Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M

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1 International Journal of Mathematics And its Applications Volume 4, Issue 4 (2016), ISSN: Available Online: International Journal of Mathematics Applications And its ISSN: International Journal of Mathematics And its Applications Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M Research Article D.Krishnaswamy 1 and S.Anusuya 1 1 Department of Mathematics, Annamalai University, Annamalai Nagar, Tamilnadu, India. Abstract: Let K be a skew field and K n n be the set of all n n matrices over K. The purpose of this paper is to give some necessary ( ) and sufficient conditions for the existence and the representations of the minkowski inverse of the block matrix A B under some conditions. C D MSC: 15A09, 65F20. Keywords: Skew fields, Block matrix, Minkowski inverse. c JS Publication. 1. Introduction Let K be a skew field, C be the complex number field, K m n be the set of all m n matrices over K, and I n be the n n identity matrix over K. For a matrix A K n n, the matrix X K n n satisfying A k XA = A K, XAX = X, (AX) = AX and (XA) = XA is called the Minkowski inverse of A and is denoted by X = A m. If A m exists if and only if rank(a) = rank(a 2 ) and R(AA m ) = R(A m ) = R(A). we denote I AA m by A π. A matrix A C n n is said to be EP, if AA = A A. A matrix A C n n is G-unitary, if AA = A A = I. The Minkowski Inverse of block matrices has numerous applications in Game Theory, matrix theory such as singular differential and difference equation. In 1979, S. campbell and C. Meyer proposed an open problem to find an explicit representation for the Drazin inverse of a 2 2 block matrix A B, where the blocks A and D are supposed to be square matrices but their sizes need not be the same. A C D simplified problem to find an explicit representation of the Drazin (group) inverse for block matrix A B (A is square, 0 C 0 is null matrix) was proposed by S. Campbell in This open problem was motivated in hoping to find general expressions for the solutions of the second order system of the differential equations Ex (t) + F x (t) + Gx(t) = 0(t 0), where E is a singular matrix. Detailed discussions of the importance of the problem can be found in [11]. 67

2 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M In this paper, we give the sufficient conditions or the necessary and sufficient conditions for the existance and the representations of the minkowski inverse for block matrix A B (A, Bn K n n ) when A and B satisfy one of the following C D conditions: (1) B m and (B π A) m exist; (2) B m and (AB π ) m exist; (3) B m exists and BAB π = 0; (4) B m exists and B π AB = Preliminaries Lemma 2.1. Let A K n n Then A has a Minkowski Inverse if and only if there exist G-unitary matrices P K n n and A 1 K n n such that A = P A1 0 P and A m = P Am 1 0 P, where rank(a) = r. Proof. Since rank(a) = r, there exists G-unitary matrices P K n n and A K n n, such that A = P A1 0 P, X = P Am 1 0 P (1) AXA = P A1 0 P.P Am 1 0 P.P A1 0 P = P A1Am 1 A 1 0 P = P A1 0 P = A (2) XAX = P Am 1 0 P.P A1 0 P.P Am 1 0 P = P Am 1 0 P = X (3) (AX) = (XA) = (P ) (A1Am 1 ) 0 P. G1 0 P A1Am 1 0 P 0 I n 1 Gm I Similarly, (XA) = (P ) (A1Am 1 ) 0 P. Therefore (AX) = (XA). Hence X = A m. = (P ) G1(A1Am 1 )G 1 0 P = Lemma 2.2 ([? ]). Let A, G K n n, ind(a) = K. Then G = A D if and only if A K GA = A K, AG = GA, rank(g) rank(a K ). Lemma 2.3. Let A B S = B π AB π, A, B X n n. (i) B m and (B π A) m exist in M then S m and M m exist in M. (ii) If B m exist in M and BAB π = 0, then M m exist if and only if (AB π ) m exist in M. Proof. Suppose rank(b) = r. Applying Lemma 2.1, there exist unitary matrix P C n n and invertible matrix B 1 k r r, 68

3 D.Krishnaswamy and S.Anusuya such that B = p B1 0 and B m = p B First we can find B π, B π = I BB m = Ir 0 p B1 0.p B I n r = Ir 0 p 0 I n r B1B = Ir 0 p I 0 0 I n r = ppr 0 p I 0 0 (pp ) n r = p Ir 0 p I 0 0 I n r B π = p 0 I n r (i) Because (B π A) m exist in M. We have rank(b π A) = rank(b π A) 2. That is B π A = p p A1 A2 0 I n r A 3 A 4 = p A1 A2 0 I n r A 3 A 4 = p A A 4 = p A 3 A 4 B π A = p A 3 A 4 Therefore rank(b π A) = rank(a 3A 4) (1) (B π A) 2 = p p A 3 A 4 A 3 A 4 = p. Therefore A 3A 4 (A 4) 2 rank(b π A) 2 = rank(a 3A 4 A 4) (2) 69

4 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M Equating (1) and (2), we get rank(a 3 A 4) = rank(a 3A 4 A 2 4) rank(a 4(A 3 A 4)) rank(a 4) and rank(a 3 A 4) rank(a 4) We have rank(a 3 A 4) = rank(a 4), so there exist a matrix x k (n r) r such that A 3 = A 4X. Because order(a 3) = (n r) r and order(a 4X) = (n r) r corresponding entries we also equal. We get rank(a 4) = rank(a 2 4). Therefore A m exist in M. Nothing that S = B π AB π = p p A1 A2 p 0 I n r A 3 A 4 0 I n r = p A1 A2 0 I n r A 3 A 4 0 I n r = p A A 4 0 I n r = p A 3 A 4 0 I n r = p = p A 4 0 A 4 S = p 0 A 4 Therefore rank(s) = rank(a 4). Similarly, S 2 = p rank(s 2 ) = rank(a 2 4). Hence rank(a 4) = rank(a 2 4). 0 A 4 A 1 A 2 B 1 0 Which implies rank(s) = rank(s 2 ). Thus rank(s m A 3 A 4 ) exist in M. Since rank(m) = rank = B 1 0 B A 4 rank By using rank(b 1) = r. Therefore rank(b 1) = r. Also rank(m) = 2r + rank(a 4) and we B

5 D.Krishnaswamy and S.Anusuya can find rank(m 2 ) By A 3 = A 4X, we get, M 2 = A B A B = A2 + B 2 AB = rank A2 + B 2 AB BA B 2 BA B 2 = rank A2 ABB m B + B B B1 2 + A 2A 3 A 2A 4 = rank A 4A 3 A 2 4 B1 2 0 rank(m 2 0 A 2 1 ) = B1 2 0 rank(m 2 ) = 2r + rank(a 2 4) and rank(m) = rank(m 2 ). That is M m exist. (ii) If BAB π = 0, then A 2 = 0.Thus AB π = p 0 A 4 A 1 0 B 1 0 B 1 0 A 3 A 4 0 A 4 rank(m) = rank = rank = 2r + rank(a 4). B 1 0 B 1 0 Similarly, rank(m 2 ) = r A2 + B 2 AB = rank A2 ABB m B1 2 0 BA 0 = rank A 4A 3 A 2 BA B 2 4. By A3 = A4X, we get, B rank(m 2 0 A 2 1 ) = B1 2 0 B1 2 0 rank(m 2 0 A 2 1 ) = B1 2 0 rank(m 2 ) = 2r + rank(a 2 4) As AB π exist in M, we can get rank(a 4)=rank(A 2 4). Thus rank(m)=rank(m 2 ). That is M m exists. Conversely assume that M m exist if and only if rank(m 2 ) = rank(m). rank(a 4) = rank(a 2 4). That is (AB π ) m exist in M. Hence Proved. 71

6 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M Lemma 2.4. Let A, B K n n, S = B π AB π suppose B m and (B π A) m exists in M then s m exist in M and the following conclusion holds: (i) B π As m A = B π A (ii) B π AS m = S m AB π (iii) BS m = S m B = B m s m = S m B m = 0. Proof. Suppose rank(b) = r, Applying Lemma 2.1 there exist G-unitary matrix p K n n and B 1 K r r such that B = p B1 0 and B m = p Bm 1 0. Let A = p A1 A2, where A 1 K r r, A 2 K r (n r), A 3 A 3 A 4 K (n r) r, A 4 K (n r) (n r). From Lemma 2.3 (i), we get S m exist in M and S m = p 0 A 4 (i) B π AS π A = p p p A1 A2 A 3 A 4 0 A m 4 A 3 A 4 = p A1 A2 A 3 A 4 0 A m 4 A 3 A 4 = p A1 A2 0 A 4A m 4 A 3 A 4 = p A1 A2 0 A 4A m 4 A 3 A 4 = p A1 A2 0 A 4A m 4 A 3 A 4 = p A 4A m 4 A 3 A 4A m 4 A 4 = p 0 + A 4A m 4 A 3 A 4A m 4 A 4 = p A 3 A 4 (ii) B π AS m = p p A 3 A 4 0 A m 4 = p A 3 A 4 0 A m 4 = p A 4A m 4 72

7 D.Krishnaswamy and S.Anusuya S m AB π = p p A1 A2 p 0 A m 4 A 3 A 4 0 I n r = p A1 A2 0 A m 4 A 3 A 4 0 I n r = p A m 4 A A m 4 A 4 0 I n r = p A m 4 A 3 A m 4 A 4 0 I n r = p A m 4 A 4 S m AB π = p 0 A m 4 A 4 Therefore B π AS m = S m AB π = p. 0 A m 4 A 4 BS m = p B1 0 p 0 A m 4 = p B1 0 0 A m 4 = p = Similarly, S m B = 0. B m S m = p Bm 1 0 p = p = 0. 0 A m 4 Therefore B m S m = 0. Similarly, we can obtain S m B m = 0. Therefore BS m = S m B = B m S m = S m B m. Hence the Lemma. 3. Main Results Theorem 3.1. Let M = A B, where A, B K n n. Suppose B m and (B π A) m exists in M then M m exist in m and M m = U11 U12, where U 21 U 22 U 11 = S m + (S m A I)BB m AB π AS m (S m A I)BB m AB π U 12 = B m S m AB m U 21 = B m B m AS m + B m A(S m A I)BB m AB π B m A(S m A I)BB m AB π AS m 73

8 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M U 22 = B m AS m AB m B m AB m S = B π AB π. Proof. The existence of M m and S m in M have given in Lemma 2.3(i). Let X = U11 U12, then MX = A B U11 U12 AU11 + BU21 = U 21 U 22 U 21 U 22 We prove X = M m, AU12 + BU22. AU 11 + BU 21 = A(S m + (S m A I)BB m AB π AS m (S m A I)BB m AB π ) + B(B m B m AS m + B m A(S m A I)BB m AB π B m A(S m A I)BB m AB π AS m ) = AS m + A(S m A I)BB m AB π AS m A(S m A I)BB m AB π ) + BB m BB m AS m + BB m A(S m A I)BB m AB π BB m A(S m A I)BB m AB π AS m ) = BB m + (I BB m )AS m + (I BB m )A(S m A I)BB m AB π AS m (I BB m )(AS m A I)BB m AB π = BB m + B π AS m + B π A(S m A I)BB m AB π AS m B π A(S m I)BB m AB π = BB m + B π AS m + B π ABB m AB π AS m B π ABB m AB π AS m B π ABB m AB π + B π ABB m AB π = B π AS m + BB m and U 11A + U 12B = (S m + (S m A I)BB m AB π AS m (S m A I)BB m AB π )A + (B m S m AB m )B = S m A + (S m A I)BB m AB π AS m A (S m A I)BB m AB π A + B m B S m AB m B = S m A + S m ABB m AB π AS m A BB m AB π AS m A S m ABB m AB π A + BB m AB π A + B m B S m AB m B = S m A(I BB m ) + S m ABB m AB π A BB m AB π A S m ABB m AB π B + BB m AB π A + B m B = S m AB π + BB m = BB m + B π AS m Therefore AU 11 + BU 21 = U 11A + U 12B AU 12 + BU 22 = A(B m S m AB m ) + B(B m AS m AB m B m AB m ) = AB m AS m AB m + BB m AS m AB m BB m AB m = AB m B π AS m AB m BB m AB m = AB m B π AB m BB m AB m = (I BB m )AB m B m AB m = 0 U 11B = (S m + (S m A I)BB m AB π AS m (S m A I)BB m AB π )B = S m B + (S m A I)BB m AB π AS m B (S m A I)BB m AB π B = (S m A I)BB m A(I BB m )B = (S m A I)(BB m AB BB m ABB m B = (S m A I)(BB m AB BB m AB = 0. 74

9 D.Krishnaswamy and S.Anusuya Therefore AU 12 + BU 22 = U 11B. Similarly, we can get, BU 11 = U 21A + U 22B = BB m AB π (I AS m ) BU 12 = U 21B = BB m. Consequently, (MX) = G1 0 BBm + B π AS m 0 G1 0 0 I BB m AB π (I AS m ) BB m 0 I = G1 0 (BBm + B π AS m ) (BB m AB π (I AS m )) G1 0 0 I 0 (BB m ) 0 I = G1(BBm + B π AS m ) G 1(BB m AB π (I AS m )) G1 0 0 (BB m ) 0 I = G1(BBm + B π AS m ) G 1 G 1(BB m AB π (I AS m )) 0 (BB m ) Therefore (MX) = (BBm + B π AS m ) G 1(BB m AB π (I AS m )). Similarly, 0 (BB m ) (XM) = (BBm + B π AS m ) G 1(BB m AB π (I AS m )) 0 (BB m ) MXM = BBm + B π AS m 0 A B BB m AB π (I AS m ) BB m BB m A + B π AS m A BB m B + B π AS m B = BB m AB π (I AS m )A + BB m B BB m AB π (I AS m )B BB m A + (I BB m )AS m A B = BB m AB π A BB m AB π AS m A + B BB m AB π B BB m AB π AS m )B BB m A + A BB m )A B = BB m AB π A BB m AB π AS m A + B BB m AB π B BB m AB π AS m B = BBm A + A BB m )A B B BB m AB BB m ABB m B = A B. Therefore MXM = M. Suppose rank(b) = r. By Lemma 2.1, there exist G-unitary matrices P K n n and invertible matrices B = P B1 0 P and B m = P Bm 1 0 P. Let A = P A1 A2 P, where A 1 K r r, A 2 K r (n r), A 3 A 3 A 4 75

10 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M K (n r) r, A 4 K (n r) (n r). Since X = U11 U12 U 21 U 22 rank(x) = rank U11 U12 U 21 U 22 = rank Sm + (S m A I)BB m AB π AS m (S m A I)BB m AB π B m S m AB m B m 0 = rank Sm B m S m AB m B m 0 B A m 4 = rank B 1 0 = 2r + rank(a m 4 ) rank(x) = 2r + rank(a m 4 ) = 2r + rank(a 4) = rank(m) From Lemma 2.3, we get X = M m. Theorem 3.2. Let M = A B, where A, B K n n, suppose B m and (AB π ) m exists in M then M m exist in M and M m = U11 U12, where U 21 U 22 U 11 = S m + S m AB π ABB m (AS m I ) B π ABB m (AS m I ) U 12 = B m S m AB m S m AB π S m AB π ABB m (AS m I)AB m + B π ABB m (I AS m )AB m U 21 = B m B m AS m U 22 = B m AS m AB m B m AB m S = B π AB π. Theorem 3.3. Let M = A B, where A, B K n n, suppose B m exists in M and BAB π = 0 then (i). M m exists in M iff (AB π ) m exists in M. (ii). If M m exists in M, then M m = U B m V, where B m AB m U = B π A(B m ) 2 (AB π ) m AB π A(B m ) 2 + (AB π ) m V = B π A(B m ) 2 AB m + (AB π ) m AB π A(B m ) 2 AB m (AB π ) m AB m + B m. 76

11 D.Krishnaswamy and S.Anusuya Proof. (i) The existence of M m has been given in Lemma 2.3 (ii) (ii) Let X = U B m V then B m AB m MX = A B U XM = U B m B m V AU + BBm = AV BB m AB m B m AB m BU BV V A B = UA + V B B m AB m B m AB π UB B m B AU + BB m = AB π A(B m ) 2 A(AB π ) m AB π A(B m ) 2 + A(AB π ) m + BB m = AB π A(B m ) 2 AB π A(B m ) 2 + A(AB π ) m + BB m = A(AB π ) m + BB m UA + V B = (B π A(B m ) 2 (AB π ) m AB π A(B m ) 2 + (AB π ) m )A + ( B π A(B m ) 2 AB m + (AB π ) m AB π A(B m ) 2 AB m (AB π ) m AB m + B m )B = B π A(B m ) 2 A (AB π ) m AB π A(B m ) 2 A + (AB π ) m A B π A(B m ) 2 AB m B (AB π ) m AB π A(B m ) 2 AB m B (AB π ) m AB m B + B m B = B π A(B m ) 2 A(I BB m ) (AB π ) m AB π A(B m ) 2 A(I BB m ) + (AB π ) m A(I BB m ) + BB m = B π A(B m ) 2 AB π (AB π ) m AB π A(B m ) 2 AB π + (AB π ) m AB π + BB m = (AB π ) m AB π + BB m = A(AB π ) m + BB m. Therefore AU + BB m = UA + V B. AV BB m AB m = A( B π A(B m ) 2 AB m + (AB π ) m AB π A(B m ) 2 AB m (AB π ) m AB m + B m ) BB m AB m = AB π A(B m ) 2 AB m + A(AB π ) m AB π A(B m ) 2 AB m A(AB π ) m AB m + AB m BB m AB m = AB π A(B m ) 2 AB m + AB π A(B m ) 2 AB m A(AB π ) m AB m (I BB m )AB m = A(AB π ) m AB m + B m AB m UB = [B π A(B m ) 2 (AB π ) m AB π A(B m ) 2 + (AB π ) m ]B = B π A(B m ) 2 B (AB π ) m AB π A(B m ) 2 B + (AB π ) m B = B π AB m B (AB π ) m AB π AB m B + (AB π ) m B = B π AB m BB m (AB π ) m AB π AB m BB m + (AB π ) m B = B π AB m (AB π ) m AB π AB m + (AB π ) m B = B π AB m A(AB π ) m AB m + (AB π ) m B = A(AB π ) m AB m + B π AB m. 77

12 Minkowski Inverse for the Range Symmetric Block Matrix with Two Identical Sub-blocks Over Skew Fields in Minkowski Space M Hence AV BB m AB m = UB. BU = B[B π A(B m ) 2 (AB π ) m AB π A(B m ) 2 + (AB π ) m ] = BB π A(B m ) 2 B(AB π ) m AB π A(B m ) 2 + B(AB π ) m = B(I B m )A(B m ) 2 = BA(B m ) 2 BB m A(B m ) 2 = BA(B m ) 2 BB m BA(B m ) 2 = BA(B m ) 2 BA(B m ) 2 = 0. B m AB π = 0. Hence BU = B m AB π. BV = B[ B π A(B m ) 2 AB m + (AB π ) m AB π A(B m ) 2 AB m (AB π ) m AB m + B m ] = BB π A(B m ) 2 AB m + B(AB π ) m AB π A(B m ) 2 AB m B(AB π ) m AB m + BB m = B(I BB m )A(B m ) 2 AB m + BB m = BA(B m ) 2 AB m + BBB m A(B m ) 2 AB m + BB m = BA(B m ) 2 AB m + BB m BA(B m ) 2 AB m + B m B = BA(B m ) 2 AB m + BA(B m ) 2 AB m + B m B = B m B. Hence BV = B m B. Consequently, (MX) = G1 0 A(ABπ ) m + BB m A(AB π ) m AB m + B π AB m 0 I 0 B m B = G1(A(ABπ ) m + BB m ) G 1 0 (A(AB π ) m AB m + B π AB m ) G 1 (B m B) = (A(AB π ) m + BB m ) 0. (A(AB π ) m AB m + B π AB m ) G 1 (B m B) Similarly, (XM) = (A(AB π ) m + BB m ) 0. (A(AB π ) m AB m + B π AB m ) G 1 (B m B) MXM = A(ABπ ) m + BB m A(AB π ) m AB m + B π AB m A B 0 B m B G1 0 0 I = A(ABπ ) m A + BB m A A(AB π ) m AB m B + B π AB m B A(AB π ) m B + BB m B BB m = A B. 78

13 D.Krishnaswamy and S.Anusuya Therefore MXM = M. Suppose rank(b) = r. By Lemma 2.1, there exist G-unitary matrices P K n n and invertible matrices B 1 K r r such that B = P B1 0 P and B m = P Bm 1 0 P. Let A = P A1 A2 P, where A 1 A 3 A 4 K r r, A 2 K r (n r), A 3 K (n r) r, A 4 K (n r) (n r). Then rank(x) = rank U B m V B m AB m = rank U Bm B m 0 = rank (ABπ ) m B m B m 0 B A m 4 = rank B 1 0 = 2r + rank(a m 4 ). Since rank(m) = 2r + rank(a 4) = rank(r). Then X = M m. Theorem 3.4. Let M = A B, where A, B K n n, suppose B m exists in M and B π AB = 0 then (i). M m exists in M iff (B π A) m exists in M. (ii). If M m exists in M, then M m = U Bm, where V B m AB m U = (B m ) 2 AB π (B m ) 2 AB π A(B π A) m + (B π A) m, V = B m B m B π + B m A(B m ) 2 AB π A(B π A) m B m A(B π A) m + B m. References [1] A.Ben-Israel and T.N.E.Greville, Generalized inverses: Theory and applications, Second ed., Springer-Verlag, New York, (2003). [2] K.P.S.Bhaskara Rao, The theory of generalized inverses over commutative rings, Taylor and Francis, London and New York, (2002). [3] R.Bru, C.Coll and N.Thome, Symmetric singular linear control systems, Appl. Math. Letters, 15(2002), [4] C.Bu, J.Zhao and J.Zheng, Group inverse for a Class 2 2 block matrices over skew fields, Applied Math. Comput., 24(2008), [5] C.Bu, J.Zhao and K.Zhang, Some results on group inverses of block matrices over skew fields, Electronic Journal of Linear Algebra, 18(2009), [6] C.Bu, M.Li, K.Zhang and L.Zheng, Group inverse for the block matrices with an invertible subblock, Appl. Math. Comput., 215(2009),

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