More on generalized inverses of partitioned matrices with Banachiewicz-Schur forms

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1 More on generalized inverses of partitioned matrices wit anaciewicz-scur forms Yongge Tian a,, Yosio Takane b a Cina Economics and Management cademy, Central University of Finance and Economics, eijing, Cina b epartment of Psycology, McGill University, Montréal, Québec, Canada bstract. Necessary and sufficient conditions are derived for a 2-by-2 partitioned matrix to ave {1}-, {1,2}-, {1,3}-, {1,4}-inverses and te Moore-Penrose inverse wit anaciewicz-scur forms. s applications, te anaciewicz-scur forms of {1}-, {1,2}-, {1,3}-, {1,4}-inverses and te Moore-Penrose inverse of a 2-by-2 partitioned Hermitian matrix are also given. Keywords: anaciewicz-scur form; partitioned matrix; generalized inverse; Moore-Penrose inverse; Hermitian matrix; Scur complement; matrix rank metod Matematics Subject Classifications: 1503; Introduction Trougout tis paper, C m n stands for te set of all m n matrices over te field of complex numbers. Te symbols, r() and R() stand for te conjugate transpose, te rank and te range (column space) of a matrix C m n, respectively;, denotes a row block matrix consisting of and. Te Moore-Penrose inverse of C m n, denoted by, is defined to be te unique matrix X C n m satisfying te following four matrix equations (1) X = (2) XX = X, (3) (X) = X, (4) (X) = X. Furter, let E = I m and F = I n stand for te two ortogonal projectors. matrix X is called an {i,..., j}-inverse of, denoted by (i,...,j), if it satisfies te i,..., jt equations. Te collection of all {i,..., j}-inverses of is denoted by { (i,...,j) }. Some frequently used generalized inverses of are (1), (1,2), (1,3) and (1,4). Let M be a 2 2 block matrix M =, (1.1) C were C m n, C m k, C C l n and C l k. If in (1.1) is square and nonsingular, ten M can be decomposed as Im 0 0 Im M = 1 C 1 I l 0 C 1. (1.2) 0 I l Tis decomposition is often called itken block-diagonalization formula in te literature, see Puntanen and Styan 9. Moreover, if bot M and are nonsingular, ten te Scur complement S = C 1 is nonsingular too, and te inverse of M can be written in te following form M 1 Im = I m 0 0 I l 0 S 1 C 1 I l = S 1 C 1 1 S 1 S 1 C 1 S 1. (1.3) Tis well-known formula is called te anaciewicz inversion formula for te inverse of a nonsingular matrix in te literature, see Puntanen and Styan 9, and can be found in most linear algebra books. Te two formulas in (1.2) and (1.3) and teir consequences are widely used in manipulating partitioned matrices and teir operations. Wen bot and M in (1.1) are singular, te s: yongge@mail.sufe.edu.cn, takane@psyc.mcgill.ca 1

2 two formulas in (1.2) and (1.3) can be extended to generalized inverses of matrices. reasonable extension of (1.3) wit generalized inverses of submatrices in (1.1) is given by N( (i,...,j), S (i,...,j) In (i,...,j) (i,...,j) 0 I m 0 0 I k 0 S (i,...,j) C (i,...,j) I l = (i,...,j) + (i,...,j) S (i,...,j) C (i,...,j) (i,...,j) S (i,...,j) S (i,...,j) C (i,...,j) S (i,...,j), (1.4) were S = C (i,...,j). Eq. (1.4) is called te anaciewicz-scur form induced from M, see aksalary and Styan 1. It can be seen from (1.4) tat te matrix N( (i,...,j), S (i,...,j) ) varies over te coice of (i,...,j) and S (i,...,j). Let {N( (i,...,j), S (i,...,j) )} denote te collection of all N( (i,...,j), S (i,...,j) ). ltoug te rigt-and side of (1.4) is obtained by replacing inverses wit generalized inverses, it is not necessarily an {i,..., j}-inverse of M. In tis case, it is of interest to investigate relations between generalized inverses of M in (1.1) and te matrix N( (i,...,j), S (i,...,j) ) in (1.4), in particular, to derive necessary and sufficient conditions for N( (i,...,j), S (i,...,j) ) to be generalized inverses of M in (1.1). some autors ave investigated te relations between M (i,...,j) and N( (i,...,j), S (i,...,j) ) for some special coices of {i,..., j}. well-known result asserts tat M = N(, S ) R() R(), R(C ) R( ), R(C) R(S) and R( ) R(S ), (1.5) were S = C ; see, e.g., 1. Oter results can be found, e.g., in 2, 4, 5, 8, 15, 16. In a recent paper 16, we considered relations between M (1) and N( (1), S (1) ) troug te matrix rank metod and obtain te following two rank formulas r M (1) N( (1), S (1) ) (1), M (1) { } = max r(m) r() r C,, r(m) r() r, 0, (1.6) max r M (1) N( (1), S (1) ) (1) M (1) = r r r r C, 2r(). (1.7) 0 C C y setting te rigt-and sides of te two rank equalities to zero, we obtain tat (a) Tere exist (1) and S (1) suc tat N( (1), S (1) ) is a {1}-inverse of M if and only if { r(m) r() + r C,, } r() + r (1.8) olds. (b) Te set inclusion {N( (1), S (1) )} {M (1) } olds if and only if 0 r = r() + r C, and r 0 0 = r() +r 0 C C, (1.9) old, or equivalently, 0 C R (E ) R 0 and R R CF old. 2

3 s an extension of te previous investigation, we derive in tis paper some rank formulas for te difference M (i,...,j) N( (i,...,j), S (i,...,j) ) (1.10) for {1,2}-, {1,3}-, {1,4}-inverses and te Moore-Penroses of matrices, and use te rank formulas to caracterize te following relations {N( (i,...,j), S (i,...,j) )} {M (i,...,j) }, (1.11) {N( (i,...,j), S (i,...,j) )} {M (i,...,j) }. (1.12) In order to establis rank equalities associated wit (1.10), we need a variety of rank formulas for partitioned matrices and generalized Scur complements. Lemma 1.1 (7) Let C m n, C m k, C C l n and C l k. Ten r, = r() + r( (1) r() + r( (1) ), (1.13) r = r() + r( C C (1) r(c) + r( C (1) C ), (1.14) C r = r() + r(c) + r (I C 0 m (1) )(I n C (1) C), (1.15) 0 r = r() + r (1) C C C (1) C (1). (1.16) Lemma 1.2 (12, 14) Let M be as given in (1.1). Ten r( C (1) r() + r C, + r + r (1) C max (1) { r( C (1) r C,, r, r C 0 0 r r 0, 0 C C (1.17) } r(), (1.18) r( C (1,2) r + r C, + r() + max{ r (1,2) 1, r 2 }, (1.19) { } max r( C (1,2) r() + r(), r C,, r, r r(), (1.20) (1,2) C r( C (1,3) r + r r 0 0, (1.21) (1,3) C C { } max r( C (1,3) r r(), r, (1.22) (1,3) C r( C (1,4) 0 r C, + r (1,4) C r, (1.23) 0 C { } max r( C (1,4) r C,, r (1,4) C r(), (1.24) r( C r C r(), (1.25) were 0 r 1 = r r r 0 0 C 0 C C 0, r 2 = r() r C r 0. 3

4 Te following lemma is derived from Lemma 1.2 by setting and C to identity matrices in (1.17), (1.19), (1.21) and (1.23). Lemma 1.3 Let C m n and C n m. Ten (1) r( (1) r( ), (1.26) (1,2) r( (1,2) max{ r( ), r() + r() r() r() }, (1.27) (1,3) r( (1,3) r( ), (1.28) (1,4) r( (1,4) r( ). (1.29) 2 Generalized inverses of partitioned matrices wit anaciewicz-scur forms We first give two rank formulas for te difference M (1,2) N( (1,2), S (1,2) ). Teorem 2.1 Let M and N( (1,2), S (1,2) ) be as given in (1.1) and (1.4), respectively, were S = C (1,2). Ten were (1,2), M (1,2) r M (1,2) N( (1,2), S (1,2) ) = max{r 1, r 2, r 3, 0}, (2.1) max r M (1,2) N( (1,2), S (1,2) ) = { s 1, s 2 }, (2.2) (1,2) M (1,2) r 1 = r(m) 2r() r(), r 2 = r(m) r() r C,, r 3 = r(m) r() r, s 1 = r r 0 C C 0 s 2 = r + r C 0 r r C, 2r(), + r(m) 2r() r() r C, r. Hence, (a) 16 Tere exist (1,2) and S (1,2) suc tat N( (1,2), S (1,2) ) is a {1, 2}-inverse of M if and only if { } r(m) 2r() + r(), r() + r, r() + r C,. (2.3) (b) Te set inclusion {N( (1,2), S (1,2) )} {M (1,2) } olds if and only if (1.9) olds, or 0 r(m r + r C, r() and r = r = r() + r() (2.4) C 0 old. Proof It was sown in 16 tat r M (1,2) N( (1,2), S (1,2) ) = r(m) r() r( C (1,2) ), (2.5) M (1,2) 4

5 so tat (1,2),M (1,2) r M (1,2) N( (1,2), S (1,2) ) = r(m) r() max (1,2) r( C (1,2) ), (2.6) max r M (1,2) N( (1,2), S (1,2) ) = r(m) r() r( C (1,2) ). (2.7) (1,2) M (1,2) (1,2) Substituting (1.19) and (1.20) into (2.6) and (2.7) gives (2.1) and (2.2). Setting te rigt-and side of (2.1) to zero leads to r 1 0, r 2 0 and r 3 0, tat is, (2.3) olds. Setting te rigt-and side of (2.2) to zero leads to s 1 0 or s 2 0. It can be seen from (1.7) tat s 1 0 is equivalent to (1.9). Note tat s 2 in (2.2) can be rewritten as a sum of tree parts ( ) ( ) 0 s 2 = + r() r() r C ( r(m) + r() r C, r + r 0 ), r() r() were eac part is nonnegative. In tis case, Setting s 2 = 0 leads to (2.4). Teorem 2.2 Let M and N( (1,3), S (1,3) ) be as given in (1.1) and (1.4). Ten were Hence, (1,3), M (1,3) r M (1,3) N( (1,3), S (1,3) ) = max{ r 1, r 2 }, (2.8) max r M (1,3) N( (1,3), S (1,3) ) = r 1 + r 3, (2.9) (1,3) M (1,3) r 1 = r, + r C, r, C r 2 = r, + r C, r() r, r 3 = r 0 0 r r(). C (a) 16 Tere exist (1,3) and S (1,3) suc tat N( (1,3), S (1,3) ) is a {1, 3}-inverse of M if and only if R() R(), R C R = {0} and r C, r old. (b) Te set inclusion {N( (1,3), S (1,3) )} {M (1,3) } olds if and only if R() R(), R C R = {0} and R(CF ) R(F ) old. Proof Te following formula r M (1,3) N( (1,3), S (1,3) ) = r, + r C, r() r( C (1,3) ) (2.10) M (1,3) was sown in 16. Substituting (1.21) and (1.22) into (2.10) gives (2.8) and (2.9). Setting te rigtand side of (2.8) to zero, we see tat tere exist (1,3) and S (1,3) suc tat N( (1,3), S (1,3) ) {M (1,3) } if and only if r, + r C, =r and r, + r C, r() + r. (2.11) C 5

6 Note tat r, + r C, r() + r C, r C Hence, te first equality in (2.11) is equivalent to R() R() and r = r, + r C,, C and te second inequality in (2.11) is equivalent to r C, r. Setting te rigt-and side of (2.9) to zero and noting tat te two terms on te rigt-and side of (2.9) are nonnegative, we obtain (b). Te following teorem can be sown similarly. Teorem 2.3 Let M and N( (1,4), S (1,4) ) be as given in (1.1) and (1.4). Ten. were (1,4), M (1,4) r M (1,4) N( (1,4), S (1,4) ) = max{ r 1, r 2 }, max r M (1,4) N( (1,4), S (1,4) ) = r 1 + r 3, (1,4) M (1,4) r 1 = r + r r C C, r 2 = r + r r() r C,, C 0 r 3 = r r C, r(). 0 C Hence, (a) 16 Tere exist (1,4) and S (1,4) suc tat N( (1,4), S (1,4) ) is a {1, 4}-inverse of M if and only if R(C ) R( ), R C R = {0} and r r C, old. (b) Te set inclusion {N( (1,4), S (1,4) )} {M (1,4) } olds if and only if R(C ) R( ), R C R = {0} and R( E ) R( E C ) old. special case of (1.4) corresponding to te Moore-Penrose inverse is given by N(, S + S C S S C S, (2.12) were S = C. Te relations between N(, S ) and {i,..., j}-inverse of M are given in te following teorems. Teorem 2.4 Let M and N(, S ) be as given in (1.1) and (2.12), respectively. Ten r M (1) N(, S ) = r M (1,2) N(, S ) = r(m) r M (1) M (1,2) C. (2.13) Hence, te following statements are equivalent: 6

7 (a) N(, S ) is a {1}-inverse of M. (b) N(, S ) is a {1, 2}-inverse of M. (c) r(m r() + r( C ). (d) r(m r C. (e) R C = R(M) and R C = R(M ). Proof It follows from (1.26) and (1.27) tat M (1) r M (1) N(, S ) = r M MN(, S )M, (2.14) M (1,2) r M (1,2) N(, S ) = max{ r M MN(, S )M, It is also easy to verify tat rn(, S ) + r(m) rmn(, S ) rn(, S )M }. (2.15) rmn(, S rn(, S )M = r() + r(s), r M MN(, S )M = r(m) r() r(s), were S = C. Substituting tese two equalities and (1.25) into (2.14) and (2.15) gives (2.13). Te equivalence of (a), (b), (c) and (d) follows from (2.13). Recall a simple fact tat r(p Q r() R( P R( ) and R(Q R(). pplying tis result to (d) gives te equivalence of (d) and (e). Teorem 2.5 Let M and N(, S ) be as given in (1.1) and (2.12), respectively. Ten r M (1,3) N(, S ) = r, + r C, r M (1,3) C. (2.16) Hence, te following statements are equivalent: (a) N(, S ) is a {1, 3}-inverse of M. (b) r C = r, + r C,. (c) r C = r, + r C,, r C, = r C, and r, = r,. (d) R C R = {0}, R() R() and R C, = R C,. Proof It follows from (1.28) tat It is easy to verify tat r M (1,3) N(, S ) = r M MN(, S ) M. (2.17) M (1,3) MN(, S E S C SS Im 0 C, I l M MN(, S ) M = M I m+l MN(, S ) = M E + E S C E S E S C E S. 7

8 Recall tat elementary block matrix operations (EMOs) don t cange te rank of a matrix. Hence we can derive by elementary block matrix operations tat ( ) r M MN(, S ) M = r M E + E S C E S E S C E S ( ) = r M E 0 0 E S E E = r E S C E S 0 = r r() r(s) (by (1.13)) 0 S C 0 0 = r r() r(s) 0 C 0 = r, + r C, r() r( C ). Tus we ave (2.16) by (1.25). lso note tat r C r, + r C, r, + r C,. pplying tis inequality to (b) leads to te equivalence of (b) and (c). Te equivalence of (c) and (d) is obvious. Te following result can be sown similarly. Teorem 2.6 Let M and N(, S ) be as given in (1.1) and (2.12), respectively. Ten r M (1,4) N(, S ) = r + r r M (1,4) C C. Hence, te following statements are equivalent: (a) N(, S ) is a {1, 4}-inverse of M. (b) r C = r + r. C (c) r C = r C + r (d) R C R, r C = r and r C = {0}, R, = R, and R(C ) R( ). = r. 3 Generalized inverses of partitioned Hermitian matrices Let M be an Hermitian matrix, and partition M as M =, (3.1) were = C m m, C m n and = C n n. Te anaciewicz-scur form induced from M is N( (i,...,j), S (i,...,j) (i,...,j) + (i,...,j) S (i,...,j) (i,...,j) (i,...,j) S (i,...,j) S (i,...,j) (i,...,j) S (i,...,j), (3.2) were S = (i,...,j). pplying te results in Section 2 to (3.1) and (3.2) gives te following results. 8

9 Teorem 3.1 Let M and N( (1), S (1) ) be as given in (3.1) and (3.2). Ten Hence, r M (1) N( (1), S (1) ) = max {r(m) r() r,, 0}, (1), M (1) max r M (1) N( (1), S (1) 0 ) = 2r (1) M (1) 0 2r() 2r,. (a) Tere exist (1) and S (1) suc tat N( (1), S (1) ) is a {1}-inverse of M if and only if r(m) r() + r,. (b) Te set inclusion {N( (1), S (1) )} {M (1) } olds if and only if 0 r 0 = r() + r,. Proof It follows from (1.6) and (1.7) by setting C =. Teorem 3.2 Let M and N( (1,2), S (1,2) ) be as given in (3.1) and (3.2), were S = (1,2). Ten were Hence, r M (1,2) N( (1,2), S (1,2) ) = max{r 1, r 2, 0}, (1,2), M (1,2) max r M (1,2) N( (1,2), S (1,2) ) = { s 1, s 2 }, (1,2) M (1,2) r 1 = r(m) 2r() r(), r 2 = r(m) r() r,, 0 s 1 = 2r 0 s 2 = 2r 0 2r, 2r(), + r(m) 2r() r() 2r,. (a) Tere exist (1,2) and S (1,2) suc tat N( (1,2), S (1,2) ) is a {1, 2}-inverse of M if and only if r(m) { 2r() + r(), r() + r, }. (3.3) (b) Te set inclusion {N( (1,2), S (1,2) )} {M (1,2) } olds if and only if (3.3) olds, or r(m 2r r() and r 0 = r() + r(). Proof It follows from Teorem 2.1 by setting C =. Teorem 3.3 Let M and N( (1,3), S (1,3) ) be as given in (3.1) and (3.2). Ten r M (1,3) N( (1,3), S (1,3) ) = max{ r 1, r 2 }, (1,3), M (1,3) max r M (1,3) N( (1,3), S (1,3) ) = r 1 + r 3, (1,3) M (1,3) 9

10 were r 1 = r, + r 2, r r 2 = r, r(), r 3 = r 0 0 r r()., Hence, (a) Tere exist (1,3) and S (1,3) suc tat N( (1,3), S (1,3) ) is {1, 3}-inverse of M if and only if 2 R() R() and R R = {0}. (b) Te set inclusion {N( (1,3), S (1,3) )} {M (1,3) } olds if and only if 2 R() R() and R R = {0}. Proof It follows from Teorem 2.2 by setting C =. special case of (3.2) corresponding to te Moore-Penrose inverse is N(, S + S S S S, (3.4) were S =. Te relations between N(, S ) and {i,..., j}-inverse of M are given in te following teorems. Teorem 3.4 Let M and N(, S ) be as given in (3.1) and (3.4), respectively. Ten r M (1) N(, S ) = r M (1,2) N(, S 3 ) = r(m) r M (1) M (1,2). Hence, te following statements are equivalent: (a) N(, S ) is a {1}-inverse of M. (b) N(, S ) is a {1, 2}-inverse of M. (c) r(m r() + r( ). 3 (d) r(m r. 2 (e) R = R(M). Proof It follows from Teorem 2.4 by setting C =. Teorem 3.5 Let M and N(, S ) be as given in (3.1) and (3.4), respectively. Ten r M (1,3) N(, S ) = r M (1,4) N(, S 3 ) = r M (1,3) M (1,4) + r r Hence, te following statements are equivalent: (a) N(, S ) is a {1, 3}-inverse of M.. 10

11 (b) N(, S ) is a {1, 4}-inverse of M. 3 (c) r = r + r. 3 (d) R R = {0}, R() R() and R, = R,. Proof It follows from Teorems 2.5 and 2.6 by setting C =. ssume te Hermitian matrix in (3.1) is nonnegative definite, tat is, tere exists a matrix U suc tat M = UU. In tis case, R() R() and R( ) R() and S = (1) = for any (1) old. Relations between {1}- and {1, 2}-inverses of M and N( (1), S (1) ) and N( (1,2), S (1,2) ) were considered by Rode 11. pplying Teorems 3.1, 3.2, 3.3 and (1.5) to te nonnegative Hermitian matrix in (3.1) gives te following result. Teorem 3.6 Let M and N( (1), S (1) ) be as given in (3.1) and (3.2), and assume M is nonnegative definite. lso let S =. Ten, (a) 11 Te set inclusion {N( (1), S (1) )} {M (1) } always olds. (b) 11 Te set inclusion {N( (1,2), S (1,2) )} {M (1,2) } always olds. (c) Te following statements are equivalent: (i) Te set inclusion {N( (1,3), S (1,3) )} {M (1,3) } olds. (ii) Te set inclusion {N( (1,4), S (1,4) )} {M (1,4) } olds. (iii) M = N(, S ). (iv) R R = {0}. n {i,..., j}-inverse of a square matrix is said to be an Hermitian {i,..., j}-inverse of and is denoted by (i,...,j), if it Hermitian. It is easy to verify tat any Hermitian matrix always as an Hermitian {i,..., j}-inverse for any give set {i,..., j}. Furter, te Hermitian anaciewicz- Scur form induced from te Hermitian matrix M in (3.1) is defined to be (i,...,j) + (i,...,j) S (i,...,j) (i,...,j) (i,...,j) S (i,...,j) N( (i,...,j), S (i,...,j) S (i,...,j) (i,...,j) S (i,...,j), (3.5) were S = (i,...,j). In order to caracterize relations between M and N( (i,...,j), S (i,...,j) ), we need to know te extremal ranks of (i,...,j) wit respect to (i,...,j), wic now are open problems. 4 Generalized inverses of a bordered Hermitian matrix Setting = 0 in (3.1) gives M =, (4.1) 0 were C m m is nonnegative definite and C m n. Tis matrix occurs widely in various problems in matrix teory, in particular, in regression analysis. Te anaciewicz-scur form induced from M is N( (i,...,j), S (i,...,j) (i,...,j) + (i,...,j) S (i,...,j) (i,...,j) (i,...,j) S (i,...,j) S (i,...,j) (i,...,j) S (i,...,j), (4.2) were S = (i,...,j). pplying te results in Section 2 to (4.1) and (4.2) gives us te following results. 11

12 Teorem 4.1 Let M and N( (1), S (1) ) be as given in (4.1) and (4.2). Ten te following statements are equivalent: (a) Tere exist (1) and S (1) suc tat N( (1), S (1) ) {M (1) }. (b) Te set inclusion {N( (1), S (1) )} {M (1) } olds. (c) Tere exist (1,3) and S (1,3) suc tat N( (1,3), S (1,3) ) {M (1,3) }. (d) Te set inclusion {N( (1,3), S (1,3) )} {M (1,3) } olds. (e) Tere exist (1,4) and S (1,4) suc tat N( (1,4), S (1,4) ) {M (1,4) }. (f) Te set inclusion {N( (1,4), S (1,4) )} {M (1,4) } olds. (g) M = N(, S ). () R() R(). Teorem 4.2 Let M and N( (1), S (1) ) be as given in (4.1) and (4.2). Ten te following statements are equivalent: (a) Tere exist (1,2) and S (1,2) suc tat N( (1,2), S (1,2) ) {M (1,2) }. (b) Te set inclusion {N( (1,2), S (1,2) )} {M (1,2) } olds. (c) r, 2r() r() or R() R(). 5 Concluding remarks If in (1.1) is square and nonsingular, ten te matrix in (1.1) can also be decomposed as Im M = 1 1 C 0 Im 0 0 I l 0 1. (5.1) C I l If bot M and are nonsingular in (1.1), ten te Scur complement T = 1 C is nonsingular, too, and te inverse of M can also be written as M 1 T = 1 T CT CT 1 1. (5.2) y symmetry, anoter type of anaciewicz-scur form induced from M is given by K( (i,...,j), T (i,...,j) T (i,...,j) T (i,...,j) (i,...,j) (i,...,j) CT (i,...,j) (i,...,j) + (i,...,j) CT (i,...,j) (i,...,j), (5.3) were T = (i,...,j) C. ltoug (1.3) and (5.2) are identical, (1.4) and (5.3) are not necessarily te same. pplying te results in te previous sections to (5.3), we can derive various conclusions on relations between M and K( (i,...,j), T (i,...,j) ). Furtermore, it is of interest to give necessary and sufficient conditions for te following equality N( (i,...,j), S (i,...,j) K( (i,...,j), T (i,...,j) ) (5.4) to old, or equalities of submatrices in tem to old. ll te results in tis paper on partitioned matrices and anaciewicz-scur forms induced from te matrices can be used to furter study various problems related to block matrices and teir generalized inverses. cknowledgements. We are grateful to anonymous referees for teir elpful comments and suggestions on an earlier version of tis paper. 12

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ELA

ELA Electronic Journal of Linear Algebra ISSN 181-81 A publication of te International Linear Algebra Society ttp://mat.tecnion.ac.il/iic/ela RANK AND INERTIA OF SUBMATRICES OF THE MOORE PENROSE INVERSE OF