Special MatricesResearch Article • DOI: 10.2478/spma-2013-0002 • Special Matrices • 2013 • 3-9

Factorizable matrices

AbstractWe study square matrices which are products of simpler fac-tors with the property that any ordering of the factors yields amatrix cospectral with the given matrix. The results general-ize those obtained previously by the authors.

KeywordsGenerator • F-matrix • generalized complementary basicmatrix

MSC: 15A23, 15A57, 15A48© 2013 Miroslav Fiedler et al., licensee Versita Sp. z o. o.This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs license (http://creativecommons.org/licenses/by-nc-nd/3.0/), which meansthat the text may be used for non-commercial purposes, provided credit is givento the author.

Miroslav Fiedler1,2∗† , Frank J. Hall3‡

1 Academy of Sciences of the Czech Republic,Inst. of Mathematics, Žitná 25 115 67 Praha 1

2 Inst. of Computer Science, Pod vodárenskou veží 2,182 07 Praha 8, Czech Republic

3 Department of Mathematics and Statistics,Georgia State University, Atlanta, GA 30303, USA

Received 7 August 2013Accepted 8 August 2013

1. IntroductionThe complementary basic matrices originated in [2] as follows. Let C1, C2, . . . , Cn−1 be 2×2 matrices. For k = 1, . . . , n−1,define n× n matrices Gk byGk =

Ik−1Ck

In−k−1

. (1)Then for permutations (i1, . . . , in−1) of (1, . . . , n− 1), products of the form

Gi1Gi2 · · ·Gin−1 (2)are called complementary basic matrices, CB-matrices for short. CB-matrices and the associated “complementary zig-zagshapes" were investigated in detail in [3].In [4] connections between the 2 × 2 matrices Ck and the resulting CB-matrix ∏

Gk were explored. In particular,properties that are inherited from the Ck to the ∏Gk were enumerated. Two situations were considered. The first is forthe ordinary real CB-matrices and the second is for the corresponding sign pattern matrices.In two more recent papers [5] and [6], an extended version of (1) and (2) was given as follows. Let A1, A2, ..., As bematrices of respective orders k1, k2, . . . , ks, ki ≥ 2 for all i. Denote n = ∑s

i=1 ki − s + 1, and form the block diagonalmatrices G1, G2, . . ., Gs as follows:G1 = [

A1 00 In−k1]

, G2 = Ik1−1 0 00 A2 00 0 In−k1−k2+1

, . . . , Gs−1 = In−ks−1−ks+1 0 00 As−1 00 0 Iks−1

, Gs = [In−ks 00 As

].

∗ E-mail: fiedler@math.cas.cz† Research supported in part by grant GACR 201/09/0473.‡ This work was done while the second author was in Prague, Czech Republic, as a visiting researcher.

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Miroslav Fiedler, Frank J. Hall

Then, for any permutation (i1, i2, . . . , is) of (1, 2, . . . , s), we can consider the productGi1Gi2 · · ·Gis . (3)

We call products of this form generalized complementary basic matrices, GCB-matrices for short. We have continuedto use the notation ∏Gk for these more general products. The diagonal blocks Ak are called distinguished blocks andthe Gk are called the generators of the ∏

Gk . The GCB-matrices form a rich class of matrices. Particularly, in [6],[8], and [9], interesting spectral and permanental properties of GCB-matrices were exhibited, and in [7], graph theoreticconnections with the GCB-matrices were explored.Notice that for a GCB-matrix ∏Gk , any two distinguished blocks Ak and Ak+1 over-lap in precisely one diagonalposition. The purpose of this paper is to extend this idea to the situation where the blocks Ak and Ak+1 can over-lapin more than one position of the matrices. The diagonal positions in a distinguished block will be called essential, theremaining ones inessential. Let us assign to the corresponding generator matrix G an interval J (the support of G) inthe ordered set of indices N = (1, 2, . . . , n) consisting of the indices corresponding to the essential positions.Suppose now that we have t ≥ 1, n×n such generators G1, G2, . . . , Gt the corresponding intervals J1, J2, . . . , Jt of whichare chained as follows: J1 contains the number 1, J2 (if t > 1) has with J1 a non-void intersection, J3 has with J2 anon-void intersection but is disjoint with J1, in general Jk+1 has a non-void intersection with Jk but is disjoint with Jk−1for k = 2, . . . , t − 1; in addition, Jt contains the number n.

Definition 1.We consider products of the form Gi1Gi2 · · ·Git , where (i1, i2, . . . , it) is a permutation of (1, 2, . . . , t), and call themfactorizable matrices, F-matrices for short.We continue to use the notation ∏

Gk for the F-matrices. Observe that for the previously defined [2] CB-matrices J1was the interval (1, 2), J2 was the interval (2, 3), ..., Jn−1 was the interval (n − 1, n). For the generalized CB-matrices(GCB-matrices for short) defined in [6] the sizes of Jk could be greater than two but the intersections Jk ∩ Jk+1 for alladmissible k had just one element.Let us note that even in our general case all indices 1, . . . , n enter in the union J1 ∪ J2 ∪ · · · ∪ Jt ; in fact, each at mosttwice.From our construction of the numbering of the generators, we derive the following observation.Observation 2.The generators Gi and Gk commute if |i− k | > 1.

Having this in mind, we introduce a simplification of the F-matrices. We move Gt in a given F-matrix Π = ∏Gk to theleft as far as possible without changing the product, ie., either to the first position, or until it meets Gt−1. In the lattercase, move the pair Gt−1Gt as far to the left as possible, and continue until such product is in front. Then take thelargest remaining index and move the corresponding G to the left, etc. In this way, we arrive at the left normal form of

the matrix∏

Gk : Π = (Gt1Gt1+1 · · ·Gt)(Gt2Gt2+1 · · ·Gt1−1) · · · (G1G2 · · ·Gtr−1), (4)as given in [6]. We note here that the parentheses are only formally indicated. However, they separate the originalproduct into so called terms - the expressions in the parentheses.Example 3.The left normal form of

G6G2G4G7G8G1G5G3 (5)is (G6G7G8)(G4G5)(G2G3)(G1);

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Factorizable matrices

it has four terms. The left normal form of the reverse product

G3G5G1G8G7G4G2G6is (G8)(G7)(G5G6)(G3G4)(G1G2)with five terms.In comparison, we could present the right normal form of the product

∏Gk by moving Gt (and everything next) to theright. It leads to the same result as if we start moving the smallest, ie., G1, to the left, etc. Thus in Example 3 the rightnormal form of the matrix (5) is (G2G1)(G4G3)(G6G5)(G7)(G8).

Remark 4.A given F-matrix ∏Gk can also be written as ΠP , where P is a permutation (i1, i2, . . . , it) of (1, 2, . . . , t). The dependenceof ΠP on P is the following: for P1 and P2 they are equal if P1 and P2 have the same precedence properties, ie., if

k ∈ {1, . . . , t − 1}, then k precedes k + 1 in P1 if and only if k precedes k + 1 in P2. This is clearly an equivalencerelation in the class of all permutations. It is immediate that two such products are equivalent if and only if they havethe same left normal form, or if they have the same right normal form. We denote in the sequel by S the set of indices kin {1, 2, . . . , t − 1} for which k precedes k + 1, and by S the set of the complementary indices. It is easily seen that Sis the set of the first indices in the terms in the right normal form (of course, except t), S is the set of the last indices interms in the left normal form (except t). So S = {2, 4, 6, 7} in (5) of Example 3. The number of elements in S is thus byone less than the number of terms in the right normal form, the number of elements in S is by one less than the numberof terms in the left normal form.Observation 5.The transpose matrix of an F-matrix is also an F-matrix. The precedence properties of the transpose are reversed.

Observation 6.In the same notation as in Definition 1, for a given system of generators, there are at most 2t−1 F-matrices none ofwhich can be identified with another product by exchange of commuting generators.

Remark 7.By the definition, every square matrix itself can be considered as an F-matrix with one generator. More interesting casesare those with at least two generators. We note that every F-matrix has a “maximum" number of possible generators ina factorization, and in a given context, also an “explicit" number of generators.2. Characteristic Properties of F-matricesTheorem 8.Let S = {G1, G2, . . . , Gt} be generators of an n × n F-matrix. Then, for every permutation P = (i1, i2, . . . , it) of(1, 2, . . . , t), the products ΠP = Gi1Gi2 · · ·Git have the same spectrum.

Proof. It suffices to show that every matrix ΠP = Gi1Gi2 · · ·Git has the same spectrum as the matrix G1G2 · · ·Gt .We use induction with respect to t. The assertion holds trivially for t = 1 and t = 2. Let S = {G1, G2, . . . , Gt} be suchsystem with t > 2 and suppose the assertion holds for all smaller t’s. Since transposition (see Observation 5) changesthe precedence properties, we can assume that t − 1 precedes t in the given permutation. Also, cyclic permutations ofthe factors do not change the spectrum so that we can assume that Gt is the last factor and Gt−1 is the next to the lastfactor. It is easy to see that the given product is then cospectral with Π = Gl1Gl2 · · ·Glt−2 (Gt−1Gt) and since the productGt−1Gt can be considered as a generator whose essential indices are formed by the union of essential indices of Gt−1and Gt , the product Π has t − 1 factors. By the induction hypothesis, Π is cospectral with G1G2 · · ·Gt−2(Gt−1Gt).

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Miroslav Fiedler, Frank J. Hall

Remark 9.The natural ordering corresponding to the identity permutation, ie., G1G2 · · ·Gt , is then in the generalized Hessenbergform in which the strict lower triangular part (ie., the block diagonal excluded) is the same as that of the sum G1 +G2 +· · · + Gt ; it thus contains many zero entries in contrast to the upper triangular part which can be completely filled bynon-zeros. For the permutation GtGt−1 · · ·G2G1, we obtain a transpose shape to the previous generalized Hessenbergform. The approach using the left normal form allows to find the structure of the matrix ∏

Gk . Indeed, each termcorresponds to one diagonal block having the generalized Hessenberg form (possibly with only one matrix G1) whereasthe product of the terms behaves like the transpose of a (big) generalized Hessenberg form whose generating matricesare the previous blocks.We have thus the result about the shape (ie., the structure of the non-zero entries) of the matrix ∏

Gk .Theorem 10.The strict upper triangular part of

∏Gk is the same as the strict upper triangular part of the sum of the terms in the

left normal form of∏

Gk , and the strict lower triangular part of∏

Gk is the same as the strict lower triangular part ofthe sum of the terms in the right normal form of

∏Gk .

Proof. The result about the strict upper triangular part follows from the argument in Remark 9. The second partfollows from the preceding if we apply it to the transpose of the reverse product and transpose the result.As we already noticed, every square matrix itself can be considered as an F-matrix with one generator. More interestingcases are those with at least two generators. We can thus ask the question of how we recognize if a square matrix is aproduct of two such generators. The result is described in the following.Theorem 11.Let p, q and n be positive integers, where p + q < n. Let A, B be n × n partitioned matrices (we could call themoverlapping)

A = [A0 00 Ip

], B = [

Iq 00 B0]

. (6)Then the product C = AB is a matrix the lower-left corner p×q submatrix of which is a zero matrix and the complementary(n− p)× (n− q) submatrix of which has rank at most n− p− q.Conversely, let an n× n matrix C have the property that its lower-left corner p× q submatrix is a zero matrix and thecomplementary (n− p)× (n− q) submatrix has rank at most n− p− q. Then C can be factorized as C = AB, where Aand B have the forms in (6).

Proof. For the first part, write A0 as a block matrix [A11 A12A21 A22

], where A22 is (n− p− q)× (n− p− q), and B0 as[

B22 B23B32 B33

], where B22 is also (n− p− q)× (n− p− q). Then the product is

AB = A11 A12B22 A12B23

A21 A22B22 A22B230 B32 B33

. (7)The zero matrix has clearly dimension p× q and the complementary matrix rank at most n− p− q.To prove the converse, write C in the partitioned form

C = [C11 C120 C22

],

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Factorizable matrices

where the zero block is p×q. By our assumption, C12 can be factorized as PQ, where P is (n− p)× (n− p−q) and Q

is (n− p− q)× (n− q). It follows that C = AB, where A and B have the form (6) with A0 = [C11 P ] and B0 = [QC22

].

Remark 12.Let us note again that even in our general case of F-matrices all indices 1, . . . , n enter in the union J1 ∪ J2 ∪ · · · ∪ Jt , infact, each at most twice. To handle F-matrices with more than two factors, it is advisable to restrict oneself to the caseof formally simple products, ie., products for which neither (1, 1) nor (n, n) belong to two generators but exactly to onegenerator. Explicitly, if Jk = (jk , jk + 1, . . . , lk ), k = 1, 2, . . . , t, then1 = j1 < j2 ≤ l1 < j3 ≤ l2 < j4 ≤ · · · < jt ≤ lt−1 < lt = n. (8)

Formally simple F-matrices with at least two factors have a certain structure and contain always some zero entries.To better understand such structure, we now characterize the set of all possible formally simple products of Boolean(0, 1) matrices with Boolean multiplication which arise if all the distinguished diagonal blocks are full matrices. It isimmediate that such product is determined by the sizes and location of the blocks (or, supports) and by the permutation.Example 13.Suppose that G1, G2, G3, and G4 are the described full Boolean generators of an 10 × 10 matrix, G1 with supportJ1 = (1, 2, 3), G2 with J2 = (2, 3, 4, 5, 6, 7), G3 with J3 = (6, 7, 8, 9) and G4 with J4 = (9, 10). Then the product

G1G3G4G2 =

1 1 1 1 1 1 1 0 0 01 1 1 1 1 1 1 0 0 01 1 1 1 1 1 1 0 0 00 1 1 1 1 1 1 0 0 00 1 1 1 1 1 1 0 0 00 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 10 0 0 0 0 0 0 0 1 1

; (9)

it has the left normal form (G3G4)(G1G2) and the right normal form G1(G3G2)G4.Observe that this product has a certain zig-zag shape. This means that if there is a zero entry in the upper triangularpart of a square matrix, then all entries to the right and up are equal to zero, too, and if there is a zero in the lowertriangular part, then all entries to the left and down are also equal to zero. Observe that the framed zero entries havethe property that they generate the zero structure (the zero structure is the minimal zig-zag structure containing thosezeros) and, in addition, the complements of the corresponding zero blocks have Boolean rank one. As claimed in Theorem10, the strict upper triangular part of (9) is the same as that of the sum of G3G4 and G1G2. Clearly, G3G4 has supportJ3 ∪ J4 = (6, 7, 8, 9, 10), G1G2 support (1, . . . , 7). There is just one framed entry (5, 8); its first index is 5 as the lastnumber before J3 ∪ J4 and the second is 8 as the first number after J1 ∪ J2. One framed entry is there in coincidence withRemark 4 since the number of terms in the left normal form being 2, the number of elements in S is 1. Similar factsfollow in the strict lower triangular part of (9).We now prove that these are characteristic properties of such Boolean matrices.Theorem 14.Suppose that G1, . . . , Gt , t ≥ 2, are n × n Boolean matrices as described above which form a formally simple systemas generators, Gk corresponding to a support Jk , k = 1, . . . , t. Let P be a permutation of (1, . . . , t). Then the matrix

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Miroslav Fiedler, Frank J. Hall

A = ∏P Gk has a zig-zag shape with t − 1 framed zeros obtained as follows. Define S as the subset of indices

{1, . . . , t−1} for which p precedes p+1 in P , and let S be the complement. For k = 1, . . . , t, let Jk = (jk , jk +1, . . . , lk ).Thus, of course, j1 = 1, lt = n. Then, the t−1 framed entries, for k = 1, . . . , t−1, have index pairs either (lk +1, jk+1−1),if k ∈ S, or (jk+1 − 1, lk + 1), if k ∈ S. In addition, the set of zeros in A is the union of blocks with corner entries in theframed positions.Conversely, suppose that A = [aik ] is an n × n Boolean matrix such that s entries in the positions (pk , qk ), none ofthem in the main diagonal and satisfying max(pk , qk ) ≤ min(pk+1, qk+1) , are corners of zero blocks in the correspondingtriangular part, and in addition, the complementary block has Boolean rank one. Then A is a Boolean product of s + 1generators forming a formally simple system, each of the generators having ones in the corresponding distinguishedblock.

Proof. To prove the first part, we use induction with respect to t. The case t = 1 being trivial, suppose that t > 1and that the assertion is correct for smaller positive t’s. Suppose first that t − 1 precedes t in P . We then can assumethat t is in P in the last position and the remaining indices form a permutation P0 with t − 1 factors. The productA = ∏

P0 Gk satisfies the induction hypothesis, of course in the class of m×m matrices, where m is the greatest indexin Jt−1. Thus A has the zig-zag shape with t − 2 framed zeros as described above. We can then apply Theorem 11 forthe product A = [

A 00 I

]×Gt . It follows that one obtains a single further framed position (lt−1 +1, jt −1) (indeed, t−1

is in S) since in (7) the upper-right corner entry of the zero block has this position.The case that t precedes t − 1 leads similarly to the case Gt

[A 00 I

] in which the zero block is in the upper triangularpart.To prove the second part, it suffices to choose j1 = 1, jk = min(pk−1, qk−1) + 1, k = 2, . . . , s + 1, lk = max(pk , qk ) − 1,k = 1, . . . , s, ls+1 = n. The corresponding Boolean product of s + 1 full Boolean generators with supports Jk =(jk , jk + 1, . . . , lk ), k = 1, . . . , s + 1, and permutation P for which k ∈ S if and only if pk > qk is then A.In the proof, we could, of course, also use Theorem 10. The result implies the following corollary the second part ofwhich is easily proved by induction and Theorem 11.Corollary 15.All products with fixed dimensions of the generators and fixed permutation have the same upper and lower bounds withinwhich the non-zero entries can occur. In addition, a necessary and sufficient condition that a matrix within these boundsis such product is that for each framed zero, the rank of the complement of the zero block having this zero as its corner,is less than or equal to the number of the diagonal entries in the complement.

3. Concluding RemarksAn n×n matrix with no zero entry cannot of course be factored as an F-matrix with more than one generator. However,even if an n× n matrix C has just one zero entry (placed in the lower left corner by permutation equivalence), we seeby Theorem 11 that C can be factored as an F-matrix AB if and only if the complementary submatrix of order n − 1is singular. In fact, the idea in the converse of Theorem 11 led the first author to factorize the companion matrix as aCB-matrix, [1].References[1] M. Fiedler, A note on companion matrices, Linear Algebra Appl. 372 (2003), 325–331.[2] M. Fiedler, Complementary basic matrices, Linear Algebra Appl. 384 (2004), 199–206.[3] M. Fiedler, Intrinsic products and factorizations of matrices, Linear Algebra Appl. 428 (2008), 5–13.[4] M. Fiedler and F. J. Hall, Some inheritance properties for complementary basic matrices, Linear Algebra Appl. 433(2010), 2060–2069.[5] M. Fiedler and F. J. Hall, G-matrices, Linear Algebra Appl. 436 (2012), 731–741.

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[6] M. Fiedler and F. J. Hall, A note on permanents and generalized complementary basic matrices, Linear Algebra Appl.436 (2012), 3553–3569.[7] M. Fiedler and F. J. Hall, Some graph theoretic properties of generalized complementary basic matrices, LinearAlgebra Appl. 438 (2013), 3365–3374.[8] M. Fiedler and F. J. Hall, Combinatorial aspects of generalized complementary basic matrices, to appear in CentralEuropean Journal of Mathematics.[9] M. Fiedler, F. J. Hall, and M. Stroev, Permanents, determinants, and generalized complementary basic matrices, toappear in Operators and Matrices.

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