# Existence and Uniqueness of the Positive Definite Solution for the Matrix EquationX=Q+A∗(X^−C)−1A

Authors: Dongjie Gao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider the nonlinear matrix equationX=Q+A∗(X^−C)−1A, where Q is positive definite, C is positive semidefinite, and X^ is the block diagonal matrix defined by X^=diag(X,X,…,X). We prove that the equation has a unique positive definite solution via
variable replacement and fixed point theorem. The basic fixed point iteration for the equation is given.


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## Body

## 1. Introduction

We consider the matrix equation(1)X=Q+A*(X^-C)-1A,
where Q is an n×n positive definite matrix, C is an mn×mn positive semidefinite matrix, A is arbitrary mn×n matrix, and X^ is the block diagonal matrix defined by X^=diag(X,X,…,X) in which X is an n×n matrix. This matrix equation is connected with certain interpolation problem (see [1]).WhenC=0, we write A as the  m×1  block matrix
(2)A=(A1A2⋮A4),
where A1,…,Am are n×n matrices. Then (1) is equal to X=Q+∑i=1mAi*X-1Ai, which has been investigated by many authors (see [2–9]). However, are few little theoretical results when C≠0. Ran and Reurings [1] have proved that (1) has a unique positive definite solution. Based on this, Sun [10] presented some perturbation results for the unique solution.Fixed point theorem is often used to discuss the existence and uniqueness of the solution. However, for (1), it is difficult to prove the uniqueness of the solution directly by using fixed point theorems. In this paper, we turn (1) into its equivalent equation via variable replacement. Then we consider its equivalent equation by using fixed point theorem. This provides a new proof for the existence and uniqueness of the positive definite solution. And this method is shown to be much easier than the way of [1]. In addition, the basic fixed point iteration for the equation is given.In this paper we use𝒞(n) to denote n×n complex 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). A* denotes the conjugate transpose of a matrix A. Let λ¯=λmax(A*A), λ_=λmin(A*A), C1¯=λmax(C1),  and  C1_=λmin(C1).


## 2. The Case ofQ^>C

In this section, we discuss (1) with Q^>C. Let S1(n)⊂P(n) be the set defined by
(3)S1(n)={X∈P(n)∣X^>C}.
We are interested in positive definite solutions of (1) in this set. A matrix X is a solution of (1) if and only if it is a fixed point of the map  G1 defined by
(4)G1(X)=Q+A*(X^-C)-1A.
Let C1=Q^-C. For Q^>C, we know that C1 is a positive definite matrix. Then (1) turns into
(5)X-Q=A*(X^-Q^+C1)-1A.
Let Y=X-Q. Then (1) eventually becomes
(6)Y=A*(Y^+C1)-1A.In the following, we consider the existence and uniqueness of the positive definite solution of (6). And then we can easily get corresponding conclusions about (1).Theorem 1.
Equation (6) has a positive semidefinite solution for any A∈𝒞(n).Proof.
A matrixY is a solution of (6) if and only if it is a fixed point of the map F defined by F(Y)=A*(Y^+C1)-1A. Note that F maps [0,F(0)] into itself, because F is order reversing. Hence it has a fixed point in the [0,F(0)]. That is to says (6) has a positive semidefinite solution.Theorem 2.
IfY is positive semidefinite solution of (6), then Y∈[0,F(0)].Proof.
ByY≥0, and  C1>0, we know that (Y^+C1)-1>0 and A*(Y^+C1)-1A≥0; then Y=A*(Y^+C1)-1A≥0. By Y≥0, we know that Y=A*(Y^+C1)-1A≤A*C1-1A=F(0). That is,  Y∈[0,F(0)].For proving the uniqueness, we first verify the following lemma.Lemma 3.
Letη(t)=(((1-t)C1_)/(C1¯+C1_-1λ¯))>0. For any Y∈[0,F(0)] and 0<t<1, one has
(7)F2(tY)≥t(1+η(t))F2(Y).Proof.
By Theorem2, we know F(Y)∈[0,F(0)] for any Y∈[0,F(0)]. Then  F2(Y)∈[0,F(0)] for any  Y∈[0,F(0)]. Let F(tY)=A*(tY^+C1)-1A, and  F1(tY)=A*(Y^+t-1C1)-1A. For any Y∈[0,F(0)] and 0<t<1, we have
(8)F2(tY)-t(1+η(t))F2(Y)=A*(F(tY)^+C1)-1A-tA*(F(Y)^+C1)-1A-tη(t)F2(Y)=tA*(F1(tY)^+tC1)-1A-tA*(F(Y)^+C1)-1A-tη(t)F2(Y)≥tA*(F(Y)^+tC1)-1A-tA*(F(Y)^+C1)-1A-tη(t)F2(Y)=tA*[(F(Y)^+tC1)-1-(F(Y)^+C1)-1]A-tη(t)F2(Y)=tA*[(F(Y)^+tC1)-1(1-t)C1(F(Y)^+C1)-1]A-tη(t)F2(Y)≥tA*[(F(Y)^+C1)-1(1-t)C1(F(Y)^+C1)-1]A-tη(t)F2(Y)≥t(1-t)C1_C1¯+C1_-1λ¯A*(F(Y)^+C1)-1A-tη(t)F2(Y)=tη(t)F2(Y)-tη(t)F2(Y)=0.Theorem 4.
Equation (6) has a unique positive definite solution Y∈[0,F(0)], and for any Y0∈[0,F(0)], the iteration
(9)Yn+1=A*(Y^n+C1)-1A,n=0,1,2,…,
converges to Y; that is,  limn→∞Yn=Y.Proof.
Consider the matrix sequence (9). Let F(Yn)=A*(Yn^+C1)-1A. We first suppose that Y0=0. Then we get
(10)0≤Y1=A*(Y0^+C1)-1A=F(0),0≤Y2=A*(Y1^+C1)-1A=F2(0)≤F(0),F2(0)≤Y3=A*(Y2^+C1)-1A=F3(0)≤F(0),F2(0)≤Y4=A*(Y3^+C1)-1A=F4(0)≤F3(0).
Analogously one can prove that
(11)0≤F2k(0)≤F2k+2(0)≤F2k+1(0)≤F2k-1(0)≤F(0).
Hence, the sequence {F2k(0),k≥0} is monotone increasing and is bounded from above by F(0). Thus, the sequence {F2k(0),k≥0} has a finite positive definite limit. Moreover, the sequence {F2k+1(0),k≥0} is monotone decreasing and is bounded from below by 0. Thus, the sequence {F2k+1(0),k≥0} has a finite positive definite limit. Let
(12)limk→∞F2k(0)=Y(1),limk→∞F2k+1(0)=Y(2);
then 0≤Y(i)≤F(0)  for i=1,2. Clearly, both  Y(1) and Y(2) are the positive fixed points of F2(Y). Then
(13)Y(1)=F2(Y(1))=A*[F(Y(1))^+C1]-1A≥A*(λ¯C1-2C1+C1)-1A≥A*(λ¯C1_-2+1)-1C1-1A=1λ¯C1_-2+1A*C1-1A≥1λ¯C1_-2+1F2(Y(2))=1λ¯C1_-2+1Y(2).
Let t0=sup{t∣Y(1)≥tY(2)}; then 0<t0<+∞. Now we prove that t0≥1. Assume, on the contrary, that 0<t0<1; then Y(1)≥t0Y(2). By Lemma 3 and monotonicity of F2(Y), we have
(14)Y(1)=F2(Y(1))≥F2(t0Y(2))≥(1+η(t0))t0F2(Y(2))=(1+η(t0))t0Y2.
Since (1+η(t0))t0>t0, then it is a contradiction to the definition of t0. Hence we have t0≥1; therefore Y(1)≥Y(2). Similarly, we can get Y(1)≤Y(2). Therefore, Y(1)=Y(2). Let Ω=[0,F(0)]. By Theorem 1, we know F(Ω)⊆Ω. Hence F2(Ω)⊆Ω. That is, F2 has only one positive fixed point in Ω. In other words, the equation Y=F2(Y) has only one positive definite solution. Hence Y(1)=Y(2). It follows that limn→∞Fn(0) is a fixed point of F2. Since the positive definite solution of F(Y)=Y solves Y=F2(Y), then F2(Y)=Y has only one positive definite solution. Hence  limn→∞F(n)(0)  is the unique fixed point of F.
For anyY0>0, we get
(15)Y1≥0,0≤Y2=A*(Y1^+C1)-1A≤A*C1-1A=F(0).
By induction, we have for any k=1,2,…(16)F2k(0)≤Y2k+1≤F2k-1(0),F2k-2(0)≤Y2k≤F2k-1(0).
Now letting k→∞ on both sides of the above inequalities, we can get
(17)limk→∞Yk=limn→∞F(n)(0)=Y.
Since (1) is equal to (6) when X=Y+Q, then we know that X is a positive definite solution of (1) if and only if Y is a positive semidefinite solution of (6). And furthermore, Y∈[0,F(0)] if and only if X∈[Q,G1(Q)]. Thus, we can get the following conclusions about (1).Theorem 5.
Equation (1) with Q^>C has a positive definite solution for any A∈𝒞(n).Theorem 6.
IfX is positive definite solution of  (1) with Q^>C, then X∈[Q,G1(Q)]⊂S1(n).Theorem 7.
LetQ∈𝒫(n) and C∈𝒫¯(mn) such that Q^>C. Then (1) has a unique solution X in S1(n) and for any X0∈[Q,G1(Q)], the iteration
(18)Xn+1=Q+A*(X^n-C)-1A,n=0,1,2,…,
converges to X; that is,  limn→∞Xn=X.


## 3. The Case ofQ^<C

In this section, we discuss (1) with Q^<C and A*(C-Q^)-1A<Q. In fact, C∈𝒫¯(n) must be a positive definite matrix in this case, because Q^<C and because Q is a positive definite matrix. We consider the matrix equation
(19)X=Q-A*(C-X^)-1A,
which is equal to (1). Let S2(n)⊂P(n) be the set defined by
(20)S2(n)={X∈P(n)∣X^<C}.
We are interested in positive definite solutions of (1) in this set. A matrix X is a solution of (1) if and only if it is a fixed point of the map G2 defined by
(21)G2(X)=Q-A*(C-X^)-1A.
Let C2=C-Q^;  For Q^<C, we have C2 being  a positive definite matrix. Then (1) turns into
(22)Q-X=A*(Q^-X^+C2)-1A.
Let Y=Q-X. Then (1) eventually becomes
(23)Y=A*(Y^+C2)-1A.
We note that (23) should have the same results as (6). Then we directly give the following conclusions without proof.Theorem 8.
Equation (1) with Q^<C and A*(C-Q^)-1A<Q has a positive definite solution for any A∈𝒞(n).Theorem 9.
IfX is a positive definite solution of (1) with Q^<C and A*(C-Q^)-1A<Q, then X∈[G2(Q),Q]⊂S2(n).Theorem 10.
LetQ∈𝒫(n) and C∈𝒫¯(mn) such that Q^<C. Then (1) has a unique solution X in S2(n), and for any X0∈[G2(Q),Q], the iteration
(24)Xn+1=Q-A*(C-X^n)-1A,n=0,1,2,…,
converges to X; that is,  limn→∞Xn=X.


## 4. Numerical Examples

We now use numerical examples to illustrate our results. All computations were performed using MATLAB, version7.01. We denoteε(X)=∥X-A*(X^-C)-1A-Q∥∞ and  use the stopping criterion ε(X)<1.0×10-10.Example  1. Consider (1) with n=2, m=3, and
(25)A=(2-210-5362-3165),Q=(20-1-113),C=(61-205219-1-13-2-2-171-120-1160-253-10902-22-204),
where the matrices Q and C satisfy Q^>C. Considering the iterative method (18) with X0=Q, after 23 iterations we get the following result:
(26)X≈X23=(27.286668734882351.124933045738341.1249330457383416.21667821374470)
and ε(X23)=∥X23-A*(X23^-C)-1A-Q∥∞=8.622080827080936×10-11. It is not difficult to verify that X∈[Q,G1(Q)]⊂S1(n).Example  2. Consider (1) with n=2, m=3, and
(27)A=(0-110-101221-11),Q=(4-1-15),C=(151-1201119-2-13-1-1-2102-222-12190-103-202301-12-1030),
where the matrices Q and C satisfy Q^<C and A*(C-Q^)-1A<Q. Considering the iterative method (24) with X0=Q, after 7 iterations one gets an approximation to the positive definite solution X. It is
(28)X≈X7=(3.48707713254602-1.33049056598352-1.330490565983524.39117111733130)
and ε(X7)=∥X23-A*(X7^-C)-1A-Q∥∞=6.447897771266753×10-11. It is not difficult to verify that X∈[G2(Q),Q]⊂S2(n).



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