minus 1198781) 119878minus1119862119860minus1= 119863minus1119862119860minus1 + 119863minus1119863119878minus1119862119860minus1minus 119863minus1 1198781119878minus1119862119860minus1 = 119878minus1119862119860minus1

119866minus1119861119863minus1 = (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119862119860minus1119861119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1 (119863 minus 1198781)119863minus1= 119860minus1119861119863minus1 + 119860minus1119861119878minus1119863119863minus1minus 119860minus1119861119878minus1 1198781119863minus1 = 119860minus1119861119878minus1

119863minus1 + 119863minus1119862119866minus1119861119863minus1 = 119863minus1+ 119863minus1119862 (119860minus1 + 119860minus1119861119878minus1119862119860minus1) 119861119863minus1

= 119863minus1 + 119863minus1119862119860minus1119861119863minus1+ 119863minus1119862119860minus1119861119878minus1119862119860minus1119861119863minus1

= 119863minus1 + 119863minus1 (119863 minus 1198781)119863minus1+ 119863minus1 (119863 minus 1198781) 119878minus1 (119863 minus 1198781)119863minus1

= 119878minus1 (27)

Thus

1198831 = [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ]

= [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732(28)

Therefore1198731 sube 1198732 sube 1198721 2 3(rArr) Since 11987311 2 3 sube 1198721 2 3 and 11987321 2 3 sube1198721 2 3 by Lemmas 4 and 6 we get that for arbitrary119860minus isin 1198601 2 3 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3the following hold 119865119860119861 = 0 119865119866119861 = 0 and 119865119863119862 = 0Moreover by11987311 2 3 sube 11987321 2 3 it follows that for any119883 isin 11987311 2 3 there exists some 119884 isin 11987321 2 3 such that119883 = 119884 Hence for any 119860minus isin 1198601 2 3 119878minus isin 1198781 2 3 thereexist119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 such that

[119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(29)

A simple computation shows that

119878119878minus = (119863 minus 119862119860minus119861) (119863minus + 119863minus119862119866minus119861119863minus)= 119863119863minus + 119863119863minus119862119866minus119861119863minus minus 119862119860minus119861119863minusminus 119862119860minus119860119866minus119861119863minus + 119862119860minus119866119866minus119861119863minus

= 119863119863minus + 119862 (119868 minus 119860minus119860)119866minus119861119863minus= 119863119863minus + 119862 (119868 minus 119860minus119860)119860minus119861119878minus = 119863119863minus

(30)

thus we get 119865119878 = 119865119863 Since119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119878minus119862119860minus119861119863minus

= 119863minus + 119878minus (119863 minus 119878)119863minus= 119863minus + 119878minus119863119863minus minus 119878minus119878119863minus = 119878minus

(31)

Mathematical Problems in Engineering 5

we get119863minus = 119878minus119878119863minus that is 119864119878119863minus = 0 Similarly

119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119863minus119862119860minus119861119878minus= 119863minus + 119863minus (119863 minus 119878) 119878minus= 119863minus + 119863minus119863119878minus minus 119863minus119878119878minus = 119878minus

(32)

which implies 119864119863119878minus = 0(ii) (lArr) Similar to (i) since 119865119860119861 = 0 and 119865119878119862 = 0we have 11987311 2 3 sube 1198721 2 3 Now we will prove that11987321 2 3 sube 11987311 2 3 Take arbitrary1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732 1 2 3 (33)

For 119866minus1 there exists 1 2 3-inverse of 119860 which we denote by119860minus1 such that 119865119860 = 1198651198661 119864119860119866minus1 = 0 and 1198641198661119860minus1 = 0 Let 119878 =119863 minus 119862119860minus1119861 A simple computation shows that

119878minus1 = 119863minus1 + 119863minus1119862119866minus1119861119863minus1 isin 119878 1 2 3 119878minus1119862119860minus1 = 119863minus1119862119866minus1 119860minus1119861119878minus1 = 119866minus1119861119863minus1

119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (34)

Hence

1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1]

= [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (35)

Therefore 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 The reversepart of the proof is similar to (i)(iii) Combining (i) and (ii) we get (iii)

The case for 1 2 4-inverses is treated completely analo-gously and the corresponding result follows by taking adjoint

Theorem 10 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 4 sube 11987321 2 4 sube 1198721 2 4 if and only iffor arbitrary 119860minus isin 1198601 2 4 119863minus isin 1198631 2 4 and 119866minus isin1198661 2 4

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(36)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(37)

(ii) 11987321 2 4 sube 11987311 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4 and 119878minus isin 1198781 2 4

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(38)

and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that

119864119860 = 119864119866119866minus119865119860 = 0119860minus119865119866 = 0

(39)

(iii) 11987311 2 4 = 11987321 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4

119862119864119860 = 0119861119864119863 = 0 (40)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that 119864119878 = 119864119863 119878minus119865119863 = 0 and 119863minus119865119878 = 0and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that 119864119860 = 119864119866 119866minus119865119860 = 0 and119860minus119865119866 = 0

Similarly we get the following results

Theorem 11 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 3 given by (7) there exists119884 isin 11987321 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119860minus isin 1198601 2 3 and 119878minus isin 1198781 2 3 and there exist 119863minus isin1198631 2 3 and 119866minus isin 1198661 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(41)

(ii) For arbitrary 119884 isin 11987321 2 3 given by (8) there exists119883 isin 11987311 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 and there exist 119860minus isin1198601 2 3 and 119878minus isin 1198781 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 5: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 5

we get119863minus = 119878minus119878119863minus that is 119864119878119863minus = 0 Similarly

119863minus + 119863minus119862119866minus119861119863minus = 119863minus + 119863minus119862119860minus119861119878minus= 119863minus + 119863minus (119863 minus 119878) 119878minus= 119863minus + 119863minus119863119878minus minus 119863minus119878119878minus = 119878minus

(32)

which implies 119864119863119878minus = 0(ii) (lArr) Similar to (i) since 119865119860119861 = 0 and 119865119878119862 = 0we have 11987311 2 3 sube 1198721 2 3 Now we will prove that11987321 2 3 sube 11987311 2 3 Take arbitrary1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1] isin 1198732 1 2 3 (33)

For 119866minus1 there exists 1 2 3-inverse of 119860 which we denote by119860minus1 such that 119865119860 = 1198651198661 119864119860119866minus1 = 0 and 1198641198661119860minus1 = 0 Let 119878 =119863 minus 119862119860minus1119861 A simple computation shows that

119878minus1 = 119863minus1 + 119863minus1119862119866minus1119861119863minus1 isin 119878 1 2 3 119878minus1119862119860minus1 = 119863minus1119862119866minus1 119860minus1119861119878minus1 = 119866minus1119861119863minus1

119860minus1 + 119860minus1119861119878minus1119862119860minus1 = 119866minus1 (34)

Hence

1198841 = [ 119866minus1 minus119866minus1119861119863minus1minus119863minus1119862119866minus1 119863minus1 + 119863minus1119862119866minus1119861119863minus1]

= [119860minus1 + 119860minus1119861119878minus1119862119860minus1 minus119860minus1119861119878minus1minus119878minus1119862119860minus1 119878minus1 ] isin 1198731 1 2 3 (35)

Therefore 11987321 2 3 sube 11987311 2 3 sube 1198721 2 3 The reversepart of the proof is similar to (i)(iii) Combining (i) and (ii) we get (iii)

The case for 1 2 4-inverses is treated completely analo-gously and the corresponding result follows by taking adjoint

Theorem 10 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) 11987311 2 4 sube 11987321 2 4 sube 1198721 2 4 if and only iffor arbitrary 119860minus isin 1198601 2 4 119863minus isin 1198631 2 4 and 119866minus isin1198661 2 4

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(36)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(37)

(ii) 11987321 2 4 sube 11987311 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4 and 119878minus isin 1198781 2 4

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(38)

and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that

119864119860 = 119864119866119866minus119865119860 = 0119860minus119865119866 = 0

(39)

(iii) 11987311 2 4 = 11987321 2 4 sube 1198721 2 4 if and only if forarbitrary 119860minus isin 1198601 2 4119863minus isin 1198631 2 4

119862119864119860 = 0119861119864119863 = 0 (40)

and for any119860minus isin 1198601 2 4 and any 119878minus isin 1198781 2 4 there exists119863minus isin 1198631 2 4 such that 119864119878 = 119864119863 119878minus119865119863 = 0 and 119863minus119865119878 = 0and for any 119863minus isin 1198631 2 4 and any 119866minus isin 1198661 2 4 thereexists 119860minus isin 1198601 2 4 such that 119864119860 = 119864119866 119866minus119865119860 = 0 and119860minus119865119866 = 0

Similarly we get the following results

Theorem 11 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 3 given by (7) there exists119884 isin 11987321 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119860minus isin 1198601 2 3 and 119878minus isin 1198781 2 3 and there exist 119863minus isin1198631 2 3 and 119866minus isin 1198661 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119865119863 = 119865119878

119864119863119878minus = 0119864119878119863minus = 0

(41)

(ii) For arbitrary 119884 isin 11987321 2 3 given by (8) there exists119883 isin 11987311 2 3 such that 119883 = 119884 isin 1198721 2 3 if and onlyif 119863minus isin 1198631 2 3 and 119866minus isin 1198661 2 3 and there exist 119860minus isin1198601 2 3 and 119878minus isin 1198781 2 3 such that

119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

6 Mathematical Problems in Engineering

119865119860 = 119865119866119864119860119866minus = 0119864119866119860minus = 0

(42)

Theorem 12 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then the following happens

(i) For arbitrary 119883 isin 11987311 2 4 given by (7) there exists119884 isin 11987321 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119860minus isin 1198601 2 4 and 119878minus isin 1198781 2 4 and there exist 119863minus isin1198631 2 4 and 119866minus isin 1198661 2 4 such that

119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0119864119878 = 119864119863

119878minus119865119863 = 0119863minus119865119878 = 0

(43)

(ii) For arbitrary 119884 isin 11987321 2 4 given by (8) there exists119883 isin 11987311 2 4 such that 119883 = 119884 isin 1198721 2 4 if and onlyif 119863minus isin 1198631 2 4 and 119866minus isin 1198661 2 4 and there exist 119860minus isin1198601 2 4 and 119878minus isin 1198781 2 4 such that

119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0119864119860 = 119864119866

119866minus119865119860 = 0119860minus119865119866 = 0

(44)

Theorem 13 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(45)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0

119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119863 = 119864119878119865119863 = 119865119878

(46)

Proof ByTheorems 11 and 12 we get

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(47)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(48)

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 7

Since 119864119878 = 119864119863 and 119864119863119863minus = 0 then 119864119878119863minus = 0 Thus

119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(49)

implies

119864119863119878minus = 0119864119878119863minus = 0119864119860119866minus = 0119864119866119860minus = 0119863minus119865119878 = 0119878minus119865119863 = 0119866minus119865119860 = 0119860minus119865119866 = 0

(50)

Now we only need to prove that

119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119878 = 119864119863119864119860 = 119864119866119865119863 = 119865119878119865119860 = 119865119866

(51)

is equivalent to (46) Denote1198661015840 = 119860minus+119860minus119861119878minus119862119860minus By119865119860119861 =0 119861119864119863 = 0 and 119865119878 = 119865119863 a simple computation shows that

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119863119878minus119862119860minus + 119861119863minus119878119878minus119862119860minus = 119860119860minus

(52)

Similarly by 119865119863119862 = 0 119862119864119860 = 0 and 119864119878 = 119864119863 we get 1198661015840119866 =119860minus119860 Moreover 1198661198661015840119866 = 119866 and 11986610158401198661198661015840 = 1198661015840 Thus 1198661015840 = 119866daggerand 119864119860 = 119864119866 and 119865119860 = 119865119866 Therefore (51) is equivalent to(46)

Corollary 14 (see [9]) Let119872 119878 and 119866 be defined by (3) (5)and (6) respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(53)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(54)

Proof ByCorollary 14 from [9] it is easy to conclude that (54)is equivalent to (46)

Corollary 15 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(55)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119862119864119860 = 0119861119864119863 = 0119864119860 = 119864119866119865119860 = 119865119866

(56)

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

8 Mathematical Problems in Engineering

Corollary 16 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872dagger = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(57)

if and only if

119860minus = 119860dagger119863minus = 119863dagger119878minus = 119878dagger119866minus = 119866dagger119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(58)

Theorem 17 Let119872 119878 and 119866 be defined by (3) (5) and (6)respectively Then

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus ]

(59)

if and only if one of the following conditions holds(i)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119878119862 = 0119862119864119860 = 0119861119864119863 = 0119861119864119878 = 0

(60)

(ii)

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119863119862 = 0119865119866119861 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(61)

Proof (i) By Lemma 8

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ] (62)

if and only if

119860minus = 119860119878minus = 119878

119865119860119861 = 0119865119878119862 = 0119862119864119860 = 0119861119864119878 = 0

(63)

Similarly

119872 = [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus] (64)

if and only if

119863minus = 119863119866minus = 119866119865119866119861 = 0119865119863119862 = 0119862119864119866 = 0119861119864119863 = 0

(65)

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 9: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 9

Since the group inverse is unique if it exists combining (63)and (65) we get

119872 = [119860minus + 119860minus119861119878minus119862119860minus minus119860minus119861119878minusminus119878minus119862119860minus 119878minus ]

= [ 119866minus minus119866minus119861119863minusminus119863minus119862119866minus 119863minus + 119863minus119862119866minus119861119863minus]

(66)

if and only if

119860minus = 119860119863minus = 119863119878minus = 119878119866minus = 119866119865119860119861 = 0119865119866119861 = 0119865119878119862 = 0119865119863119862 = 0119861119864119878 = 0119862119864119860 = 0119861119864119863 = 0119862119864119866 = 0

(67)

Now we only need to prove (67) is equivalent to (60)Denote 1198661015840 = 119860minus + 119860minus119861119878minus119862119860minus If (60) holds then

1198661198661015840 = (119860 minus 119861119863minus119862) (119860minus + 119860minus119861119878minus119862119860minus)= 119860119860minus + 119860119860minus119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus119862119860minus119861119878minus119862119860minus

= 119860119860minus + 119861119878minus119862119860minus minus 119861119863minus119862119860minusminus 119861119863minus (119863 minus 119878) 119878minus119862119860minus = 119860119860minus

1198661015840119866 = (119860minus + 119860minus119861119878minus119862119860minus) (119860 minus 119861119863minus119862)= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862119860minus119860minus 119860minus119861119878minus119862119860minus119861119863minus119862

= 119860minus119860 minus 119860minus119861119863minus119862 minus 119860minus119861119878minus119862minus 119860minus119861119878minus (119863 minus 119878)119863minus119862 = 119860minus119860

(68)

Now it is easy to get

1198661198661015840119866 = 119860119860minus (119860 minus 119861119863minus119862) = 119860 minus 119861119863minus119862 = 11986611986610158401198661198661015840 = 119860minus119860 (119860minus + 119860minus119861119878minus119862119860minus)

= 119860minus + 119860minus119861119878minus119862119860minus = 1198661015840(69)

and 1198661198661015840 = 1198661015840119866 Thus 1198661015840 = 119866 Hence 119866119866 = 119860minus119860 and119866119866 = 119860119860minus Now we get 119865119860119861 = 119865119866119861 = 0 and 119862119864119860 =119862119864119866 = 0 which means (60) implies (67) Obviously (67)implies (60) Thus (67) is equivalent to (60)(ii)The proof is similar to the proof of (i)3 Representations of Drazin Inverse in

terms of Banachiewicz-Schur Forms

TheDrazin inverse of a squarematrix has various applicationsin singular differential equations and singular differenceequations Markov chains and iterative methods Actually in1979 Campbell and Meyer [13] posed the problem of findingan explicit representation for the Drazin inverse of a complexblock matrix 119872 in terms of its blocks No formula for 119872119863has yet been offered without any restrictions upon the blocksMany papers have considered this open problem and each ofthem offered a formula for the Drazin inverse and specificconditions for the 2times2 blockmatrix119872 (see [9 11 12 14ndash22])

Wei [23] proved that 119872119863 can be expressed by[ 119860119863+119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863

] if 119860120587119862 = 0 119861119860120587 = 0 119862119878120587 = 0119878120587119861 = 0 and119863119878120587 = 0 where119860120587 = 119868minus119860119860119863 and 119878120587 = 119868minus119878119878119863

Cvetkovic-Ilic [9] generalized the result from [23] andgave the representation of119872119863 under simpler conditions thanin [23]

Sheng and Chen [12] proved that under the conditions119860120587119861 = 0 119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0 and 119878120587 = 119863120587 = 0119872119863can be represented by both of Banachiewicz-Schur forms

In the following we will present conditions more generalthan those in [9 12 23] which ensure the representation of119872119863 by the Banachiewicz-Schur form

Theorem 18 Let119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861 bethe generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(70)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (71)

Proof Denote the right side of (71) by 119883 By simple compu-tations we show that

119872119883= [[119860119860119863 minus (119868 minus 119860119860119863) 119861119878119863119862119860119863 (119868 minus 119860119860119863) 119861119878119863

(119868 minus 119878119878119863) 119862119860119863 119878119878119863 ]]

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

10 Mathematical Problems in Engineering

119883119872= [[119860119863119860 minus 119860119863119861119878119863119862 (119868 minus 119860119863119860) 119860119863119861 (119868 minus 119878119863119878)

119878119863119862 (119868 minus 119860119863119860) 119878119863119878 ]](72)

Since 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 119860120587119861119878119863 = 0 and 119878120587119862119860119863 = 0(72) reduce to

119872119883 = [119860119860119863 00 119878119878119863]

119883119872 = [119860119863119860 00 119878119863119878]

(73)

Obviously119872119883 = 119883119872 On the other hand

119883119872119883= [119860119863119860 0

0 119878119863119878][119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863

minus119878119863119862119860119863 119878119863 ]

= [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] = 119883

(74)

Now we need to prove that119872 minus1198722119873 is a nilpotent matrixSince

119872minus1198722119873 = [[(119860 minus (119868 minus 119860119860119863) 119861119878119863119862) (119868 minus 119860119863119860) (119868 minus 119860119860119863) 119861 (119868 minus 119878119863119878)

(119868 minus 119878119878119863) 119862 (119868 minus 119860119863119860) 119878 (119868 minus 119878119878119863) ]] (75)

by 119860119863119861119878120587 = 0 119878119863119862119860120587 = 0 we get119872minus1198722119873 = [

[119860(119868 minus 119860119863119860) 0

0 119878 (119868 minus 119878119878119863)]] (76)

which is nilpotent and ind(119872minus1198722119873) ⩾ maxind(119860) ind(119878)Corollary 19 (see [9]) Let 119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860119863119861119878120587 = 0119878119863119862119860120587 = 0119860120587119861119878119863 = 0119878120587119862119860119863 = 0

(77)

and if there exists a nonnegative integer 119896 such that119872119896119875 = 0then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (78)

where 119875 = [ 119860120587 00 119878120587

]Corollary 20 (see [23]) Let119872 be given by (3) and 119878 = 119863 minus119862119860119863119861 be the generalized Schur complement of 119860 in119872 If

119860120587119861 = 0119861119878120587 = 0119862119860120587 = 0

119878120587119862 = 0119863119878120587 = 0

(79)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (80)

Theorem 21 Let119872 be defined as in (3) and 119866 = 119860 minus 119861119863119863119862be the generalized Schur complement of 119863 in119872 If

119863119863119862119866120587 = 0119866119863119861119863120587 = 0119866120587119861119863119863 = 0119863120587119862119866119863 = 0

(81)

then

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (82)

Proof The proof is analogous to that of Theorem 18

Theorem 22 Let 119872 be defined by (3) and 119878 = 119863 minus 119862119860119863119861119866 = 119860 minus 119861119863119863119862 be the generalized Schur complement of 119860 in119872 and 119863 in119872 respectively If any of the conditions hold

(i) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119878120587119862119860119863 = 0119860119863119861119878120587 = 0(ii) 119860120587119861 = 0 119862119860120587 = 0 119861119863120587 = 0 119863120587119862 = 0 119866120587119861119863119863 = 0119863119863119862119866120587 = 0

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 11

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(83)

Proof (i) Denote 119883 = 119860119863 + 119860119863119861119878119863119862119860119863 Firstly we prove119866119863 = 119860119863 + 119860119863119861119878119863119862119860119863 It is easy to see119866119883 = (119860 minus 119861119863119863119862) (119860119863 + 119860119863119861119878119863119862119860119863)

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119862119860119863119861119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863 (119863 minus 119878) 119878119863119862119860119863

= 119860119860119863 + 119860119860119863119861119878119863119862119860119863 minus 119861119863119863119862119860119863minus 119861119863119863119863119878119863119862119860119863 + 119861119863119863119878119878119863119862119860119863 = 119860119860119863

(84)

by 119860120587119861 = 0 119861119863120587 = 0 and 119878120587119862119860119863 = 0 and119883119866 = (119860119863 + 119860119863119861119878119863119862119860119863) (119860 minus 119861119863119863119862)

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119862119860119863119861119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863 (119863 minus 119878)119863119863119862

= 119860119863119860 + 119860119863119861119863119863119862 minus 119860119863119861119878119863119862119860119863119860minus 119860119863119861119878119863119863119863119863119862 + 119860119863119861119878119863119878119863119863119862 = 119860119863119860

(85)

by 119862119860120587 = 0 119863120587119862 = 0 and 119860119863119861119878120587 = 0 Thus 119866119883 = 119883119866Moreover

119883119866119883 = 119860119863119860(119860119863 + 119860119863119861119878119863119862119860119863)= (119860119863 + 119860119863119861119878119863119862119860119863) = 119883

119866119896119883119866 = 119866119896119860119863119860 = 119866119896minus1 (119860 minus 119861119863119863119862)119860119863119860= 119866119896minus1 (119860 minus 119861119863119863119862) = 119866119896

(86)

by using 119862119860120587 = 0 Therefore 119866119866119863 = 119860119860119863 Since 119860120587119861 = 0and 119862119860120587 = 0 we get 119866120587119861 = 0 119862119866120587 = 0 Now we can deducethat

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ] (87)

by 119860119863119861119878120587 = 0 119862119860120587 = 0 119860120587119861 = 0 and 119878120587119862119860119863 = 0 and

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (88)

by 119862119866120587 = 0 119861119863120587 = 0 119866120587119861 = 0 and119863120587119862 = 0 Then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(89)

follows by the uniqueness of the Drazin inverse(ii) Similarly we denote 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 In thesame way we can easily get 119878119884 = 119863119863119863 by 119862119860120587 = 0 119861119863120587 = 0119863120587119862119866119863 = 0 and 119884119878 = 119863119863119863 by 119860120587119861 = 0 119862119866120587 = 0 and119866120587119861119863119863 = 0 Since119884119878119884 = 119863119863119863(119863119863 + 119863119863119862119866119863119861119863119863)

= 119863119863 + 119863119863119862119866119863119861119863119863 = 119884119878119896119884119878 = 119878119896minus1 (119863 minus 119862119860119863119861)119863119863119863 = 119878119896minus1 (119863 minus 119862119860119863119861)

= 119878119896(90)

by using 119861119863120587 = 0 we obtain 119878119863 = 119863119863 + 119863119863119862119866119863119861119863119863 Thuswe have 119878119878119863 = 119863119863119863 which implies 119861119878120587 = 0 and 119878120587119862 = 0where we used 119861119863120587 = 0 and 119863120587119862 = 0 Now by 119861119878120587 = 0119878120587119862 = 0 119860120587119861 = 0 and 119862119860120587 = 0 we get

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863119878119863119862119860119863 119878119863 ] (91)

and by 119863120587119862 = 0 119861119863120587 = 0 119866120587119861119863119863 = 0 and119863119863119862119866120587 = 0 wehave

119872119863 = [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863] (92)

Therefore

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(93)

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

12 Mathematical Problems in Engineering

Corollary 23 (see [12]) Let119872 be defined by (3) and 119878 = 119863 minus119862119860119863119861 119866 = 119860 minus 119861119863119863119862 be the generalized Schur complementof 119860 in119872 and 119863 in119872 respectively If

119860120587119861 = 0119861119863120587 = 0119862119860120587 = 0119863120587119862 = 0119878120587 = 119863120587

(94)

then

119872119863 = [119860119863 + 119860119863119861119878119863119862119860119863 minus119860119863119861119878119863minus119878119863119862119860119863 119878119863 ]

= [ 119866119863 minus119866119863119861119863119863minus119863119863119862119866119863 119863119863 + 119863119863119862119866119863119861119863119863]

(95)

Proof According to the proof of Theorem 22(ii) we have119878119863 = 119884 = 119863119863 + 119863119863119862119866119863119861119863119863 which implies 119878119878119863 = 119863119863119863by 119878119884 = 119863119863119863 Now it is sufficient to conclude that (95)holds

4 Quotient Property and the FirstSylvester Identity in terms of theGeneralized Schur Complement

Crabtree and Haynsworth in [18] showed a quotient formulafor Schur complement of amatrix After that the formula wasextended by Ostrowski in 1971 [24] and in 1973 [19]

Let

119876 = [[[119860 119861 119864119862 119863 119865119866 119867 119869

]]]

119872 = [119860 119861119862 119863]

1198722 = [119861 119864119863 119865]

1198723 = [119863 119865119867 119869]

(96)

1198724 = [119862 119863119866 119867] (97)

If 119876 and119872 are square and nonsingular then

( 119876119872) = ( (119876119860)(119872119860))= ([ 119860 119864119866 119869 ]119860 ) minus ([ 119860 119861119866 119867 ]119860 )(119872119860 )minus1 ([ 119860 119864119862 119865 ]119860 )

(98)

This is the so-called quotient property which is proven byCrabtree and Haynsworth in [18]

The first Sylvester identity is that

( 1198761198723) = ((119876119863)(1198723119863))

= (119872119863 ) minus (1198722119863 )(1198723119863 )minus1 (1198724119863 ) (99)

where1198723 and119863 are nonsingularSheng and Chen [12] studied the quotient properties and

the first Sylvester identity with respect to MP group andDrazin inverses Cvetkovic-Ilic in [9] derived the propertiesunder conditions different than those used in [12]

In this section we will give the quotient properties andthe first Sylvester identity of MP group and Drazin inversesunder the conditions which are much simpler than those of[9 12] Notice that the proofs of the following results arealmost identical to the proofs of the corresponding resultsfrom [12] We will state the theorems without proofs

We first give the quotient properties based on MP groupand Drazin inverses For convenience we denote

(119872119860 )119901= 119863 minus 119862119860dagger119861

(119872119860 )119892= 119863 minus 119862119860119861

(119872119860 )119889= 119863 minus 119862119860119863119861

(119872119863 )119901= 119860 minus 119861119863dagger119862

(119872119863 )119892= 119860 minus 119861119863119862

(119872119863 )119889= 119860 minus 119861119863119863119862

(100)

Theorem 24 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Corollary 15 then

( 119876119872)119901= ( (119876119860)119901(119872119860)119901)119901 (101)

Theorem 25 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 11(i) or (ii) then

( 119876119872)119892= ( (119876119860)119892(119872119860)119892)119892 (102)

Theorem 26 Let119876 = [ 119860 119861 119864119862 119863 119865119866 119867 119869

] and119872 = [ 119860 119861119862 119863 ] If119872 satisfiesthe conditions of Theorem 18 then

( 119876119872)119889= ( (119876119860)119889(119872119860)119889)119889 (103)

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Mathematical Problems in Engineering 13

The first Sylvester identity with respect to MP group andDrazin inverses is expressed as follows

Theorem 27 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Corollary 15 then

( 1198761198723)119901 = ((119876119863)119901(1198723119863)119901)119901

= (119872119863 )119901minus (1198722119863 )

119901(1198723119863 )

119901

dagger (1198724119863 )119901

(104)

Theorem 28 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 11(i) or (ii) then

( 1198761198723)119892 = ((119876119863)119892(1198723119863)119892)119892

= (119872119863 )119892minus (1198722119863 )

119892(1198723119863 )

119892

(1198724119863 )119892

(105)

Theorem 29 Let11987611987211987221198723 and1198724 be the forms of (97)If1198723 satisfies the conditions of Theorem 18 then

( 1198761198723)119889 = ((119876119863)119889(1198723119863)119889)119889

= (119872119863 )119889minus (1198722119863 )

119889(1198723119863 )

119889

119863 (1198724119863 )119889

(106)

5 Applications of the GeneralizedSchur Complement

The Schur complements and generalized Schur complementswere shown playing an important role in various appliedmathematical settings such as statistics matrix theoryelectrical network theory discrete-time regulator problemsophisticated techniques and many other fields

In this section we will give an application of the gener-alized Schur complement in the solution of a linear systemUsing the generalized Schur complement we can split a largersystem into two small linear systems by the following steps

Let

119872119909 = 119910 (107)

be a linear system Applying the block Gaussian eliminationto the system we have

[119860 1198610 119863 minus 119862119860dagger119861][

11990911199092] = [1199101

1199102 minus 119862119860dagger1199101] (108)

Hence we get

1198601199091 + 1198611199092 = 11991011198781199092 = 1199102 minus 119862119860dagger1199101 (109)

that is

1198601199091 = 1199101 minus 11986111990921198781199092 = 1199102 minus 119862119860dagger1199101 (110)

Now the solution of system (107) can be obtained by thetwo small linear systems above In this case the operation canbe significantly simplifiedWewill also notice that theMoore-Penrose inverse of 119860 can be replaced by other generalizedinverses such as the group inverse the Drazin inverse andthe generalized inverse of119860 or even the ordinary inverse119860minus1

In the following we will give the group inverse solutionsof the linear system

Theorem 30 Let

119872119909 = 119910 (111)

be a linear system Suppose 119872 satisfies all the conditions ofTheorem 17 partitioning 119909 and 119910 as

119909 = [11990911199092]

119910 = [11991011199102] (112)

which have appropriate sizes with 119872 If 119910 isin 119877(119872) then thesolution 119909 = 119872119910 of linear system (111) can be expressed as

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (113)

where 119878 = 119863 minus 119862119860119861Proof Since 119910 isin 119877(119872) we conclude that 119909 = 119872119910 is thesolution of linear system (111) ByTheorem 17 we can get thefollowing

119909 = 119872119910 = [119860 + 119860119861119878119862119860 minus119860119861119878minus119878119862119860 119878 ][11991011199102]

= [1198601199101 + 1198601198611198781198621198601199101 minus 1198601198611198781199102119878 (1199102 minus 1198621198601199101) ] = [11990911199092]

(114)

Now it is easy to see the solution 119909 = 119872119910 can be expressedas

1199091 = 119860 (1199101 minus 1198611199092) 1199092 = 119878 (1199102 minus 1198621198601199101) (115)

which is also the group inverse solutions of the two smalllinear systems of (110) respectively

Competing Interests

The authors declare that they have no competing interests

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

14 Mathematical Problems in Engineering

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11361009) The third author is sup-ported by Grant no 174007 of the Ministry of EducationScience and Technological Development Serbia

References

[1] I Schur ldquoPotenzreihn in innern des heitskreisesrdquo Journal fur dieReine und Angewandte Mathematik vol 147 pp 205ndash232 1917

[2] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andIts Applications vol 27 pp 173ndash186 1979

[3] D Carlson ldquoWhat are Schur complements anywayrdquo LinearAlgebra and Its Applications vol 74 pp 257ndash275 1986

[4] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[5] C-K Li and R Mathias ldquoExtremal characterizations of theSchur complement and resulting inequalitiesrdquo SIAM Reviewvol 42 no 2 pp 233ndash246 2000

[6] G Corach A Maestripieri and D Stojanoff ldquoOblique projec-tions and Schur complementsrdquo Acta Scientiarum Mathemati-carum vol 67 pp 439ndash459 2001

[7] G Corach A Maestripieri and D Stojanoff ldquoGeneralizedSchur complements and oblique projectionsrdquo Linear Algebraand Its Applications vol 341 no 1ndash3 pp 259ndash272 2002

[8] R W Cottle ldquoManifestations of the Schur complementrdquo LinearAlgebra and its Applications vol 8 no 3 pp 189ndash211 1974

[9] D S Cvetkovic-Ilic ldquoExpression of the Drazin and MP-inverseof partitionedmatrix and quotient identity of generalized Schurcomplementrdquo Applied Mathematics and Computation vol 213no 1 pp 18ndash24 2009

[10] J K Baksalary and G P Styan ldquoGeneralized inverses of par-titioned matrices in Banachiewicz-Schur formrdquo Linear Algebraand Its Applications vol 354 pp 41ndash47 2002

[11] J Benıtez and N Thome ldquoThe generalized Schur complementin group inverses and (119896 + 1) -potent matricesrdquo Linear andMultilinear Algebra vol 54 no 6 pp 405ndash413 2006

[12] X Sheng and G Chen ldquoSome generalized inverses of partitionmatrix and quotient identity of generalized Schur complementrdquoApplied Mathematics and Computation vol 196 no 1 pp 174ndash184 2008

[13] S L Campbell and C D Meyer Generalized Inverse of LinearTransformations Pitman London UK 1979

[14] N Castro-Gonzalez EDopazo and J Robles ldquoFormulas for theDrazin inverse of special block matricesrdquo Applied Mathematicsand Computation vol 174 no 1 pp 252ndash270 2006

[15] D S Cvetkovic-Ilic ldquoA note on the representation for theDrazin inverse of 2 times 2 block matricesrdquo Linear Algebra and itsApplications vol 429 no 1 pp 242ndash248 2008

[16] D S Cvetkovic-Ilic J Chen and Z Xu ldquoExplicit represen-tations of the Drazin inverse of block matrix and modifiedmatrixrdquo Linear and Multilinear Algebra vol 57 no 4 pp 355ndash364 2009

[17] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[18] D E Crabtree and E V Haynsworth ldquoAn identity for theSchur complement of a matrixrdquo Proceedings of the AmericanMathematical Society vol 22 pp 364ndash366 1969

[19] A M Ostrowski ldquoOn Schurrsquos complementrdquo Journal of Combi-natorial Theory Series A vol 14 pp 319ndash323 1973

[20] R E Hartwig and J M Shoaf ldquoGroup inverses and Drazininverses of bidiagonal and triangular Toeqlitz matricesrdquo Journalof the Australian Mathematical Society vol 24 no 1 pp 10ndash141977

[21] D S Cvetkovic-Ilic and B Zheng ldquoWeighted generalizedinverses of partitioned matrices in Banachiewicz-Schur formrdquoJournal of Applied Mathematics amp Computing vol 22 no 3 pp175ndash184 2006

[22] J Miao ldquoGeneral expressions for the Moore-Penrose inverse ofa 2 times 2 block matrixrdquo Linear Algebra and its Applications vol151 pp 1ndash15 1991

[23] Y Wei ldquoExpression for the Drazin inverse of a 2 times 2 blockmatrixrdquo Linear andMultilinear Algebra vol 45 no 2-3 pp 131ndash146 1998

[24] A Ostrowski ldquoA new proof of Haynswortus quotient formulafor Schur complementrdquo Linear Algebra and its Applications vol4 no 4 pp 389ndash392 1971

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article Representations of Generalized Inverses ...downloads.hindawi.com/journals/mpe/2016/9236281.pdf · e Schur complements and generalized Schur comple-ments were studied

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


INVERSE OFA ITS APPLICATIONSAsystematic development of the calculus of generalized inverses and their applicationsaregivenin aforthcomingbookbytheauthors,entitled Generalized Inverse

INVERSE OFA ITS APPLICATIONSAsystematic development of the calculus of generalized inverses and their applicationsaregivenin aforthcomingbookbytheauthors,entitled Generalized Inverse

Editors dacteurs en chef Jonathan Borweinv1ranick/papers/bengrev.pdftreat generalized inverses of finite matrices, while the eighth introduces gen-eralized inverses of operators between

Editors dacteurs en chef Jonathan Borweinv1ranick/papers/bengrev.pdftreat generalized inverses of finite matrices, while the eighth introduces gen-eralized inverses of operators between

Generalized inverses and solutions of linear systems · Val. 72B, No.4, October- December 1968 Generalized Inverses and Solutions of Linear Systems * John Z. Hearon** (September 28,

Generalized inverses and solutions of linear systems · Val. 72B, No.4, October- December 1968 Generalized Inverses and Solutions of Linear Systems * John Z. Hearon** (September 28,

STRUCTURE ESTIMATION FOR DISCRETE GRAPHICAL …wainwrig/Papers/LohWai13b.pdf · STRUCTURE ESTIMATION FOR DISCRETE GRAPHICAL MODELS: GENERALIZED COVARIANCE MATRICES AND THEIR INVERSES

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