AN ITERATIVE METHOD FOR THE GENERALIZED CENTRO-SYMMETRIC SOLUTION OF A LINEAR MATRIX EQUATION AXB + CY D = E. Ying-chun LI and Zhi-hong LIU

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1 Acta Universitatis Apulensis ISSN: No. 9/01 pp AN ITERATIVE METHOD FOR THE GENERALIZED CENTRO-SYMMETRIC SOLUTION OF A LINEAR MATRIX EQUATION AXB + CY D = E Ying-chun LI and Zhi-hong LIU Abstract. A matrix R n n is said to be a symmetric orthogonal matrix if = T = 1. A matrix A R n n is said to be generalized centro-symmetric (generalized central anti-symmetric )with respect to, if A = A (A = A ). In this paper, an iterative method is constructed to solve the generalized centrosymmetric solutions of a linear matrix equation AXB + CY D = E, with real pair matrices X and Y. We show when the matrix equation is consistent over generalized centro-symmetric pair matrices X and Y, for any initial pair matrices X 0 and Y 0, the generalized centro-symmetric solution can be obtained within finite iterative steps in the absence of roundoff errors, and the minimum norm of the generalized centro-symmetric solutions can be obtained by choosing a special kind of initial pair matrices. Furthermore, the optimal approximation pair solution X and Ŷ to a given matrices X and Y can be derived. 000 Mathematics Subject Classification: 15A4, 15A Introduction Throughout the paper, the notations R n n, SOR n n represent the set of all n n real and real symmetric orthogonal matrices, respectively. A B stands for the Kronecker product of matrices A and B. A T, trace(a) and R(A) denote the transpose, trace and column space of the matrix A respectively. Also vec(a) represents the vector operator. i.e. vec(a) = (a T 1,, at n ) T R mn for the matrix A = (a 1,, a n ) R m n, a i R m, i = 1,, n. We define an inner product as A, B = trace(b T A). Then the norm of a matrix A generated by this inner product is the Frobenius norm and is denoted by A. Definition 1.1. Let be some real symmetric orthogonal n n matrix, i.e. = T = 1. If A = A, then A is called a generalized centro-symmetric 335

2 matrix with respect to. CSR n n denotes the set of order n generalized centrosymmetric matrices with respect to SOR n n. The following two problems are considered in this paper. roblem I. Given matrices A R p n, B R n q, C R p m, D R m q, and E R p q, find a pair matrices X CSR n n and Y CSR m m, such that AXB + CY D = E. (1) roblem II. When roblem I is consistent, let S E denote the set of solutions of roblem I S E = {(X, Y ) CSR n n CSR m m : AXB + CY D = E}. For given pair matrices X R n n and Y R m m,find the generalized centrosymmetric pair matrices X S E and Ŷ S E such that X X + Ŷ Y = min { X X + Ŷ (X,Y ) S E Y }. () Matrix equation is one of the topics of very active research in computational mathematics, and has been widely applied in various areas. Many results have been obtained about equation (1).For example, Chu [1] gave the consistency conditions and the minmum norm solution by making use of the generalized singular value decomposition(gsvd). Huang and Zeng [] and Özgüler [3], respectively, gave the solvability conditions over a simple Artinian ring and principal ideal domain by using the generalized inverse. Shim and Chen [4],Xu, Wei and Zheng [5] presented the least square solution with the minimun norm by using the canonical correlation decomposition(ccd) and GSVD. Yuan,Liao and Lei [6] obtain a unique least squares symmetric the Kronecker product of matrices. eng and eng [7] solve the solution of equation (1) by using iterative method Sheng and Chen [8] obtained the symmetric and skew symmetric of equation (1) by using iterative method. In this paper, we will use iterative method to solve the generalized centro-symmetric solutions of equation (1). This paper is organized as follows: In Section, we introduce an iterative algorithm for solving roblem I. Then we prove several properties of Algorithm I. Also, when the linear matrix equation (1) is consistent, we show for any initial generalized centro-symmetric matrix pair X 0 and Y 0, a generalized centro-symmetric solution can be obtained within finite iteration steps, and also show that if the initial matrix is chosen as X 0 = A T HB T + A T HB T and Y 0 = C T HD T + C T HD T, where 336

3 H is arbitrary, then the generalized centro-symmetric X and Y obtained by the iterative method is the minimum norm solution. The optimal approximation generalized centro-symmetric solution to given generalized centro-symmetric matrix pair X and Y in the solution set of the linear matrix equation (1) is obtained in Section 3..The iterative algorithm for solving roblem I In this section, we will construct an iterative method to solve roblem I. Then, some basic properties of the introduced iterative method are described. Finally, we show that it is convergent. Algorithm I. Step 1. Input matrices A, B, C, D, E and X 0 CSR n n, Y 0 CSR m m Step. R 0 = E AX 0 B CY 0 D; 0 = A T R 0 B T ; 0 s = ; Q 0 = C T R 0 D T ; Q s 0 = Q 0 + Q 0 ; k = 0. Step 3. If R k = 0, then stop; else, k = k + 1. Step 4. Calculate X k = X k 1 + Y k = Y k 1 + R k 1 s k 1 + Q s k 1 s k 1 ; R k 1 s k 1 + Q s k 1 Qs k 1 ; R k = E AX k B CY k D = R k 1 R k 1 k 1 s + Q s (A s k 1 k 1 B + CQs k 1 D);. 337

4 k = A T R k B T ; Q k = C T R k D T ; s k = k + k Q s k = Q k + Q k tr( k s k 1 ) + tr(q kq s k 1 ) s k 1 + Q s k 1 s k 1 ; tr( k s k 1 ) + tr(q kq s k 1 ) s k 1 + Q s k 1 Q s k 1. Step 5. Go to step 3. Remark.1 Obviously, k s, Qs k CSRn n p for k = 0, 1,, from Algorithm I. Lemma. [8] Let A, B R n n, then we have A, B = B, A = A T, B T = B T, A T. and X k CSR n n,y k CSR m m Lemma.3 Let SOR n n. i.e. = T = 1, A R n n,b CSR n n, then A + A, B = A, B. roof. A, B = tr(b T A) = tr[b T ( A + A + A A = tr[ BT (A + A ) ] + tr( BT A ) A tr(bt ) = A + A, B + tr( BT A T ) tr( B T T A ) = A + A = A + A, B + tr( BT A ) tr(bt A ), B. Lemma.4 Assume that the linear matrix equation (1) is consistent and (X, Y ) is one of its solutions, then, for any initial generalized centro-symmetric matrix pair (X 0, Y 0 ), the sequences {X i }, {Y i }, {R i }, {i s} and {Qs i } generalized by Algorithm I satisfy i s, X X i + Q s i, Y Y i = R i, (i = 0, 1,, ). (3) )] 338

5 roof. We prove the conclusion by induction and notice that X, X 0, Y, Y 0 are all generalized centro-symmetric matrices. When i = 0, we have 0 s, X X 0 + Q s 0, Y Y 0 = 0 + 0, X X 0 + Q 0 + Q 0, Y Y 0 = 0, X X 0 + Q 0, Y Y 0 = A T R 0 B T, X X 0 + C T R 0 D T, Y Y 0 = R 0, A(X X 0 )B + R 0, C(Y Y 0 )D = R 0, A(X X 0 )B + C(Y Y 0 )D = R 0, R 0 = R 0. Assume that (3) holds for i = t(for t 0), that is s t, X X t + Q s t, Y Y t = R t, then for i = t + 1, we have s t+1, X X t+1 + Q s t+1, Y Y t+1 = t+1 + t+1 + Q t+1 + Q t+1 = t+1 + t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t s t, X X t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t Q s t, Y Y t+1, X X t+1 + Q t+1 + Q t+1, Y Y t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t [ s t, X X t+1 + Q s t, Y Y t+1 ] = t+1, X X t+1 + Q t+1, Y Y t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t s t, X X t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t Q s t, Y Y t+1 = A T R t+1 B T, X X t+1 + C T R t+1 D T, Y Y t+1 tr( t+1t s ) + tr(q t+1 Q s t) t s + Q s s t t, X R t X t t s + Q s s t t tr( t+1t s ) + tr(q t+1 Q s t) t s + Q s Q s t, Y R t Y t t t s + Q s t Qs t 339

6 = R t+1 tr( t+1t s ) + tr(q t+1 Q s t) t s + Q s [ s t t, X X t t s R t, t s + Q s s t t ] tr( t+1t s ) + tr(q t+1 Q s t) t s + Q s [ Q s t, Y Y t Q s R t t, t t s + Q s t Qs t ] = R t+1 tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t R t + tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t R t s t + Q s t [ s t, s t + Q s t, Q s t ] = R t+1. By the principle of induction, the conclusion (3) holds for all i = 0, 1,,. Remark.5 From the formulae of s i and Q s i i in Algorithm I and Lemma.4, we know that if the linear matrix equation (1) is consistent, then, R i = 0 if and only if i = 0 and Q i = 0. This result implies that if there exists a positive number k such that s k = Qs k = 0 but R k 0, then the linear matrix equation (1) has no generalized centro-symmetric solution. Hence, the solvability of the linear matrix equation (1) can be determined automatically by Algorithm I in the absence of roundoff errors. Lemma.6 Assume that the linear matrix equation (1) is consistent and the sequences {R i }, {i s} and {Qs i }, where R i 0(i = 0, 1,,, k) generated by Algorithm I. Then we have R i, R j = 0, s i, s j + Q s i, Q s j = 0, (i j, i, j = 0, 1,, k) (4) roof. From Lemma. we know that A, B = B, A holds for all matrices A and B in R p q, we only prove that the conclusion holds for all 0 i < j k. Using induction and two steps are required. Step 1. Show that R i, R i+1 = 0 and i s, i+1 s + Qs i, Qs i+1 = 0 for all i = 0, 1,,, k. To prove this conclusion, we also use induction. 340

7 and For i = 0, we have R 0, R 1 = R 0, R 0 = R 0 = R 0 = R 0 = 0. s 0, s 1 + Q s 0, Q s 1 = s 0, R 0 s 0 + Q s 0 (A s 0 B + CQ s 0D) R 0 s 0 + Q s 0 [ R 0, A s 0 B + R 0, CQ s 0D ] R 0 s 0 + Q s 0 [ 0, s 0 + Q 0, Q s 0 ] R 0 s 0 + Q s 0 ( s 0 + Q s 0 ) + Q s 0, Q 1 + Q 1 = s 0, tr( 1 s 0 ) + tr(q 1Q s 0 ) s 0 + Q s 0 s 0 tr( 1 s 0 ) + tr(q 1Q s 0 ) s 0 + Q s 0 Q s 0 + Q s 0, Q 1 + Q 1 tr( 1 s 0 ) + tr(q 1Q s 0 ) s 0 + Q s 0 [ s 0, s 0 + Q s 0, Q s 0 ] = tr( 1 s 0 ) + tr(q 1 Q s 0) [tr( 1 s 0 ) + tr(q 1 Q s 0)] = 0. Assume (4) holds for all i t(for0 < t < k), then R t, R t+1 = R t, R t = R t = R t = R t R t s t + Q s t (A s t B + CQ s td) R t s t + Q s t [ AT R t B T, s t + C T R t D T, Q s t ] R t s t + Q s t [ t, s t + Q t, Q s t ] R t t s + Q s [ t + t, t t s + Q t + Q t, Q s t ] 341

8 = R t = R t = R t R t = 0 and R t t s + Q s t + t t R t t s + Q s Q t + Q t t R t s t + Q s t [ s t, s t + Q s t, Q s t ] s t, s t+1 + Q s t, Q s t+1 = s t, t+1 + t+1 + Q s t, Q t+1 + Q t+1 = s t, t+1 + t+1 tr( t s t 1 ) + tr(q tq s t 1 ) s t 1 + Q s t 1 s t 1, s t tr( t s t 1 ) + tr(q tq s t 1 ) s t 1 + Q s t 1 Q s t 1, Q s t tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t s t tr( t+1 s t ) + tr(q t+1 Q s t) s t + Q s t Q s t + Q s t, Q t+1 + Q t+1 tr( t+1t s ) + tr(q t+1 Q s t) t s + Q s [ s t t, t s + Q s t, Q s t ] = tr( t+1 s t ) + tr(q t+1 Q s t) [tr( t+1 s t ) + tr(q t+1 Q s t)] = 0. By the principle of induction, R i, R i+1 = 0 and i s, i+1 s + Qs i, Qs i+1 = 0 hold for all i = 0, 1,, k. Step. Assume that R i, R i+l = 0 and i s, i+l s + Qs i, Qs i+l = 0 hold for all 0 i k and 1 l k, show that R i, R i+l+1 = 0 and i s, i+l+1 s + Qs i, Qs i+l+1 = 0. R i+l s R i, R i+l+1 = R i, R i+l i+l s + Q s (Ai+l B + i+l CQs i+l D) R i+l = R i, R i+l i+l s + Q s [ R i, A s i+l i+l B + R i, Ci+l s D ] R i+l = i+l s + Q s [ i, s i+l i+l + Q i, Q s i+l ] R i+l = i+l s + Q s [ i + i, i+l i+l s + Q i + Q i, Q s i+l ] R i+l = i+l s + Q s i + i tr( ii 1 s ) + tr(q iq s i 1 ) i+l i 1 s + Q s i 1, s s i 1 i+l 34

9 and R i+l i+l s + Q s Q i + Q i tr( ii 1 s ) + tr(q iq s i 1 ) i+l i 1 s + Q s Q s i 1, Q s i 1 i+l R i+l s = i+l s + Q s [ i, i+l s + i+l Qs i, Q s i+l ] = 0 s i, s i+l+1 + Qs i, Q s i+l+1 = s i, i+l+1 + i+l+1 + Q s i, Q i+l+1 + Q i+l+1 tr( i+l+1i+l s ) + tr(q i+l+1q s i+l ) i+l s + Q s i+l s i+l tr( i+l+1i+l s ) + tr(q i+l+1q s i+l ) i+l s + Q s Q s i+l i+l = i s, i+l+1 + i+l+1 + Q s i, Q i+l+1 + Q i+l+1 tr( i+l+1i+l s ) + tr(q i+l+1q s i+l ) s i+l + Q s i+l = i s, i+l+1 + i+l+1 [ s i, s i+l + Qs i, Q s i+l ] + Q s i, Q i+l+1 + Q i+l+1 = s i, i+l+1 + Q s i, Q i+l+1 = s i, A T R i+l+1 B T + Q s i, C T R i+l+1 D T = A s i B, R i+l+1 + CQ s i D, R i+l+1 = s i + Q s i R s i [ A(X i+1 X i )B, R i+l+1 + C(X i+1 X i )D, R i+l+1 ] = s i + Q s i R s i [ R i R i+1, R i+l+1 ] = 0. From Step 1 and Step, the conclusion (4) holds by the principle of induction. Remark.7 Lemma.6 implies that if the linear matrix equation (1) is consistent, then, for any initial generalized centro-symmetric matrix pair X 0 and Y 0, a solution can be obtained within at most pq iteration steps. Since the R 0, R 1, are orthogonal each other in the finite dimension matrix space R p q, it is certain that there exists a positive number k pq such that R k = 0. Lemma.8 [9] Let the linear system Ax = b be consistent, if x is a solution, satisfied x R(A ), then x is the unique minimum normal solution of it. 343

10 Lemma.9 In Algorithm I, if we choose the X 0 = A T HB T + A T HB T and Y 0 = C T HD T + C T HD T, where H is an arbitrary matrix in R p q, then the sequences of {X k } and {Y k } generated by Algorithm I have the following properties X k = A T Hk B T + A T HT k B T and Y k = A T Ĥ k B T + A T Ĥ T k BT, where H k and Ĥ k is some matrix in R p q. Consider the following system of matrix equations { AXB + CY D = E A X B + C Y D = E Obviously, the solvability of the above system of matrix equation is equivalent to roblem I. The system of matrix equations (5) is equivalent to ( B T A D T ) ( ) ( ) C B T A D T vec(x) vec(e) =. C vec(y ) vec(e) Now suppose H R n n is obviously matrices, we have (5) ( A vec T HB T + A T HB T C T HD T + C T HD T ) = = ( B A T B A T ) ( ) D C T D vec(h) C T vec(h) ( B T A D T ) T ( C B T A D T vec(h) C vec(h) ( ( B T A D R T ) ) T C B T A D T C ) Obviously, if we consider X 0 = A T HB T + A T HB T, Y 0 = C T HD T + C T HD T, then all Y k, generated by Algorithm I satisfy ( ) ( ( vec(xk ) B T A D R T C vec(y k ) B T A D T C ) T ). Hence by Lemma.8, if we take an initial matrices X 0 = A T HB T + A T HB T, Y 0 = C T HD T + C T HD T, then X and Y generated by Algorithm I are the least Frobenius norm generalized centro-symmetric solution. By using the above conclusions, we can prove the following theorem. 344

11 Theorem.10 Suppose that roblem I is consistent. If we take initial matrices X 0 = A T HB T + A T HB T, Y 0 = C T HD T + C T HD T, where H R n n is arbitrary, or more especially X 0,Y 0, then the solutions X and Y are the least Frobenius norm generalized centro-symmetric solution of roblem I. 3.The Solution of roblem II We assume that X R n n and Y R m m in roblem II, it is well known that a generalized centro-symmetric matrix and a generalized central anti-symmetric are orthogonal each other, for any X CSR n n and Y CSR m m, we have that X X + Y Y = X ( X + X = X X + X X+ X + X X + X X ) + Y ( Y + Y + Y Y + Y + Y Y ) + Y Y Denote X = and Y = Y + Y, when the linear matrix equation (1) is consistent, the solution set S E of the matrix equation (1) is no-empty, then AXB + CY D = E A(X X)B + C(Y Y )D = E AXB CY D. Let X = X X, Ỹ = Y Y and Ẽ = E AXB CY D, then the matrix nearness roblem II is the equivalent to find the minimum norm solution of the pair of matrix equations A XB + C XD = Ẽ. (6) By using Algorithm I, and let the initial matrix X 0 = A T HB T + A T HB T, Ỹ0 = C T HD T + C T HD T, where H is an arbitrary matrix in R p q, more specially, let X 0 = 0 and Ỹ0 = 0, we can obtain the unique minimum norm solution X and Ỹ of linear matrix equation (6). Once the above matrix X and Ỹ are obtained, the unique generalized centro-symmetric solution pair of the matrix nearness roblem II can be obtained. In this case, X and Ŷ can be expressed X = X +X = X + and Ŷ = Ỹ + Y = Ỹ + Y + Y, respectively.. X+ X Acknowledgements: The present investigation was supported by the Scientific Research Fund of Yunnan rovincial Education Department under Grant 09C006 of eoples Republic of China and the Important Course Construction of Honghe University(ZDKC1003). 345

12 References [1] K.E.Chu, Singular value and generalized value decomposition and the solution of linear matrix equations, Linear Algebra and it Application, 87,(1987), [] L..Huang and Q.G.Zeng, The matrix equation AXB + CY D = E over a simple artinium ring, Linear and Multilinear Algebra, 38,(1995),5-3. [3] A.B.Özgüler, The equation AXB + CY D = E over a principal ideal domain, SIAM Journal on Matrix Analysis and Applications 38,(1991), [4] S.Y.Shim and Y.Chen, Least squares solution of matrix equation AXB + CY D = E, SIAM Journal on Matrix Analysis and Applications, 3,(003), [5] G.Xu, M.Wei and D.Zhang, On solution of matrix equation AXB+CY D = F, Linear Algebra and it Application, 79,(1998), [6] S.Yuan, A.Liao and Y.Lei, Least squares symmetric solution of the matrix equation AXB + CY D = E with the least norm, Mathematica Numerica Sinica. 9,(007), [7] Z.eng and Y.eng, An efficient iterative method for solving the matrix equation AXB + CY D = E, Numerical Linear Algebra with Applications. 13(6),(006), [8] X..Sheng,G.L.Chen, An iterative method for the symmetric and skew symmetric solutions of a linear matrix equation AXB + CY D = E, Journal of Computational and Applied Mathematics. 3,(010), [9] A.Ben-Israel and T.N.E.Greville, Generalized Inverse:Theory and Application, nd edition, Springer Verlag, New york, 003. Ying-chun LI and Zhi-hong LIU Department of Mathematics Honghe University Mengzi , Yunnan,.R.China liyingchunmath@163.com, liuzhihongmath@163.com 346