Abstract and Applied Analysis

Volume 2014 (2014), Article ID 705830, 9 pages

http://dx.doi.org/10.1155/2014/705830

Research Article

The Iteration Solution of Matrix Equation Subject to a Linear Matrix Inequality Constraint

Na Huang and Changfeng Ma

School of Mathematics and Computer Science, Fujian Normal University, Fuzhou 350007, China

Received 1 May 2014; Accepted 22 July 2014; Published 27 August 2014

Academic Editor: Ivan Ivanov

Copyright © 2014 Na Huang and Changfeng Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We propose a feasible and effective iteration method to find solutions to the matrix equation subject to a matrix inequality constraint , where means that the matrix is nonnegative. And the global convergence results are obtained. Some numerical results are reported to illustrate the applicability of the method.

1. Introduction

In this paper, we consider the following problem:

where , , , , , and are known constant matrixes and is unknown matrix.

The solutions to the linear matrix equation with special structures have been widely studied, for example, symmetric solutions (see [1–5]), -symmetric solutions (see [6]), -symmetric solutions (see [7, 8]), bisymmetric solutions (see [9–12]), centrosymmetric solutions (see [13]), and other general solutions (see [14–23]). Some iterative methods to solve a pair of linear matrix equations have been studied as well (see [24–31]).

However, very little research has been done on the solutions to a matrix equation subjected to a matrix inequality constraint. In 2012, Peng et al. (see [32]) proposed a feasible and effective algorithm to find solutions to the matrix equation subjected to a matrix inequality constraint based on the polar decomposition in Hilbert space. Next year, Li et al. (see [33]) used a similar way to study the bisymmetric solutions to the same problem. Motivated and inspired by the work mentioned above, in this paper, we consider the solutions of the matrix equation over linear inequality constraint. We use the theory on the analytical solution of matrix equation to transform the problem into a matrix inequality smallest nonnegative deviation problem. And then, combined with the polar decomposition theory, an iterative method for solving this transformed problem is proposed. Meanwhile, the global convergence results are obtained. Some numerical results are reported and indicate that the proposed method is quite effective.

Throughout this paper, we use the following notation: for , we write , , and to denote the transpose, the Moore-Penrose generalized inverse, and the Frobenius norm of the matrix , respectively. For any , , we write if . denotes the Kronecker product defined as . For the matrix , denotes the operator defined as . For , is a matrix with th entry equal to . Obviously, . The inner product in space is defined as

Hence is a inner product space, and the norm of a matrix generated by this inner product space is the Frobenius norm.

This paper is organized as follows. In Section 2, we transform problem (1) into a matrix inequality smallest nonnegative deviation problem. Then we study the existence of the solutions for problem (1) in Section 3. The iterative method to the transformed problem and convergence analysis are presented in Section 4. Section 5 shows some numerical experiments. Finally, we conclude this paper in Section 6.

2. Transforming the Original Problem

In this section, we use the theory on the analytical solution of matrix equation to transform problem (1) into a matrix inequality smallest nonnegative deviation problem. Firstly, we present the following lemma about the analytical solution of matrix equation .

Lemma 1 (see Theorem 1.21 in [34]). Given , , and , the matrix equation is solvable for in if and only if . Moreover, if the matrix equation is solvable, then the general solutions can be expressed as

where is an arbitrary matrix.

Assume that ; that is, assume that the matrix equation is solvable. Substituting (3) into the second inequality of (1), we get

By simple calculation, we have

Hence (3) is a solution of (1) if and only if the matrix in (1) satisfies (5). However, inequality (5) may be unsolvable; namely,

where . In this case, we can find such that

is solvable (if (5) is solvable, then ). Obviously, there exist many satisfied (7). Here we need to find a such that , where , satisfied (7). Thus we consider the following smallest nonnegative deviation of the matrix inequality, which is also a quadratic programming problem:

If and solve (8) with , then solves (7), and (3) is a solution of (1). If and solve (8) , then solves (7) in the smallest nonnegative deviation sense, and (3) is a solution of the matrix equation over the nonnegative smallest deviation constraint of the inequality . Conversely, if solves (7), then and solve (8). So, to find satisfied (1), we only need to solve the smallest nonnegative deviation problem (8).

Suppose that and solve (8); then

On the other hand, if a pair of matrices and solves the smallest nonnegative deviation problem (8), then there exists a nonnegative matrix satisfying

Consequently, the smallest nonnegative deviation problem (8) is equivalent to the following optimization problem:

Eliminating from (11) yields the following optimization problem:

Suppose that a pair solves (12); then

which together with the matrices and solves (11), and hence and solve (8). This allows one to determine whether or not (3) is a solution of (1). Therefore, to solve (1), we first solve optimization problem (12). Our iteration method proposed below will take advantage of these equivalent forms of (1).

3. The Solution of the Problem

To illustrate the existence of the solutions , , and of (8), (11), and (12), we give the following theorem in the first place.

Theorem 2 (see [35, 36]). Let be a closed convex cone (i.e., is closed convex set and , for all and ). Let be the polar cone of ; that is,

Then for all , has unique polar decomposition of the form

Theorem 2 implies that is the projection of onto and is the projection of onto .

For problem (12), we give the following two matrix sets:

Now we will prove that is a closed convex cone and .

Lemma 3. The matrix set is a closed convex cone in the Hilbert space .

Proof. For all , there exists and such that . By the definition of Kronecker product, we have

where the second equality follows from the definition of Moore-Penrose generalized inverse, and

It is easy to see that . As and are arbitrary, is arbitrary as well. Let and , where ; then . Thus is equivalent to the following set:

By the result in [37], we know that the set is a closed convex cone in the Hilbert space . Hence is a closed convex cone in the Hilbert space .

Lemma 4. The matrix set is the polar cone of the matrix set .

Proof. By the definition of the polar cone, we get

So we just need to prove . Firstly, we prove .

For all and , we have

Thus . Then .

Now we prove . For all , if , then . So there exists a positive real number and such that

Let , where :

This contradicts the assumption .

Theorem 5. Assume that the matrices and solve (12). Define matrices and as

Then , , , and ; namely, and are the polar decomposition of .

Proof. As and solve (12), we have . Then

Thus .

By Lemmas 3 and 4 and Theorem 2, we get that has unique polar decomposition with and ; that is, there exists unique and such that and .

Consider optimization problem (12). The objective function in (12) is

Since , solve (12), , and , we have . Thus . Then . This completes the proof.

Remark 6. Theorem 5 implies that if a pair of matrices and solves optimization problem (12), then and are the projections of onto and , respectively. Conversely, by Theorem 2 we get that has unique polar decomposition of the form , , and . By the definition of , there exists and such that and . Moreover, , , and solve optimization problem (11). Thus problem (11) is solvable.

By the above analysis, we get the following theorem immediately.

Theorem 7. Problem (1) is solvable if and only if and , where and are the solutions of optimization problem (12).

4. Iterative Method and Convergence Analysis

In this section, we present an iteration method to solve (1) and give the convergence analysis. We are now in a position to give our algorithm to compute the solutions , , and of (8), (11), and (12).

Algorithm 8 (an iteration method for (1)).

Step 0. Input matrices , , , , , and . Choose the initial matrix . Compute , , and . Take the stopping criterion . Set .

Step 1. Find a solution of the least squares problem

Step 2. Update the sequences

Step 3. If or , then stop; otherwise, set and go to Step 1.

Next we give the following lemma.

Lemma 9. is the direct sum of and ; that is,

where

Proof. Obviously, and are linear subspaces of . By orthogonal decomposition theorem in the space, we obtain , where is the orthogonal complement space of . So we just need to prove .

We prove firstly. For all and , we have

where the third equality follows from the definition of Moore-Penrose generalized inverse. Hence ; namely, .

Then we prove . For all and , we have . By the same way, we get . As is arbitrary, we take . Then

So ; that is, . Thus .

From the above, we get . Therefore, .

Now we present the convergence theorem.

Theorem 10. Let be the unique polar decomposition of . Then

Proof. Since matrix solves (27), we have . This together with Algorithm 8 yields

where the first inequality follows from and the second inequality follows from nonexpansive property of the projection. This implies that the sequence is monotonically decreasing and is bounded from below. So there exists a constant such that . Furthermore, is bounded. Then has at least one cluster point. Next we show that any cluster point of the sequence is equal to . Consequently, converges toward to .

Let be any cluster point of the sequence . Without loss of generality, we suppose . Obviously, . It follows from Lemma 9 that has unique orthogonal decomposition of the form

Moreover, by (27), satisfies . Thus :

Let ; we get . This together with yields that

So . Therefore

Since , as well. This together with follows that .

Since

we get . By Lemma 3 and , we gain . By the definition of , there exists and such that . Let . Then is the unique polar decomposition of . Hence and . Furthermore,

This completes the proof of the theorem.

5. Numerical Experiments

In this section, we present two numerical examples to illustrate the efficiency and the performance of Algorithm 8. Firstly, we consider least squares problem (27) in Algorithm 8.

By the definition of Kronecker product, least squares problem (27) in Algorithm 8 is equivalent to

where and . It is well known that the normal equation of problem (41) is

Notice that

and therefore (42) is equal to

Taking the definition of and into the above equation, we get

which is the normal equation of problem (27). Let

and . Then problem (27) is equivalent to , which can be solved by the modified conjugate gradient method (see [33]).

Algorithm 11 (modified conjugate gradient method).

Step 0. Input matrices , , , , , , and . Choose the initial matrix . Compute , , , and . Take the stopping criterion . Set .

Step 1. If , then stop; otherwise, set and go to Step 2.

Step 2. Update the sequences

Step 3. Return to Step 1.

In our experiment, all computations were done using the PC with Pentium Dual-Core CPU E5800 @2.40 GHz. All the programming is implemented in MATLAB R2011b. The initial matrix in Algorithm 8 is taken as the null matrix and the termination criterion is or .

Example 12. Matrices , , , , , and are given as follows:

Computing by Algorithm 8, we have ,

and a solution to inequality (5) as follows:

It follows from and that problem (1) is solvable. By substituting into (3), we obtain a solution to problem (1) as follows:

Furthermore, denote

We have the following iterative error curve in Figure 1.

Figure 1: and for Example 12.

Example 13. Given matrices , , and being the same as in Example 12, and are identity matrices, and is given as follows:

Following Algorithm 8, we get ,

and a solution to inequality (5) as follows:

It follows from and that problem (1) is solvable. By substituting into (3), we obtain a solution to problem (1) as follows:

Furthermore, we have the following iterative error curve in Figure 2.

Figure 2: and for Example 13.

6. Conclusion

In this paper, we propose Algorithm 8 to find solutions to the matrix equation subject to a matrix inequality constraint . And the global convergence results are obtained. For least squares problem (27) in Algorithm 8, we use the modified conjugate gradient method to solve it. Numerical results also confirm the good theoretical properties of our approach.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Authors’ Contribution

All authors contributed equally and significantly to the writing of this paper. All authors read and approved the final paper.

Acknowledgments

The project is supported by the National Natural Science Foundation of China (Grant nos. 11071041, 11201074), Fujian Natural Science Foundation (Grant no. 2013J01006), the University Special Fund Project of Fujian (Grant no. JK2013060), and R&D of Key Instruments and Technologies for Deep Resources Prospecting (the National R&D Projects for Key Scientific Instruments) under Grant no. ZDYZ2012-1-02-04.

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