An iterative method for the skew-symmetric solution and the optimal approximate solution of the matrix equation AXB =C

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1 Journal of Computational and Applied Mathematics 1 008) An iterative method for the skew-symmetric solution and the optimal approximate solution of the matrix equation AXB =C Guang-Xin Huang a Feng Yin b KeGuo a a College of Information and Management Chengdu University of echnology Chengdu PR China b Department of Mathematics Sichuan University of Science and Engineering Zigong PR China Received 7 May 006; received in revised form 4 August 006 Abstract In this paper an iterative method is constructed to solve the linear matrix equation AXB = C over skew-symmetric matrix X. By the iterative method the solvability of the equation AXB = C over skew-symmetric matrix can be determined automatically. When the equation AXB = C is consistent over skew-symmetric matrix X for any skew-symmetric initial iterative matrix X 1 the skew-symmetric solution can be obtained within finite iterative steps in the absence of roundoff errors. he unique least-norm skew-symmetric iterative solution of AXB = C can be derived when an appropriate initial iterative matrix is chosen. A sufficient and necessary condition for whether the equation AXB = C is inconsistent is given. Furthermore the optimal approximate solution of AXB = C for a given matrix X 0 can be derived by finding the least-norm skew-symmetric solution of a new corresponding matrix equation A XB = C. Finally several numerical examples are given to illustrate that our iterative method is effective. 006 Elsevier B.V. All rights reserved. MSC: 65F15; 65F0 Keywords: Iterative method; Skew-symmetric solution; Least-norm skew-symmetric solution; Optimal approximate solution 1. Introduction In this paper the following notations are considered and used. Let R m n denote the set of all m n real matrices SSR n n denote the set of all n n real skew-symmetric matrices in R n n. We denote by the superscripts and + the transpose and Moore Penrose generalized inverse of matrices respectively. In matrix space R m n define inner product as: A B =trb A) for all A B R m n A represents the Frobenius norm of A. RA) represents the column space of A. vec ) represents the vector operator. i.e. veca) = a 1 a...a n ) R mn for the matrix A = a 1 a...a n ) R m n a i R m i= 1...n. A B stands for the Kronecker product of matrices A and B. For more notations we refer to 3. In this paper we will consider the following two problems. his work is supported by the Key Science Fund and the Youngth Foundation of Chengdu University of echnology. Corresponding authors. addresses: huangx@cdut.edu.cn G.-X. Huang) fyin@suse.edu.cn F. Yin) /$ - see front matter 006 Elsevier B.V. All rights reserved. doi: /j.cam

2 3 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) Problem I. For given matrices A R m n B R n p C R m p find matrix X SSR n n such that AXB = C. 1) Problem II. When Problem I is consistent let S E denote the set of skew-symmetric solutions of Problem I for a given matrix X 0 R n n find X S E such that X X 0 = min X S E X X 0. ) In fact Problem II is to find the optimal approximately skew-symmetric solution to a given matrix X 0 R n n. Problem I has been considered in the case of special solution structures e.g. symmetric triangular or diagonal solution X.We refer to Dai 5 Eric Chu Don 4 Magnus 6 Morris and Odell 8 Bjerhammer 1 for more details. Mitra 7 considered common solutions to a pair of linear matrix equations A 1 XB 1 = C 1 A XB = C.In these literatures the problem was discussed by using matrices decomposition such as the singular value decomposition SVD) the generalized SVD GSVD) the quotient SVD QSVD) and the canonical correlation decomposition CCD). However these methods are difficult to be applied to solve Problem II and the representation of solutions of Problems I and II are complicated. Huang and Yin 111 recently solve the constrained inverse eigenproblem and associated approximation problem for anti-hermitian R-symmetric matrices and the matrix inverse problem and its optimal approximation problem for R-symmetric matrices. Recently Peng et al. 9 have constructed an iteration method to solve the linear matrix equation AXB = C over symmetric matrix X. Peng 10 has shown an iterative method to solve the minimum Frobenius norm residual problem: min AXB C with unknown symmetric matrix X. In this paper we will consider the skew-symmetric solution of the linear matrix equation AXB = C we construct an iterative method by which the solvability of Problem I can be determined automatically the solution can be obtained within finite iterative steps when Problem I is consistent and the solution of Problem II can be obtained by finding the least-norm skew-symmetric solution of a new matrix equation A XB = C. his paper is organized as follows. In Section we will solve Problem I by construct an iterative method i.e. for an arbitrary initial matrix X 1 SSR n n if there exists a positive integer k such that R k = 0 and Q k = 0 then Problem I is inconsistent where R k and Q k k = 1...) are defined in Algorithm.1. If Problem I is consistent then for an arbitrary initial matrix X 1 SSR n n we can obtain a solution X SSR n n of Problem I within finite iterative steps in the absence of roundoff errors. Let X 1 = A H B BHA where H is an arbitrary matrix in R p m or more especially let X 1 =0 SSR n n we can obtain the unique least-norm solution of Problem I. hen in Section 3 we give the optimal approximate solution of Problem II by finding the least-norm skew-symmetric solution of a corresponding new matrix equation. In Section 4 several numerical examples are given to illustrate the application of our iterative methods.. he solution of Problem I In this section we will first introduce an iterative method to solve Problem I then prove that it is convergent. Algorithm.1. Step 1: Input matrices A R m n B R n p C R m p ; Step : Choose an arbitrary matrix X 1 SSR n n compute R 1 = C AX 1 B P 1 = A R 1 B Q 1 = 1 P 1 P 1 ) k := 1; Step 3: If R 1 = 0 orr 1 = 0 and Q 1 = 0 then stop; Else go to step 4; Step 4: Compute X k+1 = X k + R k Q k Q k

3 R k+1 = C AX k+1 B G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) P k+1 = A R k+1 B Q k+1 = 1 P k+1 P k+1 ) + trp k+1q k ) Q k Q k ; Step 5: If R k+1 = 0 orr k+1 = 0 and Q k+1 = 0 then stop; Else let k :=k + 1 go to step 4. Obviously it can be seen that Q i SSR n n X i SSR n n where i = Lemma.1. For the sequences {R i } {P i } and {Q i } generated in Algorithm.1 we have trr i+1 R j ) = trr i R j ) + R i Proof. Noting that Q i = Q i by Algorithm.1 we have Q i trq ip j ). 3) trri+1 R j ) = trc AX i+1 B) R j = tr C AX i + R i ) Q i Q i)b R j = tr C AX i B) R j R i Q i AQ ib) R j = trr i R j ) R i = trr i R j ) R i = trr i R j ) R i = trr i R j ) + R i Q i trb Q i A R j ) Q i trq i A R j B ) Q i trq i P j ) Q i trq ip j ). Lemma.. For the sequences {R i } and {Q i } generated by Algorithm.1 and k we have trr i R j ) = 0 trq i Q j ) = 0 ij = 1...ki = j. 4) Proof. Since trr i R j ) = trr j R i) and trq i Q j ) = trq j Q i) for all i j = 1...k we only need prove that trr i R j ) = 0 trq i Q j ) = 0 for all 1 j <i k. We prove the conclusion by induction and two steps are required. Step 1: we will show that trr i+1 R i) = 0 trq i+1 Q i) = 0 i = 1...k 1. 5) o prove this conclusion we also use induction.

4 34 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) For i = 1 by Lemma.1 and Algorithm.1 noting that Q 1 = Q 1 it follows that trr R 1) = trr1 R 1) + R 1 Q 1 trq 1P 1 ) = R 1 + R 1 Q 1 tr Q 1 P 1 = R 1 + R 1 Q 1 tr ) + Q 1P 1 ) ) Q 1 P 1 + P 1 Q 1 ) = R 1 + R 1 Q 1 tr P 1 P1 Q 1 = R 1 + R 1 Q 1 trq 1Q 1 ) = R 1 R 1 Q 1 trq 1 Q 1) = 0 and ) trq Q P P 1) = tr + trp Q 1 ) Q 1 Q 1 Q 1 ) P P = tr Q 1 + trp Q 1 ) Q 1 trq 1 Q 1) ) P Q 1 + Q 1 P ) = tr + trp Q 1 ) ) P Q 1 + P Q 1 = tr + trp Q 1 ) = 0. Assume 5) holds for i = s 1 i.e. trr s R s 1) = 0 trq s Q s 1) = 0. Noting that Q s = Q s by Lemma.1 we have trrs+1 R s) = trrs R s) + R s Q s trq sp s ) = R s R s Q s trq s P s) = R s R s Q s tr Q s P s + P s Q s ) = R s R s Q s tr Q P s P ) s s = R s R s Q s tr Q s Q s trp sq s 1 ) Q s 1 ) Q s Q s = R s R s Q s tr = R s R s Q s trq s Q s) = 0 Q s 1 ) trp sq s 1 ) Q s 1 trq s Q s 1)

5 and G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) ) Ps+1 trq s+1 Q P s) = tr s+1 + trp s+1q s ) Q s Q s Q s = tr P ) s+1 Ps+1 + trp s+1q s ) Q s Q s Q s = tr P s+1q s Ps+1 Q s + trp s+1q s ) Q s trq s Q s) P s+1 Q s + Q s P s+1 ) = tr + trp s+1 Q s ) P s+1 Q s + P s+1 Q s ) = tr + trp s+1 Q s ) = trp s+1 Q s ) + trp s+1 Q s ) = 0. ) Hence 5) holds for i = s. herefore 5) holds by the principle of induction. Step : Assume that trr s R j ) = 0 trq s Q j ) = 0j = 1...s 1 then we show that trr s+1 R j ) = 0 trq s+1 Q j ) = 0 j = 1...s. 6) In fact by Lemma.1 we have trrs+1 R j ) = trrs R j ) + R s Q s trq sp j ) = R s Q s tr Qs P j + Q sp j ) = R s Q s tr P j P ) j Q s = R s Q Q s tr s = R s Q s = R s Q s Q j trp j Q j 1 ) Q j 1 Q j 1 trq s Q j ) trp j Q j 1 ) ) Q j 1 trq sq j 1 ) trq s Q j ) + trp j Q j 1 ) Q j 1 trq s Q j 1) = 0 and Ps+1 trq s+1 Q P j ) = tr = tr s+1 P s+1 P s+1 = tr P s+1 P s+1 ) + trp s+1q s ) Q s Q s Q j ) + trp s+1q s ) Q s Q s Q j Q j ) + trp s+1q s ) Q s trq s Q j )

6 36 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) = 1 trp s+1q j ) + 1 trp s+1 Q j ) = 1 trp s+1q j ) + 1 trp s+1 Q j ) = 1 trp s+1q j ) + 1 trq j P s+1) = 1 trp s+1q j ) 1 trq j P s+1 ) = trp s+1 Q j ) = trq j P s+1 ). Noting that trr s+1 R j ) = 0 trr s+1 R j+1) = 0 by Lemma.1 we have trq s+1 Q j ) = trq j P s+1 ) = Q j R j trr j R s+1) trr j+1 R s+1) = Q j R j trr s+1 R j ) trrs+1 R j+1)=0. By the principle of induction 6) holds. Noting that 4) is implied in steps 1 and by the principle of induction we complete the proof. Lemma.3. Suppose X be an arbitrary solution of Problem I i.e. A XB = C and X = X then tr X X k )Q k = R k k = ) where the sequences {X k } {R k } {Q k } are generated by Algorithm.1. Proof. We proof the conclusion by induction. For k = 1 tr X X 1 )Q 1 =tr X X 1 ) P 1 P1 X X 1 )P 1 = tr + X X 1 ) P1 X X 1 )P 1 = tr + P 1 X X 1 ) = tr X X 1 )P 1 = tr X X 1 )A C AX 1 B)B = trbc AX 1 B) A X X 1 ) = trbc AX 1 B) A X X 1 ) = tra X X 1 )BC AX 1 B) = tra XB AX 1 B)R 1 = trc AX 1 B)R 1 = trr 1 R 1 ) = trr 1 R 1) = R 1.

7 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) Assume 7) holds for k = s. Since tr X X s+1 )Q s =tr X X s R s Q s Q s)q s = tr X X s )Q s R s = R s + R s = 0 then by Algorithm.1 we have tr X X s+1 )Q s+1 =tr X X s+1 ) = tr Q s trq sq s ) Q s trq s Q s) Ps+1 P s+1 ) + trp s+1q s ) Q s Q s X X s+1 )P s+1 + X X s+1 ) Ps+1 + trp s+1q s ) Q s tr X X s+1 )Q s X X s+1 )P s+1 + P s+1 X X s+1 ) = tr = tr X X s+1 )P s+1 = tr X X s+1 )A R s+1 B = tr X X s+1 )A C AX s+1 B)B = trbc AX s+1 B) A X X s+1 ) = trbc AX s+1 B) A X X s+1 ) = tra X X s+1 )BC AX s+1 B) = tra XB AX s+1 B)C AX s+1 B) = trc AX s+1 B)C AX s+1 B) = trr s+1 R s+1 ) = trr s+1 R s+1) = R s+1. herefore 7) holds for k = s + 1. By the principle of induction the proof is completed. heorem.1. Suppose that Problem I is consistent then for an arbitrary initial matrix X 1 SSR n n a solution of Problem I can be obtained with finite iteration steps in the absence of roundoff errors and the minimum of the steps marked t 0 is within minmp n ). Proof. If R i = 0i = 1...mp by Lemma.3 we have Q i = 0 i = 1...mp then we can compute X mp+1 R mp+1 by Algorithm.1. By Lemma. we have trr mp+1 R i) = 0 i = 1...mp

8 38 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) and trr i R j ) = 0 ij = 1...mp i = j. It can be seen that the set of R 1 R...R mp is an orthogonal basis of the matrix space R m p which implies that R mp+1 = 0 i.e.x mp+1 is a solution of Problem I. When Problem I is consistent we can verify that the solution of Problem I can be obtained within t 0 iterative steps where t 0 = minmp n ). In fact if n mp and if R i = 0i = 1...n then Q i = 0i = 1...n and we can compute X n +1R n +1Q n +1 by Algorithm.1. Similar to the previous proof we have Q n +1 = 0 and then by Lemma.3 we have R n +1 = 0 i.e. X n +1 is a solution of Problem I. From heorem.1 we can easily obtain the following result. heorem.. he necessary and sufficient conditions of the inconsistency of Problem I is that there exists a positive integer k such that R k = 0 and Q k = 0 in the process of Algorithm.1. Proof. Sufficiency: If there exists a positive integer k such that R k = 0 and Q k = 0 by Lemma.3 it is easy to see that Problem I is inconsistent. Necessity: he inconsistency of Problem I implies that R i = 0 for all positive integer i. IfQ i = 0 for all positive integer i then Problem I has solutions by heorem.1 which contradict to the inconsistency of Problem I. herefore there exists a positive integer number k such that R k = 0 and Q k = 0. From heorem.1 and. we get the conditions when Algorithm.1 can be terminated. o show the least-norm skew-symmetric solution of Problem I we first introduce the following result. Lemma.4 See Peng et al. 9 Lemma.4). Suppose that the consistent system of linear equation My = b has a solution y 0 RM ) then y 0 is the least-norm solution of the system of linear equations. By Lemma.4 the following result can be obtained. heorem.3. Suppose that Problem I is consistent. If we choose the initial iterative matrix X 1 = A H B BHA where H is an arbitrary matrix in R p m especially let X 1 = 0 SSR n n we can obtain the unique least-norm skew-symmetric solution of Problem I within finite iterative steps in the absence of roundoff errors by using Algorithm.1. Proof. By Algorithm.1 and heorem.1 if we let X 1 = A H B BHA where H is an arbitrary matrix in R p m we can obtain the solution X of Problem I within finite iterative steps in the absence of roundoff errors and the solution X can be represented that X = A Y B BYA. In the sequel we will prove that X is just the least-norm solution of Problem I. Consider the following system of matrix equations: { AXB = C B XA = C. 8) If Problem I has a solution X 0 SSR n n then X 0 = X 0AX 0 B = C and B X 0 A = AX 0 B) = AX 0 B) = AX 0 B) = C. Hence the systems of matrix equations 8) also has a solution X 0.

9 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) Conversely if the systems of matrix equations 8) has a solution X R n n such that AXB = C B XA = C let X 0 = X X )/ then X 0 SSR n n and AX 0 B = 1 AX X )B = 1 AXB AX B) = 1 AXB B XA ) = 1 C C ) =C. herefore X 0 is a solution of Problem I. So the solvability of Problem I is equivalent to that of the systems of matrix equations 8) and the solution of Problem I must be the solution of the systems of matrix equations 8). Let S E denote the set of all solutions of the systems of matrix equations 8) then we know that S E S E where S E is the set of all solutions of Problem I. In order to prove that X is the least-norm solution of Problem I it is enough to prove that X is the least-norm solution of the systems of matrix equations 8). Denote vecx) = xvecx ) = x vecy ) = y 1 vecy ) = y vecc) = c 1 vecc ) = c then the systems of matrix equations 8) is equivalent to the systems of linear equations B A x = A B Noting that c1 c x = veca Y B BYA). 9) = B A )y 1 A B)y B y1 A ) = B A A B R y A B by Lemma.4 we know that X is the least-norm solution of the systems of linear equations 9). Since vector operator is isomorphic X is the unique least-norm solution of the systems of matrix equations 8) then X is the unique least-norm solution of Problem I. 3. he solution of Problem II In this section we will show that the optimal approximate solution of Problem II for a given matrix can be derived by finding the least-norm skew-symmetric solution of a new corresponding matrix equation A XB = C. For a given matrix X 0 R n n since symmetric matrix and a skew-symmetric matrix are orthogonal each other we have ) X X 0 = X X0 + X0 + X 0 X0 for any X SSR n n. = X X ) 0 X0 = X X 0 X0 X 0 + X 0 X 0 + X0 +

10 40 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) When Problem I is consistent the set of solutions of Problem I denoted by S E equation is not empty then linear AXB = C is equivalent to the following equation: A X X ) 0 X0 B = C A X 0 X0 B. Let X = X X 0 X0 )/ C = C AX 0 X0 )/B then Problem II is equivalent to finding the least-norm skew-symmetric solution X of the matrix equation A XB = C. 10) By using Algorithm.1 let initially iterative matrix X 1 = A H B BHA or more especially let X 1 = 0 R n n we can obtain the unique least-norm solution X of the matrix equation 10) then we can obtain the solution X of Problem II and X can be represented that X = X + X 0 X 0 )/. 4. Examples for the iterative methods In this section we will show several numerical examples to illustrate our results. All the tests are performed by MALAB Example 1. Consider the skew-symmetric solution of the equation AXB = C where A = B = C = We will find the skew-symmetric solution of the matrix equation AXB = C by using Algorithm.1. Because of the influence of the error of calculation the residual R i is usually unequal to zero in the process of the iteration where i = For any chosen positive number ε however small enough e.g. ε = e 010 whenever R k <ε stop the iteration and X k is regarded to be a solution of the matrix equation AXB = C. Choose an initially iterative

11 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) matrix X 1 SSR 5 5 such as X 1 = by Algorithm.1 we have X 14 = R 14 =3.646e 011 <ε. So we obtain a skew-symmetric solution of the matrix equation AXB = C as follows: X = Let X 1 = by Algorithm.1 we have X 14 = R 14 =9.8875e 011 <ε.

12 4 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) So we obtain a skew-symmetric solution of the matrix equation AXB = C as follows: X = Example. Consider the least-norm solution of the equation AXB = C in Example 1. Let H = and X 1 = A H B BHA. By using Algorithm.1 we have X 17 = R 17 =8.116e 011 <ε. So we obtain the least-norm solution of the matrix equation AXB = C as follows: X = Example 3. Consider the skew-symmetric solution of the equation AXB = C where A = B = C =

13 G.-X. Huang et al. / Journal of Computational and Applied Mathematics 1 008) Let ε = e 005 and initial matrix X 1 = 0. By using Algorithm.1 we have R 6 =1.0408e Q 6 =1.9143e 008 <ε. herefore there is no skew-symmetric solution for the matrix equation AXB = C by heorem.. Example 4. Let S E denote the set of all skew-symmetric solutions of the matrix equation AXB =C where the matrices A B and C are mentioned in Example 1. Suppose X 0 = we will find X S E such that X X 0 = min X S E X X 0 i.e. find the optimal approximate solution to the matrix X 0 in S E. Let X 0 =X 0 X0 )/ X =X X 0 C =C A X 0 B by the method mentioned in Section 3 we can obtain the least-norm skew-symmetric solution X of the matrix equation A XB = C by choosing the initial iteration matrix X 1 = 0 and X is that X 14 = R 14 =7.0170e 011 <ε= e 010 and X = X 14 + X0 = Acknowledgements he authors are grateful to the anonymous referee for the constructive and helpful comments and Prof. Lothar Reichel for the communication. References 1 A. Bjerhammer Rectangular reciprocal matrices with special reference to geodetic calculations Kung. ekn. Hogsk. Handl. Stockholm ) K.W. Eric Chu Symmetric solutions of linear matrix equations by matrix decompositions Linear Algebra Appl ) G.H. Golub C.F. Van Loan Matrix Computations John Hopkins University Press Baltimore MD F.J. Henk Don On the symmetric solution of a linear matrix equation Linear Algebra Appl ) 1 7.

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