_{1}

^{*}

Consider the nonlinear matrix equation
*X*-
*A*X*
^{p}
*A*-
*B*X*
^{-q}
*B*=
*I *(0<
*p*,
*q<*

1). By using the fixed point theorem for mixed monotone operator in a normal cone, we prove that the equation with
0<*p*,*q<*

1

always has the unique positive definite solution. Two different iterative methods are given, including the basic fixed point iterative method and the multi-step stationary iterative method. Numerical examples show that the iterative methods are feasible and effective.

We consider the matrix equation

X − A * X p A − B * X − q B = I , 0 < p , q < 1 (1.1)

where I is an n × n identity matrix, A , B are arbitrary m n × n matrices. A * , B * 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 [

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 P ( n ) . 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 H ( n ) to denote n × n Hermitian matrices, P ( n ) to denote n × n positive definite matrices, and P ¯ ( n ) to denote n × n positive semidefinite matrices. For X , Y ∈ P ( n ) , we write X ≥ Y ( X > Y ) if X − Y is positive semidefinite(definite).

In this section, we will prove that the equation X − A * X p A − B * X − q B = I ( 0 < p , q < 1 ) 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. P 0 to denote the interior points set of P. A cone is said to be a solid cone if P 0 ≠ ϕ .

Definition 2.1. [

Definition 2.2. [

Γ ( x 1 , y 1 ) ≥ Γ ( x 2 , y 2 ) , ∀ x 1 , x 2 , y 1 , y 2 ∈ D with x 1 ≤ x 2 and y 2 ≤ y 1 .

An element x * is called a fixed point of Γ if Γ ( x * , x * ) = x * .

Lemma 2.1. [

Γ ( t x , 1 t y ) ≥ t β Γ ( x , y )

holds for all x , y ∈ P 0 . Then Γ has exactly one fixed point x * in P 0 . And for any x 0 , y 0 ∈ P 0 , we have

lim n → ∞ x n = lim n → ∞ y n = x * ,

where

x n = Γ ( x n − 1 , y n − 1 ) , y n = Γ ( y n − 1 , x n − 1 ) , n = 0 , 1 , 2 , ⋯

Lemma 2.2. [

Lemma 2.3. [

We define the spectral norm ‖ ⋅ ‖ in H ( n ) , then H ( n ) is a real Banach space. We all know that P ¯ ( n ) is a cone in H ( n ) and the interior points set is P ( n ) . Since the spectral norm is monotone, then we have that the set P ¯ ( n ) is normal cone from Definition 2.1. We define F by F ( X , Y ) = I + A * X p A + B * Y − q B , then we have the following theorem.

Theorem 2.1. Eqution (1.1) always has a unique positive definite solution X. And for any X 0 , Y 0 ∈ P 0 , we have

lim n → ∞ X n = lim n → ∞ Y n = X * ,

where

X n = F ( X n − 1 , Y n − 1 ) , Y n = F ( Y n − 1 , X n − 1 ) , n = 0 , 1 , 2 , ⋯

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 ∀ X , Y , Z , W ∈ P ( n ) with X ≥ Y , W ≥ Z , by Lemma 2.3, we have

1) F ( X , Y ) = I + A * X p A + B * Y − q B ∈ P ( n ) , that is F : P ( n ) × P ( n ) → P ( n ) ;

2) F ( X , Z ) = I + A * X p A + B * Z − q B ≥ I + A * Y p A + B * W − q B = F ( Y , W ) ,

which means that the operator F is mixed monotone;

3) Let β = max { p , q } , then we have 0 < β < 1 . For ∀ t ∈ ( 0,1 ) , we get

F ( t X , 1 t Y ) = I + A * t p X p A + B * ( 1 t ) − q Y − q B = I + A * t p X p A + B * t q Y − q B ≥ t β I + t β A * X p A + t β B * Y − q B = t β F ( X , Y ) .

According to Lemma 2.1, we get that the operator F has a unique fixed point X in P ( n ) , which is the unique positive definite solution of Equation (1.1).

Furthermore, for any X 0 , Y 0 ∈ P 0 , we have

lim n → ∞ X n = lim n → ∞ Y n = X * ,

where

X n = F ( X n − 1 , Y n − 1 ) , Y n = F ( Y n − 1 , X n − 1 ) , n = 0 , 1 , 2 , ⋯

,

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 X n = F ( X n − 1 , Y n − 1 ) , Y n = F ( Y n − 1 , X n − 1 ) , n = 0 , 1 , 2 , ⋯ converge to the unique positive definite solution of Equation (1.1). Let X 0 = Y 0 , we have X n = Y n = F ( X n − 1 , X n − 1 ) = I + A * X n − 1 p A + B * X n − 1 − q B . Consider the above iteration, which is the basic fixed point iterative method. The following are the algorithm.

Take X 0 = I

X n + 1 = I + A * X n p A + B * X n − q B , n = 0 , 1 , 2 , ⋯ (3.1)

We then consider a kind of multi-step stationary iterative method.

Theorem 3.1. For arbitrary initial matrix X 1 , X 2 ∈ P ( n ) , the matrix sequence

X n + 1 = I + A * X n − 1 p A + B * X n − q B , n = 2 , 3 , ⋯ (3.2)

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, X 1 and X 2 , there exist a positive number 0 < α ≤ 1 satisfy:

α X ≤ X 1 , X 2 ≤ α − 1 X . (3.3)

For any positive integer k, there exists a unique nonnegative integer K such that k = 2 K + 1 or k = 2 K + 2 . We will use mathematical induction to prove the following inequality

α β K X ≤ X k ≤ ( α − 1 ) β K X , k = 1,2, ⋯ (3.4)

where β = max { p , q } . When K = 0 , that is k = 1 , 2 , the inequality (3.4) holds following (3.3). When K = n − 1 , that is k = 2 n − 1 , 2 n , Assume (3.4) is true. That is

α β n − 1 X ≤ X 2 n − 1 , X 2 n ≤ ( α − 1 ) β n − 1 X . (3.5)

Now we need to prove (3.4) is true for K = n , at this time k = 2 n + 1 , 2 n + 2 . From (3.5) and β = max { p , q } , we get that

α β n X p ≤ α β n − 1 p X p ≤ X 2 n − 1 p ≤ ( α − 1 ) β n − 1 p X p ≤ ( α − 1 ) β n X p ,

α β n X − q ≤ α β n − 1 p X − q = ( α − 1 ) β n − 1 ( − q ) X − q ≤ X 2 n − q ≤ α β n − 1 ( − q ) X ( − q ) = ( α − 1 ) β n − 1 q X − q ≤ ( α − 1 ) β n X − q ,

From 0 < α ≤ 1 , it follows that 0 < α β n ≤ 1 , ( α − 1 ) β n ≥ 1 . Then

X 2 n + 1 = I + A * X 2 n − 1 p A + B * X 2 n − q B ≥ α β n I + A * α β n X p A + B * α β n X − q B = α β n [ I + A * X p A + B * X − q B ] = α β n X

and

X 2 n + 1 = I + A * X 2 n − 1 p A + B * X 2 n − q B ≤ ( α − 1 ) β n I + A * ( α − 1 ) β n X p A + B * ( α − 1 ) β n X − q B = ( α − 1 ) β n [ I + A * X p A + B * X − q B ] = ( α − 1 ) β n X .

Therefore

α β n X ≤ X 2 n + 1 ≤ ( α − 1 ) β n X . (3.6)

From 0 < q ≤ 1 , (3.6) implies

α β n X − q ≤ α β n q X − q ≤ X 2 n + 1 − q ≤ ( α − 1 ) β n q X − q ≤ ( α − 1 ) β n X − q .

From (3.4), we have

α β n X p ≤ α β n − 1 p X p ≤ X 2 n p ≤ ( α − 1 ) β n − 1 p X p ≤ ( α − 1 ) β n X p .

Then

X 2 n + 2 = I + A * X 2 n p A + B * X 2 n + 1 − q B ≥ α β n I + A * α β n X p A + B * α β n X − q B = α β n [ I + A * X p A + B * X − q B ] = α β n X

and

X 2 n + 2 = I + A * X 2 n p A + B * X 2 n + 1 − q B ≤ ( α − 1 ) β n I + A * ( α − 1 ) β n X p A + B * ( α − 1 ) β n X − q B = ( α − 1 ) β n [ I + A * X p A + B * X − q B ] = ( α − 1 ) β n X .

Therefore

α β n X ≤ X 2 n + 2 ≤ ( α − 1 ) β n X .

Now we have proved that (3.4) is true for K = n , at this time k = 2 n + 1 , 2 n + 2 .

Hence we get that the inequality (3.4) holds for any positive integer k. Let k → ∞ , then K → ∞ , we have

α β K → 1, ( α − 1 ) β K → 1.

Therefore,

X k → X .

We now use numerical examples to illustrate our results. All computations were performed using MATLAB, version 7.01. we denote ε ( X ) = ‖ X − I − A * X p A − B * X − q B ‖ ∞ , and use the stopping criterion ε ( X ) < 1.0 × 10 − 10 .

Example 4.1. Consider Equation (1.1) with p = 1 3 , q = 1 2 and

A = ( 0.1953 0.2310 0.1835 0.1796 ) , B = ( 0.9501 0.6068 0.2311 0.4860 ) .

We use the basic fixed point iterative method (3.1) to solve Equation (1.1). Let X 0 = I , after 14 iterations we get the following result:

X ≈ X 14 = ( 1.7834 0.5628 0.5628 1.5167 )

and ε ( X 14 ) = ‖ X 14 − I − A * X 14 p A − B * X 14 − q B ‖ ∞ = 4.0992 × 10 − 11 .

Considering the multi-step stationary iterative method (3.2) with X 1 = X 2 = I , after 19 iterations we get the following result:

X ≈ X 19 = ( 1.7834 0.5628 0.5628 1.5167 )

and ε ( X 19 ) = ‖ X 19 − I − A * X 19 p A − B * X 19 − q B ‖ ∞ = 3.1824 × 10 − 11 .

Example 4.2. Consider Eq.(1.1) with p = 1 2 , q = 1 3 and

A = ( 0.2913 0.5185 0.6154 0.2621 0.4214 0.7919 0.0565 0.0447 0.8218 ) , B = ( 0.9382 0.5355 0.3936 0.4763 0.0969 0.9579 0.9057 0.0013 0.8929 ) ,

Considering the basic fixed point iterative method (3.1) with X 0 = I , after 13 iterations we get the following result:

X ≈ X 13 = ( 2.3605 0.8354 1.8154 0.8354 2.0042 1.6716 1.8154 1.6716 6.3423 )

and ε ( X 13 ) = ‖ X 13 − I − A * X 13 p A − B * X 13 − q B ‖ ∞ = 9.7863 × 10 − 11 .

Considering the multi-step stationary iterative method (3.2) with X 1 = I , X 2 = 2 I , after 35 iterations we get the following result:

X ≈ X 35 = ( 2.3605 0.8354 1.8154 0.8354 2.0042 1.6716 1.8154 1.6716 6.3423 )

and ε ( X 35 ) = ‖ X 35 − I − A * X 35 p A − B * X 35 − q B ‖ ∞ = 5.8041 × 10 − 11 .

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).

Gao, D.J. (2017) Iterative Methods for Solving the Nonlinear Matrix Equation Advances in Linear Algebra & Matrix Theory, 7, 72-78. https://doi.org/10.4236/alamt.2017.73007