An Iterative Algorithm for the Generalized Reflexive Solution of the Matrix Equations 𝐴𝑋𝐵=𝐸, 𝐶𝑋𝐷=𝐹

Abstract

An iterative algorithm is constructed to solve the linear matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 over generalized reflexive matrix 𝑋. When the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 is consistent over generalized reflexive matrix 𝑋, for any generalized reflexive initial iterative matrix 𝑋₁, the generalized reflexive solution can be obtained by the iterative algorithm within finite iterative steps in the absence of round-off errors. The unique least-norm generalized reflexive iterative solution of the matrix equation pair can be derived when an appropriate initial iterative matrix is chosen. Furthermore, the optimal approximate solution of 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 for a given generalized reflexive matrix 𝑋₀ can be derived by finding the least-norm generalized reflexive solution of a new corresponding matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 with 𝐸=𝐸−𝐴𝑋₀𝐵,𝐹=𝐹−𝐶𝑋₀𝐷. Finally, several numerical examples are given to illustrate that our iterative algorithm is effective.

1. Introduction

Let ℛ^𝑚×𝑛 denote the set of all 𝑚-by-𝑛 real matrices. 𝐼_𝑛 denotes the 𝑛 order identity matrix. Let 𝑃∈ℛ^𝑚×𝑚 and 𝑄∈ℛ^𝑛×𝑛 be two real generalized reflection matrices, that is, 𝑃^𝑇=𝑃,𝑃²=𝐼_𝑚,𝑄^𝑇=𝑄,𝑄²=𝐼_𝑛. A matrix 𝐴∈ℛ^𝑚×𝑛 is called generalized reflexive matrix with respect to the matrix pair (𝑃,𝑄) if 𝑃𝐴𝑄=𝐴. For more properties and applications on generalized reflexive matrix, we refer to [1, 2]. The set of all 𝑚-by-𝑛 real generalized reflexive matrices with respect to matrix pair (𝑃,𝑄) is denoted by ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄).We denote by the superscripts 𝑇 the transpose of a matrix. In matrix space ℛ^𝑚×𝑛, define inner product as tr(𝐵^𝑇𝐴)=trace(𝐵^𝑇𝐴) for all 𝐴,𝐵∈ℛ^𝑚×𝑛;  ‖𝐴‖ represents the Frobenius norm of 𝐴. ℛ(𝐴) represents the column space of 𝐴. vec(⋅) represents the vector operator; that is, vec(𝐴)=(𝑎^𝑇1,𝑎^𝑇2,…,𝑎^𝑇𝑛)^𝑇∈ℛ^𝑚𝑛 for the matrix 𝐴=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℛ^𝑚×𝑛,𝑎_𝑖∈𝑅^𝑚,𝑖=1,2,…,𝑛. 𝐴⊗𝐵 stands for the Kronecker product of matrices 𝐴 and 𝐵.

In this paper, we will consider the following two problems.

Problem 1. For given matrices 𝐴∈ℛ^𝑝×𝑚,𝐵∈ℛ^𝑛×𝑞,𝐶∈ℛ^𝑠×𝑚, 𝐷∈ℛ^𝑛×𝑡,𝐸∈ℛ^𝑝×𝑞,𝐹∈ℛ^𝑠×𝑡, find matrix 𝑋∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄) such that 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹.(1.1)

Problem 2. When Problem 1 is consistent, let 𝑆_𝐸 denote the set of the generalized reflexive solutions of Problem 1. For a given matrix 𝑋₀∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄), find 𝑋∈𝑆_𝐸 such that ‖‖𝑋−𝑋₀‖‖=min_{𝑋∈𝑆𝐸}‖‖𝑋−𝑋₀‖‖.(1.2)

The matrix equation pair (1.1) may arise in many areas of control and system theory. Dehghan and Hajarian [3] presented some examples to show a motivation for studying (1.1). Problem 2 occurs frequently in experiment design; see for instance [4]. In recent years, the matrix nearness problem has been studied extensively (e.g., [3, 5–19]).

Research on solving the matrix equation pair (1.1) has been actively ongoing for last 40 or more years. For instance, Mitra [20, 21] gave conditions for the existence of a solution and a representation of the general common solution to the matrix equation pair (1.1). Shinozaki and Sibuya [22] and vander Woude[23] discussed conditions for the existence of a common solution to the matrix equation pair (1.1). Navarra et al. [11] derived sufficient and necessary conditions for the existence of a common solution to (1.1). Yuan [18] obtained an analytical expression of the least-squares solutions of (1.1) by using the generalized singular value decomposition (GSVD) of matrices. Recently, some finite iterative algorithms have also been developed to solve matrix equations. Deng et al. [24] studied the consistent conditions and the general expressions about the Hermitian solutions of the matrix equations (𝐴𝑋,𝑋𝐵)=(𝐶,𝐷) and designed an iterative method for its Hermitian minimum norm solutions. Li and Wu [25] gave symmetric and skew-antisymmetric solutions to certain matrix equations 𝐴₁𝑋=𝐶₁,𝑋𝐵₃=𝐶₃ over the real quaternion algebra 𝐻. Dehghan and Hajarian [26] proposed the necessary and sufficient conditions for the solvability of matrix equations 𝐴₁𝑋𝐵₁=𝐷₁,𝐴₁𝑋=𝐶₁,𝑋𝐵₂=𝐶₂ and 𝐴₁𝑋=𝐶₁,𝑋𝐵₂=𝐶₂,𝐴₃𝑋=𝐶₃,𝑋𝐵₄=𝐶₄ over the reflexive or antireflexive matrix 𝑋 and obtained the general expression of the solutions for a solvable case. Wang [27, 28] gave the centrosymmetric solution to the system of quaternion matrix equations 𝐴₁𝑋=𝐶₁,𝐴₃𝑋𝐵₃=𝐶₃. Wang [29] also solved a system of matrix equations over arbitrary regular rings with identity. For more studies on iterative algorithms on coupled matrix equations, we refer to [6, 7, 15–17, 19, 30–34]. Peng et al. [13] presented iterative methods to obtain the symmetric solutions of (1.1). Sheng and Chen [14] presented a finite iterative method when (1.1) is consistent. Liao and Lei [9] presented an analytical expression of the least-squares solution and an algorithm for (1.1) with the minimum norm. Peng et al. [12] presented an efficient algorithm for the least-squares reflexive solution. Dehghan and Hajarian [3] presented an iterative algorithm for solving a pair of matrix equations (1.1) over generalized centrosymmetric matrices. Cai and Chen [35] presented an iterative algorithm for the least-squares bisymmetric solutions of the matrix equations (1.1). However, the problem of finding the generalized reflexive solutions of matrix equation pair (1.1) has not been solved. In this paper, we construct an iterative algorithm by which the solvability of Problem 1 can be determined automatically, the solution can be obtained within finite iterative steps when Problem 1 is consistent, and the solution of Problem 2 can be obtained by finding the least-norm generalized reflexive solution of a corresponding matrix equation pair.

This paper is organized as follows. In Section 2, we will solve Problem 1 by constructing an iterative algorithm; that is, if Problem 1 is consistent, then for an arbitrary initial matrix 𝑋₁∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄), we can obtain a solution 𝑋⁻∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄) of Problem 1 within finite iterative steps in the absence of round-off errors. Let 𝑋₁=𝐴^𝑇𝐻𝐵^𝑇+𝐶^𝑇𝐻𝐷^𝑇+𝑃𝐴^𝑇𝐻𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐻𝐷^𝑇𝑄, where 𝐻∈ℛ^𝑝×𝑞,𝐻∈ℛ^𝑠×𝑡 are arbitrary matrices, or more especially, letting 𝑋₁=𝟎∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄), we can obtain the unique least norm solution of Problem 1. Then in Section 3, we give the optimal approximate solution of Problem 2 by finding the least norm generalized reflexive solution of a corresponding new matrix equation pair. In Section 4, several numerical examples are given to illustrate the application of our iterative algorithm.

2. The Solution of Problem 1

In this section, we will first introduce an iterative algorithm to solve Problem 1 and then prove that it is convergent. The idea of the algorithm and it’s proof in this paper are originally inspired by those in [13]. The idea of our algorithm is also inspired by those in [3]. When 𝑃=𝑄,𝑅=𝑆,𝑋^𝑇=𝑋 and 𝑌^𝑇=𝑌, the results in this paper reduce to those in [3].

Algorithm 2.1. Step 1. Input matrices 𝐴∈ℛ^𝑝×𝑚,𝐵∈ℛ^𝑛×𝑞,𝐶∈ℛ^𝑠×𝑚, 𝐷∈ℛ^𝑛×𝑡,𝐸∈ℛ^𝑝×𝑞,𝐹∈ℛ^𝑠×𝑡, and two generalized reflection matrix 𝑃∈ℛ^𝑚×𝑚,𝑄∈ℛ^𝑛×𝑛.
Step 2. Choose an arbitrary matrix 𝑋₁∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄). Compute 𝑅₁=⎛⎜⎜⎝𝐸−𝐴𝑋¹0𝐵𝐹−𝐶0𝑋₁𝑃₁𝐷⎞⎟⎟⎠,=12𝐴^𝑇𝐸−𝐴𝑋₁+𝑃𝐶_𝑇𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇+𝑃𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇𝑄𝐹−𝐶𝑋₁𝐷𝐷^𝑇𝑄,(2.1)𝑘∶=1.
Step 3. If 𝑅₁=𝟎, then stop. Else go to Step 4.
Step 4. Compute 𝑋_𝑘+1=𝑋_𝑘+‖‖𝑅^𝑘‖‖²‖‖𝑃_𝑘‖‖²𝑅_𝑘+1𝑃_𝑘,=⎛⎜⎜⎝𝐸−𝐴𝑋^𝑘+10𝐵𝐹−𝐶0𝑋_𝑘+1𝐷⎞⎟⎟⎠=𝑅^𝑘−‖‖𝑅^𝑘‖‖²‖‖𝑃_𝑘‖‖²⎛⎜⎜⎝𝐴𝑃0_𝑘𝐵𝐶𝑃0_𝑘𝑃_𝑘+1𝐷⎞⎟⎟⎠,=12𝐴^𝑇𝐸−𝐴𝑋_𝑘+1+𝑃𝐶_𝑇𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑘+1𝐷𝐷^𝑇+𝑃𝐴^𝑇𝐸−𝐴𝑋_𝑘+1𝐵𝐵^𝑇𝑄𝐹−𝐶𝑋_𝑘+1𝐷𝐷^𝑇𝑄+‖‖𝑅^𝑘+1‖‖²‖‖𝑅_𝑘‖‖²𝑃_𝑘.(2.2)
Step 5. If 𝑅_𝑘+1=𝟎, then stop. Else, letting 𝑘∶=𝑘+1, go to Step 4.

Obviously, it can be seen that 𝑃_𝑖∈𝑅^{𝑟𝑚×𝑛}(𝑃,𝑄),𝑋_𝑖∈𝑅^{𝑟𝑚×𝑛}(𝑃,𝑄), where 𝑖=1,2,….

Lemma 2.2. For the sequences {𝑅_𝑖} and {𝑃_𝑖} generated in Algorithm 2.1, one has tr𝑅^𝑇𝑖+1𝑅_𝑗=tr𝑅^𝑇𝑖𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²tr𝑃^𝑇𝑖𝑃_𝑗+‖‖𝑅^𝑖‖‖²‖‖𝑅_𝑗‖‖²‖‖𝑃_𝑖‖‖²‖‖𝑅_𝑗−1‖‖²tr𝑃_𝑇𝑖+1tr𝑃^𝑇𝑖𝑃_𝑗−1,𝑃_𝑗=‖‖𝑃^𝑗‖‖²‖‖𝑅_𝑗‖‖²tr𝑅^𝑇𝑖+1𝑅_𝑗−tr𝑅^𝑇𝑖+1𝑅_𝑗+1+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²tr𝑃^𝑇𝑖𝑃_𝑗.(2.3)

Proof. By Algorithm 2.1, we have tr𝑅^𝑇𝑖+1𝑅_𝑗=tr⎛⎜⎜⎜⎝⎛⎜⎜⎝𝑅_𝑖−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²⎛⎜⎜⎝𝐴𝑃0_𝑖𝐵𝐶𝑃0_𝑖𝐷⎞⎟⎟⎠⎞⎟⎟⎠^𝑇𝑅_𝑗=tr𝑅_𝑇𝑖⎞⎟⎟⎟⎠𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²tr⎛⎜⎜⎝⎛⎜⎜⎝𝐵^𝑇𝑃0^𝑇𝑖𝐴^𝑇𝐷_𝑇0𝑃^𝑇𝑖𝐶^𝑇=tr𝑅_𝑇𝑖⎞⎟⎟⎠𝑅_𝑗⎞⎟⎟⎠𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²tr⎛⎜⎜⎝⎛⎜⎜⎝𝐵^𝑇𝑃0^𝑇𝑖𝐴^𝑇𝐷_𝑇0𝑃^𝑇𝑖𝐶^𝑇⎞⎟⎟⎠⎛⎜⎜⎝𝐸−𝐴0𝑋_𝑗𝐵𝐹−𝐶0𝑋_𝑗=tr𝑅_𝑇𝑖𝐷⎞⎟⎟⎠⎞⎟⎟⎠𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²=tr𝑅_𝑇𝑖tr𝐵^𝑇𝑃^𝑇𝑖𝐴^𝑇𝐸−𝐴𝑋_𝑗𝐵+𝐷^𝑇𝑃^𝑇𝑖𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²=tr𝑅_𝑇𝑖tr𝑃^𝑇𝑖𝐴^𝑇𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²×tr𝑃^𝑇𝑖𝐴^𝑇𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇𝑄+𝐴_𝑇2𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇+−𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇𝑄−𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇𝑄=tr𝑅_𝑇𝑖2𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²×tr𝑃^𝑇𝑖𝐴^𝑇𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑗𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑗𝐷𝐷^𝑇𝑄=tr𝑅_𝑇𝑖2𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²tr𝑃^𝑇𝑖𝑃^𝑗−‖‖𝑅^𝑗‖‖²‖‖𝑅_𝑗−1‖‖²𝑃_𝑗−1=tr𝑅_𝑇𝑖𝑅_𝑗−‖‖𝑅^𝑖‖‖²‖‖𝑃_𝑖‖‖²tr𝑃^𝑇𝑖𝑃_𝑗+‖‖𝑅^𝑖‖‖²‖‖𝑅_𝑗‖‖²‖‖𝑃_𝑖‖‖²‖‖𝑅_𝑗−1‖‖²tr𝑃_𝑇𝑖+1tr𝑃^𝑇𝑖𝑃_𝑗−1,𝑃_𝑗=tr𝐴^𝑇𝐸−𝐴𝑋_𝑖+1𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑖+1𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑖+1𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑖+1𝐷𝐷^𝑇𝑄2+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²𝑃_𝑖^𝑇𝑃_𝑗=tr𝐴_𝑇⎞⎟⎟⎠𝐸−𝐴𝑋_𝑖+1𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑖+1𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑖+1𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑖+1𝐷𝐷^𝑇𝑄2^𝑇𝑃_𝑗+‖‖𝑅_𝑖+1⎞⎟⎟⎠‖‖²‖‖𝑅_𝑖‖‖²=tr𝑃_𝑇𝑗tr𝑃^𝑇𝑖𝑃_𝑗𝐴^𝑇𝐸−𝐴𝑋_𝑖+1𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑖+1𝐷𝐷^𝑇+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²=tr𝐸−𝐴𝑋_𝑖+1tr𝑃^𝑇𝑖𝑃_𝑗𝐵^𝑇𝐴𝑃_𝑗𝐵+𝐹−𝐶𝑋_𝑖+1𝐷^𝑇𝐶𝑃_𝑗𝐷+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²=tr⎛⎜⎜⎝⎛⎜⎜⎝𝐸−𝐴𝑋_𝑖+1tr𝑃^𝑇𝑖𝑃_𝑗0𝐵^𝑇𝐹−𝐶𝑋0_𝑖+1𝐷^𝑇⎞⎟⎟⎠⎛⎜⎜⎝𝐴𝑃0_𝑗𝐵𝐶𝑃0_𝑗+‖‖𝑅_𝑖+1𝐷⎞⎟⎟⎠⎞⎟⎟⎠‖‖²‖‖𝑅_𝑖‖‖²=‖‖𝑃_𝑗tr𝑃^𝑇𝑖𝑃_𝑗‖‖²‖‖𝑅_𝑗‖‖²tr𝑅^𝑇𝑖+1𝑅_𝑗−𝑅_𝑗+1+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²=‖‖𝑃_𝑗tr𝑃^𝑇𝑖𝑃_𝑗‖‖²‖‖𝑅_𝑗‖‖²tr𝑅^𝑇𝑖+1𝑅_𝑗−tr𝑅^𝑇𝑖+1𝑅_𝑗+1+‖‖𝑅^𝑖+1‖‖²‖‖𝑅_𝑖‖‖²tr𝑃^𝑇𝑖𝑃_𝑗.(2.4) This completes the proof.

Lemma 2.3. For the sequences {𝑅_𝑖} and {𝑃_𝑖} generated by Algorithm 2.1, and 𝑘≥2, one has tr𝑅^𝑇𝑖𝑅_𝑗=0,tr𝑃^𝑇𝑖𝑃_𝑗=0,𝑖,𝑗=1,2,…,𝑘,𝑖≠𝑗.(2.5)

Proof. Since tr(𝑅^𝑇𝑖𝑅_𝑗)=tr(𝑅^𝑇𝑗𝑅_𝑖) and tr(𝑃^𝑇𝑖𝑃_𝑗)=tr(𝑃^𝑇𝑗𝑃_𝑖) for all 𝑖,𝑗=1,2,…,𝑘, we only need to prove that tr(𝑅^𝑇𝑖𝑅_𝑗)=0,tr(𝑃^𝑇𝑖𝑃_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 tr𝑅^𝑇𝑖+1𝑅_𝑖=0,tr𝑃^𝑇𝑖+1𝑃_𝑖=0,𝑖=1,2,…,𝑘−1.(2.6) To prove this conclusion, we also use induction.
For 𝑖=1, by Algorithm 2.1 and the proof of Lemma 2.2, we have that tr𝑅^𝑇2𝑅₁=tr⎛⎜⎜⎜⎝⎛⎜⎜⎝𝑅₁−‖‖𝑅¹‖‖²‖‖𝑃₁‖‖²⎛⎜⎜⎝𝐴𝑃0₁𝐵𝐶𝑃0₁𝐷⎞⎟⎟⎠⎞⎟⎟⎠^𝑇𝑅₁=tr𝑅_𝑇1⎞⎟⎟⎟⎠𝑅₁−‖‖𝑅¹‖‖²‖‖𝑃₁‖‖²×tr𝑃^𝑇1𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋₁𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇𝑄=‖‖𝑅2₁‖‖²−‖‖𝑅¹‖‖²‖‖𝑃₁‖‖²tr𝑃_𝑇2=0,tr𝑃^𝑇1𝑃₁𝑃₁=‖‖𝑃¹‖‖²‖‖𝑅₁‖‖²tr𝑅^𝑇2𝑅₁−tr𝑅^𝑇2𝑅₂+‖‖𝑅²‖‖²‖‖𝑅₁‖‖²‖‖𝑃₁‖‖²=0.(2.7)
Assume (2.6) holds for 𝑖=𝑠−1, that is, tr(𝑅^𝑇𝑠𝑅_𝑠−1)=0,tr(𝑃^𝑇𝑠𝑃_𝑠−1)=0. When 𝑖=𝑠, by Lemma 2.2, we have that tr𝑅^𝑇𝑠+1𝑅_𝑠=tr𝑅^𝑇𝑠𝑅_𝑠−‖‖𝑅^𝑠‖‖²‖‖𝑃_𝑠‖‖²tr𝑃^𝑇𝑠𝑃_𝑠+‖‖𝑅^𝑠‖‖⁴‖‖𝑃_𝑠‖‖²‖‖𝑅_𝑠−1‖‖²=‖‖𝑅_𝑠tr𝑃^𝑇𝑠𝑃_𝑠−1‖‖²−‖‖𝑅_𝑠‖‖²+‖‖𝑅^𝑠‖‖⁴‖‖𝑃_𝑠‖‖²‖‖𝑅_𝑠−1‖‖²tr𝑃_𝑇𝑠+1=0,tr𝑃^𝑇𝑠𝑃_𝑠−1𝑃_𝑠=‖‖𝑃^𝑠‖‖²‖‖𝑅_𝑠‖‖²tr𝑅^𝑇𝑠+1𝑅_𝑠−tr𝑅^𝑇𝑠+1𝑅_𝑠+1+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²=−‖‖𝑃_𝑠tr𝑃^𝑇𝑠𝑃_𝑠‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑅_𝑠+1‖‖²+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑃_𝑠‖‖²=0.(2.8) Hence, (2.6) holds for 𝑖=𝑠. Therefor, (2.6) holds by the principle of induction.
Step 2. Assuming that tr(𝑅^𝑇𝑠𝑅_𝑗)=0,tr(𝑃^𝑇𝑠𝑃_𝑗)=0,𝑗=1,2,…,𝑠−1, then we show that tr𝑅^𝑇𝑠+1𝑅_𝑗=0,tr𝑃^𝑇𝑠+1𝑃_𝑗=0,𝑗=1,2,…,𝑠.(2.9) In fact, by Lemma 2.2 we have tr𝑅^𝑇𝑠+1𝑅_𝑗=tr𝑅^𝑇𝑠𝑅_𝑗−‖‖𝑅^𝑠‖‖²‖‖𝑃_𝑠‖‖²tr𝑃^𝑇𝑠𝑃_𝑗+‖‖𝑅^𝑠‖‖²‖‖𝑅_𝑗‖‖²‖‖𝑃_𝑠‖‖²‖‖𝑅_𝑗−1‖‖²=0.tr𝑃^𝑇𝑠𝑃_𝑗−1(2.10) From the previous results, we have tr(𝑅^𝑇𝑠+1𝑅_𝑗+1)=0. By Lemma 2.2 we have that tr𝑃^𝑇𝑠+1𝑃_𝑗=‖‖𝑃^𝑗‖‖²‖‖𝑅_𝑗‖‖²tr𝑅^𝑇𝑠+1𝑅_𝑗−tr𝑅^𝑇𝑠+1𝑅_𝑗+1+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²=‖‖𝑃_𝑗tr𝑃^𝑇𝑠𝑃_𝑗‖‖²‖‖𝑅_𝑗‖‖²=0.tr𝑅^𝑇𝑠+1𝑅_𝑗−tr𝑅^𝑇𝑠+1𝑅_𝑗+1(2.11)
By the principle of induction, (2.9) holds. Note that (2.5) is implied in Steps 1 and 2 by the principle of induction. This completes the proof.

Lemma 2.4. Supposing 𝑋⁻ is an arbitrary solution of Problem 1, that is, 𝐴𝑋⁻𝐵=𝐸 and 𝐶𝑋⁻𝐷=𝐹, then tr𝑋⁻−𝑋_𝑘^𝑇𝑃_𝑘=‖‖𝑅^𝑘‖‖²,𝑘=1,2,…,(2.12) where the sequences {𝑋_𝑘}, {𝑅_𝑘}, and {𝑃_𝑘} are generated by Algorithm 2.1.

Proof. We proof the conclusion by induction.
For 𝑘=1, we have that tr𝑋⁻−𝑋₁^𝑇𝑃₁=tr⁻𝑋−𝑋₁^𝑇12𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇=tr+𝑃₋𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇𝑄=tr𝑋₋−𝑋₁^𝑇𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇𝑋−𝑋₁^𝑇𝐴^𝑇𝐸−𝐴𝑋₁𝐵𝐵^𝑇+𝑋⁻−𝑋₁^𝑇𝐶^𝑇𝐹−𝐶𝑋₁𝐷𝐷^𝑇=tr𝐸−𝐴𝑋₁=tr⎛⎜⎜⎝⎛⎜⎜⎝𝐸−𝐴𝑋₁𝐵^𝑇𝐴𝑋⁻−𝑋₁𝐵+𝐹−𝐶𝑋¹𝐷^𝑇𝐶𝑋⁻−𝑋₁𝐷0𝐵^𝑇𝐹−𝐶0𝑋₁𝐷^𝑇⎞⎟⎟⎠⎛⎜⎜⎝𝐴𝑋⁻−𝑋₁0𝐵𝐶₋0=tr⎛⎜⎜⎝⎛⎜⎜⎝𝐸−𝐴𝑋₁𝑋−𝑋₁𝐷⎞⎟⎟⎠⎞⎟⎟⎠0𝐵^𝑇𝐹−𝐶0𝑋₁𝐷^𝑇⎞⎟⎟⎠⎛⎜⎜⎝𝐸−𝐴0𝑋₁𝐵𝐹−𝐶0𝑋₁=tr𝑅_𝑇1𝐷⎞⎟⎟⎠⎞⎟⎟⎠𝑅₁=‖‖𝑅₁‖‖².(2.13)
Assume (2.12) holds for 𝑘=𝑠. By Algorithm 2.1, we have that tr𝑋⁻−𝑋_𝑠+1^𝑇𝑃_𝑠+1=tr₋𝑋−𝑋_𝑠+1^𝑇×𝐴^𝑇𝐸−𝐴𝑋_𝑠+1𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑠+1𝐷𝐷^𝑇+𝑃𝐴_𝑇2𝐸−𝐴𝑋_𝑠+1𝐵𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐹−𝐶𝑋_𝑠+1𝐷𝐷^𝑇𝑄2+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²𝑃_𝑠=tr₋𝑋−𝑋_𝑠+1^𝑇𝐴^𝑇𝐸−𝐴𝑋_𝑠+1𝐵𝐵^𝑇+𝐶^𝑇𝐹−𝐶𝑋_𝑠+1𝐷𝐷^𝑇+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²𝑃_𝑠=tr⎛⎜⎜⎝⎛⎜⎜⎝𝐸−𝐴𝑋_𝑠+10𝐵^𝑇𝐹−𝐶𝑋0_𝑠+1𝐷^𝑇⎞⎟⎟⎠⎛⎜⎜⎝𝐸−𝐴0𝑋_𝑠+1𝐵𝐹−𝐶0𝑋_𝑠+1+‖‖𝑅_𝑠+1𝐷⎞⎟⎟⎠⎞⎟⎟⎠‖‖²‖‖𝑅_𝑠‖‖²tr𝑋⁻−𝑋_𝑠+1^𝑇𝑃_𝑠=tr𝑅_𝑇𝑠+1𝑅_𝑠+1+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²tr𝑋⁻−𝑋_𝑠^𝑇𝑃_𝑠−‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑃_𝑠‖‖²=‖‖𝑅_𝑠+1tr𝑃^𝑇𝑠𝑃_𝑠‖‖²+‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑅_𝑠‖‖²−‖‖𝑅^𝑠+1‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑅_𝑠‖‖²‖‖𝑃_𝑠‖‖²‖‖𝑃_𝑠‖‖²=‖‖𝑅_𝑠+1‖‖².(2.14) Therefore, (2.12) holds for 𝑘=𝑠+1. By the principle of induction, (2.12) holds. This completes the proof.

Theorem 2.5. Supposing that Problem 1 is consistent, then for an arbitrary initial matrix 𝑋₁∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄), a solution of Problem 1 can be obtained with finite iteration steps in the absence of round-off errors.

Proof. If 𝑅_𝑖≠𝟎,𝑖=1,2,…,𝑝𝑞+𝑠𝑡, by Lemma 2.4 we have 𝑃_𝑖≠𝟎,𝑖=1,2,…,𝑝𝑞+𝑠𝑡, then we can compute 𝑋_{𝑝𝑞+𝑠𝑡+1},𝑅_{𝑝𝑞+𝑠𝑡+1} by Algorithm 2.1.
By Lemma 2.3, we have trtr𝑅𝑅^{𝑇𝑝𝑞+𝑠𝑡+1}_𝑇𝑖𝑅_𝑖=0,𝑖=1,2,…,𝑝𝑞+𝑠𝑡,𝑅_𝑗=0,𝑖,𝑗=1,2,…,𝑝𝑞+𝑠𝑡,𝑖≠𝑗.(2.15) Therefore, 𝑅₁,𝑅₂,…,𝑅_{𝑝𝑞+𝑠𝑡} is an orthogonal basis of the matrix space 𝑆=⎧⎪⎨⎪⎩𝑊∣𝑊=⎛⎜⎜⎝𝑊¹0𝑊0₄⎞⎟⎟⎠,𝑊₁∈ℛ^𝑝×𝑞,𝑊₄∈ℛ^𝑠×𝑡⎫⎪⎬⎪⎭,(2.16) which implies that 𝑅_{𝑝𝑞+𝑠𝑡+1}=𝟎; that is, 𝑋_{𝑝𝑞+𝑠𝑡+1} is a solution of Problem 1. This completes the proof.

To show the least norm generalized reflexive solution of Problem 1, we first introduce the following result.

Lemma 2.6 (see [8, Lemma  2.4]). Supposing that the consistent system of linear equation 𝑀𝑦=𝑏 has a solution 𝑦₀∈𝑅(𝑀^𝑇), then 𝑦₀ is the least norm solution of the system of linear equations.

By Lemma 2.6, the following result can be obtained.

Theorem 2.7. Suppose that Problem 1 is consistent. If one chooses the initial iterative matrix 𝑋₁=𝐴^𝑇𝐻𝐵^𝑇+𝐶^𝑇𝐻𝐷^𝑇+𝑃𝐴^𝑇𝐻𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐻𝐷^𝑇𝑄, where 𝐻∈ℛ^𝑝×𝑞,𝐻∈ℛ^𝑠×𝑡 are arbitrary matrices, especially, let 𝑋₁=𝟎∈ℛ^{𝑟𝑚×𝑛}, one can obtain the unique least norm generalized reflexive solution of Problem 1 within finite iterative steps in the absence of round-off errors by using Algorithm 2.1.

Proof. By Algorithm 2.1 and Theorem 2.5, if we let 𝑋₁=𝐴^𝑇𝐻𝐵^𝑇+𝐶^𝑇𝐻𝐷^𝑇+𝑃𝐴^𝑇𝐻𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐻𝐷^𝑇𝑄, where 𝐻∈ℛ^𝑝×𝑞,𝐻∈ℛ^𝑠×𝑡 are arbitrary matrices, we can obtain the solution 𝑋^∗ of Problem 1 within finite iterative steps in the absence of round-off errors, and the solution 𝑋^∗ can be represented that 𝑋^∗=𝐴^𝑇𝐺𝐵^𝑇+𝐶^𝑇𝐺𝐷^𝑇+𝑃𝐴^𝑇𝐺𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐺𝐷^𝑇𝑄.
In the sequel, we will prove that 𝑋^∗ is just the least norm solution of Problem 1.
Consider the following system of matrix equations: 𝐴𝐶𝐴𝑃𝐶𝑃𝑋𝑋𝑋𝑋𝑄𝑄𝐷𝐵𝐷𝐵====𝐸𝐹𝐸,,𝐹.,(2.17)
If Problem 1 has a solution 𝑋₀∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄), then 𝐴𝑋₀𝑃𝑋₀𝑄=𝑋₀,𝐵=𝐸,𝐶𝑋₀𝐷=𝐹.(2.18) Thus 𝐴𝑃𝑋₀𝑄𝐵=𝐸,𝐶𝑃𝑋₀𝑄𝐷=𝐹.(2.19) Hence, the systems of matrix equations (2.17) also have a solution 𝑋₀.
Conversely, if the systems of matrix equations (2.17) have a solution 𝑋∈𝑅^𝑚×𝑛, let 𝑋₀=(𝑋+𝑃𝑋𝑄)/2, then 𝑋₀∈𝑅^{𝑟𝑚×𝑛}(𝑃,𝑄), and 𝐴𝑋₀𝐵=12𝐴𝑋+𝑃𝑋𝑄𝐵=12𝐴𝑋𝐵+𝐴𝑃𝑋𝑄𝐵=1𝐶𝑋₀2(𝐸+𝐸)=𝐸,𝐷=12𝐶𝑋+𝑃𝑋𝑄𝐷=12𝐶𝑋𝐷+𝐶𝑃𝑋𝑄𝐷=12(𝐹+𝐹)=𝐹.(2.20) Therefore, 𝑋₀ is a solution of Problem 1.
So the solvability of Problem 1 is equivalent to that of the systems of matrix equations (2.17), and the solution of Problem 1 must be the solution of the systems of matrix equations (2.17).
Letting 𝑆^𝐸 denote the set of all solutions of the systems of matrix equations (2.17), then we know that 𝑆_𝐸⊂𝑆^𝐸, where 𝑆_𝐸 is the set of all solutions of Problem 1. In order to prove that 𝑋^∗ is the least-norm solution of Problem 1, it is enough to prove that 𝑋^∗ is the least-norm solution of the systems of matrix equations (2.21). Denoting vec(𝑋)=𝑥,vec(𝑋^∗)=𝑥^∗,vec(𝐺)=𝑔₁,vec(𝐺)=𝑔₂,vec(𝐸)=𝑒,vec(𝐹)=𝑓, then the systems of matrix equations (2.17) are equivalent to the systems of linear equations ⎛⎜⎜⎜⎜⎜⎜⎝𝐷𝐵^𝑇⊗𝐴_𝑇𝐵⊗𝐶_𝑇𝐷𝑄⊗𝐴𝑃_𝑇𝑄⊗𝐶𝑃⎞⎟⎟⎟⎟⎟⎟⎠𝑥=⎛⎜⎜⎜⎜⎜⎜⎝𝑓𝑓𝑒𝑒⎞⎟⎟⎟⎟⎟⎟⎠.(2.21) Noting that 𝑥^∗==v𝐵ec⊗𝐴𝐴^𝑇_𝑇𝐺𝐵^𝑇+𝐶^𝑇𝐺𝐷^𝑇+𝑃𝐴^𝑇𝐺𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐺𝐷^𝑇𝑄𝑔₁+𝐷⊗𝐶^𝑇𝑦₂+𝑄𝐵⊗𝑃𝐴^𝑇𝑔₁+𝑄𝐷⊗𝑃𝐶^𝑇𝑔₂=𝐵⊗𝐴^𝑇𝐷⊗𝐶^𝑇𝑄𝐵⊗𝑃𝐴^𝑇𝑄𝐷⊗𝑃𝐶^𝑇⎛⎜⎜⎜⎜⎜⎜⎝𝑔¹𝑔₂𝑔₁𝑔₂=⎛⎜⎜⎜⎜⎜⎜⎝𝐵_𝑇⎞⎟⎟⎟⎟⎟⎟⎠𝐷⊗𝐴_𝑇𝐵⊗𝐶_𝑇𝐷𝑄⊗𝐴𝑃_𝑇𝑄⊗𝐶𝑃⎞⎟⎟⎟⎟⎟⎟⎠^𝑇⎛⎜⎜⎜⎜⎜⎜⎝𝑔₁𝑔₂𝑔₁𝑔₂⎞⎟⎟⎟⎟⎟⎟⎠∈ℛ⎛⎜⎜⎜⎜⎜⎜⎜⎝⎛⎜⎜⎜⎜⎜⎜⎝𝐷𝐵^𝑇⊗𝐴_𝑇𝐵⊗𝐶_𝑇𝐷𝑄⊗𝐴𝑃_𝑇𝑄⊗𝐶𝑃⎞⎟⎟⎟⎟⎟⎟⎠^𝑇⎞⎟⎟⎟⎟⎟⎟⎟⎠,(2.22) by Lemma 2.6 we know that 𝑋^∗ is the least norm solution of the systems of linear equations (2.21). Since vector operator is isomorphic and 𝑋^∗ is the unique least norm solution of the systems of matrix equations (2.17), then 𝑋^∗ is the unique least norm solution of Problem 1.

3. The Solution of Problem 2

In this section, we will show that the optimal approximate solution of Problem 2 for a given generalized reflexive matrix can be derived by finding the least norm generalized reflexive solution of a new corresponding matrix equation pair 𝐴𝑋𝐵=𝐸, 𝐶𝑋𝐷=𝐹.

When Problem 1 is consistent, the set of solutions of Problem 1 denoted by 𝑆_𝐸 is not empty. For a given matrix 𝑋₀∈ℛ^{𝑟𝑚×𝑛}(𝑃,𝑄) and 𝑋∈𝑆_𝐸, we have that the matrix equation pair (1.1) is equivalent to the following equation pair: 𝐶𝐴𝑋𝑋𝐷𝐵==𝐸𝐹,,(3.1) where 𝑋=𝑋−𝑋₀,𝐸=𝐸−𝐴𝑋₀𝐵,𝐹=𝐹−𝐶𝑋₀𝐷. Then Problem 2 is equivalent to finding the least norm generalized reflexive solution 𝑋^∗ of the matrix equation pair (3.1).

By using Algorithm 2.1, let initially iterative matrix 𝑋₁=𝐴^𝑇𝐻𝐵^𝑇+𝐶^𝑇𝐻𝐷^𝑇+𝑃𝐴^𝑇𝐻𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐻𝐷^𝑇𝑄, or more especially, letting 𝑋₁=𝟎∈𝑅^{𝑟𝑚×𝑛}(𝑃,𝑄), we can obtain the unique least norm generalized reflexive solution 𝑋^∗ of the matrix equation pair (3.1); then we can obtain the generalized reflexive solution 𝑋 of Problem 2, and 𝑋 can be represented that 𝑋=𝑋^∗+𝑋₀.

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 MATLAB 7.8.

Example 4.1. Consider the generalized reflexive solution of the equation pair (1.1), where 𝐶𝐴==⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−16292⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝35−120385𝐹−𝐸131961−520=306452=2⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−−−−−321−−49424853824214925636242065690142876323942542387324−24−−−227637661−−−77919−−−−3−4213261412438−−−1191−−−−77−−−0755−−13950211103791311119180109698139245119⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠−−−,41111,1241022642140953813649159756364819340932𝐵68𝐷−−=−−−−−−1=−21111251331⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝223170⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝418876−−95778−−4440482503167314612319345−−−−−−−−−5703−−2512312581685461233831740488217342−90681498090186308−296⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠0809⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠2,−−.12225−−98462−134516−1123⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠7456,⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,(4.1) Let 𝑃=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝0001000001−000011000001000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,𝑄=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−0000100010−0000100100−00001⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.2)
We will find the generalized reflexive solution of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 by using Algorithm 2.1. It can be verified that the matrix equation pair is consistent over generalized reflexive matrix and has a solution with respect to 𝑃,𝑄 as follows: 𝑋^∗=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−1551371−193824−−−16186194832−−1173155⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠∈ℛ^𝑟5×5(𝑃,𝑄).(4.3)
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=1,2,…. For any chosen positive number 𝜀, however small enough, for example, 𝜀=1.0000𝑒−010, whenever ‖𝑅_𝑘‖<𝜀, stop the iteration, and 𝑋_𝑘 is regarded to be a generalized reflexive solution of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹. Choose an initially iterative matrix 𝑋₁∈ℛ^𝑟5×5(𝑃,𝑄), such as 𝑋₁=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−115369−11184024−−−6181611148420−−196351⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.4) By Algorithm 2.1, we have 𝑋₁₇=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..1.00.00.000000000‖‖00000𝑅00−193824...00.0.00000000000000000−−−16186..00...000000000000000000148392....00.000000000000000000−−1173155.0....00000000000000000₁₇00⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,‖‖=3.2286𝑒−011<𝜀.(4.5) So we obtain a generalized reflexive solution of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 as follows: 𝑋⁻=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..100..00.0000000000000000−193824...00.0.00000000000000000−−−16618..00...000000000000000000194832....000.00000000000000000−−1171355..0...0000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.6) The relative error of the solution and the residual are shown in Figure 1, where the relative error re_𝑘=‖𝑋_𝑘−𝑋^∗‖/‖𝑋^∗‖ and the residual 𝑟_𝑘=‖𝑅_𝑘‖.

Figure 1 

The relative error of the solution and the residual for Example 4.1 with 𝑋1≠𝟎.

Letting 𝑋₁=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝0000000000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,(4.7) by Algorithm 2.1, we have 𝑋₁₇=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..1.00.00.000000000‖‖00000𝑅00−193824...00.0.00000000000000000−−−16186..00...000000000000000000148392....00.000000000000000000−−1173155.0....00000000000000000₁₇00⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,‖‖=3.1999𝑒−011<𝜀.(4.8) So we obtain a generalized reflexive solution of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 as follows: 𝑋⁻=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..100..00.0000000000000000−193824...00.0.00000000000000000−−−16618..00...000000000000000000194832....000.00000000000000000−−1171355..0...0000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.9) The relative error of the solution and the residual are shown in Figure 2.

Figure 2 

The relative error of the solution and the residual for Example 4.1 with 𝑋1=𝟎.

Example 4.2. Consider the least norm generalized reflexive solution of the matrix equation pair in Example 4.1. Let 𝐻𝑋=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−101201−−000111−100122−010011−201003⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,𝐻=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−−012011−101011−−001212−−010112−02₁3003⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,=𝐴^𝑇𝐻𝐵^𝑇+𝐶^𝑇𝐻𝐷^𝑇+𝑃𝐴^𝑇𝐻𝐵^𝑇𝑄+𝑃𝐶^𝑇𝐻𝐷^𝑇𝑄.(4.10) By using Algorithm 2.1, we have 𝑋₁₉=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..1.00.00.000000000‖‖00000𝑅00−193824...00.0.00000000000000000−−−16186..00...000000000000000000148392....00.000000000000000000−−1173155.0....00000000000000000₁₉00⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,‖‖=6.3115𝑒−011<𝜀.(4.11) So we obtain the least norm generalized reflexive solution of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 as follows: 𝑋^∗=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..100..0.00000000000000000−189324...000..00000000000000000−−−16618..00...000000000000000000194832....000.00000000000000000−−1171355..0...0000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.12) The relative error of the solution and the residual are shown in Figure 3.

Figure 3 

The relative error of the solution and the residual for Example 4.2.

Example 4.3. Let 𝑆_𝐸 denote the set of all generalized reflexive solutions of the matrix equation pair in Example 4.1. For a given matrix, 𝑋₀=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−−1010310−−63197−−01111−913671101300⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠∈ℛ^𝑟5×5(𝑃,𝑄),(4.13) we will find 𝑋∈𝑆_𝐸, such that ‖‖𝑋−𝑋₀‖‖=min_{𝑋∈𝑆𝐸}‖‖𝑋−𝑋₀‖‖.(4.14) That is, find the optimal approximate solution to the matrix 𝑋₀ in 𝑆_𝐸.

Letting 𝑋=𝑋−𝑋₀,𝐸=𝐸−𝐴𝑋₀𝐵,𝐹=𝐹−𝐶𝑋₀𝐷, by the method mentioned in Section 3, we can obtain the least norm generalized reflexive solution 𝑋^∗ of the matrix equation pair 𝐴𝑋𝐵=𝐸,𝐶𝑋𝐷=𝐹 by choosing the initial iteration matrix 𝑋₁=𝟎, and 𝑋^∗ is that 𝑋^∗17=‖‖⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝𝑅−83631....10000.0000000000000000−1153150..0.0..000000000000000000−−−72782..00...000000000000000000−−1131550...0..0000000000000000000−−−131683.0....0000000000000000₁₇000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠,𝑋=𝑋_∗17‖‖=3.0690𝑒−011<𝜀=1.0000𝑒−010,+𝑋₀=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝−−155137..1.00.00.0000000000000000−193824...00.0.00000000000000000−−−16861..00...000000000000000000148392....00.000000000000000000−−1173155.0....0000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠.(4.15) The relative error of the solution and the residual are shown in Figure 4, where the relative error re_𝑘=‖𝑋_𝑘+𝑋₀−𝑋^∗‖/‖𝑋^∗‖ and the residual 𝑟_𝑘=‖𝑅_𝑘‖.

Figure 4 

The relative error of the solution and the residual for Example 4.3.

Acknowledgments

The authors are very much indebted to the anonymous referees and our editors for their constructive and valuable comments and suggestions which greatly improved the original manuscript of this paper. This work was partially supported by the Research Fund Project (Natural Science 2010XJKYL018), Opening Fund of Geomathematics Key Laboratory of Sichuan Province (scsxdz2011005), Natural Science Foundation of Sichuan Education Department (12ZB289) and Key Natural Science Foundation of Sichuan Education Department (12ZA008).