Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The...

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Research Article The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs Xiaogen Chen School of Information Science and Technology, Zhanjiang Normal University, Zhanjiang, Guangdong 524048, China Correspondence should be addressed to Xiaogen Chen; [email protected] Received 26 February 2014; Revised 5 May 2014; Accepted 6 May 2014; Published 14 May 2014 Academic Editor: Chein-Shan Liu Copyright © 2014 Xiaogen Chen. 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. Let ≥1 be an integer, let B 2 denote the Brualdi-Li matrix of order 2, and let LB 2 denote the Laplacian matrices of Brualdi- Li tournament digraphs. We obtain the eigenvalues and eigenvectors of LB 2 . 1. Introduction e Laplacian spectral theory is currently not only a hot research direction of spectral graph theory but also one of the research directions of combined matrix theory. e Laplacian spectrum of a graph is of importance in graph theory, matrix theory, and the definite solution of partial differential equa- tions. It also has applications in quantum chemistry, biology, and complex network. erefore, it has important theoretical and practical values to study the Laplacian eigenvalues and eigenvectors of graphs. e Laplacian spectrum of graph has attracted the attention of researchers; see [1–5] and so on. We follow [1, 6] for terminology and notations. Let be a digraph of order with vertex set () and arc set (), where () = {V 1 , V 2 ,..., V }. e adjacency matrix of is the (0, 1) matrix () = ( ) of order , where =1 if there is an arc from V to V and =0 otherwise. e digraph is called the associated digraph of matrix (). Let () = diag( 1 , 2 ,..., ), the diagonal matrix with vertex outdegrees 1 , 2 ,..., of V 1 , V 2 ,..., V . () = () − () is called the Laplacian matrix of the digraph . e characteristic polynomial of the adjacency matrix (), that is, (, ) = ((), ) = det( − ()), is called the characteristic polynomial of the digraph . e equation det( − ()) = 0 has complex roots and these roots are called the eigenvalues of (). Suppose the distinct eigenvalues of () are denoted by 1 , 2 ,..., ∈ C with corresponding algebraic multiplicities 1 , 2 ,..., , where is a nonnegative integer = 1, 2, . . . , . = ( 1 2 ⋅⋅⋅ 1 2 ⋅⋅⋅ ) is called the Laplacian spectrum of digraph . e Laplacian spectral radius of is the largest modulus of an eigenvalue of (), denoted by (()). e symbol C will denote the complex field. Let ∈ C be an eigenvalue of matrix . ere is vector ̸ =0 satisfying = , and then is called the eigenvectors of matrix corresponding to . A tournament matrix of order is a (0, 1) matrix satisfying the equation + = − , where is the all ones matrix, is the identity matrix, and is the transpose of . Let B 2 =( + ), (1) where is strictly upper triangular tournament matrix (all of whose entries above the main diagonal are equal to one); that is, =( 0 1 1 ⋅⋅⋅ 1 0 0 1 ⋅⋅⋅ 1 . . . ddd . . . 0 0 ⋅⋅⋅ 0 1 0 0 ⋅⋅⋅ 0 0 ) × (2) is the tournament matrix of order 2. e matrix B 2 has been dubbed by the Brualdi-Li matrix. e associated digraph of matrix B 2 is called the Brualdi-Li tournament digraph. In 1983 Brualdi and Li Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 940834, 7 pages http://dx.doi.org/10.1155/2014/940834

Transcript of Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The...

Page 1: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

Research ArticleThe Spectrum and Eigenvectors of the Laplacian Matrices ofthe Brualdi-Li Tournament Digraphs

Xiaogen Chen

School of Information Science and Technology Zhanjiang Normal University Zhanjiang Guangdong 524048 China

Correspondence should be addressed to Xiaogen Chen oamc163com

Received 26 February 2014 Revised 5 May 2014 Accepted 6 May 2014 Published 14 May 2014

Academic Editor Chein-Shan Liu

Copyright copy 2014 Xiaogen ChenThis is an open access article distributed under theCreative CommonsAttribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Let 119898 ge 1 be an integer letB2119898

denote the Brualdi-Li matrix of order 2119898 and letLB2119898

denote the Laplacianmatrices of Brualdi-Li tournament digraphs We obtain the eigenvalues and eigenvectors ofLB

2119898

1 Introduction

The Laplacian spectral theory is currently not only a hotresearch direction of spectral graph theory but also one of theresearch directions of combinedmatrix theoryThe Laplacianspectrum of a graph is of importance in graph theory matrixtheory and the definite solution of partial differential equa-tions It also has applications in quantum chemistry biologyand complex networkTherefore it has important theoreticaland practical values to study the Laplacian eigenvalues andeigenvectors of graphs The Laplacian spectrum of graph hasattracted the attention of researchers see [1ndash5] and so on Wefollow [1 6] for terminology and notations

Let 119866 be a digraph of order 119899 with vertex set 119881(119866) andarc set 119864(119866) where 119881(119866) = V

1 V2 V

119899 The adjacency

matrix of 119866 is the (0 1) matrix 119860(119866) = (119886119894119895

) of order 119899where 119886

119894119895= 1 if there is an arc from V

119894to V119895and 119886

119894119895= 0

otherwise The digraph 119866 is called the associated digraph ofmatrix 119860(119866) Let 119863(119866) = diag(119889

1 1198892 119889

119899) the diagonal

matrix with vertex outdegrees 1198891 1198892 119889

119899of V1 V2 V

119899

119871(119866) = 119863(119866) minus 119860(119866) is called the Laplacian matrix of thedigraph 119866 The characteristic polynomial of the adjacencymatrix 119860(119866) that is 119875(119866 120582) = 119875(119860(119866) 120582) = det(120582119868 minus 119860(119866))is called the characteristic polynomial of the digraph 119866 Theequation det(120582119868 minus 119860(119866)) = 0 has 119899 complex roots andthese roots are called the eigenvalues of 119860(119866) Suppose thedistinct eigenvalues of 119871(119866) are denoted by 120582

1 1205822 120582

119903isin C

with corresponding algebraic multiplicities 1205831205821

1205831205822

120583120582119903

where 120583

120582119894

is a nonnegative integer 119894 = 1 2 119903

119871119878 = (12058211205822sdotsdotsdot 120582119903

12058312058211205831205822sdotsdotsdot 120583120582119903

) is called the Laplacian spectrum ofdigraph 119866 The Laplacian spectral radius of 119866 is the largestmodulus of an eigenvalue of 119871(119866) denoted by 120588(119871(119866)) Thesymbol C will denote the complex field Let 120582 isin C bean eigenvalue of matrix 119860 There is vector 119909 = 0 satisfying119860119909 = 120582119909 and then 119909 is called the eigenvectors of matrix 119860

corresponding to 120582A tournament matrix of order 119899 is a (0 1) matrix 119879

119899

satisfying the equation 119879119899

+ 119879119905

119899= 119869119899

minus 119868119899 where 119869

119899is the all

ones matrix 119868119899is the identity matrix and 119879

119905

119899is the transpose

of 119879119899 Let

B2119898

= (

119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (1)

where 119880119898is strictly upper triangular tournament matrix (all

of whose entries above the main diagonal are equal to one)that is

119880119898

= (

0 1 1 sdot sdot sdot 1

0 0 1 sdot sdot sdot 1

d d d

0 0 sdot sdot sdot 0 1

0 0 sdot sdot sdot 0 0

)

119898times119898

(2)

is the tournament matrix of order 2119898The matrix B

2119898has been dubbed by the Brualdi-Li

matrix The associated digraph of matrix B2119898

is calledthe Brualdi-Li tournament digraph In 1983 Brualdi and Li

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 940834 7 pageshttpdxdoiorg1011552014940834

2 Journal of Applied Mathematics

conjectured that themaximal spectral radius for tournamentsof order 2119898 is attained by the Brualdi-Li matrix [7] Thisconjecture has been confirmed in [8] The properties of theBrualdi-Li matrix have been investigated in [9ndash14]

In this paper we obtain the spectrum and eigenvectorsof the Laplacian matrices of the Brualdi-Li tournamentdigraphs

Theorem 1 Let 119898 ge 1 be an integer and let SLB2119898

bethe Laplacian spectrum of the Brualdi-Li tournament digraphThen

SLB2119898

= (

0 119898 120582119896

120582119896

1 lfloor

119898 minus 1

2

rfloor + lfloor

119898 + 1

2

rfloor 1 1

) (3)

where lfloor119886rfloor is the floor of number 119886 120582119896

= 119898 minus 119894 cot ((120587 +

2119896120587)2119898) and 120582119896is the conjugate complex number of 120582

119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

Theorem 2 Let 119898 ge 1 be an integer and let 120585 = (V119908 ) = 0

be the eigenvector of LB2119898

corresponding to 120582 where V =

(V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905 then

(1) if 120582 = 119898 then120585119898

= (0

1) is the eigenvector of LB

2corresponding to

120582 = 119898 = 1120585119898

= (1 minus1 1 minus1)119905 is the eigenvector of LB

4

corresponding to 120582 = 119898 = 2120585119898

= sum119898

119896=2119897119896120585119898119896

is the eigenvector of LB2119898

corre-sponding to 120582 = 119898 gt 2where 119897

119896is an arbitrary constant 119896 = 2 3 119898

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(4)

(2) if 120582 = 119898 then

V119896

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

)

119908119896

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

)

119896 = 1 2 sdot sdot sdot 119898

(5)

Corollary 3 Let 119898 ge 1 be an integer and let 120588(LB2119898

)

be the Laplacian spectral radius of the Brualdi-Li tournamentdigraph Then

120588 (LB2119898

) = radic1198982

+ cot2 120587

2119898

(6)

Proof ByTheorem 1

120588 (LB2119898

) = max0le119896lelfloor1198982rfloorminus1

0 119898

10038161003816100381610038161003816100381610038161003816

119898 plusmn 119894cot120587 + 2119896120587

2119898

10038161003816100381610038161003816100381610038161003816

= radic1198982

+ cot2 120587

2119898

(7)

Corollary 4 Let 119898 ge 1 be an integer then

(1) if 119898 gt 1 is odd thenLB2119898

is not diagonalizable

(2) if 119898 is even or 119898 = 1 thenLB2119898

is diagonalizable

2 Some Lemmas

Fundamental Theorem of Algebra Every nonzero single-variable degree 119899 polynomial with complex coefficients hascounted with multiplicity exactly 119899 rootsComplex Conjugate RootTheorem If 119875 is a polynomial in onevariable with real coefficients and 119886 + 119887119894 is a root of 119875 with 119886

and 119887 real numbers then its complex conjugate 119886 minus 119887119894 is alsoa root of 119875 where 119894

2= minus1

The symbol LB2119898

denotes the Laplacian matrix ofthe Brualdi-Li tournament digraph Clearly LB

2119898=

((119898minus1)119868

119898minus119880119898minus119880119905

119898

minus119868119898minus119880119905

119898119898119868119898minus119880119898

)

Lemma 5 Let 119898 gt 1 be an integer and 119883 = (1 119909 1199092

119909119898minus1

)119905 where 119909 = 1 is real variable Then

(1) 119883119905119880119898

= minus

1

1 minus 119909

119883119905

+

1

1 minus 119909

1119905119898

(2) 119883119905119880119905

119898=

119909

1 minus 119909

119883119905

minus

119909119898

1 minus 119909

1119905119898

(8)

Proof Consider

(1) 119883119905119880119898

= (1 119909 1199092 119909

119898minus1) 119880119898

= (0

0

sum

119896=0

119909119896

1

sum

119896=0

119909119896

2

sum

119896=0

119909119896

3

sum

119896=0

119909119896

119898minus3

sum

119896=0

119909119896

119898minus2

sum

119896=0

119909119896)

= (

1 minus 1

1 minus 119909

1 minus 119909

1 minus 119909

1 minus 1199092

1 minus 119909

1 minus 1199093

1 minus 119909

1 minus 119909119898minus2

1 minus 119909

1 minus 119909119898minus1

1 minus 119909

)

= minus

1

1 minus 119909

119883119905

+

1

1 minus 119909

1119905119898

Journal of Applied Mathematics 3

(2) 119883119905119880119905

119898

= (1 119909 1199092 119909

119898minus1) 119880119905

119898

= (

119898minus1

sum

119896=1

119909119896

119898minus1

sum

119896=2

119909119896

119898minus1

sum

119896=119898minus2

119909119896

119898minus1

sum

119896=119898minus1

119909119896 0)

= (

119909 minus 119909119898

1 minus 119909

1199092

minus 119909119898

1 minus 119909

119909119898minus2

minus 119909119898

1 minus 119909

119909119898minus1

minus 119909119898

1 minus 119909

119909119898

minus 119909119898

1 minus 119909

)

=

119909

1 minus 119909

119883119905

minus

119909119898

1 minus 119909

1119905119898

(9)

Lemma6 Let119898 gt 1 be an integer120582 = 119898 let120582 isin C be an arbi-trary eigenvalue ofLB

2119898 and let 120585 = (

V119908 ) = 0 be the eigenvec-

tor of LB2119898

corresponding to 120582 where V = (V1 V2 V

119898)119905

119908 = (1199081 1199082 119908

119898)119905 Let 119883 = (1 119909 119909

2 119909

119898minus1)119905 119891(119909) =

sum119898

119896=1V119896119909119896minus1 and 119892(119909) = sum

119898

119896=1119908119896119909119896minus1 where 119909 is real variable

Then

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(10)

where 119886 = 1119905119898V 119887 = 1119905

119898119908

Proof If 120582( = 119898) is an eigenvalue with eigenvector 120585 = (V119908 )

ofLB2119898 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (11)

Therefore

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119880119905

119898119908 = 120582V

(minus119868119898

minus 119880119905

119898) V + (119898119868

119898minus 119880119898

) 119908 = 120582119908

(12)

We have

119883119905

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119883119905119880119905

119898119908 = 120582119883

119905V

119883119905

(minus119868119898

minus 119880119905

119898) V + 119883

119905(119898119868119898

minus 119880119898

) 119908 = 120582119883119905119908

(13)

According to Lemma 5 we have

((119898 minus 1 minus 120582) (1 minus 119909) + 1) 119891 (119909) minus 119909119892 (119909) = 119886 minus 119887119909119898

minus119891 (119909) + ((119898 minus 120582) (1 minus 119909) + 1) 119892 (119909) = 119887 minus 119886119909119898

(14)

Notice that this equation holds for 119909 = 1 too Consider

119863 = det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 minus119909

minus1 (119898 minus 120582) (1 minus 119909) + 1)

= ((119898 minus 1 minus 120582) (1 minus 119909) + 1) ((119898 minus 120582) (1 minus 119909) + 1) minus 119909

= (119898 minus 120582) (1 minus 119909) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909)

119863119891

= det(

119886 minus 119887119909119898

minus119909

119887 minus 119886119909119898

(119898 minus 120582) (1 minus 119909) + 1)

= (119886 minus 119887119909119898

) ((119898 minus 120582) (1 minus 119909) + 1) + 119909 (119887 minus 119886119909119898

)

= 119886 (119898 + 1 minus 120582) + (119887 minus 119886 (119898 minus 120582)) 119909

minus 119887 (119898 + 1 minus 120582) 119909119898

+ (119887 (119898 minus 120582) minus 119886) 119909119898+1

= (1 minus 119909) (119886 (119898 + 1 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

119863119892

= det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 119886 minus 119887119909119898

minus1 119887 minus 119886119909119898)

= (119887 minus 119886119909119898

) ((119898 minus 1 minus 120582) (1 minus 119909) + 1) + (119886 minus 119887119909119898

)

= 119886 + 119887 (119898 minus 120582) minus 119887 (119898 minus 1 minus 120582) 119909

minus (119887 + 119886 (119898 minus 120582)) 119909119898

+ 119886 (119898 minus 1 minus 120582) 119909119898+1

= (1 minus 119909) (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

(15)

By Cramerrsquos rule

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(16)

We are done

4 Journal of Applied Mathematics

Lemma 7 Under the assumptions and in the notation ofLemma 6

119886 + 119887 = 1119905119898V + 1119905119898

119908 = 0 119886 = 1119905119898V = 0 119887 = 1119905

119898119908 = 0

(17)

Proof In Lemma 6(1) by setting 119909 = 1 we have

119886 = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + (119886 + 119887) (119898 minus 1) + (119886 minus 119887 (119898 minus 120582))

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

(18)

Since 120582 = 119898 it follows that

(119886 + 119887) (1 minus 120582) = 2119886 (19)

As we all know that the eigenvalues of a real skew-symmetricmatrix are all pure imaginary or nonzero hence det((1 minus

119898)119868119898

+ 119880 minus 119880119905) = 0 where 119898 gt 1 Since

det (1119868119898

minus LB2119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119880119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

(2 minus 119898) 119868119898

+ 119880119898

+ 119880119905

119898

119868119898

+ 119880119905

119898(2 minus 119898) 119868

119898+ 119880119898

+ 119880119905

119898

)

= de119905 (

(1 minus 119898) 119868119898

+ 119880 minus 119880119905

0

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119869119898

)

= det ((1 minus 119898) 119868119898

+ 119880 minus 119880119905) det ((1 minus 119898) 119868

119898+ 119869119898)

= (1 minus 119898)119898minus1 det ((1 minus 119898) 119868

119898+ 119880 minus 119880

119905) = 0

(20)

hence 120582 = 1 Notice that (119886 + 119887)(1 minus 120582) = 2119886 and if 119886 + 119887 = 0then 119886 = 0 and 119887 = (119886 + 119887) minus 119886 = 0 By Lemma 6

119891 (119909) =

119898

sum

119896=1

V119896119909119896minus1

equiv 0

119892 (119909) =

119898

sum

119896=1

119908119896119909119896minus1

equiv 0

(21)

for arbitrary real variable 119909 It is not possible Hence 119886 + 119887 = 0It is easy to see that 119886 = 0 and 119887 = 0

Lemma 8 (see [15]) If A is a square matrix then

(1) for every eigenvalue of A the geometric multiplicity isless than or equal to the algebraic multiplicity

(2) A is diagonalizable if and only if the geometric mul-tiplicity of every eigenvalue is equal to the algebraicmultiplicity

3 Proof of Theorem 1

Let 120582 isin C be an arbitrary eigenvalue of LB2119898 and let

120585 = (V119908 ) = 0 be the eigenvector of LB

2119898corresponding to

120582 where V = (V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905

For 119898 = 1 then B2

= (0 0

1 0) LB

2= (0 0

minus1 1) Obviously

1205821

= 0 1205822

= 1 are eigenvalues of LB2and 120583

0= 1205831

= 1Theorem 1 holds

For 119898 gt 1 using the previous assumptions and notationobviously 119875(LB

2119898 0) = det(0119868

2119898minus LB

2119898) = 0 by

definition 120582 = 0 is an eigenvalue ofLB2119898 1205830

ge 1Note that 119875(LB

2119898 119898) = det(119898119868

2119898minus LB

2119898) = 0 and

by definition 120582 = 119898 is an eigenvalue of LB2119898 By simple

calculation the rank ofmatrix1198981198682119898

minusLB2119898

is equal to119898+1By Lemma 8(1) 120583

119898ge 2119898 minus (119898 + 1) = 119898 minus 1

For 120582 = 119898 by Lemma 7 we set 119886 = 1119905119898V 119886 + 119887 = 1119905

119898V +

1119905119898

119908 = 1 hence 119887 = 1 minus 1119905119898V = 1 minus 119886 In Lemma 6(1) by

setting 119909 = 1 we have

119886 = 1119905119898V = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + 119898 minus 1 + 119886 + (119886 minus 1) (119898 minus 120582)

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

=

(119898 minus 120582) (2119886 minus 1) + 119898 + 2119886 minus 1

2 (119898 minus 120582)

(22)

Hence

119886 =

1 minus 120582

2

(23)

Denoting 1198910

= (119898 + 1 minus 120582)119886 1198911

= 1198912

= sdot sdot sdot = 119891119898minus1

= 1 119891119898

=

119886 + (119886 minus 1)(119898 minus 120582) 119888 = (119898 minus 120582)(119898 + 1 minus 120582) and 119889 = (119898 minus 1 minus

120582)(119898 + 1 minus 120582) we have

119891 (119909)

=

119898

sum

119896=1

V119896119909119896minus1

= ((119898 + 1 minus 120582) 119886 + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

+ (119886 + (119886 minus 1) (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119891119895119889119896minus119895

) 119909119896

(24)

It must be that

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

119896 = 1 2 119898 (25)

Therefore

119886 =

119898

sum

119896=1

V119896

=

1

119888

119898

sum

119896=1

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

119896minus1

sum

119895=1

119889119896minus1minus119895

)

Journal of Applied Mathematics 5

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

119888(1 minus 119889)2

times (1198910

(1 minus 119889119898

) (1 minus 119889) + 119898 (1 minus 119889) minus 1 + 119889119898

)

=

1

119888(1 minus 119889)2

times ((1 minus 1198910

(1 minus 119889)) 119889119898

+ (1198910

+ 119898) (1 minus 119889) minus 1)

(26)

That is 119886119888(1 minus 119889)2

= (1 minus 1198910(1 minus 119889))119889

119898+ (1198910

+ 119898)(1 minus 119889) minus 1Since 119886 = (1minus120582)2119891

0= (119898+1minus120582)119886 119888 = (119898minus120582)(119898+1minus120582)

and 119889 = (119898 minus 1 minus 120582)(119898 + 1 minus 120582) then

(1 minus 1198910

(1 minus 119889)) 119889119898

= 119886119888(1 minus 119889)2

minus (1198910

+ 119898) (1 minus 119889) + 1

(1 minus

2119886 (119898 + 1 minus 120582)

119898 + 1 minus 120582

) 119889119898

=

4119886 (119898 minus 120582)

119898 + 1 minus 120582

minus

2 ((119898 + 1 minus 120582) 119886 + 119898)

119898 + 1 minus 120582

+ 1

120582119889119898

= minus120582

(27)

Denote 120579119896

= (120587 + 2119896120587)2119898 119896 = 0 1 2 119898 minus 1 As 120582 = 0 120582 minus

119898 = 0 we have 119889 = 1198902119894120579119896 that is (119898minus1minus120582)(119898+1minus120582) = 119890

2119894120579119896

where 1198942

= minus1Therefore

120582 = 120582119896

= 119898 minus

1 + 1198902119894120579119896

1 minus 1198902119894120579119896

= 119898 minus

119894 sin 2120579119896

1 minus cos 2120579119896

= 119898 minus 119894cot120579119896

(28)

where 1 minus 1198902119894120579119896

= 0 119896 = 0 1 2 119898 minus 1Note that if 119898 is odd by 119889

119898= ((119898minus1minus120582)(119898+1minus120582))

119898=

minus1 then (119898 minus 1 minus 120582)(119898 + 1 minus 120582) = minus1 hence 120582 = 119898 It is aneigenvalue ofLB

2119898

Furthermore 120583120582119896

ge 1 Note that 120582119896

= 120582119898minus1minus119896

119896 =

0 1 2 lfloor1198982rfloor minus 1To sum up by fundamental theorem of algebra and

complex conjugate root theorem for 119898 ge 1 we obtain thefollowing conclusions

If 119898 is odd then 1205830

= 120583120582119896

= 120583120582119896

= 1 and 120583119898

= 119898 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorIf119898 is even then120583

0= 120583120582119896

= 120583120582119896

= 1 and120583119898

= 119898minus1 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorConsider 119896 = 0 1 2 lfloor1198982rfloor minus 1 We complete theproof of Theorem 1

4 Proofs of Theorem 2 and Corollary 4

It is easy to see that the distinct eigenvalues of LB2119898

are0 120582119896

= 119898 minus 119894cot((120587 + 2119896120587)2119898) 120582119896 119898 with corresponding

algebraic multiplicities 1 1 1 lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloor 119896 =

1 2 lfloor1198982rfloor minus 1For 120582 = 119898 = 1LB

2= (0 0

minus1 1) and obviously 120585

119898= (0

1)

is the eigenvector ofLB2corresponding to 120582 = 119898 = 1

For 120582 = 119898 = 2

LB4

= (

1 minus1 0 0

0 1 minus1 0

minus1 0 2 minus1

minus1 minus1 0 2

) (29)

By simple calculation 120585119898

= (1 minus1 1 minus1)119905 is the eigenvec-

tor ofLB4corresponding to 120582 = 119898 = 2

For 120582 = 119898 gt 2 let 120585119898

= (V119908 ) = 0 be the eigenvector

of LB2119898

corresponding to 119898 and then LB2119898

120585119898

= 119898120585119898

expands to

(

119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (

V119908

) = 0 (30)

Equation (30) is equivalent to the following equation

(

119868119898

(119868 + 119880119898

)minus1

119880119905

119898

0119898

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898

) (

V119908

) = 0 (31)

By simple calculation

(119868 + 119880119898

)minus1

119880119905

119898= (

minus119868119898minus1

0

1119905119898minus1

0)

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898= (

119869119898minus1

1119898minus1

0 0)

(32)

We obtain the solution of (30) as follows

120585119898

=

119898

sum

119896=2

119897119896120585119898119896

(33)

where 119897119896is an arbitrary constant 119896 = 2 3 119898 and

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(34)

For 120582 = 119898 let 120585 = (V119908 ) = 0 be the eigenvector of LB

2119898

corresponding to 120582( = 119898) where V = (V1 V2 V

119898)119905 119908 =

(1199081 1199082 119908

119898)119905 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (35)

By Lemma 7 1119905119898V + 1119905119898

119908 = 1 we put 1119905119898V + 1119905119898

119908 = 1 ByTheorem 1 119886 = 1119905

119898V = (1 minus 120582)2 119887 = 1119905

119898119908 = 1 minus 119886

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

2 Journal of Applied Mathematics

conjectured that themaximal spectral radius for tournamentsof order 2119898 is attained by the Brualdi-Li matrix [7] Thisconjecture has been confirmed in [8] The properties of theBrualdi-Li matrix have been investigated in [9ndash14]

In this paper we obtain the spectrum and eigenvectorsof the Laplacian matrices of the Brualdi-Li tournamentdigraphs

Theorem 1 Let 119898 ge 1 be an integer and let SLB2119898

bethe Laplacian spectrum of the Brualdi-Li tournament digraphThen

SLB2119898

= (

0 119898 120582119896

120582119896

1 lfloor

119898 minus 1

2

rfloor + lfloor

119898 + 1

2

rfloor 1 1

) (3)

where lfloor119886rfloor is the floor of number 119886 120582119896

= 119898 minus 119894 cot ((120587 +

2119896120587)2119898) and 120582119896is the conjugate complex number of 120582

119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

Theorem 2 Let 119898 ge 1 be an integer and let 120585 = (V119908 ) = 0

be the eigenvector of LB2119898

corresponding to 120582 where V =

(V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905 then

(1) if 120582 = 119898 then120585119898

= (0

1) is the eigenvector of LB

2corresponding to

120582 = 119898 = 1120585119898

= (1 minus1 1 minus1)119905 is the eigenvector of LB

4

corresponding to 120582 = 119898 = 2120585119898

= sum119898

119896=2119897119896120585119898119896

is the eigenvector of LB2119898

corre-sponding to 120582 = 119898 gt 2where 119897

119896is an arbitrary constant 119896 = 2 3 119898

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(4)

(2) if 120582 = 119898 then

V119896

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

)

119908119896

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

)

119896 = 1 2 sdot sdot sdot 119898

(5)

Corollary 3 Let 119898 ge 1 be an integer and let 120588(LB2119898

)

be the Laplacian spectral radius of the Brualdi-Li tournamentdigraph Then

120588 (LB2119898

) = radic1198982

+ cot2 120587

2119898

(6)

Proof ByTheorem 1

120588 (LB2119898

) = max0le119896lelfloor1198982rfloorminus1

0 119898

10038161003816100381610038161003816100381610038161003816

119898 plusmn 119894cot120587 + 2119896120587

2119898

10038161003816100381610038161003816100381610038161003816

= radic1198982

+ cot2 120587

2119898

(7)

Corollary 4 Let 119898 ge 1 be an integer then

(1) if 119898 gt 1 is odd thenLB2119898

is not diagonalizable

(2) if 119898 is even or 119898 = 1 thenLB2119898

is diagonalizable

2 Some Lemmas

Fundamental Theorem of Algebra Every nonzero single-variable degree 119899 polynomial with complex coefficients hascounted with multiplicity exactly 119899 rootsComplex Conjugate RootTheorem If 119875 is a polynomial in onevariable with real coefficients and 119886 + 119887119894 is a root of 119875 with 119886

and 119887 real numbers then its complex conjugate 119886 minus 119887119894 is alsoa root of 119875 where 119894

2= minus1

The symbol LB2119898

denotes the Laplacian matrix ofthe Brualdi-Li tournament digraph Clearly LB

2119898=

((119898minus1)119868

119898minus119880119898minus119880119905

119898

minus119868119898minus119880119905

119898119898119868119898minus119880119898

)

Lemma 5 Let 119898 gt 1 be an integer and 119883 = (1 119909 1199092

119909119898minus1

)119905 where 119909 = 1 is real variable Then

(1) 119883119905119880119898

= minus

1

1 minus 119909

119883119905

+

1

1 minus 119909

1119905119898

(2) 119883119905119880119905

119898=

119909

1 minus 119909

119883119905

minus

119909119898

1 minus 119909

1119905119898

(8)

Proof Consider

(1) 119883119905119880119898

= (1 119909 1199092 119909

119898minus1) 119880119898

= (0

0

sum

119896=0

119909119896

1

sum

119896=0

119909119896

2

sum

119896=0

119909119896

3

sum

119896=0

119909119896

119898minus3

sum

119896=0

119909119896

119898minus2

sum

119896=0

119909119896)

= (

1 minus 1

1 minus 119909

1 minus 119909

1 minus 119909

1 minus 1199092

1 minus 119909

1 minus 1199093

1 minus 119909

1 minus 119909119898minus2

1 minus 119909

1 minus 119909119898minus1

1 minus 119909

)

= minus

1

1 minus 119909

119883119905

+

1

1 minus 119909

1119905119898

Journal of Applied Mathematics 3

(2) 119883119905119880119905

119898

= (1 119909 1199092 119909

119898minus1) 119880119905

119898

= (

119898minus1

sum

119896=1

119909119896

119898minus1

sum

119896=2

119909119896

119898minus1

sum

119896=119898minus2

119909119896

119898minus1

sum

119896=119898minus1

119909119896 0)

= (

119909 minus 119909119898

1 minus 119909

1199092

minus 119909119898

1 minus 119909

119909119898minus2

minus 119909119898

1 minus 119909

119909119898minus1

minus 119909119898

1 minus 119909

119909119898

minus 119909119898

1 minus 119909

)

=

119909

1 minus 119909

119883119905

minus

119909119898

1 minus 119909

1119905119898

(9)

Lemma6 Let119898 gt 1 be an integer120582 = 119898 let120582 isin C be an arbi-trary eigenvalue ofLB

2119898 and let 120585 = (

V119908 ) = 0 be the eigenvec-

tor of LB2119898

corresponding to 120582 where V = (V1 V2 V

119898)119905

119908 = (1199081 1199082 119908

119898)119905 Let 119883 = (1 119909 119909

2 119909

119898minus1)119905 119891(119909) =

sum119898

119896=1V119896119909119896minus1 and 119892(119909) = sum

119898

119896=1119908119896119909119896minus1 where 119909 is real variable

Then

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(10)

where 119886 = 1119905119898V 119887 = 1119905

119898119908

Proof If 120582( = 119898) is an eigenvalue with eigenvector 120585 = (V119908 )

ofLB2119898 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (11)

Therefore

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119880119905

119898119908 = 120582V

(minus119868119898

minus 119880119905

119898) V + (119898119868

119898minus 119880119898

) 119908 = 120582119908

(12)

We have

119883119905

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119883119905119880119905

119898119908 = 120582119883

119905V

119883119905

(minus119868119898

minus 119880119905

119898) V + 119883

119905(119898119868119898

minus 119880119898

) 119908 = 120582119883119905119908

(13)

According to Lemma 5 we have

((119898 minus 1 minus 120582) (1 minus 119909) + 1) 119891 (119909) minus 119909119892 (119909) = 119886 minus 119887119909119898

minus119891 (119909) + ((119898 minus 120582) (1 minus 119909) + 1) 119892 (119909) = 119887 minus 119886119909119898

(14)

Notice that this equation holds for 119909 = 1 too Consider

119863 = det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 minus119909

minus1 (119898 minus 120582) (1 minus 119909) + 1)

= ((119898 minus 1 minus 120582) (1 minus 119909) + 1) ((119898 minus 120582) (1 minus 119909) + 1) minus 119909

= (119898 minus 120582) (1 minus 119909) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909)

119863119891

= det(

119886 minus 119887119909119898

minus119909

119887 minus 119886119909119898

(119898 minus 120582) (1 minus 119909) + 1)

= (119886 minus 119887119909119898

) ((119898 minus 120582) (1 minus 119909) + 1) + 119909 (119887 minus 119886119909119898

)

= 119886 (119898 + 1 minus 120582) + (119887 minus 119886 (119898 minus 120582)) 119909

minus 119887 (119898 + 1 minus 120582) 119909119898

+ (119887 (119898 minus 120582) minus 119886) 119909119898+1

= (1 minus 119909) (119886 (119898 + 1 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

119863119892

= det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 119886 minus 119887119909119898

minus1 119887 minus 119886119909119898)

= (119887 minus 119886119909119898

) ((119898 minus 1 minus 120582) (1 minus 119909) + 1) + (119886 minus 119887119909119898

)

= 119886 + 119887 (119898 minus 120582) minus 119887 (119898 minus 1 minus 120582) 119909

minus (119887 + 119886 (119898 minus 120582)) 119909119898

+ 119886 (119898 minus 1 minus 120582) 119909119898+1

= (1 minus 119909) (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

(15)

By Cramerrsquos rule

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(16)

We are done

4 Journal of Applied Mathematics

Lemma 7 Under the assumptions and in the notation ofLemma 6

119886 + 119887 = 1119905119898V + 1119905119898

119908 = 0 119886 = 1119905119898V = 0 119887 = 1119905

119898119908 = 0

(17)

Proof In Lemma 6(1) by setting 119909 = 1 we have

119886 = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + (119886 + 119887) (119898 minus 1) + (119886 minus 119887 (119898 minus 120582))

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

(18)

Since 120582 = 119898 it follows that

(119886 + 119887) (1 minus 120582) = 2119886 (19)

As we all know that the eigenvalues of a real skew-symmetricmatrix are all pure imaginary or nonzero hence det((1 minus

119898)119868119898

+ 119880 minus 119880119905) = 0 where 119898 gt 1 Since

det (1119868119898

minus LB2119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119880119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

(2 minus 119898) 119868119898

+ 119880119898

+ 119880119905

119898

119868119898

+ 119880119905

119898(2 minus 119898) 119868

119898+ 119880119898

+ 119880119905

119898

)

= de119905 (

(1 minus 119898) 119868119898

+ 119880 minus 119880119905

0

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119869119898

)

= det ((1 minus 119898) 119868119898

+ 119880 minus 119880119905) det ((1 minus 119898) 119868

119898+ 119869119898)

= (1 minus 119898)119898minus1 det ((1 minus 119898) 119868

119898+ 119880 minus 119880

119905) = 0

(20)

hence 120582 = 1 Notice that (119886 + 119887)(1 minus 120582) = 2119886 and if 119886 + 119887 = 0then 119886 = 0 and 119887 = (119886 + 119887) minus 119886 = 0 By Lemma 6

119891 (119909) =

119898

sum

119896=1

V119896119909119896minus1

equiv 0

119892 (119909) =

119898

sum

119896=1

119908119896119909119896minus1

equiv 0

(21)

for arbitrary real variable 119909 It is not possible Hence 119886 + 119887 = 0It is easy to see that 119886 = 0 and 119887 = 0

Lemma 8 (see [15]) If A is a square matrix then

(1) for every eigenvalue of A the geometric multiplicity isless than or equal to the algebraic multiplicity

(2) A is diagonalizable if and only if the geometric mul-tiplicity of every eigenvalue is equal to the algebraicmultiplicity

3 Proof of Theorem 1

Let 120582 isin C be an arbitrary eigenvalue of LB2119898 and let

120585 = (V119908 ) = 0 be the eigenvector of LB

2119898corresponding to

120582 where V = (V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905

For 119898 = 1 then B2

= (0 0

1 0) LB

2= (0 0

minus1 1) Obviously

1205821

= 0 1205822

= 1 are eigenvalues of LB2and 120583

0= 1205831

= 1Theorem 1 holds

For 119898 gt 1 using the previous assumptions and notationobviously 119875(LB

2119898 0) = det(0119868

2119898minus LB

2119898) = 0 by

definition 120582 = 0 is an eigenvalue ofLB2119898 1205830

ge 1Note that 119875(LB

2119898 119898) = det(119898119868

2119898minus LB

2119898) = 0 and

by definition 120582 = 119898 is an eigenvalue of LB2119898 By simple

calculation the rank ofmatrix1198981198682119898

minusLB2119898

is equal to119898+1By Lemma 8(1) 120583

119898ge 2119898 minus (119898 + 1) = 119898 minus 1

For 120582 = 119898 by Lemma 7 we set 119886 = 1119905119898V 119886 + 119887 = 1119905

119898V +

1119905119898

119908 = 1 hence 119887 = 1 minus 1119905119898V = 1 minus 119886 In Lemma 6(1) by

setting 119909 = 1 we have

119886 = 1119905119898V = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + 119898 minus 1 + 119886 + (119886 minus 1) (119898 minus 120582)

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

=

(119898 minus 120582) (2119886 minus 1) + 119898 + 2119886 minus 1

2 (119898 minus 120582)

(22)

Hence

119886 =

1 minus 120582

2

(23)

Denoting 1198910

= (119898 + 1 minus 120582)119886 1198911

= 1198912

= sdot sdot sdot = 119891119898minus1

= 1 119891119898

=

119886 + (119886 minus 1)(119898 minus 120582) 119888 = (119898 minus 120582)(119898 + 1 minus 120582) and 119889 = (119898 minus 1 minus

120582)(119898 + 1 minus 120582) we have

119891 (119909)

=

119898

sum

119896=1

V119896119909119896minus1

= ((119898 + 1 minus 120582) 119886 + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

+ (119886 + (119886 minus 1) (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119891119895119889119896minus119895

) 119909119896

(24)

It must be that

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

119896 = 1 2 119898 (25)

Therefore

119886 =

119898

sum

119896=1

V119896

=

1

119888

119898

sum

119896=1

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

119896minus1

sum

119895=1

119889119896minus1minus119895

)

Journal of Applied Mathematics 5

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

119888(1 minus 119889)2

times (1198910

(1 minus 119889119898

) (1 minus 119889) + 119898 (1 minus 119889) minus 1 + 119889119898

)

=

1

119888(1 minus 119889)2

times ((1 minus 1198910

(1 minus 119889)) 119889119898

+ (1198910

+ 119898) (1 minus 119889) minus 1)

(26)

That is 119886119888(1 minus 119889)2

= (1 minus 1198910(1 minus 119889))119889

119898+ (1198910

+ 119898)(1 minus 119889) minus 1Since 119886 = (1minus120582)2119891

0= (119898+1minus120582)119886 119888 = (119898minus120582)(119898+1minus120582)

and 119889 = (119898 minus 1 minus 120582)(119898 + 1 minus 120582) then

(1 minus 1198910

(1 minus 119889)) 119889119898

= 119886119888(1 minus 119889)2

minus (1198910

+ 119898) (1 minus 119889) + 1

(1 minus

2119886 (119898 + 1 minus 120582)

119898 + 1 minus 120582

) 119889119898

=

4119886 (119898 minus 120582)

119898 + 1 minus 120582

minus

2 ((119898 + 1 minus 120582) 119886 + 119898)

119898 + 1 minus 120582

+ 1

120582119889119898

= minus120582

(27)

Denote 120579119896

= (120587 + 2119896120587)2119898 119896 = 0 1 2 119898 minus 1 As 120582 = 0 120582 minus

119898 = 0 we have 119889 = 1198902119894120579119896 that is (119898minus1minus120582)(119898+1minus120582) = 119890

2119894120579119896

where 1198942

= minus1Therefore

120582 = 120582119896

= 119898 minus

1 + 1198902119894120579119896

1 minus 1198902119894120579119896

= 119898 minus

119894 sin 2120579119896

1 minus cos 2120579119896

= 119898 minus 119894cot120579119896

(28)

where 1 minus 1198902119894120579119896

= 0 119896 = 0 1 2 119898 minus 1Note that if 119898 is odd by 119889

119898= ((119898minus1minus120582)(119898+1minus120582))

119898=

minus1 then (119898 minus 1 minus 120582)(119898 + 1 minus 120582) = minus1 hence 120582 = 119898 It is aneigenvalue ofLB

2119898

Furthermore 120583120582119896

ge 1 Note that 120582119896

= 120582119898minus1minus119896

119896 =

0 1 2 lfloor1198982rfloor minus 1To sum up by fundamental theorem of algebra and

complex conjugate root theorem for 119898 ge 1 we obtain thefollowing conclusions

If 119898 is odd then 1205830

= 120583120582119896

= 120583120582119896

= 1 and 120583119898

= 119898 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorIf119898 is even then120583

0= 120583120582119896

= 120583120582119896

= 1 and120583119898

= 119898minus1 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorConsider 119896 = 0 1 2 lfloor1198982rfloor minus 1 We complete theproof of Theorem 1

4 Proofs of Theorem 2 and Corollary 4

It is easy to see that the distinct eigenvalues of LB2119898

are0 120582119896

= 119898 minus 119894cot((120587 + 2119896120587)2119898) 120582119896 119898 with corresponding

algebraic multiplicities 1 1 1 lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloor 119896 =

1 2 lfloor1198982rfloor minus 1For 120582 = 119898 = 1LB

2= (0 0

minus1 1) and obviously 120585

119898= (0

1)

is the eigenvector ofLB2corresponding to 120582 = 119898 = 1

For 120582 = 119898 = 2

LB4

= (

1 minus1 0 0

0 1 minus1 0

minus1 0 2 minus1

minus1 minus1 0 2

) (29)

By simple calculation 120585119898

= (1 minus1 1 minus1)119905 is the eigenvec-

tor ofLB4corresponding to 120582 = 119898 = 2

For 120582 = 119898 gt 2 let 120585119898

= (V119908 ) = 0 be the eigenvector

of LB2119898

corresponding to 119898 and then LB2119898

120585119898

= 119898120585119898

expands to

(

119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (

V119908

) = 0 (30)

Equation (30) is equivalent to the following equation

(

119868119898

(119868 + 119880119898

)minus1

119880119905

119898

0119898

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898

) (

V119908

) = 0 (31)

By simple calculation

(119868 + 119880119898

)minus1

119880119905

119898= (

minus119868119898minus1

0

1119905119898minus1

0)

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898= (

119869119898minus1

1119898minus1

0 0)

(32)

We obtain the solution of (30) as follows

120585119898

=

119898

sum

119896=2

119897119896120585119898119896

(33)

where 119897119896is an arbitrary constant 119896 = 2 3 119898 and

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(34)

For 120582 = 119898 let 120585 = (V119908 ) = 0 be the eigenvector of LB

2119898

corresponding to 120582( = 119898) where V = (V1 V2 V

119898)119905 119908 =

(1199081 1199082 119908

119898)119905 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (35)

By Lemma 7 1119905119898V + 1119905119898

119908 = 1 we put 1119905119898V + 1119905119898

119908 = 1 ByTheorem 1 119886 = 1119905

119898V = (1 minus 120582)2 119887 = 1119905

119898119908 = 1 minus 119886

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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Differential EquationsInternational Journal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

Journal of Applied Mathematics 3

(2) 119883119905119880119905

119898

= (1 119909 1199092 119909

119898minus1) 119880119905

119898

= (

119898minus1

sum

119896=1

119909119896

119898minus1

sum

119896=2

119909119896

119898minus1

sum

119896=119898minus2

119909119896

119898minus1

sum

119896=119898minus1

119909119896 0)

= (

119909 minus 119909119898

1 minus 119909

1199092

minus 119909119898

1 minus 119909

119909119898minus2

minus 119909119898

1 minus 119909

119909119898minus1

minus 119909119898

1 minus 119909

119909119898

minus 119909119898

1 minus 119909

)

=

119909

1 minus 119909

119883119905

minus

119909119898

1 minus 119909

1119905119898

(9)

Lemma6 Let119898 gt 1 be an integer120582 = 119898 let120582 isin C be an arbi-trary eigenvalue ofLB

2119898 and let 120585 = (

V119908 ) = 0 be the eigenvec-

tor of LB2119898

corresponding to 120582 where V = (V1 V2 V

119898)119905

119908 = (1199081 1199082 119908

119898)119905 Let 119883 = (1 119909 119909

2 119909

119898minus1)119905 119891(119909) =

sum119898

119896=1V119896119909119896minus1 and 119892(119909) = sum

119898

119896=1119908119896119909119896minus1 where 119909 is real variable

Then

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(10)

where 119886 = 1119905119898V 119887 = 1119905

119898119908

Proof If 120582( = 119898) is an eigenvalue with eigenvector 120585 = (V119908 )

ofLB2119898 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (11)

Therefore

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119880119905

119898119908 = 120582V

(minus119868119898

minus 119880119905

119898) V + (119898119868

119898minus 119880119898

) 119908 = 120582119908

(12)

We have

119883119905

((119898 minus 1) 119868119898

minus 119880119898

) V minus 119883119905119880119905

119898119908 = 120582119883

119905V

119883119905

(minus119868119898

minus 119880119905

119898) V + 119883

119905(119898119868119898

minus 119880119898

) 119908 = 120582119883119905119908

(13)

According to Lemma 5 we have

((119898 minus 1 minus 120582) (1 minus 119909) + 1) 119891 (119909) minus 119909119892 (119909) = 119886 minus 119887119909119898

minus119891 (119909) + ((119898 minus 120582) (1 minus 119909) + 1) 119892 (119909) = 119887 minus 119886119909119898

(14)

Notice that this equation holds for 119909 = 1 too Consider

119863 = det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 minus119909

minus1 (119898 minus 120582) (1 minus 119909) + 1)

= ((119898 minus 1 minus 120582) (1 minus 119909) + 1) ((119898 minus 120582) (1 minus 119909) + 1) minus 119909

= (119898 minus 120582) (1 minus 119909) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909)

119863119891

= det(

119886 minus 119887119909119898

minus119909

119887 minus 119886119909119898

(119898 minus 120582) (1 minus 119909) + 1)

= (119886 minus 119887119909119898

) ((119898 minus 120582) (1 minus 119909) + 1) + 119909 (119887 minus 119886119909119898

)

= 119886 (119898 + 1 minus 120582) + (119887 minus 119886 (119898 minus 120582)) 119909

minus 119887 (119898 + 1 minus 120582) 119909119898

+ (119887 (119898 minus 120582) minus 119886) 119909119898+1

= (1 minus 119909) (119886 (119898 + 1 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

119863119892

= det(

(119898 minus 1 minus 120582) (1 minus 119909) + 1 119886 minus 119887119909119898

minus1 119887 minus 119886119909119898)

= (119887 minus 119886119909119898

) ((119898 minus 1 minus 120582) (1 minus 119909) + 1) + (119886 minus 119887119909119898

)

= 119886 + 119887 (119898 minus 120582) minus 119887 (119898 minus 1 minus 120582) 119909

minus (119887 + 119886 (119898 minus 120582)) 119909119898

+ 119886 (119898 minus 1 minus 120582) 119909119898+1

= (1 minus 119909) (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

(15)

By Cramerrsquos rule

(1) 119891 (119909)

= ((119898 + 1 minus 120582) 119886 + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

+ (119886 minus 119887 (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(2) 119892 (119909)

= (119886 + 119887 (119898 minus 120582) + (119886 + 119887) (119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

)

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

(16)

We are done

4 Journal of Applied Mathematics

Lemma 7 Under the assumptions and in the notation ofLemma 6

119886 + 119887 = 1119905119898V + 1119905119898

119908 = 0 119886 = 1119905119898V = 0 119887 = 1119905

119898119908 = 0

(17)

Proof In Lemma 6(1) by setting 119909 = 1 we have

119886 = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + (119886 + 119887) (119898 minus 1) + (119886 minus 119887 (119898 minus 120582))

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

(18)

Since 120582 = 119898 it follows that

(119886 + 119887) (1 minus 120582) = 2119886 (19)

As we all know that the eigenvalues of a real skew-symmetricmatrix are all pure imaginary or nonzero hence det((1 minus

119898)119868119898

+ 119880 minus 119880119905) = 0 where 119898 gt 1 Since

det (1119868119898

minus LB2119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119880119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

(2 minus 119898) 119868119898

+ 119880119898

+ 119880119905

119898

119868119898

+ 119880119905

119898(2 minus 119898) 119868

119898+ 119880119898

+ 119880119905

119898

)

= de119905 (

(1 minus 119898) 119868119898

+ 119880 minus 119880119905

0

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119869119898

)

= det ((1 minus 119898) 119868119898

+ 119880 minus 119880119905) det ((1 minus 119898) 119868

119898+ 119869119898)

= (1 minus 119898)119898minus1 det ((1 minus 119898) 119868

119898+ 119880 minus 119880

119905) = 0

(20)

hence 120582 = 1 Notice that (119886 + 119887)(1 minus 120582) = 2119886 and if 119886 + 119887 = 0then 119886 = 0 and 119887 = (119886 + 119887) minus 119886 = 0 By Lemma 6

119891 (119909) =

119898

sum

119896=1

V119896119909119896minus1

equiv 0

119892 (119909) =

119898

sum

119896=1

119908119896119909119896minus1

equiv 0

(21)

for arbitrary real variable 119909 It is not possible Hence 119886 + 119887 = 0It is easy to see that 119886 = 0 and 119887 = 0

Lemma 8 (see [15]) If A is a square matrix then

(1) for every eigenvalue of A the geometric multiplicity isless than or equal to the algebraic multiplicity

(2) A is diagonalizable if and only if the geometric mul-tiplicity of every eigenvalue is equal to the algebraicmultiplicity

3 Proof of Theorem 1

Let 120582 isin C be an arbitrary eigenvalue of LB2119898 and let

120585 = (V119908 ) = 0 be the eigenvector of LB

2119898corresponding to

120582 where V = (V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905

For 119898 = 1 then B2

= (0 0

1 0) LB

2= (0 0

minus1 1) Obviously

1205821

= 0 1205822

= 1 are eigenvalues of LB2and 120583

0= 1205831

= 1Theorem 1 holds

For 119898 gt 1 using the previous assumptions and notationobviously 119875(LB

2119898 0) = det(0119868

2119898minus LB

2119898) = 0 by

definition 120582 = 0 is an eigenvalue ofLB2119898 1205830

ge 1Note that 119875(LB

2119898 119898) = det(119898119868

2119898minus LB

2119898) = 0 and

by definition 120582 = 119898 is an eigenvalue of LB2119898 By simple

calculation the rank ofmatrix1198981198682119898

minusLB2119898

is equal to119898+1By Lemma 8(1) 120583

119898ge 2119898 minus (119898 + 1) = 119898 minus 1

For 120582 = 119898 by Lemma 7 we set 119886 = 1119905119898V 119886 + 119887 = 1119905

119898V +

1119905119898

119908 = 1 hence 119887 = 1 minus 1119905119898V = 1 minus 119886 In Lemma 6(1) by

setting 119909 = 1 we have

119886 = 1119905119898V = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + 119898 minus 1 + 119886 + (119886 minus 1) (119898 minus 120582)

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

=

(119898 minus 120582) (2119886 minus 1) + 119898 + 2119886 minus 1

2 (119898 minus 120582)

(22)

Hence

119886 =

1 minus 120582

2

(23)

Denoting 1198910

= (119898 + 1 minus 120582)119886 1198911

= 1198912

= sdot sdot sdot = 119891119898minus1

= 1 119891119898

=

119886 + (119886 minus 1)(119898 minus 120582) 119888 = (119898 minus 120582)(119898 + 1 minus 120582) and 119889 = (119898 minus 1 minus

120582)(119898 + 1 minus 120582) we have

119891 (119909)

=

119898

sum

119896=1

V119896119909119896minus1

= ((119898 + 1 minus 120582) 119886 + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

+ (119886 + (119886 minus 1) (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119891119895119889119896minus119895

) 119909119896

(24)

It must be that

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

119896 = 1 2 119898 (25)

Therefore

119886 =

119898

sum

119896=1

V119896

=

1

119888

119898

sum

119896=1

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

119896minus1

sum

119895=1

119889119896minus1minus119895

)

Journal of Applied Mathematics 5

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

119888(1 minus 119889)2

times (1198910

(1 minus 119889119898

) (1 minus 119889) + 119898 (1 minus 119889) minus 1 + 119889119898

)

=

1

119888(1 minus 119889)2

times ((1 minus 1198910

(1 minus 119889)) 119889119898

+ (1198910

+ 119898) (1 minus 119889) minus 1)

(26)

That is 119886119888(1 minus 119889)2

= (1 minus 1198910(1 minus 119889))119889

119898+ (1198910

+ 119898)(1 minus 119889) minus 1Since 119886 = (1minus120582)2119891

0= (119898+1minus120582)119886 119888 = (119898minus120582)(119898+1minus120582)

and 119889 = (119898 minus 1 minus 120582)(119898 + 1 minus 120582) then

(1 minus 1198910

(1 minus 119889)) 119889119898

= 119886119888(1 minus 119889)2

minus (1198910

+ 119898) (1 minus 119889) + 1

(1 minus

2119886 (119898 + 1 minus 120582)

119898 + 1 minus 120582

) 119889119898

=

4119886 (119898 minus 120582)

119898 + 1 minus 120582

minus

2 ((119898 + 1 minus 120582) 119886 + 119898)

119898 + 1 minus 120582

+ 1

120582119889119898

= minus120582

(27)

Denote 120579119896

= (120587 + 2119896120587)2119898 119896 = 0 1 2 119898 minus 1 As 120582 = 0 120582 minus

119898 = 0 we have 119889 = 1198902119894120579119896 that is (119898minus1minus120582)(119898+1minus120582) = 119890

2119894120579119896

where 1198942

= minus1Therefore

120582 = 120582119896

= 119898 minus

1 + 1198902119894120579119896

1 minus 1198902119894120579119896

= 119898 minus

119894 sin 2120579119896

1 minus cos 2120579119896

= 119898 minus 119894cot120579119896

(28)

where 1 minus 1198902119894120579119896

= 0 119896 = 0 1 2 119898 minus 1Note that if 119898 is odd by 119889

119898= ((119898minus1minus120582)(119898+1minus120582))

119898=

minus1 then (119898 minus 1 minus 120582)(119898 + 1 minus 120582) = minus1 hence 120582 = 119898 It is aneigenvalue ofLB

2119898

Furthermore 120583120582119896

ge 1 Note that 120582119896

= 120582119898minus1minus119896

119896 =

0 1 2 lfloor1198982rfloor minus 1To sum up by fundamental theorem of algebra and

complex conjugate root theorem for 119898 ge 1 we obtain thefollowing conclusions

If 119898 is odd then 1205830

= 120583120582119896

= 120583120582119896

= 1 and 120583119898

= 119898 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorIf119898 is even then120583

0= 120583120582119896

= 120583120582119896

= 1 and120583119898

= 119898minus1 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorConsider 119896 = 0 1 2 lfloor1198982rfloor minus 1 We complete theproof of Theorem 1

4 Proofs of Theorem 2 and Corollary 4

It is easy to see that the distinct eigenvalues of LB2119898

are0 120582119896

= 119898 minus 119894cot((120587 + 2119896120587)2119898) 120582119896 119898 with corresponding

algebraic multiplicities 1 1 1 lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloor 119896 =

1 2 lfloor1198982rfloor minus 1For 120582 = 119898 = 1LB

2= (0 0

minus1 1) and obviously 120585

119898= (0

1)

is the eigenvector ofLB2corresponding to 120582 = 119898 = 1

For 120582 = 119898 = 2

LB4

= (

1 minus1 0 0

0 1 minus1 0

minus1 0 2 minus1

minus1 minus1 0 2

) (29)

By simple calculation 120585119898

= (1 minus1 1 minus1)119905 is the eigenvec-

tor ofLB4corresponding to 120582 = 119898 = 2

For 120582 = 119898 gt 2 let 120585119898

= (V119908 ) = 0 be the eigenvector

of LB2119898

corresponding to 119898 and then LB2119898

120585119898

= 119898120585119898

expands to

(

119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (

V119908

) = 0 (30)

Equation (30) is equivalent to the following equation

(

119868119898

(119868 + 119880119898

)minus1

119880119905

119898

0119898

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898

) (

V119908

) = 0 (31)

By simple calculation

(119868 + 119880119898

)minus1

119880119905

119898= (

minus119868119898minus1

0

1119905119898minus1

0)

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898= (

119869119898minus1

1119898minus1

0 0)

(32)

We obtain the solution of (30) as follows

120585119898

=

119898

sum

119896=2

119897119896120585119898119896

(33)

where 119897119896is an arbitrary constant 119896 = 2 3 119898 and

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(34)

For 120582 = 119898 let 120585 = (V119908 ) = 0 be the eigenvector of LB

2119898

corresponding to 120582( = 119898) where V = (V1 V2 V

119898)119905 119908 =

(1199081 1199082 119908

119898)119905 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (35)

By Lemma 7 1119905119898V + 1119905119898

119908 = 1 we put 1119905119898V + 1119905119898

119908 = 1 ByTheorem 1 119886 = 1119905

119898V = (1 minus 120582)2 119887 = 1119905

119898119908 = 1 minus 119886

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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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

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Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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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 4: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

4 Journal of Applied Mathematics

Lemma 7 Under the assumptions and in the notation ofLemma 6

119886 + 119887 = 1119905119898V + 1119905119898

119908 = 0 119886 = 1119905119898V = 0 119887 = 1119905

119898119908 = 0

(17)

Proof In Lemma 6(1) by setting 119909 = 1 we have

119886 = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + (119886 + 119887) (119898 minus 1) + (119886 minus 119887 (119898 minus 120582))

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

(18)

Since 120582 = 119898 it follows that

(119886 + 119887) (1 minus 120582) = 2119886 (19)

As we all know that the eigenvalues of a real skew-symmetricmatrix are all pure imaginary or nonzero hence det((1 minus

119898)119868119898

+ 119880 minus 119880119905) = 0 where 119898 gt 1 Since

det (1119868119898

minus LB2119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119880119898

)

= det(

(2 minus 119898) 119868119898

+ 119880119898

(2 minus 119898) 119868119898

+ 119880119898

+ 119880119905

119898

119868119898

+ 119880119905

119898(2 minus 119898) 119868

119898+ 119880119898

+ 119880119905

119898

)

= de119905 (

(1 minus 119898) 119868119898

+ 119880 minus 119880119905

0

119868119898

+ 119880119905

119898(1 minus 119898) 119868

119898+ 119869119898

)

= det ((1 minus 119898) 119868119898

+ 119880 minus 119880119905) det ((1 minus 119898) 119868

119898+ 119869119898)

= (1 minus 119898)119898minus1 det ((1 minus 119898) 119868

119898+ 119880 minus 119880

119905) = 0

(20)

hence 120582 = 1 Notice that (119886 + 119887)(1 minus 120582) = 2119886 and if 119886 + 119887 = 0then 119886 = 0 and 119887 = (119886 + 119887) minus 119886 = 0 By Lemma 6

119891 (119909) =

119898

sum

119896=1

V119896119909119896minus1

equiv 0

119892 (119909) =

119898

sum

119896=1

119908119896119909119896minus1

equiv 0

(21)

for arbitrary real variable 119909 It is not possible Hence 119886 + 119887 = 0It is easy to see that 119886 = 0 and 119887 = 0

Lemma 8 (see [15]) If A is a square matrix then

(1) for every eigenvalue of A the geometric multiplicity isless than or equal to the algebraic multiplicity

(2) A is diagonalizable if and only if the geometric mul-tiplicity of every eigenvalue is equal to the algebraicmultiplicity

3 Proof of Theorem 1

Let 120582 isin C be an arbitrary eigenvalue of LB2119898 and let

120585 = (V119908 ) = 0 be the eigenvector of LB

2119898corresponding to

120582 where V = (V1 V2 V

119898)119905 119908 = (119908

1 1199082 119908

119898)119905

For 119898 = 1 then B2

= (0 0

1 0) LB

2= (0 0

minus1 1) Obviously

1205821

= 0 1205822

= 1 are eigenvalues of LB2and 120583

0= 1205831

= 1Theorem 1 holds

For 119898 gt 1 using the previous assumptions and notationobviously 119875(LB

2119898 0) = det(0119868

2119898minus LB

2119898) = 0 by

definition 120582 = 0 is an eigenvalue ofLB2119898 1205830

ge 1Note that 119875(LB

2119898 119898) = det(119898119868

2119898minus LB

2119898) = 0 and

by definition 120582 = 119898 is an eigenvalue of LB2119898 By simple

calculation the rank ofmatrix1198981198682119898

minusLB2119898

is equal to119898+1By Lemma 8(1) 120583

119898ge 2119898 minus (119898 + 1) = 119898 minus 1

For 120582 = 119898 by Lemma 7 we set 119886 = 1119905119898V 119886 + 119887 = 1119905

119898V +

1119905119898

119908 = 1 hence 119887 = 1 minus 1119905119898V = 1 minus 119886 In Lemma 6(1) by

setting 119909 = 1 we have

119886 = 1119905119898V = 119891 (1)

=

(119898 + 1 minus 120582) 119886 + 119898 minus 1 + 119886 + (119886 minus 1) (119898 minus 120582)

(119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582))

=

(119898 minus 120582) (2119886 minus 1) + 119898 + 2119886 minus 1

2 (119898 minus 120582)

(22)

Hence

119886 =

1 minus 120582

2

(23)

Denoting 1198910

= (119898 + 1 minus 120582)119886 1198911

= 1198912

= sdot sdot sdot = 119891119898minus1

= 1 119891119898

=

119886 + (119886 minus 1)(119898 minus 120582) 119888 = (119898 minus 120582)(119898 + 1 minus 120582) and 119889 = (119898 minus 1 minus

120582)(119898 + 1 minus 120582) we have

119891 (119909)

=

119898

sum

119896=1

V119896119909119896minus1

= ((119898 + 1 minus 120582) 119886 + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

+ (119886 + (119886 minus 1) (119898 minus 120582)) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119891119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119891119895119889119896minus119895

) 119909119896

(24)

It must be that

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

119896 = 1 2 119898 (25)

Therefore

119886 =

119898

sum

119896=1

V119896

=

1

119888

119898

sum

119896=1

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

119896minus1

sum

119895=1

119889119896minus1minus119895

)

Journal of Applied Mathematics 5

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

119888(1 minus 119889)2

times (1198910

(1 minus 119889119898

) (1 minus 119889) + 119898 (1 minus 119889) minus 1 + 119889119898

)

=

1

119888(1 minus 119889)2

times ((1 minus 1198910

(1 minus 119889)) 119889119898

+ (1198910

+ 119898) (1 minus 119889) minus 1)

(26)

That is 119886119888(1 minus 119889)2

= (1 minus 1198910(1 minus 119889))119889

119898+ (1198910

+ 119898)(1 minus 119889) minus 1Since 119886 = (1minus120582)2119891

0= (119898+1minus120582)119886 119888 = (119898minus120582)(119898+1minus120582)

and 119889 = (119898 minus 1 minus 120582)(119898 + 1 minus 120582) then

(1 minus 1198910

(1 minus 119889)) 119889119898

= 119886119888(1 minus 119889)2

minus (1198910

+ 119898) (1 minus 119889) + 1

(1 minus

2119886 (119898 + 1 minus 120582)

119898 + 1 minus 120582

) 119889119898

=

4119886 (119898 minus 120582)

119898 + 1 minus 120582

minus

2 ((119898 + 1 minus 120582) 119886 + 119898)

119898 + 1 minus 120582

+ 1

120582119889119898

= minus120582

(27)

Denote 120579119896

= (120587 + 2119896120587)2119898 119896 = 0 1 2 119898 minus 1 As 120582 = 0 120582 minus

119898 = 0 we have 119889 = 1198902119894120579119896 that is (119898minus1minus120582)(119898+1minus120582) = 119890

2119894120579119896

where 1198942

= minus1Therefore

120582 = 120582119896

= 119898 minus

1 + 1198902119894120579119896

1 minus 1198902119894120579119896

= 119898 minus

119894 sin 2120579119896

1 minus cos 2120579119896

= 119898 minus 119894cot120579119896

(28)

where 1 minus 1198902119894120579119896

= 0 119896 = 0 1 2 119898 minus 1Note that if 119898 is odd by 119889

119898= ((119898minus1minus120582)(119898+1minus120582))

119898=

minus1 then (119898 minus 1 minus 120582)(119898 + 1 minus 120582) = minus1 hence 120582 = 119898 It is aneigenvalue ofLB

2119898

Furthermore 120583120582119896

ge 1 Note that 120582119896

= 120582119898minus1minus119896

119896 =

0 1 2 lfloor1198982rfloor minus 1To sum up by fundamental theorem of algebra and

complex conjugate root theorem for 119898 ge 1 we obtain thefollowing conclusions

If 119898 is odd then 1205830

= 120583120582119896

= 120583120582119896

= 1 and 120583119898

= 119898 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorIf119898 is even then120583

0= 120583120582119896

= 120583120582119896

= 1 and120583119898

= 119898minus1 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorConsider 119896 = 0 1 2 lfloor1198982rfloor minus 1 We complete theproof of Theorem 1

4 Proofs of Theorem 2 and Corollary 4

It is easy to see that the distinct eigenvalues of LB2119898

are0 120582119896

= 119898 minus 119894cot((120587 + 2119896120587)2119898) 120582119896 119898 with corresponding

algebraic multiplicities 1 1 1 lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloor 119896 =

1 2 lfloor1198982rfloor minus 1For 120582 = 119898 = 1LB

2= (0 0

minus1 1) and obviously 120585

119898= (0

1)

is the eigenvector ofLB2corresponding to 120582 = 119898 = 1

For 120582 = 119898 = 2

LB4

= (

1 minus1 0 0

0 1 minus1 0

minus1 0 2 minus1

minus1 minus1 0 2

) (29)

By simple calculation 120585119898

= (1 minus1 1 minus1)119905 is the eigenvec-

tor ofLB4corresponding to 120582 = 119898 = 2

For 120582 = 119898 gt 2 let 120585119898

= (V119908 ) = 0 be the eigenvector

of LB2119898

corresponding to 119898 and then LB2119898

120585119898

= 119898120585119898

expands to

(

119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (

V119908

) = 0 (30)

Equation (30) is equivalent to the following equation

(

119868119898

(119868 + 119880119898

)minus1

119880119905

119898

0119898

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898

) (

V119908

) = 0 (31)

By simple calculation

(119868 + 119880119898

)minus1

119880119905

119898= (

minus119868119898minus1

0

1119905119898minus1

0)

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898= (

119869119898minus1

1119898minus1

0 0)

(32)

We obtain the solution of (30) as follows

120585119898

=

119898

sum

119896=2

119897119896120585119898119896

(33)

where 119897119896is an arbitrary constant 119896 = 2 3 119898 and

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(34)

For 120582 = 119898 let 120585 = (V119908 ) = 0 be the eigenvector of LB

2119898

corresponding to 120582( = 119898) where V = (V1 V2 V

119898)119905 119908 =

(1199081 1199082 119908

119898)119905 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (35)

By Lemma 7 1119905119898V + 1119905119898

119908 = 1 we put 1119905119898V + 1119905119898

119908 = 1 ByTheorem 1 119886 = 1119905

119898V = (1 minus 120582)2 119887 = 1119905

119898119908 = 1 minus 119886

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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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 5: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

Journal of Applied Mathematics 5

=

1

119888

119898

sum

119896=1

(1198910119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

119888(1 minus 119889)2

times (1198910

(1 minus 119889119898

) (1 minus 119889) + 119898 (1 minus 119889) minus 1 + 119889119898

)

=

1

119888(1 minus 119889)2

times ((1 minus 1198910

(1 minus 119889)) 119889119898

+ (1198910

+ 119898) (1 minus 119889) minus 1)

(26)

That is 119886119888(1 minus 119889)2

= (1 minus 1198910(1 minus 119889))119889

119898+ (1198910

+ 119898)(1 minus 119889) minus 1Since 119886 = (1minus120582)2119891

0= (119898+1minus120582)119886 119888 = (119898minus120582)(119898+1minus120582)

and 119889 = (119898 minus 1 minus 120582)(119898 + 1 minus 120582) then

(1 minus 1198910

(1 minus 119889)) 119889119898

= 119886119888(1 minus 119889)2

minus (1198910

+ 119898) (1 minus 119889) + 1

(1 minus

2119886 (119898 + 1 minus 120582)

119898 + 1 minus 120582

) 119889119898

=

4119886 (119898 minus 120582)

119898 + 1 minus 120582

minus

2 ((119898 + 1 minus 120582) 119886 + 119898)

119898 + 1 minus 120582

+ 1

120582119889119898

= minus120582

(27)

Denote 120579119896

= (120587 + 2119896120587)2119898 119896 = 0 1 2 119898 minus 1 As 120582 = 0 120582 minus

119898 = 0 we have 119889 = 1198902119894120579119896 that is (119898minus1minus120582)(119898+1minus120582) = 119890

2119894120579119896

where 1198942

= minus1Therefore

120582 = 120582119896

= 119898 minus

1 + 1198902119894120579119896

1 minus 1198902119894120579119896

= 119898 minus

119894 sin 2120579119896

1 minus cos 2120579119896

= 119898 minus 119894cot120579119896

(28)

where 1 minus 1198902119894120579119896

= 0 119896 = 0 1 2 119898 minus 1Note that if 119898 is odd by 119889

119898= ((119898minus1minus120582)(119898+1minus120582))

119898=

minus1 then (119898 minus 1 minus 120582)(119898 + 1 minus 120582) = minus1 hence 120582 = 119898 It is aneigenvalue ofLB

2119898

Furthermore 120583120582119896

ge 1 Note that 120582119896

= 120582119898minus1minus119896

119896 =

0 1 2 lfloor1198982rfloor minus 1To sum up by fundamental theorem of algebra and

complex conjugate root theorem for 119898 ge 1 we obtain thefollowing conclusions

If 119898 is odd then 1205830

= 120583120582119896

= 120583120582119896

= 1 and 120583119898

= 119898 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorIf119898 is even then120583

0= 120583120582119896

= 120583120582119896

= 1 and120583119898

= 119898minus1 =

lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloorConsider 119896 = 0 1 2 lfloor1198982rfloor minus 1 We complete theproof of Theorem 1

4 Proofs of Theorem 2 and Corollary 4

It is easy to see that the distinct eigenvalues of LB2119898

are0 120582119896

= 119898 minus 119894cot((120587 + 2119896120587)2119898) 120582119896 119898 with corresponding

algebraic multiplicities 1 1 1 lfloor(119898 minus 1)2rfloor + lfloor(119898 + 1)2rfloor 119896 =

1 2 lfloor1198982rfloor minus 1For 120582 = 119898 = 1LB

2= (0 0

minus1 1) and obviously 120585

119898= (0

1)

is the eigenvector ofLB2corresponding to 120582 = 119898 = 1

For 120582 = 119898 = 2

LB4

= (

1 minus1 0 0

0 1 minus1 0

minus1 0 2 minus1

minus1 minus1 0 2

) (29)

By simple calculation 120585119898

= (1 minus1 1 minus1)119905 is the eigenvec-

tor ofLB4corresponding to 120582 = 119898 = 2

For 120582 = 119898 gt 2 let 120585119898

= (V119908 ) = 0 be the eigenvector

of LB2119898

corresponding to 119898 and then LB2119898

120585119898

= 119898120585119898

expands to

(

119868119898

+ 119880119898

119880119905

119898

119868119898

+ 119880119905

119898119880119898

) (

V119908

) = 0 (30)

Equation (30) is equivalent to the following equation

(

119868119898

(119868 + 119880119898

)minus1

119880119905

119898

0119898

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898

) (

V119908

) = 0 (31)

By simple calculation

(119868 + 119880119898

)minus1

119880119905

119898= (

minus119868119898minus1

0

1119905119898minus1

0)

119880119898

minus (119868119898

minus 119880119905

119898) (119868 + 119880

119898)minus1

119880119905

119898= (

119869119898minus1

1119898minus1

0 0)

(32)

We obtain the solution of (30) as follows

120585119898

=

119898

sum

119896=2

119897119896120585119898119896

(33)

where 119897119896is an arbitrary constant 119896 = 2 3 119898 and

(1205851198982

1205851198983

120585119898119898

) =(

(

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

1119905119898minus2

1

minus119868119898minus2

0

0 minus1

)

)2119898times(119898minus1)

(34)

For 120582 = 119898 let 120585 = (V119908 ) = 0 be the eigenvector of LB

2119898

corresponding to 120582( = 119898) where V = (V1 V2 V

119898)119905 119908 =

(1199081 1199082 119908

119898)119905 thenLB

2119898120585 = 120582120585 expands to

(

(119898 minus 1) 119868119898

minus 119880119898

minus119880119905

119898

minus119868119898

minus 119880119905

119898119898119868119898

minus 119880119898

) (

V119908

) = 120582 (

V119908

) (35)

By Lemma 7 1119905119898V + 1119905119898

119908 = 1 we put 1119905119898V + 1119905119898

119908 = 1 ByTheorem 1 119886 = 1119905

119898V = (1 minus 120582)2 119887 = 1119905

119898119908 = 1 minus 119886

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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 6: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

6 Journal of Applied Mathematics

Denote 1198920

= 119886 + (1 minus 119886)(119898 minus 120582) 1198921

= 1198922

= sdot sdot sdot = 119892119898minus1

= 1and 119892

119898= minus119886(119898 minus 1 minus 120582) By Lemma 6(2) we have

119892 (119909)

=

119898

sum

119896=1

119908119896119909119896minus1

= (119886 + (1 minus 119886) (119898 minus 120582) + 119909 + 1199092

+ sdot sdot sdot + 119909119898minus1

minus119886 (119898 minus 1 minus 120582) 119909119898

)

times ((119898 minus 120582) ((119898 + 1 minus 120582) minus (119898 minus 1 minus 120582) 119909))minus1

=

1

119888

1

(1 minus 119889119909)

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

119889119896119909119896

119898

sum

119896=0

119892119896119909119896

=

1

119888

infin

sum

119896=0

(

119896

sum

119895=0

119892119895119889119896minus119895

) 119909119896

(36)

It must be that

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

119896 = 1 2 119898 (37)

Hence

119908119896

=

1

119888

119896minus1

sum

119895=0

119892119895119889119896minus1minus119895

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

1 minus 120582 + (1 + 120582) (119898 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 + 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896

) 119896 = 1 2 119898

(38)

In the proof of Theorem 1 we have

V119896

=

1

119888

119896minus1

sum

119895=0

119891119895119889119896minus1minus119895

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

119896minus2

sum

119895=0

119889119895)

=

1

119888

(

(119898 + 1 minus 120582) (1 minus 120582)

2

119889119896minus1

+

1 minus 119889119896minus1

1 minus 119889

)

=

1

2 (119898 minus 120582)

(1 minus 120582(

119898 minus 1 minus 120582

119898 + 1 minus 120582

)

119896minus1

) 119896 = 1 2 119898

(39)

We complete the proof of Theorem 2

Let 120582 be an eigenvalue of LB2119898 and 120583

120582and ]

120582are

the algebraic multiplicity and the geometric multiplicitycorresponding to 120582 respectively

If 119898 = 1 by Theorem 1 and Theorem 2 and 1205830

= ]0

=

1205831

= ]1

= 1 thenLB2is diagonalizable

If 119898 gt 1 byTheorem 1 1205830

= 1 120583119898

= lfloor(119898 minus 1)2rfloor + lfloor(119898 +

1)2rfloor and 120583120582119896

= 120583120582119896

= 1 where 120582119896

= 119898minus119894cot((120587+2119896120587)2119898)119896 = 0 1 2 sdot sdot sdot lfloor1198982rfloor minus 1

ByTheorem 2 ]0

= 1 ]119898

= 119898 minus 1 and ]120582119896

= ]120582119896

= 1 Wehave the following

if 119898 gt 1 is odd then 119898 = 120583119898

= ]119898

= 119898 minus 1

if 119898 is even then 120583120582

= ]120582 where 120582 = 0 119898 120582

119896 120582119896

119896 = 0 1 2 lfloor1198982rfloor minus 1

By Lemma 8(2) Corollary 4 holds

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the Natural Science Foundation ofGuangdong Province China (no S2013010016994)

References

[1] F R K Chung Spectral Graph Theory vol 92 AMS Publica-tions 1997

[2] R A Brualdi ldquoSpectra of digraphsrdquo Linear Algebra and ItsApplications vol 432 no 9 pp 2181ndash2213 2010

[3] F Chung ldquoLaplacians and the Cheeger inequality for directedgraphsrdquo Annals of Combinatorics vol 9 no 1 pp 1ndash19 2005

[4] W N Anderson and T D Morley Jr ldquoEigenvalues of theLaplacian of a graphrdquo Linear and Multilinear Algebra vol 18no 2 pp 141ndash145 1985

[5] R Agaev and P Chebotarev ldquoOn the spectra of nonsymmetricLaplacian matricesrdquo Linear Algebra and Its Applications vol399 pp 157ndash168 2005

[6] L Bolian Combinatorial Matrix Theory Science Press BeijingChina 2nd edition 2006

[7] R A Brualdi andQ Li ldquoProblem 31rdquoDiscreteMathematics vol43 pp 1133ndash1135 1983

[8] S W Drury ldquoSolution of the conjecture of Brualdi and LirdquoLinear Algebra and Its Applications vol 436 no 9 pp 3392ndash3399 2012

[9] J Burk and M J Tsatsomeros ldquoOn the Brualdi-Li matrix andits Perron eigenspacerdquo Electronic Journal of Linear Algebra vol23 pp 212ndash230 2012

[10] S Kirkland ldquoA note on the sequence of Brualdi-Li matricesrdquoLinear Algebra and Its Applications vol 248 pp 233ndash240 1996

[11] S Kirkland ldquoHypertournament matrices score vectors andeigenvaluesrdquo Linear and Multilinear Algebra vol 30 no 4 pp261ndash274 1991

[12] S J Kirkland and B L Shader ldquoTournament matrices withextremal spectral propertiesrdquo Linear Algebra and Its Applica-tions vol 196 pp 1ndash17 1994

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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 7: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

Journal of Applied Mathematics 7

[13] S Kirkland ldquoPerron vector bounds for a tournament matrixwith applications to a conjecture of Brualdi and Lirdquo LinearAlgebra and Its Applications vol 262 pp 209ndash227 1997

[14] X Chen ldquoOn some properties for the sequence of Brualdi-Limatricesrdquo Journal of Applied Mathematics vol 2013 Article ID985654 5 pages 2013

[15] ZHeruiAdvancedAlgebra AdvancedEducationPress BeijingChina 1983

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 8: Research Article The Spectrum and Eigenvectors of the Laplacian … · 2019. 5. 10. · The Spectrum and Eigenvectors of the Laplacian Matrices of the Brualdi-Li Tournament Digraphs

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


Eigenvalues/Eigenvectors & Discriminant Analysis

Eigenvalues/Eigenvectors & Discriminant Analysis

Eigenvalues and Eigenvectors - WordPress.com ·  · 2017-03-24Eigenvalues and Eigenvectors ... and today it has applications in such ... 5.1 Eigenvalues and Eigenvectors 295 EXAMPLE

Eigenvalues and Eigenvectors - WordPress.com ·  · 2017-03-24Eigenvalues and Eigenvectors ... and today it has applications in such ... 5.1 Eigenvalues and Eigenvectors 295 EXAMPLE

24. Eigenvectors, spectral theorems

24. Eigenvectors, spectral theorems

Lecture04 - Eigenvalues and Eigenvectors

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On the Spectrum of the Penrose Laplacianpersonal.denison.edu/~meim/SMI/MikeChristineJulieHailee_Poster.pdf · On the Spectrum of the Penrose Laplacian Michael Dairyko, Christine Hoffman,

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Mesh Segmentation Using Laplacian Eigenvectors and ... · Mesh Segmentation Using Laplacian Eigenvectors and Gaussian Mixtures Avinash Sharma, Radu Horaud, David Knossow and Etienne

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On the Laplacian Coefficients and Laplacian-Like Energy of ...downloads.hindawi.com/journals/jam/2012/404067.pdf · On the Laplacian Coefficients and Laplacian-Like Energy of Unicyclic

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A. Candelieri, F. Archetti, E. Messina clustering to enrich analysis with the relationship between scenarios Data Affinity Graph Laplacian, Spectrum & Eigenvalues 2nd smallest eigenvectors

A. Candelieri, F. Archetti, E. Messina clustering to enrich analysis with the relationship between scenarios Data Affinity Graph Laplacian, Spectrum & Eigenvalues 2nd smallest eigenvectors

Eigenvalues and Eigenvectors: An Introduction - … ·  · 2012-06-26Eigenvalues and Eigenvectors: An ... have focused considerable attention on "eigenvalues" and "eigenvectors"-their

Eigenvalues and Eigenvectors: An Introduction - … ·  · 2012-06-26Eigenvalues and Eigenvectors: An ... have focused considerable attention on "eigenvalues" and "eigenvectors"-their

Eigenvectors,%Eigenvalues,%and% Finite%Strain% · 9.%eigenvectors,%eigenvalues,%and%finite%strain% 1 = (a+) ...

Eigenvectors,%Eigenvalues,%and% Finite%Strain% · 9.%eigenvectors,%eigenvalues,%and%finite%strain% 1 = (a+) ...

Laplacian Patch-Based Image Synthesis - KAISTvclab.kaist.ac.kr/cvpr2016p2/CVPR2016_LaplacianInpainting_supp.pdf · Laplacian Patch-Based Image Synthesis ... with Other Laplacian-Based

Laplacian Patch-Based Image Synthesis - KAISTvclab.kaist.ac.kr/cvpr2016p2/CVPR2016_LaplacianInpainting_supp.pdf · Laplacian Patch-Based Image Synthesis ... with Other Laplacian-Based

► Image Features for Neural Network Quantitative shape descriptions of a first-order Laplacian pyramid coefficients histogram combined with the power spectrum.

► Image Features for Neural Network Quantitative shape descriptions of a first-order Laplacian pyramid coefficients histogram combined with the power spectrum.

Lecture8, Eigenvalues and Eigenvectors

Lecture8, Eigenvalues and Eigenvectors

Lec 13: Low Dimension Embedding · Laplacian Eigenmap direct data embedding without explicit projection input data -> affinity graph -> graph Laplacian eigenvectors - > embedding

Lec 13: Low Dimension Embedding · Laplacian Eigenmap direct data embedding without explicit projection input data -> affinity graph -> graph Laplacian eigenvectors - > embedding

Eigenvalues and Eigenvectors Note2

Eigenvalues and Eigenvectors Note2

Lesson14: Eigenvalues And Eigenvectors

Lesson14: Eigenvalues And Eigenvectors

ULTRASONIC SIGNAL PROCESSING: SYSTEM IDENTIFICATION …ecasp.ece.iit.edu/publications/Thesis/SaniiePhDDissertation.pdf · a) RF puLse with a sinc function potler spectrum. b) LapLacian

ULTRASONIC SIGNAL PROCESSING: SYSTEM IDENTIFICATION …ecasp.ece.iit.edu/publications/Thesis/SaniiePhDDissertation.pdf · a) RF puLse with a sinc function potler spectrum. b) LapLacian

Harmonic Functions and the Spectrum of the Laplacian on the

Harmonic Functions and the Spectrum of the Laplacian on the

Find eigenvalues and eigenvectors

Find eigenvalues and eigenvectors