An iterative method for symmetric solutions and optimal approximation solution of the system of matrix equations A1XB1 = C1, A2XB2 = C2

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Abstract

The symmetric solutions of the system of matrix equations A1XB1 = C1, A2XB2 = C2 are too difficult to be obtained by applying matrices decomposition. In this paper, an iterative method is applied to solve this problem. With it, the solvability of this system of matrix equations can be determined automatically, when this system of matrix equations is consistent, its solution can be obtained within finite iterative steps, and its least-norm solution can be obtained by choosing a special kind of initial iterative matrix, furthermore, its optimal approximation solution to a given matrix can be derived by finding the least-norm symmetric solution of a new system of matrix equations A1X˜B1=C˜1,A2X˜B2=C˜2. Finally, numerical examples are given.

Introduction

Let Rm×n denote the set of all m × n real matrices, Rm = Rm×1, SRn×n denote the set of all symmetric matrices in Rn×n. For a matrix A  Rm×n, ∥A∥ represents its Frobenius norm, R(A) represents its column space, vec(·) represents the vec operator, i.e. vec(A)=a1T,a2T,,anTT for the matrix A = (a1, a2,  , an)  Rm×n, ai  Rm, i = 1, 2,  , n. A  B stands for the Kronecker product of matrices A and B.

In this paper, we consider the following problems.

Problem I

Given A1Rm1×n, B1Rn×p1, C1Rm1×p1, A2Rm2×n, B2Rn×p2, C2Rm2×p2, find X  SRn×n, such thatA1XB1=C1,A2XB2=C2.

Problem II

When Problem I is consistent, let SE denote the set of solutions of Problem I, for given X0  Rn×n, find XˆSE, such thatXˆ-X0=minXSEX-X0.In fact, Problem II is to find the optimal approximation solution to the given matrix X0.

Research on solving systems of linear matrix equations has been actively ongoing for past years. For e.g., Mitra [1], [2] has provided conditions for the existence of a solution and a representation of the general common solution to the matrix equations AX = C, XB = D and the matrix equations A1XB1 = C1, A2XB2 = C2. Also, Navarra et al. [3] studied a representation of the general common solution to the matrix equations A1XB1 = C1, A2XB2 = C2, Bhimasankaram [4] considered the linear matrix equations AX = B, CX = D and EXF = G, van der Woude [5] obtained the existence of a common solution X to the matrix equations AiXBj = Cij, (i, j)  Γ. Till now, the problem to obtain symmetric solutions of the system of matrix equations A1XB1 = C1, A2XB2 = C2 has not been solved.

In this paper, an iterative method is constructed to solve the system of matrix equations A1XB1 = C1, A2XB2 = C2 over symmetric X. With it, the solvability of the system of matrix equations can be determined automatically, when the system of matrix equations is consistent, its symmetric solution can be obtained within finite iterative steps, and its least-norm solution can be obtained by choosing a suitable initial iterative matrix, furthermore, its optimal approximation solution to a given matrix can be derived by finding the least-norm symmetric solution of a new system of matrix equation A1X˜B1=C˜1,A2X˜B2=C˜2.

Section snippets

An iterative method for solving Problem I

The conjugate gradients method is an efficient method to solve linear systems Ax = b, where A is symmetric and positive definite, x  Rn (see [6], [7], [8]). For nonsymmetric matrix A, we can use general conjugate gradients methods in [9], [10], [11]. Similarly, we construct an iterative method to obtain the symmetric solutions of the system of matrix equations A1XB1 = C1, A2XB2 = C2.

  • Step 1:

    Choose an arbitrary matrix X1  SRn×n, computeR1=C1-A1X1B100C2-A2X1B2,P1=A1T(C1-A1X1B1)B1T+A2T(C2-A2X1B2)B2T,Q1=12(P1+P1T).

The solution of Problem II

We assume that X0  SRn×n in Problem II, this is no loss of generality since that a symmetric matrix and a skew-symmetric matrix are orthogonal each other, for X  SE  SRn×n, we have thatX-X02=X-X0+X0T2+X0-X0T22=X-X0+X0T2-X0-X0T22=X-X0+X0T22+X0-X0T22.When Problem I is consistent, the set of solutions of Problem I denoted by SE is no empty, for X  SE, by A1XB1 = C1, A2XB2 = C2, we have thatA1(X-X0)B1=C1-A1X0B1,A2(X-X0)B2=C2-A2X0B2.

Let X˜=X-X0,C˜1=C1-A1X0B1,C˜2=C2-A2X0B2, then Problem II is equivalent to

Example

Example 1

Suppose the system of matrix equations A1XB1 = C1, A2XB2 = C2, where A1  R5×6, B1  R6×7, C1  R5×7, A2  R6×6, B2  R6×4, C2  R6×4, X  R6×6, andA1=2-3-1402-1-324-510-2350-17-8-21354-22-8-12-106,A2=1-134072-315-205-4130115-1181221731-29830-65-641010026,B1=30-45-12-2-23-420-1-3-56-30-23-1-23-1305-219-15-3124-6-4-99-9-10-6-122,B2=-20-41-14-46-32-61-810-22214-26522-3066-75,C1=-2952882574311-96-815738143-56-51818-79-645597-22-189-130374-1-715723-15938-67489-210950-912374644418140-154,C2=-115158-210-47-819990-2154

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Research supported by the National Natural Science Foundation of China (10571047).

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