The Moore-Penrose inverse of a free matrix

Electronic Journal of Linear Algebra Volume 16 Article 19 007 The Moore-Penrose inerse of a free matrix Thomas Britz sonofnob@gmailcom Follow this and additional works at: http://repositoryuwyoedu/ela Recommended Citation Britz, Thomas (007), "The Moore-Penrose inerse of a free matrix", Electronic Journal of Linear Algebra, Volume 16 DOI: https://doiorg/101001/1081-8101196 This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact scholcom@uwyoedu

THE MOORE-PENROSE INVERSE OF A FREE MATRIX THOMAS BRITZ Abstract A matrix is free, or generic, if its nonzero entries are algebraically independent Necessary and sufficient combinatorial conditions are presented for a complex free matrix to hae a free Moore-Penrose inerse These conditions extend preiously known results for square, nonsingular free matrices The result used to proe this characterization relates the combinatorial structure of a free matrix to that of its Moore-Penrose inerse Also, it is proed that the bipartite graph or, equialently, the zero pattern of a free matrix uniquely determines that of its Moore-Penrose inerse, and this mapping is described explicitly Finally, it is proed that a free matrix contains at most as many nonzero entries as does its Moore-Penrose inerse Key words Free matrix, Moore-Penrose inerse, Bipartite graph, Directed graph AMS subject classifications 05C50, 15A09 1 The main results A set of complex numbers S is algebraically independent oer the rational numbers Q if p(s 1,,s n ) 0 wheneer s 1,,s n are distinct elements of S and p(x 1,,x n ) is a nonzero polynomial with rational coefficients Lemma 11 Let S 1 and S be finite sets of complex numbers so that each element of S may be written as a rational form in elements of S 1, and conersely If S 1 is algebraically independent, then S is algebraically independent if and only if S 1 S Proof For each finite set of complex numbers S,let dt Q S denote the maximal cardinality of an algebraically independent subset of S and note that dt Q S dt Q Q(S) Each element of S may be written as a rational form in entries of S 1,so S Q(S 1 ) Similarly, S 1 Q(S ); thus, Q(S 1 )Q(S ) If S 1 is algebraically independent,then S 1 dt Q S 1 dt Q Q(S 1 )dt Q Q(S )dt Q S,and the lemma follows A matrix with complex entries is free,or generic,if the multiset of nonzero entries is algebraically independent These nonzero entries may be iewed as indeterminants oer the rational numbers; see [5,Chap 6] Free matrices hae been used to represent objects from transersal theory,extremal poset theory,electrical network theory,and other combinatorial areas; see [,Chap 9] for a partial oeriew The adantage of such representations is that they allow methods from linear algebra to be applied to combinatorial problems,most often ia the connection gien in Theorem 1 below A nonzero partial diagonal of a matrix A is a collection of nonzero entries no two of which lie in the same row or column,and the term rank of A is the maximal size of a nonzero partial diagonal in A Theorem 1 [4,6] The term rank of a free matrix equals its rank Note that the rank and term rank are identical for each submatrix of a free matrix Receied by the editors 7 June 007 Accepted for publication 8 August 007 Handling Editor: Stephen J Kirkland School of Mathematics and Statistics, Uniersity of New South Wales, Sydney, NSW 05, Australia (britz@unsweduau) Supported by a Villum Kann Rasmussen postdoctoral grant and an ARC Discoery Grant 08

The Moore-Penrose Inerse of a Free Matrix 09 For any complex m n matrix A [a ij ],the Moore-Penrose inerse A is the unique matrix that satisfies the following four properties [7,8]: A AA A AA A A (A A) T A A (AA ) T AA If A is a square,nonsingular matrix,then A A 1 Hence,Moore-Penrose inersion generalizes standard matrix inersion For more information on the Moore-Penrose inerse,see [1] and the extensie bibliography therein This article describes combinatorial properties of the Moore-Penrose inerses of free matrices The first two main results,theorems 1 and 14 below,describe how the combinatorial structure of a free matrix A relates to that of the Moore-Penrose inerse A Stronger but more technical results are proed in Sections and For each integer r 1,let A[1 : r] denote the leading principal (ie upper left) r r submatrix of A Furthermore, let B(A) be the bipartite graph with ertices U V and edges { {u i, j } : u i U, j V, a ij 0 } where U {u 1,,u m } and V { 1,, n } are disjoint sets If m n,then let D(A) be the digraph with ertices W {w 1,,w n } and arcs E {(w i,w j ) : a ij 0} Thus,for instance, D(I n ) denotes the digraph with arcs {(w, w) :w W } For any digraph D,let D be the transitie closure of D For (di)graphs D 1 and D,let the notation D 1 D indicate that D 1 is a sub-(di)graph of D Theorem 1 If A is a free matrix, then B(A) uniquely determines B(A ) The aboe result follows immediately from Theorem in Section Theorem 14 Let A be a free matrix of rank r for which D(I r ) D(A[1 : r]) Then A [a ij ] and A [α ij ] can be written as C F S X A and A with C A[1:r], D(I r ) D(C) D(C) D(S), B(F ) B(Y T ),andb(g) B(X T ) A proof of Theorem 14 is gien in Section Corollary 15 The Moore-Penrose inerse A of a free matrix A has at least as many nonzero entries as A In the nonsingular case,theorem 14 has the following simple form Theorem 16 For any nonsingular n n free matrix A such that D(I n ) D(A), the digraph D(A 1 ) is equal to the transitie closure D(A) Proof SinceA 1 is a polynomial in A,we hae D(A 1 ) D(A) D(I n )D(A) Conersely,Theorem 14 asserts that D(A) D(A 1 ) Hence, D(A 1 )D(A) Remark 17 No generality is lost by requiring that D(I r ) D(A[1 : r]) in Theorem 14 Indeed for each free matrix A of rank r,choose any permutation matrices P and Q for which A : PAQsatisfies this condition,and apply the theorem to A Similarly,no generality is lost by requiring that D(I n ) D(A) in Theorem 16 The final main result is expressed as Theorem 18 below It proides necessary and sufficient conditions for the Moore-Penrose inerse A of a free matrix A to be free,and thereby generalizes [,Theorem 1]

10 T Britz Theorem 18 Let A be a free matrix of rank r and let P and Q be permutation matrices for which A : PAQ satisfies D(I r ) D(A [1 : r]) Write A C F and A S X where C A [1 : r] and S are r r matrices The following statements are equialent: 1 A is free; A and A hae an equal number of nonzero (or zero) entries; D(C) D(S), B(F )B(Y T ), B(G) B(X T ),andz 0 Proof Conditions 1 and are equialent by Lemma 11 Since A is free if and only if A is free, A is free if and only if A and A hae equal numbers of zero entries By Theorem 14,this is true if and only if condition is satisfied Corollary 19 If A is a free matrix, then the bipartite graph B(A) determines whether the Moore-Penrose inerse A is also free Proof By Theorem 1, B(A) determines B(A ) and thus whether A and A contain equal numbers of zero entries Now apply Theorem 18 Remark 110 Let A be a free matrix represented as in Theorems 14 and 18 If A is free,then,by these theorems,d(c) D(C) D(S) Example 111 The aboe results are illustrated by Figures 11 and 1,each of which shows a free matrix A and its inerse A Theentriesa ij and α ij represent the nonzero entries of A and A,respectiely; for instance,α 11 a 11 /(a 11 + a 41 ) in Figure 11 The matrices are represented as in Theorems 14 and 18,and the (di)graphs associated to them in these theorems are also shown Note that in each figure D(I r ) D(C) D(C) D(S), B(F ) B(Y T ),and B(G) B(X T ),as asserted by Theorem 14 Note also that A contains at least as many nonzero entries as A,as asserted by Corollary 15 Finally,note that the Moore-Penrose inerse A in Figure 11 is not free but that A in Figure 1 is indeed free By Theorem 18,this may be seen by comparing the numbers of zero entries in A and A or by checking whether the equalities D(C) D(S), B(F )B(Y T ),and B(G) B(X T )hold The structure of the Moore-Penrose inerse of a free matrix Let Q k,l be the family of ordered k-subsets of (1,,l) Consider γ (γ 1,,γ k ) Q k,l For each i γ,let γ i denote the ordered set γ\{i}; alsoforeachi/ γ,let (i; γ) denote the ordered set (i, γ 1,,γ k ) If A is an m n matrix and γ,δ are ordered subsets of (1,,m)and(1,,n),respectiely,then let A[γ δ] denotethe γ δ matrix whose (i, j)th entry equals a γiδ j Note that A[1 : k] :A[1,,k 1,,k] Note also that A[γ δ] is the submatrix of A with rows γ and columns δ,whereas A[i; γ j; δ] has rows and columns ordered (i; γ) and(j; δ),respectiely Theorem 1 [7] Let A be a complex m n matrix A with rank r and let A [α ij ] denote the Moore-Penrose inerse of A Then det A[γ δ]det A[j; γ i; δ] α ij γ Q r 1,m,j / γ δ Q r 1,n,i/ δ ρ Q r,m τ Q r,n det A[ρ τ]det A[ρ τ]

The Moore-Penrose Inerse of a Free Matrix 11 A» C F 6 4 a 11 a 1 0 a 14 0 a a 0 0 0 a 0 0 a 4 0 0 7 5 A» S X α 11 α 1 α 1 α 14 6 0 α α α 4 7 4 0 α α α 4 5 α 41 α 4 α 4 α 44 w 1 w w w w 1 w w w 1 w w w w 1 D(I ) D(C) D(C) D(S) 4 1 4 1 u 1 u u u 4 u 1 u u u 4 B(F ) B(G) B(X T ) B(Y T ) Fig 11 The combinatorial structure of a free matrix and its Moore-Penrose inerse A» C F a 11 a 1 a 1 0 0 0 a 0 0 0 6 0 0 a 0 0 7 4 0 0 0 a 44 a 45 5 0 0 0 0 0» S X A α 11 α 1 α 0 0 0 α 0 0 0 6 0 0 α 0 0 7 4 0 0 0 α 44 0 5 0 0 0 α 54 0 w 1 w w w 4 w 1 w w w 4 4 u 1 u u u 4 1 4 u 4 D(I 4) D(C) D(C) D(S) B(F )B(Y T ) B(G) B(X T ) Fig 1 A free matrix whose Moore-Penrose inerse is also free Throughout the remainder of this section,let A [a ij ] denote a free m n matrix of rank r and let A [α ij ] denote its Moore-Penrose inerse Remark If r 0,thenA A T 0,andB(A) andb(a )haenoedges If r 1,then,by Theorem 1,rows and columns of A may be swapped so that A or A T has the form [ 0] for a nonzero ector and a possibly empty zero matrix,so

1 T Britz A A / Hence, {u i, j } is an edge of B(A ) if and only if {u j, i } is an edge of B(A) Theorem below describes how the bipartite graph of a free matrix A uniquely determines the bipartite graph of the Moore-Penrose inerse A A matching in a graph is a collection of edges no two of which are incident Theorem {u i, j } is an edge of B(A ) if and only if B(A) contains a path p from i to u j of length s +1 with s 0 and a matching with r s 1 edges, none of which is incident to p Proof Ifr 0orr 1,then the proof follows triially from Remark,so suppose that r Assume that α ij 0 By Theorem 1,there are sequences γ Q r 1,m and δ Q r 1,n so that j / γ, i/ δ,and det A[γ δ]det A[j; γ i; δ] 0 Then B(A[γ δ]) and B(A[j; γ i; δ]) each contains a matching with r 1andr edges, respectiely,say E 1 and E Let G denote the bipartite graph with edges E 1 E and delete from G each isolated edge containing neither i nor u j In G,ertices i and u j hae degree 1 and all other ertices hae degree,so G is the disjoint union of some path p from i to u j and some cycles Hence, B(A[j; γ i; δ]) contains p and at least one (perhaps empty) matching E that coers all ertices not in p Thus, there is an integer s 0sothatp has length s +1 and so thate contains r s 1edges Both p and E are contained in B(A) and satisfy the conditions stated in the theorem Now assume that B(A) contains a path p from i to u j i u j1 i1 u j i u js is u j of length s +1 with s 0 and at least one matching with r s 1edges,say { {uj t, i t } : t 1,,r s 1},none of which is incident to p Order the sets {j 1,,j s } {j t : t 1,,r s 1} and {i 1,,i s } {i t : t 1,,r s 1} to form increasing sequences, γ and δ,respectiely Then {a j1i 1,,a jsi s } {a j t i t : t 1,,r s 1} and {a j1i,a ji 1,,a jsi s 1,a jis } {a j t i t : t 1,,r s 1} are nonzero partial diagonals of A[γ δ ]anda[j; γ i; δ ],respectiely Therefore, S det A[γ δ]det A[j; γ i; δ] γ Q r 1,m,j / γ δ Q r 1,n,i/ δ is a sum of signed monomials of degree r 1,one of which is a j1i 1 a jsi s a j 1 i 1 a j r s 1 i r s 1 a j 1ia ji 1 a jsi s 1 a jis a j 1 i 1 a j r s 1 i r s 1 Since A is free,these monomials do not anish,so S 0 By Theorem 1, α ij 0

The Moore-Penrose Inerse of a Free Matrix 1 Proposition 4 Let {u j, i } be an edge of B(A) Then {u i, j } is an edge of B(A ) if and only if {u j, i } is contained in some matching of B(A) with r edges If {u j, i } is contained in eery such matching, then α ij 1 a ji If{u i, j } is contained in no such matching, and E is such a matching, then E contains edges {u i, ie } and {u je, j } for which {u ie, je } is an edge of B(A ) Proof If {u j, i } is contained in some matching E in B(A) withr edges,then u j i is a path that together with the matching E {u j, i } satisfies the conditions in Theorem,so {u i, j } is an edge of B(A ) Conersely,suppose that {u i, j } is an edge of B(A ) By Theorem, B(A) contains a path p from i to u j of length s +1, i u j1 i1 u j i u js is u j, as well as a matching E with r s 1 edges,none of which are incident to p Then E { {u j, i }, {u j1, i1 },,{u js, is } } is a matching in B(A) withr edges, including {u j, i } If {u j, i } is in eery matching in B(A) with r edges,then det A[j; γ i; δ] a ji det A[γ δ] for all γ Q r 1,m and δ Q r 1,n with j / γ, i/ δ By Theorem 1, α ij aji a jia ji 1 a ji If E has no edges incident to u i or j,then E {u i, j } is a matching in B(A) withr +1 edges,so,by Theorem 1,the rank of A is at least r+1,a contradiction Thus,E contains at least one such edge,say {u i, ie } If E does not also contain an edge {u je, j },then (E {u i, ie }) {u i, j } is a matching of B(A) withr edges,one of which is {u i, j },a contradiction Thus,E contains edges {u i, ie } and {u je, j } for some ertices ie and u je Then ie u i j u je is a path from u je to ie of length s +1 with s 1andE {u i, ie } {u je, j } is a matching in B(A) withr s 1 r edges,none of which contain u je, u i, j,or ie By Theorem, {u ie, je } is an edge of B(A ) Then {u je, ie } is not an edge of B(A) since this would imply that ( E {u je, j } {u i, ie } ) {u i, j } {u je, ie } is amatchinginb(a) withr edges,one of which is {u i, j } Example 5 To illustrate the results in this section,let A be as follows: 4 a11 a1 0 a 5 0 a " 1 a 1 a a 11 a 11 c 0 a c a 1 a a 11 c a c # 1 u 1 u u 1 u 1 u A A B(A) B(A ) Here, c a + a and the rank of A is r The graph B(A) contains the path 1 u 1 u of length s +1withs 1 Thenr s 10andTheorem implies that {u 1, } is an edge of B(A ),as is indeed the case The graph B(A) contains precisely two matchings with r edges,namely E 1 { {u 1, 1 }, {u, } } and E { {u 1, 1 }, {u, } } Thus,the edge {u 1, 1 } is contained in all matchings in B(A) withr edgesandthe edge {u 1, } is contained in none of these By Proposition 4, α 11 1 a 11, α 1 0, and the edges {u 1, } and {u 1, } are edges of B(A )

14 T Britz Proof of Theorem 14 Theorem 14 follows triially from the next result Theorem 1 Let A be a free matrix of rank r for which D(I r ) D(A[1 : r]) Then A [a ij ] and A [α ij ] canbewrittenas C F A and A S X where C A[1 : r] Furthermore, let i, j, k be positie integers with max{i, j} r<k, and suppose that there is at least one path in D(C) from w i to w j Then 1 α ij 0; if a ik 0,thenα kj 0; if a kj 0,thenα ik 0; 4 if a jk 0 or a ki 0, then paths from w i to w j in D(C) are also paths in D(S T ) Proof WriteA and A as C F S X A and A G H where C A[1 : r] ands are r r matrices The r 0andr 1 cases follow easily from Remark so suppose that r Since C contains a diagonal of r nonzero entries,theorem 1 implies that H 0 Let w i w i1 w i w js 1 w j be a (perhaps triial) path p from w i to w j in D(C) To proe statement 1,first note that D(I r ) D(C) It follows that the graph B(A) contains the path i u i i1 u i1 i u i is 1 u is 1 j u j from i to u j and the matching { {ut, t } : t {1,,r} {i, i 1,i,,i s 1,j} } By Theorem, B(A ) contains the edge {u i, j },ie,α ij 0 Similarly,if a ik 0,thenB(A) contains the path k u i i1 u i1 i u i is 1 u is 1 j u j from k to u j and the same matching as before By Theorem, α kj 0,which proes statement Statement follows from statement by transposing A To proe statement 4,suppose that a jk 0 Then { {ui, i1 }, {u i1, i },,{u is 1, j }, {u j, k } } { {u t, t } : t {1,,r} {i, i 1,i,,i s 1,j} } is a matching E in B(A) withr edges By Proposition 4, {u j, i } is an edge of B(A ) for each edge {u i, j } of E Therefore,(w j,w i )isanarcofd(s) foreach arc (w i,w j )ofp,so p is a path in D(S T ) The a ki 0 case is proed similarly Acknowledgments I gratefully thank Carsten Thomassen for his adice and encouragement and also thank the anonymous referee for good suggestions

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