We consider the matrix equation
where is an identity matrix, are arbitrary matrices. , are used to denote the conjugate transpose of the matrix A and B, separately.
Equation (1.1) often arises in dynamic programming, control theory, stochastic filtering, statistics, and so on, see  . Equation (1.1) has been investigated in some special cases  -  . Reurings  showed that the matrix equation and always has a unique positive definite solution. Recently, Gao  provides a new proof for the uniqueness of the positive definite solution of the matrix equation via the fixed point theorem. The method is shown much easier than the way of  . Based on this, we study the combination of these two kinds of equations (i.e. Equation (1.1)).
In this paper, we prove that the Equation (1.1) always has the unique positive definite solution by using the fixed point theorem for mixed monotone operator in the normal cone . Furthermore, two different iterative methods are given, including the basic fixed point iteration method and multi-step stationary iterative method. In addition, numerical examples show that the iterative methods are feasible and effective.
In this paper, we use to denote Hermitian matrices, to denote positive definite matrices, and to denote positive semidefinite matrices. For , we write if is positive semidefinite(definite).
2. Uniqueness of Positive Definite Solution
In this section, we will prove that the equation always has the unique positive definite solution. In the process, we need to use the fixed point theorem for mixed monotone operator in a normal cone. So we first introduce the relevant theory through several definitions and lemmas. In this paper, we always use P to denote a solid cone of a real Banach space E. to denote the interior points set of P. A cone is said to be a solid cone if .
Definition 2.1.  . A cone is said to be normal if there exists a constant such that implies . That is the norm is semimonotone.
Definition 2.2.  . The operator is said to be a mixed monotone operator if is increasing in and decreasing in , that is,
An element is called a fixed point of if .
Lemma 2.1.  . Let P be a normal and solid cone of a real Banach space E and be a mixed monotone operator. Assume that for all , there exists such that
holds for all . Then has exactly one fixed point in . And for any , we have
Lemma 2.2.  . A cone P is normal if and only if , and imply .
Lemma 2.3.  . If and , then .
We define the spectral norm in , then is a real Banach space. We all know that is a cone in and the interior points set is . Since the spectral norm is monotone, then we have that the set is normal cone from Definition 2.1. We define F by , then we have the following theorem.
Theorem 2.1. Eqution (1.1) always has a unique positive definite solution X. And for any , we have
Proof. Consider that the solution of Equation (1.1) is a fixed point of F. Now we will prove that the operator F satisfies the conditions of Lemma 2.1.
For with , by Lemma 2.3, we have
1) , that is ;
which means that the operator F is mixed monotone;
3) Let , then we have . For , we get
According to Lemma 2.1, we get that the operator has a unique fixed point in , which is the unique positive definite solution of Equation (1.1).
Furthermore, for any , we have
3. Iterative Methods
In this section, we propose two different iterative methods for solving Equation (1.1), including the basic fixed point iterative method and multi-step stationary iterative method.
According to Theorem 2.1, sequences converge to the unique positive definite solution of Equation (1.1). Let , we have . Consider the above iteration, which is the basic fixed point iterative method. The following are the algorithm.
We then consider a kind of multi-step stationary iterative method.
Theorem 3.1. For arbitrary initial matrix , the matrix sequence
converges to the unique positive definite solution X of Equation (1.1).
Proof. Since X is the unique positive definite solution of Equation (1.1). For X, and , there exist a positive number satisfy:
For any positive integer k, there exists a unique nonnegative integer K such that or . We will use mathematical induction to prove the following inequality
where . When , that is , the inequality (3.4) holds following (3.3). When , that is , Assume (3.4) is true. That is
Now we need to prove (3.4) is true for , at this time . From (3.5) and , we get that
From , it follows that , . Then
From , (3.6) implies
From (3.4), we have
Now we have proved that (3.4) is true for , at this time .
Hence we get that the inequality (3.4) holds for any positive integer k. Let , then , we have
4. Numerical Examples
We now use numerical examples to illustrate our results. All computations were performed using MATLAB, version 7.01. we denote , and use the stopping criterion .
Example 4.1. Consider Equation (1.1) with and
We use the basic fixed point iterative method (3.1) to solve Equation (1.1). Let , after 14 iterations we get the following result:
Considering the multi-step stationary iterative method (3.2) with , after 19 iterations we get the following result:
Example 4.2. Consider Eq.(1.1) with and
Considering the basic fixed point iterative method (3.1) with , after 13 iterations we get the following result:
Considering the multi-step stationary iterative method (3.2) with , after 35 iterations we get the following result:
Sincere thanks to the members of ALAMT for their professional performance. The work was supported by the Natural Science Foundation of Shandong Province of China (ZR2015AL017, ZR2014AM032), the Science Reseach Foundation of Heze University (XYKJ07).
 Duan, X.F. and Liao, A.P. (2009) On Hermitian Positive Definite Solution of the Matrix Equation . Journal of Computational and Applied Mathematics, 229, 27-36.