The iterative methods for solving nonlinear matrix equation $X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B=Q$

Abstract

In this paper, we study the matrix equation $X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B=Q$, where A and B are square matrices, and Q is a positive definite matrix, and propose the iterative methods for finding positive definite solutions of the matrix equation. Also, general convergence results for the basic fixed point iteration for these equations are given. Some numerical examples are presented to show the usefulness of the iterations.

MSC:65F10, 65F30, 65H10, 15A24.

1 Introduction

In this paper, we consider the matrix equation

$$X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B=Q,$$

(1)

where A and B are square matrices, Q is a positive definite matrix. It is easy to see that matrix equation (1) can be reduced to

$$X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B=I,$$

(2)

where I is the identity matrix. Trying to solve special linear systems [1] leads to solving nonlinear matrix equations of the above types as follows.

For a linear system $Mx=f$ with

$$M=\left(\begin{array}{ccc}Q& 0& A\\ 0& Q& B\\ A^{\star }& B^{\star }& Q\end{array}\right)$$

positive definite, we rewrite $M=\tilde{M}+K$, where

$$\tilde{M}=\left(\begin{array}{ccc}X& 0& A\\ 0& X& B\\ A^{\star }& B^{\star }& Q\end{array}\right),K=\left(\begin{array}{ccc}Q-X& 0& 0\\ 0& Q-X& 0\\ 0& 0& 0\end{array}\right).$$

Moreover, we decompose $\tilde{M}$ to the LU decomposition

$$\tilde{M}=\left(\begin{array}{ccc}X& 0& A\\ 0& X& B\\ A^{\star }& B^{\star }& Q\end{array}\right)=\left(\begin{array}{ccc}I& 0& 0\\ 0& I& 0\\ A^{\star }{X}^{-1}& B^{\star }{X}^{-1}& I\end{array}\right)\left(\begin{array}{ccc}X& 0& A\\ 0& X& B\\ 0& 0& X\end{array}\right).$$

A decomposition of $\tilde{M}$ exists if and only if X is a positive definite solution of matrix equation (1). Solving the linear system $\tilde{M}y=f$ is equivalent to solving two linear systems with a lower and upper block triangular system matrix. To compute the solution of $Mx=f$ from y, the Woodbury formula [2] can be applied.

The matrix equation $X+A^{\star }{X}^{-1}A=Q$ has been studied extensively by many authors [3–14]. Several conditions for the existence of positive definite solutions and some iterations to find maximal positive definite solutions for these equations were discussed. Apparently, matrix equation (1) generalizes the matrix equation $X+A^{\star }{X}^{-1}A=Q$.

Matrix equation (2) was studied in [15], and based on some conditions, the authors proved that matrix equation (2) has positive definite solutions. They also proposed two iterative methods to find the Hermitian positive definite solutions of matrix equation (2). They did not analyze the convergence rate of proposed algorithms.

In this paper, we propose two algorithms. We will show that Algorithm (7) is more accurate than Algorithm (3) pointed out in [15]. Also, Algorithm (10) needs less operation in comparison with Algorithm (3). The following notations are used throughout the rest of the paper. The notation $A\ge 0$ ($A>0$) means that A is Hermitian positive semidefinite (positive definite). For Hermitian matrices A and B, we write $A\ge B$ ($A>B$) if $A-B\ge 0$ (>0). Similarly, by ${\lambda }_1(A)$ and ${\lambda }_n$ we denote, respectively, the maximal and the minimal eigenvalues of A. The norm used in this paper is the spectral norm of the matrix A, i.e., $\parallel A\parallel ={({\lambda }_1(A^{\star }A))}^{\frac{1}{2}}$.

2 Fixed point theorems

Lemma 1 [8]

If C and P are Hermitian matrices of the same order with $P>0$, then $CPC+P^{-1}\ge 2C$.

In [15] an algorithm that avoids matrix inversion for every iteration, called inversion-free variant of the basic fixed point iteration, and a theorem to find a Hermitian positive definite solution of matrix equation (2) were proposed as follows.

Algorithm 1 [15]

Let

$$\{\begin{array}{c}{X}_0=Y_0=I,\hfill \\ X_{n+1}=I-A^{\star }{Y}_{n}A-B^{\star }{Y}_{n}B,\hfill \\ Y_{n+1}=2Y_n-Y_n{X}_n{Y}_n,n=0,1,2,\dots .\hfill \end{array}$$

(3)

Theorem 1 [15]

Assume that matrix equation (2) has a positive definite solution. Then Algorithm (3) defines a monotonically decreasing matrix sequence $\{X_n\}$ converging to the positive definite matrix X which is a solution of matrix equation (2). Also, the sequence $\{Y_n\}$ defined in Algorithm (3) defines a monotonically increasing sequence converging to $X^{-1}$.

Although it is not mentioned in the previous theorem that the sequence $\{X_n\}$ converges to the maximal Hermitian positive definite solution of equation (2), during the proof of the theorem in [15], it is obvious. So, $X=X_{\mathrm{\infty }}$, where $X_{\mathrm{\infty }}$ is the maximal positive definite solution of matrix equation (2) in Theorem 1.

The problem of convergence rate for Algorithm (3) was not considered in [15]. We now establish the following result.

Theorem 2 If matrix equation (2) has a positive definite solution, for Algorithm (1) and any $ϵ>0$, we have

$$\parallel Y_{n+1}-X_{\mathrm{\infty }}^{-1}\parallel \le {(\parallel A{X}_{\mathrm{\infty }}^{-1}\parallel +\parallel B{X}_{\mathrm{\infty }}^{-1}\parallel +ϵ)}^2\parallel Y_{n-1}-X_{\mathrm{\infty }}^{-1}\parallel$$

(4)

and

$$\parallel X_{n+1}-X_{\mathrm{\infty }}\parallel \le ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel$$

(5)

for all n large enough.

Proof From Algorithm (3), we have

$$\begin{array}{rcl}{Y}_{n+1}& =& 2Y_n-Y_n(I-A^{\star }{Y}_{n-1}A-B^{\star }{Y}_{n-1}B)Y_n\\ =& 2Y_n-Y_n^2+Y_n{A}^{\star }(X_{\mathrm{\infty }}^{-1}+Y_{n-1}-X_{\mathrm{\infty }}^{-1})A{Y}_n+Y_n{B}^{\star }(X_{\mathrm{\infty }}^{-1}+Y_{n-1}-X_{\mathrm{\infty }}^{-1})B{Y}_n\\ =& 2Y_n-Y_n{X}_{\mathrm{\infty }}{Y}_n+Y_n{A}^{\star }(Y_{n-1}-X_{\mathrm{\infty }}^{-1})A{Y}_n+Y_n{B}^{\star }(Y_{n-1}-X_{\mathrm{\infty }}^{-1})B{Y}_n.\end{array}$$

Thus

$$\begin{array}{rcl}{X}_{\mathrm{\infty }}^{-1}-Y_{n+1}& =& X_{\mathrm{\infty }}^{-1}-Y_n+Y_n{X}_{\mathrm{\infty }}{Y}_n-Y_n+Y_n{A}^{\star }(X_{\mathrm{\infty }}^{-1}-Y_{n-1})A{Y}_n+Y_n{B}^{\star }(X_{\mathrm{\infty }}^{-1}-Y_{n-1})B{Y}_n\\ =& (X_{\mathrm{\infty }}^{-1}-Y_n)X_{+}(X_{\mathrm{\infty }}^{-1}-Y_n)+Y_n{A}^{\star }(X_{\mathrm{\infty }}^{-1}-Y_{n-1})A{Y}_n+Y_n{B}^{\star }(X_{\mathrm{\infty }}^{-1}-Y_{n-1})B{Y}_n.\end{array}$$

Now, since $\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel \le \parallel Y_{n-1}-X_{\mathrm{\infty }}^{-1}\parallel$ and $lim{Y}_n=X_{\mathrm{\infty }}^{-1}$, inequality (4) follows. Also, inequality (5) is true since

$$X_{n+1}-X_{\mathrm{\infty }}=A^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)A+B^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)B.$$

(6)

This completes the proof. □

The above proof shows that Algorithm (3) should be modified as follows to improve the preceding convergence properties.

Algorithm 2 Let

$$\{\begin{array}{c}{X}_0=Y_0=I,\hfill \\ Y_{n+1}=2Y_n-Y_n{X}_n{Y}_n,\hfill \\ X_{n+1}=I-A^{\star }{Y}_{n+1}A-B^{\star }{Y}_{n+1}B,n=0,1,2,\dots .\hfill \end{array}$$

(7)

Theorem 3 Assume that matrix equation (2) has a positive definite solution. Then Algorithm (7) defines a monotonically decreasing matrix sequence $\{X_n\}$ converging to $X_{\mathrm{\infty }}$ which is the maximal Hermitian positive definite solution of equation (2). Also, the sequence $\{Y_n\}$ defined in Algorithm (7) defines a monotonically increasing sequence converging to $X_{\mathrm{\infty }}^{-1}$.

Proof Let $X_{+}$ be a positive definite solution of matrix equation (2). It is clear that

$$X_0\ge X_1\ge \cdots \ge X_n\ge X_{+},Y_0\le Y_1\le \cdots \le Y_n\le X_{+}^{-1}$$

(8)

is true for $n=1$. Assume (8) is true for $n=k$. By Lemma 1, we have that

$$Y_{k+1}=2Y_k-Y_k{X}_k{Y}_k\le X_K^{-1}\le X_{+}^{-1}.$$

Therefore,

$$X_{k+1}=I-A^{\star }{Y}_{k+1}A\ge I-A^{\star }{X}_{+}^{-1}A=X_{+}.$$

Since $Y_k\le X_{k-1}^{-1}\le X_k^{-1}$, we have $Y_k^{-1}\ge X_k$. Thus

$$Y_{k+1}-Y_k=Y_k(Y_k^{-1}-X_k)Y_k\ge 0$$

and

$$X_{k+1}-X_k=-A^{\star }(Y_{k+1}-Y_k)A\le 0.$$

We have now proved (8) for $n=k+1$. Therefore, (8) is true for all n, and ${lim}_{n\to \mathrm{\infty }}{X}_n$ and ${lim}_{n\to \mathrm{\infty }}{Y}_n$ exist. So, we have $lim{X}_n=X_{\mathrm{\infty }}$ and $lim{Y}_n=X_{\mathrm{\infty }}^{-1}$. □

Similar to Theorem 2, we can state the following theorem.

Theorem 4 If matrix equation (2) has a positive definite solution for Algorithm (7) and any $ϵ>0$, then we have

$$\parallel Y_{n+1}-X_{\mathrm{\infty }}^{-1}\parallel \le {(\parallel A{X}_{\mathrm{\infty }}^{-1}\parallel +\parallel B{X}_{\mathrm{\infty }}^{-1}\parallel +ϵ)}^2\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel$$

and

$$\parallel X_n-X_{\mathrm{\infty }}\parallel \le ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel$$

(9)

for all n large enough.

Now, we can see that Algorithm (7) can be faster than Algorithm (3) from the estimates in Theorems 2 and 4.

Algorithm 3 Take

$$\{\begin{array}{c}{X}_0=I,Y_0=I,\hfill \\ Y_{n+1}=(I-X_n)Y_n+I,\hfill \\ X_{n+1}=I-A^{\star }{Y}_{n+1}A-B^{\star }{Y}_{n+1}B,n=0,1,2,\dots .\hfill \end{array}$$

(10)

Algorithm (10) requires only five matrix multiplications per step, whereas Algorithm (3) requires six matrix multiplications per step.

Theorem 5 If matrix equation (2) has a positive definite solution and the two sequences $\{X_n\}$ and $\{Y_n\}$ are determined by Algorithm (10), then $\{X_n\}$ is monotone decreasing and converges to the maximal Hermitian positive definite solution $X_{\mathrm{\infty }}$. Also, the sequence $\{Y_n\}$ defined in Algorithm (10) is a monotonically increasing sequence converging to $X_{\mathrm{\infty }}^{-1}$.

Proof Let $X_{+}$ be a positive definite solution of equation (2). We prove that

$$X_0\ge X_1\ge \cdots \ge X_n\ge X_{+}$$

(11)

and

$$Y_0\le Y_1\le \cdots \le Y_n\le X_{+}^{-1}.$$

(12)

Since $X_{+}$ is a solution of matrix equation (1), $X_0=I\ge X_{+}$. Also, we have $Y_0=Y_1=I$, and so

$$X_1=I-A^{\star }A-B^{\star }B\ge I-A^{\star }{X}_{+}^{-1}A-B^{\star }{X}_{+}^{-1}B=X_{+},$$

i.e., $X_0\ge X_1\ge X_{+}$.

For the sequence $\{Y_n\}$, since $I\le X_{+}^{-1}$, $Y_0=Y_1=I\le X_{+}^{-1}$.

Thus inequalities (11) and (12) are true for $n=1$. Now, assume that inequalities (11) and (12) are true for $n=k$, i.e.,

$$X_0\ge X_1\ge \cdots \ge X_k\ge X_{+}$$

and

$$Y_0\le Y_1\le \cdots \le Y_k\le X_{+}^{-1}.$$

We show that inequalities (11) and (12) are true for $n=k+1$. We have

$$Y_{k+1}=(I-X_k)Y_k+I\ge (I-X_{k-1})Y_{k-1}+I=Y_k$$

and

$$Y_{k+1}=(I-X_k)Y_k+I\le (I-X_{+})X_{+}^{-1}+I=X_{+}^{-1},$$

i.e., $Y_k\le Y_{k+1}\le X_{+}^{-1}$. Then

$$X_k-X_{k+1}=A^{\star }(Y_{k+1}-Y_k)A+B^{\star }(Y_{k+1}-Y_k)B,$$

and $X_{k+1}\le X_k$, since $Y_k\le Y_{k+1}$. On the other hand,

$$\begin{array}{rcl}{X}_{k+1}& =& I-A^{\star }{Y}_{k+1}A-B^{\star }{Y}_{k+1}B\\ \ge & I-A^{\star }{X}_{+}^{-1}A-B^{\star }{X}_{+}^{-1}B\\ =& X_{+},\end{array}$$

(13)

i.e., $X_k\ge X_{k+1}\ge X_{+}$.

Then the above inequalities are true for all n, also ${lim}_{n\to \mathrm{\infty }}{X}_n$ and ${lim}_{n\to \mathrm{\infty }}{Y}_n$ exist. By taking limit on Algorithm (10), we have ${lim}_{n\to \mathrm{\infty }}{X}_n=X_{\mathrm{\infty }}$ and ${lim}_{n\to \mathrm{\infty }}{Y}_n=X_{\mathrm{\infty }}^{-1}$, where $X_{\mathrm{\infty }}$ is the maximal positive definite solution of matrix equation (2). □

By Algorithm (10), we have $I-X_n{Y}_n=Y_{n+1}-Y_n$. Then, for small $ϵ>0$, $\parallel I-X_n{Y}_n\parallel$ can be one stopping condition.

Theorem 6 If matrix equation (2) has a positive definite solution and after n iterative steps of Algorithm (10), the inequality $\parallel I-X_n{Y}_n\parallel <ϵ$ implies

$$\parallel X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I\parallel \le ϵ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel X_{\mathrm{\infty }}^{-1}\parallel .$$

Proof Since

$$\begin{array}{l}{X}_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I=X_n-X_{n+1}+A^{\star }(X_n^{-1}-Y_{n+1})A+B^{\star }(X_n^{-1}-Y_{n+1})B\\ \phantom{X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I}=A^{\star }(Y_{n+1}-X_n^{-1}+X_n^{-1}-Y_n)A\\ \phantom{X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I=}+B^{\star }(Y_{n+1}-X_n^{-1}+X_n^{-1}-Y_n)B\\ \phantom{X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I=}+A^{\star }(X_n^{-1}-Y_{n+1})A+B^{\star }(X_n^{-1}-Y_{n+1})B\\ \phantom{X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I}=A^{\star }{X}_n^{-1}(I-X_n{Y}_n)A+B^{\star }{X}_n^{-1}(I-X_n{Y}_n)B,\\ \parallel X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I\parallel \le ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel X_{\mathrm{\infty }}^{-1}\parallel \parallel I-X_n{Y}_n\parallel \\ \phantom{\parallel X_n+A^{\star }{X}_n^{-1}A+B^{\star }{X}_n^{-1}B-I\parallel }\le ϵ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel X_{\mathrm{\infty }}^{-1}\parallel .\end{array}$$

(14)

This completes the proof. □

Theorem 7 If $X_n>0$ for every n, then matrix equation (2) has a Hermitian positive definite solution.

Proof Since $X_n>0$ for every n, the proof of the monotonicity of $\{Y_n\}$ and $\{X_n\}$ noted in Theorem 5 remains valid. Therefore, the sequence $\{X_n\}$ is monotone decreasing and bounded from below by the zero matrix. Then ${lim}_{n\to \mathrm{\infty }}{X}_n=X$ exists. We claim that the sequence $\{Y_n\}$ is bounded above. Suppose that it does not hold. Then, for every $m>0$, there exists $n_m$ such that $mI<Y_{n_m}$. Since each $X_n$ is positive definite for every n, we have

$$A^{\star }{Y}_{n}A+B^{\star }{Y}_{n}B<I\text{for every }n.$$

Furthermore, since A or B are nonsingular for every $m>0$, we have

$$m(A^{\star }A+B^{\star }B)<A^{\star }{Y}_{n_m}A+B^{\star }{Y}_{n_m}B<I.$$

By [[16], Lemma 1.2],

$$m(A^{\star }A+B^{\star }B)<I\text{for every }m,$$

which is a contradiction. Then the sequence $\{Y_n\}$ is bounded above and convergent. Suppose that ${lim}_{n\to \mathrm{\infty }}{Y}_n=Y$. As $Y_0=I$ and $\{Y_n\}$ is monotone increasing, $Y\le I$. Taking limit in Algorithm (10) implies that

$$\begin{array}{c}Y=(I-X)Y+I,\hfill \\ X=I-A^{\star }YA-B^{\star }YB.\hfill \end{array}$$

Since $Y\le I$, $X=Y^{-1}>0$, and hence $X=I-A^{\star }{X}^{-1}A-B^{\star }{X}^{-1}B$. Then matrix equation (2) has a positive definite solution. □

Theorem 8 If matrix equation (2) has a positive definite solution and $\parallel A\parallel <\frac{1}{2}$ and $\parallel B\parallel <\frac{1}{2}$, then the sequence $\{X_n\}$ defined in Algorithm (10) satisfies

$$\parallel Y_{n+1}-X_{\mathrm{\infty }}^{-1}\parallel \le (\parallel A{X}_{\mathrm{\infty }}^{-1}\parallel +\parallel B{X}_{\mathrm{\infty }}^{-1}\parallel )\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel$$

(15)

and

$$\parallel X_{n+1}-X_{\mathrm{\infty }}\parallel \le ({\parallel A\parallel }^2+{\parallel B\parallel }^2)\parallel Y_n-X_{\mathrm{\infty }}^{-1}\parallel$$

(16)

for all n large enough.

Proof From Algorithm (10), we have

$$\begin{array}{rcl}{Y}_{n+1}& =& (I-X_n)Y_n+I\\ =& A^{\star }{Y}_{n}A{Y}_n+B^{\star }{Y}_{n}B{Y}_n+I\\ =& A^{\star }(Y_n+X_{\mathrm{\infty }}^{-1}-X_{\mathrm{\infty }}^{-1})A{Y}_n+B^{\star }(Y_n+X_{\mathrm{\infty }}^{-1}-X_{\mathrm{\infty }}^{-1})B{Y}_n+I\\ =& A^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})A{Y}_n+B^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})B{Y}_n+A^{\star }{X}_{\mathrm{\infty }}^{-1}A{Y}_n+B^{\star }{X}_{\mathrm{\infty }}^{-1}B{Y}_n+Y_n-Y_n+I\\ =& A^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})A{Y}_n+B^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})B{Y}_n-(I-A^{\star }{X}_{\mathrm{\infty }}^{-1}A-B^{\star }{X}_{\mathrm{\infty }}^{-1}B)Y_n+Y_n+I\\ =& A^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})A{Y}_n+B^{\star }(Y_n-X_{\mathrm{\infty }}^{-1})B{Y}_n-X_{\mathrm{\infty }}{Y}_n+Y_n+I.\end{array}$$

Thus

$$\begin{array}{rcl}{X}_{\mathrm{\infty }}^{-1}-Y_{n+1}& =& X_{\mathrm{\infty }}^{-1}+A^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)A{Y}_n+B^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)B{Y}_n+X_{\mathrm{\infty }}{Y}_n-Y_n-I\\ =& A^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)A{Y}_n+B^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)B{Y}_n+(I-X_{\mathrm{\infty }})(X_{\mathrm{\infty }}^{-1}-Y_n)\\ =& A^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)A{Y}_n+B^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)B{Y}_n\\ +A^{\star }{X}_{\mathrm{\infty }}^{-1}A(X_{\mathrm{\infty }}^{-1}-Y_n)+B^{\star }{X}_{\mathrm{\infty }}^{-1}B(X_{\mathrm{\infty }}^{-1}-Y_n).\end{array}$$

Therefore, we have

$$\begin{array}{rcl}\parallel X_{\mathrm{\infty }}^{-1}-Y_{n+1}\parallel & \le & (\parallel A^{\star }\parallel \parallel A{Y}_n\parallel +\parallel B^{\star }\parallel \parallel B{Y}_n\parallel +\parallel A^{\star }{X}_{\mathrm{\infty }}^{-1}A\parallel +\parallel B^{\star }{X}_{\mathrm{\infty }}^{-1}B\parallel )\parallel X_{\mathrm{\infty }}^{-1}-Y_n\parallel \\ \le & ((\parallel A{Y}_n\parallel +\parallel X_{\mathrm{\infty }}^{-1}A\parallel )\parallel A^{\star }\parallel +(\parallel B{Y}_n\parallel +\parallel X_{\mathrm{\infty }}^{-1}B\parallel )\parallel B^{\star }\parallel )\parallel X_{\mathrm{\infty }}^{-1}-Y_n\parallel .\end{array}$$

Now, since ${lim}_{n\to \mathrm{\infty }}{Y}_n=X_{\mathrm{\infty }}^{-1}$, inequality (15) follows. Also, inequality (16) is true since

$$X_{n+1}-X_{\mathrm{\infty }}=A^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)A+B^{\star }(X_{\mathrm{\infty }}^{-1}-Y_n)B.$$

(17)

This completes the proof. □

3 Numerical examples

In this section, we present some numerical examples to show the effectiveness of the new inversion-free variant of the basic fixed point iteration methods. Hermitian positive definite solutions of matrix equation (2) for different matrices A and B are computed. We will compare the suggested algorithms, Algorithm (7) and Algorithm (10), by Algorithm (3). All programs were written in MATLAB.

Example 1 Consider equation (2) with

$$A=\frac{1}{200}\left(\begin{array}{ccc}0.2& -0.1& 0.3\\ 0.56& 0.3& -0.7\\ 0.2& 0.5& 0.6\end{array}\right),B=\frac{1}{20}\left(\begin{array}{ccc}0.46& -0.01& 0.020\\ -0.15& -0.488& -0.060\\ 0.04& -0.01& -0.120\end{array}\right).$$

Algorithm (3) needs six iterations to get the solution

$$\begin{array}{c}X=\left(\begin{array}{ccc}0.999400612248657& -0.000176704506272& -0.000028208026789\\ -0.000176704506272& 0.999395021004650& -0.000077249011425\\ -0.000028208026789& -0.000077249011425& 0.999930483901903\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=1.3885\mathrm{e}-013.\hfill \end{array}$$

Algorithm (7) needs six iterations to get the solution

$$\begin{array}{c}X=\left(\begin{array}{ccc}0.999400612248567& -0.000176704506276& -0.000028208026792\\ -0.000176704506276& 0.999395021004514& -0.000077249011443\\ -0.000028208026792& -0.000077249011443& 0.999930483901898\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=6.7763\mathrm{e}-021.\hfill \end{array}$$

We can easily see that Algorithm (7) is more accurate than Algorithm (3).

Algorithm (10) needs six iterations to get the solution

$$\begin{array}{c}X=\left(\begin{array}{ccc}0.999400612248567& -0.000176704506276& -0.000028208026792\\ -0.000176704506276& 0.999395021004514& -0.000077249011443\\ -0.000028208026792& -0.000077249011443& 0.999930483901898\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=6.7763\mathrm{e}-021.\hfill \end{array}$$

Example 2 Consider equation (2) with

$$A=\frac{1}{820}\left(\begin{array}{ccccc}41& 15& 23& 35& 66\\ 25& 12& 27& 45& 21\\ 23& 27& 28& 16& 24\\ 15& 45& 16& 52& 65\\ 66& 21& 24& 65& 35\end{array}\right),B=\frac{1}{830}\left(\begin{array}{ccccc}23& 21& 23& 25& 32\\ 21& 45& 60& 42& 33\\ 23& 24& 34& 18& 17\\ 13& 42& 18& 44& 30\\ 32& 33& 26& 30& 26\end{array}\right).$$

Algorithm (3) after 21 iterations gives the solution

$$\begin{array}{c}X=\left(\begin{array}{ccccc}0.98393799066& -0.01161748103& -0.01233926321& -0.01833845539& -0.01633619168\\ -0.01161748103& 0.98497686219& -0.01315828865& -0.01745583944& -0.01639741581\\ -0.01233926321& -0.01315828865& 0.98561286596& -0.01623773649& -0.01467582916\\ -0.01833845539& -0.01745583944& -0.01623773649& 0.97439947749& -0.02237728728\\ -0.01633619168& -0.01639741581& -0.01467582916& -0.02237728728& 0.97634558763\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=3.7975\mathrm{e}-013.\hfill \end{array}$$

Algorithm (7) after 21 iterations gives the solution

$$\begin{array}{c}X=\left(\begin{array}{ccccc}0.98393799066& -0.01161748103& -0.01233926321& -0.01833845539& -0.01633619168\\ -0.01161748103& 0.98497686219& -0.01315828865& -0.01745583944& -0.01639741581\\ -0.01233926321& -0.01315828865& 0.98561286596& -0.01623773649& -0.01467582916\\ -0.01833845539& -0.01745583944& -0.01623773649& 0.97439947749& -0.02237728728\\ -0.01633619168& -0.01639741581& -0.01467582916& -0.02237728728& 0.97634558763\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=6.0963\mathrm{e}-018.\hfill \end{array}$$

Algorithm (10) after 21 iterations gives the solution

$$\begin{array}{c}X=\left(\begin{array}{ccccc}0.98393799066& -0.01161748103& -0.01233926321& -0.01833845539& -0.01633619168\\ -0.01161748103& 0.98497686219& -0.01315828865& -0.01745583944& -0.01639741581\\ -0.01233926321& -0.01315828865& 0.98561286596& -0.01623773649& -0.01467582916\\ -0.01833845539& -0.01745583944& -0.01623773649& 0.97439947749& -0.02237728728\\ -0.01633619168& -0.01639741581& -0.01467582916& -0.02237728728& 0.97634558763\end{array}\right),\hfill \\ {\parallel X+A^{\star }{X}^{-1}A+B^{\star }{X}^{-1}B-I\parallel }_{\mathrm{\infty }}=6.1220\mathrm{e}-018.\hfill \end{array}$$

We can see that Algorithm (10) needs to find a Hermitian positive definite solution.