# On the Positive Definite Solutions of a Nonlinear Matrix Equation

Authors: Panpan Liu; Shugong Zhang; Qingchun Li
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

The positive definite solutions of the nonlinear
matrix equationXs+A∗f(X)A=Q are discussed. A necessary and sufficient condition for the existence of positive definite solutions for this equation is derived. Then, the uniqueness of the Hermitian positive definite solution is studied based on an iterative method proposed in this paper. Lastly the perturbation analysis
for this equation is discussed.


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## Body

## 1. Introduction

Denote the set of alln×n positive definite matrices by P(n). In this paper, we consider the matrix equation
(1)Xs+A*f(X)A=Q,
where A is nonsingular, Q is a Hermitian positive definite matrix, s is a positive real number, f is a continuous map from  P(n) into P(n), and f is either monotone (meaning that 0≤X≤Y implies that f(X)≤f(Y)) or antimonotone (meaning that 0≤X≤Y implies that f(X)≥f(Y)).Nonlinear matrix equation of the form (1) often arises in dynamic programming, control theory, stochastic filtering, statistics, and so on. In recent years, many authors have been much interested in studying this class of matrix equations [1–10].Equation (1) has been investigated in some special cases. For the case s=1, Ran and Reurings [1] derived some sufficient conditions for the existence and uniqueness of a positive definite solution of (1). In addition, an iterative method for obtaining Hermitian positive definite solutions of (1) with f(X)=X-t is proposed by Yueting [10]. Liu and Gao [9] proved the existence of the symmetric positive definite solutions of (1) with f(X)=±X-t and Q=I. Many other authors investigated (1) for particular choices of the map f [2–4, 6–8].In Section2, we will derive a necessary and sufficient condition for the existence of positive definite solutions of (1). In Section 3, we will propose an iterative method and investigate the uniqueness of the Hermitian positive definite solution. Finally, in Section 4, we will discuss the perturbation analysis of (1).The following notations are used throughout this paper. For a positive definite matrixA, λmax(A) and λmin(A) stand for the maximal and minimal eigenvalues of matrix A, respectively. A* is the conjugate transpose of the matrix A, and A-* is the inversion of A*. ∥A∥ denotes the spectral norm of A. A>0(A≥0) denotes that A is a positive definite (semidefinite) matrix, and A>B(A≥B) means that A-B>0(A-B≥0). The notation LA,B denotes the line segment joining A and B; that is,
(2)LA,B={tA+(1-t)B∣t∈[0,1]}.


## 2. On the Positive Definite Solutions of (1)

In this section, we will derive a necessary and sufficient condition for the existence of positive definite solutions of (1).Lemma 1 (see [8]).
Assume thatA≥B≥0(A>B≥0); if r∈(0,1], then Ar≥Br(Ar>Br), and if r∈[-1,0), then Ar≤Br(Ar<Br).If there is unique Hermitian positive definite matrixT, such that f(X)=T2,X∈P(n), then we denote that T=f1/2(X).Theorem 2.
Assume thatf is a continuous map from P(n) into HP(n), where HP(n) denotes the set of all n×n Hermitian positive definite matrices. Then, (1) has a Hermitian positive definite solution if and only if there is a nonsingular matrix W, such that W*W=WW*, and A=(f1/2(W*W))-1ZQ1/2, where
(3)(Q-1/2)*(Ws)*(Ws)Q-1/2+Z*Z=I.
In this case, (1) has a Hermitian positive definite solution X=W*W.Proof.
IfX is a Hermitian positive definite solution of (1), then there is unique Hermitian positive definite matrix W, such that X=W2 (see [11]). So, f(X)=f(W*W)∈HP(n), and therefore there is unique Hermitian positive definite matrix T, such that f(W*W)=T2. Substituting X=W2=W*W into (1) gives
(4)(W*W)s+A*f(W*W)A=Q.
Then, we have
(5)(Ws)*(Ws)+A*T*TA=Q.Q is Hermitian positive definite, and so
(6)(Q-1/2)*(Ws)*(Ws)Q-1/2+(Q-1/2)*A*T*TAQ-1/2=I;
that is
(7)[WsQ-1/2TAQ-1/2]*[WsQ-1/2TAQ-1/2]=I.
Let Z=TAQ-1/2. Then, A=T-1ZQ1/2=(f1/2(W*W))-1ZQ1/2, and by (7) we know that (Q-1/2)*(Ws)*(Ws)Q-1/2+Z*Z=I.
Conversely, ifA=(f1/2(W*W))-1ZQ1/2, let X=W*W. Then,
(8)Xs+A*f(X)A=(W*W)s+(Q1/2)*Z*(f1/2(W*W))-*×f(W*W)(f1/2(W*W))-1ZQ1/2=(Ws)*Ws+(Q1/2)*Z*ZQ1/2=(Q1/2)*×((WsQ-1/2)*(WsQ-1/2)+Z*Z)Q1/2=Q.
So,X=W*W is a Hermitian positive definite solution of (1).Ifs=1, then the restriction W*W=WW* in Theorem 2 can be omitted.


## 3. Iterative Method

In order to discuss an iterative method for solving (1), we assume that for a given matrix B, the equation f(X)=B always has a positive definite solution and its solution is easy to obtain. We are interested in the inverse iteration, and consider the following iterative method:
(9)Xk+1=f-1(A-*(Q-Xks)A-1),k=0,1,2,…
In this section, we assume that A, Q, f in (1) satisfy f-1(A-*QA-1)≤Q1/s.Theorem 3.
Suppose thatf-1 exists and that f is antimonotone. Let s=1. Equation (1) has a positive definite solution in the interval (0,λmax(Q)I) if and only if there is a number χ∈(0,λmax(Q)), such that Xk≤χI for all k. Moreover, in this case, the iteration (9) with X0=0 converges to the smallest positive definite solution of (1).Proof.
Sincef-1 exists and f is anti-monotone, then f-1 is also anti-monotone. Assume that there is a number χ as in the theorem. Since X0=0, one has
(10)X1=f-1(A-*QA-1)≥X0=0.
Furthermore, we get
(11)X2=f-1(A-*(Q-X1)A-1)≥f-1(A-*(Q-X0)A-1)=X1.
Now, if Xk≥Xk-1, we have
(12)Xk+1=f-1(A-*(Q-Xk)A-1)≥f-1(A-*(Q-Xk-1)A-1)=Xk.
Then, the sequence {Xk} is a monotonically nondecreasing sequence and bounded above by some positive definite matrix χI. Consequently, the sequence {Xk} converges to a positive definite matrix X, which is a solution of (1); that is,
(13)X=f-1(A-*(Q-X)A-1).
Conversely, let (1) have a positive definite solution X∈(0,λmax(Q)I), and let χ∈(0,λmax(Q)) be the largest eigenvalue of X. In order to prove that Xk≤χI for every Xk generated from (9), we will prove that Xk≤X. Consider
(14)X0=0<X,X1=f-1(A-*(Q-X0)A-1)≤f-1(A-*(Q-X)A-1)=X.
Assume that Xk≤X, for some fixed k, and then we have
(15)Xk+1=f-1(A-*(Q-Xk)A-1)≤f-1(A-*(Q-X)A-1)=X.
So, Xk+1∈(0,X). Then, the theorem is proved.Lemma 4 (see [1]).
Letf:U→M(n)(U⊂M(n)open) be differentiable at any point of U. Then,
(16)∥f(X)-f(Y)∥≤supZ∈LX,Y∥𝒟f(Z)∥∥X-Y∥,
for all X,Y∈U.Lemma 5.
Suppose thatf-1 exists and that f is anti-monotone. If (1) with s≥1 has a Hermitian positive definite solution X, then X∈[f-1(A-*QA-1),Q1/s].Proof.
Sincef-1 exists and f is anti-monotone, then f-1 is also anti-monotone. Assume that (1) has a Hermitian positive definite solution X. Since f maps into P(n), we have f(X)>0 and A*f(X)A≥0. Therefore, Xs=Q-A*f(X)A≤Q. By Lemma 1 we know that X≤Q1/s. On the other hand, we have A*f(X)A≤Q, f(X)≤A-*QA-1. Then, we obtain X≥f-1(A-*QA-1) because f-1 is anti-monotone.Theorem 6.
Suppose thatf-1 exists and that f is anti-monotone, and suppose that f, f-1 are differentiable at any point of Ω1=[f-1(A-*QA-1),Q1/s] and Ω2=[f(Q1/s),A-*QA-1], respectively. Let
(17)M1=supZ∈Ω1∥𝒟f(Z)∥,M2=supX∈Ω2∥𝒟f-1(X)∥,
and a=M1M2∥A-1∥2∥A∥2.(i)
If (1) with s>1 has a Hermitian positive definite solution X and a<1, then X is the unique solution of (1).(ii)
Assume that there is a closed setΩ⊆Ω1 satisfying that g:Ω→Ω and g(X)=f-1(A-*(Q-Xs)A-1); if a<1, then (1) with s>1 has a unique solution X in Ω. Furthermore, one considers the iterative method (9) with X0∈Ω. The sequence {Xk} in (9) converges to the unique solution X; moreover,
(18)∥Xn-X∥≤an1-a∥X1-X0∥,∥Xn-X∥≤a1-a∥Xn-Xn-1∥.Proof.
(i) Assume thatX and X¯ are two different Hermitian positive definite solutions of (1), and by Lemma 5, X,X¯∈[f-1(A-*QA-1),Q1/s]=Ω1; then, f(X),f(X¯)∈[f(Q1/s),A-*QA-1]=Ω2. Let T1=A-*(Q-Xs)A-1,T2=A-*(Q-X¯s)A-1, and
(19)M2=supX∈Ω2∥𝒟f-1(X)∥.
From Lemma 4,
(20)∥X-X¯∥=∥f-1(A-*(Q-Xs)A-1)-f-1(A-*(Q-X¯s)A-1)∥≤supZ∈LT1,T2∥𝒟f-1(Z)∥∥A-*(Q-Xs)A-1-A-*(Q-X¯s)A-1∥=supZ∈Lf(X),f(X¯)∥𝒟f-1(Z)∥∥A-*(X¯s-Xs)A-1∥≤supZ∈Ω2∥𝒟f-1(Z)∥∥A-*(X¯s-Xs)A-1∥≤M2∥A-1∥2∥X¯s-Xs∥.
Let
(21)M1=supZ∈Ω1∥𝒟f(Z)∥.
Then,
(22)∥X¯s-Xs∥=∥Q-A*f(X¯)A-Q+A*f(X)A∥≤∥A∥2∥f(X)-f(X¯)∥≤∥A∥2supZ∈LX,X¯∥𝒟f(Z)∥∥X-X¯∥≤∥A∥2supZ∈Ω1∥𝒟f(Z)∥∥X-X¯∥≤∥A∥2M1∥X-X¯∥.
By a<1 we have
(23)∥X-X¯∥≤M2∥A-1∥2∥A∥2M1∥X-X¯∥=a∥X-X¯∥<∥X-X¯∥,
which is a contradiction; so, X=X¯. That is, X is the unique solution of (1).
(ii) LetY1, Y2∈Ω⊆Ω1. Then, f(g(Y1)),f(g(Y2))∈Ω2; from Lemma 4,
(24)∥g(Y1)-g(Y2)∥=∥f-1(A-*(Q-Y1s)A-1)-f-1(A-*(Q-Y2s)A-1)∥≤supZ∈Lf(g(Y1)),f(g(Y2))∥𝒟f-1(Z)∥∥A-*(Y2s-Y1s)A-1∥≤supZ∈Ω2∥𝒟f-1(Z)∥∥A-1∥2∥Y2s-Y1s∥≤M2∥A-1∥2∥Q-A*f(Y2)A-Q+A*f(Y1)A∥≤M2∥A-1∥2∥A∥2∥f(Y1)-f(Y2)∥≤M2∥A-1∥2∥A∥2supZ∈LY1,Y2∥𝒟f(Z)∥∥Y1-Y2∥≤M2∥A-1∥2∥A∥2supZ∈Ω1∥𝒟f(Z)∥∥Y1-Y2∥≤M2∥A-1∥2∥A∥2M1∥Y1-Y2∥=a∥Y1-Y2∥.
The interval Ω is a complete metric space because it is a closed subset of P(n). And a<1; so, g is a contraction on Ω. Then, it follows from contractive mapping principle that the map g has a unique fixed point X in Ω. Furthermore, the sequence {Xk} in (9) converges to the unique solution of (1); moreover,
(25)∥Xn-X∥≤an1-a∥X1-X0∥,∥Xn-X∥≤a1-a∥Xn-Xn-1∥.


## 4. Perturbation Analysis

Let(26)X~s+A~*f(X~)A~=Q~
be the perturbation equation of (1). Let A, A~ be nonsingular matrices and Q, Q~ be positive definite, and 0<s<1. Suppose that f-1:P(n)→P(n) exists and f is anti-monotone.Lemma 7 (see [12]).
If0<r<1, the operators X, Y satisfy X≥aI and Y≥aI for some positive number a; then,
(27)∥Xr-Yr∥≤rar-1∥X-Y∥.Theorem 8.
LetX, X~ be the positive definite solutions of (1) and its perturbation equation (26), respectively. Map f-1 is differentiable at any point of Ω4 with
(28)Ω4={Y∈f([f-1(A~-*Q~A~-1),Q~1/s])αX+(1-α)Y∣α∈[0,1],X∈f([f-1(A-*QA-1),Q1/s]),Y∈f([f-1(A~-*Q~A~-1),Q~1/s])}.
Let
(29)M4=supZ∈Ω4∥𝒟f-1(Z)∥.
If M4∥A-1∥∥A~-1∥sλ~s-1<1, one has
(30)∥X-X~∥∥X∥≤Tλmin(f-1(A-*QA-1))(M4-1-∥A-1∥∥A~-1∥sλ~s-1),
where T=∥A-1-A~-1∥(∥Q-λ~sI∥∥A-1∥+∥Q~-λ~sI∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥, λ~=min{λmin(f-1(A-*QA-1)),λmin(f-1(A~-*Q~A~-1))}.Proof.
X, X~ are the positive definite solutions of (1) and (26), respectively. Let T1=A-*(Q-Xs)A-1, T2=A~-*(Q~-X~s)A~-1. From Lemma 4,
(31)∥X-X~∥=∥-f-1(A~-*(Q~-X~s)A~-1)f-1(A-*(Q-Xs)A-1)-f-1(A~-*(Q~-X~s)A~-1)∥≤supZ∈LT1,T2∥𝒟f-1(Z)∥∥-f-1(A~-*(Q~-X~s)A~-1)A-*(Q-Xs)A-1-A~-*(Q~-X~s)A~-1∥.
By Lemma 5,
(32)X∈[f-1(A-*QA-1),Q1/s],X~∈[f-1(A~-*Q~A~-1),Q~1/s].
Let
(33)Ω4={Y∈f([f-1(A~-*Q~A~-1),Q~1/s])αX+(1-α)Y∣α∈[0,1],X∈f([f-1(A-*QA-1),Q1/s]),Y∈f([f-1(A~-*Q~A~-1),Q~1/s])},M4=supZ∈Ω4∥𝒟f-1(Z)∥.
Then, we have
(34)supZ∈LT1,T2∥𝒟f-1(Z)∥=supZ∈Lf(X),f(X~)∥𝒟f-1(Z)∥≤supZ∈Ω4∥𝒟f-1(Z)∥≤M4.
Notice that
(35)X≥f-1(A-*QA-1)≥λmin(f-1(A-*QA-1))I,X~≥f-1(A~-*Q~A~-1)≥λmin(f-1(A~-*Q~A~-1))I.
Let λ~=min{λmin(f-1(A-*QA-1)), λmin(f-1(A~-*Q~A~-1))}. Then, X, X~≥λ~I. By Lemma 1, Xs, X~s≥λ~sI since 0<s<1. And from Lemma 7,
(36)∥A-*(Q-Xs)A-1-A~-*(Q~-X~s)A~-1∥=∥A-*(Q-Xs)(A-1-A~-1)+A-*(Q-Xs-Q~+X~s)A~-1+(A-*-A~-*)(Q~-X~s)A~-1∥≤∥A-1∥∥Q-Xs∥∥A-1-A~-1∥+∥A-1∥∥A~-1∥(∥Q-Q~∥+∥X~s-Xs∥)+∥A-1-A~-1∥∥A~-1∥∥Q~-X~s∥=∥A-1-A~-1∥(∥Q-Xs∥∥A-1∥+∥Q~-X~s∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥+∥A-1∥∥A~-1∥∥X~s-Xs∥≤∥A-1-A~-1∥(∥Q-λ~sI∥∥A-1∥+∥Q~-λ~sI∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥+∥A-1∥∥A~-1∥sλ~s-1∥X~-X∥.
Therefore,
(37)∥X-X~∥≤M4(∥A-1-A~-1∥×(∥Q-λ~sI∥∥A-1∥+∥Q~-λ~sI∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥+∥A-1∥∥A~-1∥sλ~s-1∥X~-X∥).
That is,
(38)(M4-1-∥A-1∥∥A~-1∥sλ~s-1)∥X~-X∥≤∥A-1-A~-1∥(∥Q-λ~sI∥∥A-1∥+∥Q~-λ~sI∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥.
If M4∥A-1∥∥A~-1∥sλ~s-1<1, by X≥λmin(f-1(A-*QA-1))I we have
(39)∥X-X~∥∥X∥≤Tλmin(f-1(A-*QA-1))(M4-1-∥A-1∥∥A~-1∥sλ~s-1),
where
(40)T=∥A-1-A~-1∥(∥Q-λ~sI∥∥A-1∥+∥Q~-λ~sI∥∥A~-1∥)+∥A-1∥∥A~-1∥∥Q-Q~∥.



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