# On the Hermitian Positive Definite Solutions of Nonlinear Matrix EquationXs+A∗X−t1A+B∗X−t2B=Q

Authors: Aijing Liu; Guoliang Chen
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

Nonlinear matrix equationXs+A∗X−t1A+B∗X−t2B=Q has many applications in engineering; control theory; dynamic programming; ladder networks; stochastic filtering; statistics and so forth. In this paper, the Hermitian positive definite solutions of nonlinear matrix equation Xs+A∗X−t1A+B∗X−t2B=Q are considered, where Q is a Hermitian positive definite matrix, A, B are nonsingular complex matrices, s is a positive number, and 0<ti≤1, i=1,2. Necessary and sufficient conditions for the existence of Hermitian positive definite solutions are derived. A sufficient condition for the existence of a unique Hermitian positive definite solution is given. In addition, some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions are presented. Finally, an iterative method is proposed to compute the maximal Hermitian positive definite solution, and numerical example is given to show the efficiency of the proposed iterative method.


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## Body

## 1. Introduction

We consider the nonlinear matrix equation(1.1)Xs+A*X-t1A+B*X-t2B=Q,
where Q is an n×n Hermitian positive definite matrix, A,B are n×n nonsingular complex matrices, s is a positive number, and 0<ti≤1, i=1,2. Here A* stands for the conjugate transpose of the matrix A.Nonlinear matrix equations with the form of (1.1) have many applications in engineering; control theory; dynamic programming; ladder networks; stochastic filtering; statistics and so forth. The solutions of practical interest are their Hermitian positive definite (HPD) solutions. The existence of HPD solutions of (1.1) has been investigated in some special cases. Long et al. [1] studied (1.1) when s=1, t1=t2=1. In addition, there have been many papers considering the Hermitian positive solutions of(1.2)Xs+A*X-tA=Q.
For instance, the authors [2–5] studied (1.2) when s=1, t=1. In Hasanov [6, 7], the authors investigated (1.2) when s=1, t∈(0,1]. Then Peng et al. [8] proposed iterative methods for the extremal positive definite solutions of (1.2) for s=1 with two cases: 0<t≤1 and t≥1. Cai and Chen [9, 10] studied (1.2) with two cases: s and t are positive integers, and s≥1, 0<t≤1 or 0<s≤1, t≥1 respectively.In this paper, we study the HPD solutions of (1.1). The paper is organized as follows. In Section 2, we derive necessary and sufficient conditions for the existence of HPD solutions of (1.1) and give a sufficient condition for the existence of a unique HPD solution of (1.1). We also present some necessary conditions and sufficient conditions for the existence of HPD solutions of (1.1). Then in Section 3, we propose an iterative method for obtaining the maximal HPD solution of (1.1). We give a numerical example in Section 4 to show the efficiency of the proposed iterative method.We start with some notations which we use throughout this paper. The symbolCm×n denotes the set of m×n complex matrices. We write D>0(D≥0) if the matrix D is positive definite(semidefinite). If D-E is positive definite(semidefinite), then we write D>E(D≥E). We use λ1(D) and λn(D) to denote the maximal and minimal eigenvalues of a matrix D. We use ∥D∥ and ∥D∥F to denote the spectral and Frobenius norm of a matrix D, and we also use ∥b∥ to denote l2-norm of a vector b. We use XS and XL to denote the minimal and maximal HPD solution of (1.1), that is, for any HPD solution X of (1.1), then XS≤X≤XL. The symbol I denotes the n×n identity matrix. The symbol ρ(D) denotes the spectral radius of D. Let [D,E]={X∣D≤X≤E} and (D,E)={X∣D<X<E}. For matrices D=(d1,d2,…,dn)=(dij) and E, D⊗E=(dijE) is a Kronecker product and vec(D) is a vector defined by vec(D)=(d1T,d2T,…,dnT)T.


## 2. Solvability Conditions and Properties of the HPD Solutions

In this section, we will derive the necessary and sufficient conditions for (1.1) to have an HPD solution and give a sufficient condition for the existence of a unique HPD solution of (1.1). We also will present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of (1.1).Lemma 2.1 (see [11]).
IfD≥E>0(or D>E>0), then Dp≥Ep>0(or Dp>Ep>0) for all p∈(0,1], and Ep≥Dp>0(or Ep>Dp>0) for all p∈[-1,0).Lemma 2.2 (see [12]).
LetD and E be positive operators on a Hilbert space such that 0<m1I≤D≤M1I, 0<m2I≤E≤M2I,  and 0<D≤E. Then
(2.1)Dq≤(M1m1)q-1Eq,Dq≤(M2m2)q-1Eq
hold for any q≥1.Lemma 2.3 (see [13]).
Letf(x)=xt(ζ-xs), ζ>0, x≥0. Then(1)
f is increasing on [0,((t/(s+t))ζ)1/s] and decreasing on [((t/(s+t))ζ)1/s,+∞);(2)
fmax=f(((t/(s+t))ζ)1/s)=(s/(s+t))(t/(s+t))t/sζ(t/s)+1.Lemma 2.4 (see [14]).
IfD and E are Hermitian matrices of the same order with E>0, then DED+E-1≥2D.Lemma 2.5 (see [15]).
If0<θ≤1, and D and E are positive definite matrices of the same order with D,E≥bI>0, then ∥Dθ-Eθ∥≤θbθ-1∥D-E∥ and ∥D-θ-E-θ∥≤θb-(θ+1)∥D-E∥. Here ∥·∥ stands for one kind of matrix norm.Lemma 2.6 (see [5]).
LetD and E be two arbitrary compatible matrices. Then ρ(D*E-E*D)≤ρ(D*D+E*E).Theorem 2.7.
Equation (1.1) has an HPD solution if and only if A,B can factor as
(2.2)A=(L*L)t1/2sN1,B=(L*L)t2/2sN2,
where L is a nonsingular matrix and (LQ-1/2N1Q-1/2N2Q-1/2) is column orthonormal.Proof.
If (1.1) has an HPD solution, then Xs>0. Let Xs=L*L be the Cholesky factorization, where L is a nonsingular matrix. Then (1.1) can be rewritten as
(2.3)Q-1/2L*LQ-1/2+Q-1/2A*(L*L)-t1/2s(L*L)-t1/2sAQ-1/2+Q-1/2B*(L*L)-t2/2s(L*L)-t2/2sBQ-1/2=I.
Let N1=(L*L)-t1/2sA,N2=(L*L)-t2/2sB, then A=(L*L)t1/2sN1, B=(L*L)t2/2sN2. Moreover, (2.3) turns into
(2.4)Q-1/2L*LQ-1/2+Q-1/2N1*N1Q-1/2+Q-1/2N2*N2Q-1/2=I,
that is,
(2.5)(LQ-1/2N1Q-1/2N2Q-1/2)*(LQ-1/2N1Q-1/2N2Q-1/2)=I,
which means that (LQ-1/2N1Q-1/2N2Q-1/2) is column orthonormal.
Conversely, ifA,B have the decompositions as (2.2), let X=(L*L)1/s, then X is an HPD matrix, and it follows from (2.2) and (2.4) that
(2.6)Xs+A*X-t1A+B*X-t2B=L*L+N1*N1+N2*N2=Q1/2(Q-1/2L*LQ-1/2+Q-1/2N1*N1Q-1/2+Q-1/2N2*N2Q-1/2)Q1/2=Q.
Hence (1.1) has an HPD solution.Theorem 2.8.
Equation (1.1) has an HPD solution if and only if there exist a unitary matrix V∈Cn×n, a column-orthonormal matrix U=(U1U2)∈C2n×n(in which U1,U2∈Cn×n), and diagonal matrices C>0 and S≥0 with C2+S2=I such that
(2.7)A=(Q1/2V*C2VQ1/2)t1/2sU1SVQ1/2,B=(Q1/2V*C2VQ1/2)t2/2sU2SVQ1/2.Proof.
If (1.1) has an HPD solution, we have by Theorem 2.7 that the matrix (LQ-1/2N1Q-1/2N2Q-1/2) is column orthonormal. According to the CS decomposition theorem (Theorem  3.8 in [16]), there exist unitary matrices P=(P100P2)∈C3n×3n (in which P1∈Cn×n,P2∈C2n×2n), V∈Cn×n, such that
(2.8)(P100P2)(LQ-1/2N1Q-1/2N2Q-1/2)V*=(CS0),
where C=diag(cosθ1,…,cosθn),S=diag(sinθ1,…,sinθn), and 0≤θ1≤⋯≤θn≤π/2. Thus the diagonal matrices C,S≥0 and C2+S2=I. Furthermore, noting that L is nonsingular, by (2.8), we have
(2.9)C=P1LQ-1/2V*>0,(2.10)P2(N1N2)Q-1/2V*=(S0).
Equation (2.10) is equivalent to (N1N2)=P2*(S0)VQ1/2. Let P2* be partitioned as P2*=(U1U3U2U4), in which Ui∈Cn×n, i=1,2,3,4, then we have(2.11)(N1N2)=(U1U3U2U4)(S0)VQ1/2=(U1SVQ1/2U2SVQ1/2),
from which it follows that N1=U1SVQ1/2, N2=U2SVQ1/2. By (2.9), we have L=P1*CVQ1/2. Then by (2.2), we have
(2.12)A=(L*L)t1/2sN1=(Q1/2V*C2VQ1/2)t1/2sU1SVQ1/2,B=(L*L)t2/2sN2=(Q1/2V*C2VQ1/2)t2/2sU2SVQ1/2.
Conversely, assume thatA,B have the decomposition (2.7). Let X=(Q1/2V*C2VQ1/2)1/s, which is an HPD matrix. Then it is easy to verify that X is an HPD solution of (1.1).Theorem 2.9.
If (1.1) has an HPD solution X, then X∈(M¯,N¯),  where
(2.13)M¯=12((μ1ν1)(1-t1)/t1(AQ-1A*)1/t1+(μ2ν2)(1-t2)/t2(BQ-1B*)1/t2),N¯=(Q-A*Q-t1/sA-B*Q-t2/sB)1/s,
in which μ1 and ν1 are the minimal and maximal eigenvalues of AQ-1A* respectively, μ2 and ν2 are the minimal and maximal eigenvalues of BQ-1B*, respectively.Proof.
LetX be an HPD solution of (1.1), then it follows from 0<Xs<Q and Lemma 2.1 that X-ti>Q-ti/s, i=1,2. Hence
(2.14)Xs=Q-A*X-t1A-B*X-t2B<Q-A*Q-t1/sA-B*Q-t2/sB.
Thus we have
(2.15)X<(Q-A*Q-t1/sA-B*Q-t2/sB)1/s=N¯.
On the other hand, from A*X-t1A<Q, it follows that
(2.16)Q-1/2A*X-t1/2X-t1/2AQ-1/2<I,X-t1/2AQ-1A*X-t1/2<I,AQ-1A*<Xt1.
Let μ1 and ν1 be the minimal and maximal eigenvalues of AQ-1A*, respectively. Since 1/t1≥1, and μ1I≤AQ-1A*≤ν1I, by Lemma 2.2, we get (μ1/ν1)(1-t1)/t1(AQ-1A*)1/t1<X.
Similarly, we have(μ2/ν2)(1-t2)/t2(BQ-1B*)1/t2<X, in which μ2 and ν2 are the minimal and maximal eigenvalues of BQ-1B*, respectively.
Hence we haveX>1/2((μ1/ν1)(1-t1)/t1(AQ-1A*)1/t1+(μ2/ν2)(1-t2)/t2(BQ-1B*)1/t2)=M¯.Theorem 2.10.
IfA*X-t1A+B*X-t2B≤Q-M¯s for all X∈[M¯,Q1/s], and
(2.17)p=1s(t1λn-(s+t1)(M¯)‖A‖F2+t2λn-(s+t2)(M¯)‖B‖F2)<1,
where M¯ is defined by (2.13),  then (1.1) has a unique HPD solution.Proof.
By the definition ofM¯, we have M¯>0. Hence λn(M¯)>0.
We consider the mapF(X)=(Q-A*X-t1A-B*X-t2B)1/s and let X∈Ω={X∣M¯≤X≤Q1/s}.Obviously, Ω is a convex, closed, and bounded set and F(X) is continuous on Ω.
By the hypothesis of the theorem, we have(2.18)Q1/s≥(Q-A*X-t1A-B*X-t2B)1/s≥(Q-Q+M¯s)1/s=M¯,
that is, M¯≤F(X)≤Q1/s. Hence F(Ω)⊆Ω.
For arbitraryX,Y∈Ω, we have
(2.19)A*X-t1A+B*X-t2B≤Q-M¯s,A*Y-t1A+B*Y-t2B≤Q-M¯s.
Hence
(2.20)F(X)=(Q-A*X-t1A-B*X-t2B)1/s≥(Q-Q+M¯s)1/s=M¯≥λn(M¯)I,F(Y)=(Q-A*Y-t1A-B*Y-t2B)1/s≥(Q-Q+M¯s)1/s=M¯≥λn(M¯)I.
From (2.20), it follows that
(2.21)‖F(X)s-F(Y)s‖F=‖∑i=0s-1F(X)i(F(X)-F(Y))F(Y)s-1-i‖F=‖vec⁡[∑i=0s-1F(X)i(F(X)-F(Y))F(Y)s-1-i]‖=‖∑i=0s-1vec⁡[F(X)i(F(X)-F(Y))F(Y)s-1-i]‖=‖∑i=0s-1(F(Y)s-1-i⊗F(X)i)vec(F(X)-F(Y))‖≥∑i=0s-1λns-1(M¯)‖vec⁡(F(X)-F(Y))‖=sλns-1(M¯)‖F(X)-F(Y)‖F.
According to the definition of the mapF, we have
(2.22)F(X)s-F(Y)s=(Q-A*X-t1A-B*X-t2B)-(Q-A*Y-t1A-B*Y-t2B)=A*(Y-t1-X-t1)A+B*(Y-t2-X-t2)B.
Combining (2.21) and (2.22), we have by Lemma 2.5 that
(2.23)‖F(X)-F(Y)‖F≤1sλns-1(M¯)‖F(X)s-F(Y)s‖F=1sλns-1(M¯)‖A*(Y-t1-X-t1)A+B*(Y-t2-X-t2)B‖F≤1sλns-1(M¯)(‖A‖F2‖Y-t1-X-t1‖F+‖B‖F2‖Y-t2-X-t2‖F)≤1sλns-1(M¯)(t1λn-(t1+1)(M¯)‖A‖F2+t2λn-(t2+1)(M¯)‖B‖F2)‖Y-X‖F=1s(t1λn-(s+t1)(M¯)‖A‖F2+t2λn-(s+t2)(M¯)‖B‖F2)‖X-Y‖F=p‖X-Y‖F.
Sincep<1, we know that the map F(X) is a contraction map in Ω. By Banach fixed point theorem, the map F(X) has a unique fixed point in Ω and this shows that (1.1) has a unique HPD solution in [M¯,Q1/s].Theorem 2.11.
If (1.1) has an HPD solution X, then
(2.24)λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)≤(ts+t)t/sss+t,X≤α̂Q1/s,
where t=min{t1,t2}, and α̂ is a solution of the equation
(2.25)yt(1-ys)=λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)
in [(t/(s+t))1/s,1].Proof.
Consider the sequence defined as follows:(2.26)α0=1,αk+1=(1-λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)αkt)1/s,k=0,1,2,….
Let X be an HPD solution of (1.1), then
(2.27)X=(Q-A*X-t1A-B*X-t2B)1/s<Q1/s=α0Q1/s.
Assuming that X<αkQ1/s, then by Lemma 2.1, we have
(2.28)Xs=Q-A*X-t1A-B*X-t2B<Q-A*(αkQ1/s)-t1A-B*(αkQ1/s)-t2B<Q-A*Q-t1/sA+B*Q-t2/sBαkt=Q1/2(I-Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2αkt)Q1/2≤Q1/2(1-λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)αkt)Q1/2=αk+1sQ.
Therefore X<αk+1Q1/s. Then by the principle of induction, we get X<αkQ1/s,k=0,1,2,….
Noting that the sequenceαk is monotonically decreasing and positive, hence αk is convergent. Let limk→∞αk=α̂, then α̂=(1-λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)/α̂t)1/s, that is, α̂ is a solution of the equation yt(1-ys)=λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2).
Consider the functionf(y)=yt(1-ys), since
(2.29)max⁡y∈[0,1]=f((ts+t)1/s)=(ts+t)t/sss+t,
from which it follows that λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)≤(t/(s+t))t/s(s/(s+t)).
Next we will prove thatα̂∈[(t/(s+t))1/s,1]. Obviously, α̂≤1. On the other hand, for the sequence αk, since α0=1>(t/(s+t))1/s, we may assume that αk>(t/(s+t))1/s without loss of generality. Then
(2.30)αk+1=(1-λn(Q-1/2A*Q-t1/sAQ-1/2+Q-1/2B*Q-t2/sBQ-1/2)αkt)1/s≥(1-1αkt(ts+t)t/sss+t)1/s>(1-1(t/(s+t))t/s(ts+t)t/sss+t)1/s=(ts+t)1/s.
Hence αk>(t/(s+t))1/s, k=0,1,2,…. So α̂=limk→∞αk≥(t/(s+t))1/s.
Consequently, we haveα̂∈[(t/(s+t))1/s,1].
This completes the proof.Theorem 2.12.
If (1.1) has an HPD solution, then
(2.31)(ρ(A))2≤ss+t1(t1s+t1)t1/s(ρ(Q))(t1/s)+1,(2.32)(ρ(B))2≤ss+t2(t2s+t2)t2/s(ρ(Q))(t2/s)+1.Proof.
For any eigenvalueλ(A) of A, let x be a corresponding eigenvector. Multiplying left side of (1.1) by x* and right side by x, we have
(2.33)x*Xsx+x*A*X-t1Ax+x*B*X-t2Bx=x*Qx,
which yields
(2.34)x*Xsx+|λ(A)|2x*X-t1x+x*B*X-t2Bx=x*Qx.
Since X>0, there exists an unitary matrix U such that X=U*ΛU, where Λ=diag(η1,…,ηn)>0. Then (2.34) turns into the following form:
(2.35)x*U*ΛsUx+|λ(A)|2x*U*Λ-t1Ux≤x*Qx.
Let y=(y1,y2,…,yn)T=Ux, then (2.35) reduces to
(2.36)y*Λsy+|λ(A)|2y*Λ-t1y≤y*UQU*y,
from which we obtain
(2.37)|λ(A)|2≤y*(UQU*-Λs)yy*Λ-t1y≤y*(λ1(Q)I-Λs)yy*Λ-t1y=∑i=1nyi2(λ1(Q)-ηis)∑i=1nyi2ηi-t1.
Form Lemma 2.3, we know that
(2.38)ηit1(λ1(Q)-ηis)≤ss+t1(t1s+t1)t1/sλ1(t1/s)+1(Q),
that is,
(2.39)(λ1(Q)-ηis)≤ss+t1(t1s+t1)t1/sλ1(t1/s)+1(Q)ηi-t1.
Noting that y≠0, we get
(2.40)∑i=1nyi2(λ1(Q)-ηis)≤ss+t1(t1s+t1)t1/sλ1(t1/s)+1(Q)∑i=1nyi2ηi-t1.
Consequently,
(2.41)|λ(A)|2≤∑i=1nyi2(λ1(Q)-ηis)∑i=1nyi2ηi-t1≤ss+t1(t1s+t1)t1/sλ1(t1/s)+1(Q).
Then(ρ(A))2≤(s/(s+t1))(t1/(s+t1))t1/sλ1(t1/s)+1(Q).
SinceQ>0, clearly denote λ1(Q)=ρ(Q), and the last inequality implies directly (2.31).
The proof of (2.32) is similar to that of (2.31), thus it is omitted here.Theorem 2.13.
IfQ≤I and (1.1) has an HPD solution, then
(2.42)ρ(As/t1+(A*)s/t1)≤ρ(Q),ρ(As/t1-(A*)s/t1)≤ρ(Q),(2.43)ρ(Bs/t2+(B*)s/t2)≤ρ(Q),ρ(Bs/t2-(B*)s/t2)≤ρ(Q).Proof.
If (1.1) has an HPD solution, we have by Theorem 2.7 that
(2.44)A=(L*L)t1/2sN1,B=(L*L)t2/2sN2,
and the matrix (LQ-1/2N1Q-1/2N2Q-1/2) is column orthonormal. From which we have
(2.45)L*L+N1*N1+N2*N2=Q.
Hence,
(2.46)Q-(As/t1+(A*)s/t1)=L*L+N1*N1+N2*N2-(L*L)1/2N1s/t1-(N1*)s/t1(L*L)1/2=((L*L)1/2-N1s/t1)*((L*L)1/2-N1s/t1)+N2*N2+(N1*N1-(N1*)s/t1N1s/t1)≥0.
Similarly, we haveQ+(As/t1+(A*)s/t1)≥0.
Thus,-Q≤(As/t1+(A*)s/t1)≤Q. Hence ρ(As/t1+(A*)s/t1)≤ρ(Q).
On the other hand, by Lemma2.6 and (2.2), we get
(2.47)ρ(As/t1-(A*)s/t1)=ρ((L*L)1/2N1s/t1-(N1*)s/t1(L*L)1/2)≤ρ(L*L+(N1*)s/t1N1s/t1)≤ρ(L*L+N1*N1)≤ρ(Q).
The proof of (2.43) is similar to that of (2.42).Ift1=t2, we denote t=t1=t2. Then (1.1) turns into(2.48)Xs+A*X-tA+B*X-tB=Q.
Consider the following equations:(2.49)xs+t-λn(Q)xt+λ1(A*A)+λ1(B*B)=0,(2.50)xs+t-λ1(Q)xt+λn(A*A)+λn(B*B)=0.We assume thatA,B, and Q satisfy(2.51)λ1(A*A)+λ1(B*B)<ss+tξ*tλn(Q),
where  ξ*=((t/(s+t))λn(Q))1/s. By (2.51) and Lemma 2.3, we know that (2.49) has two positive real roots α2<β1. We also get that (2.50) has two positive real roots α1<β2. It is easy to prove that (2.52)0<α1≤α2<ξ*<β1≤β2<λ11/s(Q).We define matrix sets as follows:(2.53)φ1={X=X*∣0<X<α1I},φ2={X=X*∣α1I≤X≤α2I},φ3={X=X*∣α2I<X<β1I},φ4={X=X*∣β1I≤X≤β2I},φ5={X=X*∣β2I<X<λ11/s(Q)I}.Theorem 2.14.
Suppose thatA,B, and Q satisfy (2.51), that is,
(2.54)λ1(A*A)+λ1(B*B)<ss+t(ts+t)t/sλn(t/s)+1(Q).
Then(i)
Equation (2.48) has a unique HPD solution in φ4;(ii)
Equation (2.48) has no HPD solution in φ1,φ3,φ5.Proof.
Consider the mapG(X)=(Q-A*X-tA-B*X-tB)1/s, which is continuous on φ4. Obviously, φ4 is a convex, closed, and bounded set. If X∈φ4,
(2.55)λ1s(G(X))=λ1(G(X)s)=λ1(Q-A*X-tA-B*X-tB)≤λ1(Q)-λn(A*A)+λn(B*B)λ1t(X)≤λ1(Q)-λn(A*A)+λn(B*B)β2t=β2s.
Hence, we have λ1(G(X))<β2. One has
(2.56)λns(G(X))=λn(G(X)s)=λn(Q-A*X-tA-B*X-tB)≥λn(Q)-λ1(A*A)+λ1(B*B)λnt(X)≥λn(Q)-λ1(A*A)+λ1(B*B)β1t=β1s.
Hence, we have λn(G(X))<β1.
Thus,G(X) maps φ4 into itself.
For arbitraryX,Y∈φ4, similar to (2.21) and (2.22), we have
(2.57)‖G(X)s-G(Y)s‖F≥sβ1s-1‖G(X)-G(Y)‖F,G(X)s-G(Y)s=A*(Y-t-X-t)A+B*(Y-t-X-t)B.
Combining (2.57), we have by Lemma 2.5 and (2.49)
(2.58)‖G(X)-G(Y)‖F≤1sβ1s-1‖G(X)s-G(Y)s‖F=1sβ1s-1‖A*(Y-t-X-t)A+B*(Y-t-X-t)B‖F≤1sβ1s-1(‖A‖22+‖B‖22)‖Y-t-X-t‖F≤1sβ1s-1(λ1(A*A)+λ1(B*B))tβ1-(t+1)‖Y-X‖F=tsλ1(A*A)+λ1(B*B)β1s+t‖X-Y‖F=ts(λn(Q)β1s-1)‖X-Y‖F<‖X-Y‖F.
Thus, we know that the mapG(X) is a contraction map in φ4. By Banach fixed point theorem, the map G(X) has a unique fixed point in φ4 and this shows that (2.48) has a unique HPD solution in φ4.
AssumeX is the HPD solution of (2.48), then
(2.59)λ1s(X)=λ1(Xs)=λ1(Q-A*X-tA-B*X-tB)≤λ1(Q)-λn(A*A)+λn(B*B)λ1t(X),
that is, λ1s+t(X)-λ1(Q)λ1t(X)+λn(A*A)+λn(B*B)≤0. So, α1≤λ1(X)≤β2, thus (2.48) has no HPD solution in φ1,φ5. 							(2.60)λns(X)=λn(Xs)=λn(Q-A*X-tA-B*X-tB)≥λn(Q)-λ1(A*A)+λ1(B*B)λnt(X),
that is, λns+t(X)-λn(Q)λnt(X)+λ1(A*A)+λ1(B*B)≥0. So, λn(X)≤α2 or λn(X)≥β1, thus (2.48) has no HPD solution in φ3.
This completes the proof.


## 3. Iterative Method for the Maximal HPD Solution

In this section, we consider the iterative method for obtaining the maximal HPD solutionXL of (1.1). We propose the following algorithm which avoids calculating matrix inversion in the process of iteration.Algorithm 1.
Step 1.
Input initial matrices:(3.1)X0=γQ1/s,Y0=γ+12γQ-1/s,
where γ∈(α̂,1), and α̂ is defined in Theorem 2.11.Step 2.
Fork=0,1,2,…,  compute
(3.2)Yk+1=Yk(2I-XkYk),Xk+1=(Q-A*Yk+1t1A-B*Yk+1t2B)1/s.Theorem 3.1.
If (1.1) has an HPD solution, then it has the maximal one XL. Moreover, to the sequences Xk and Yk generated by Algorithm 1, one has
(3.3)X0>X1>X2>…,lim⁡k→∞Xk=XL;Y0<Y1<Y2<⋯,lim⁡k→∞Yk=XL-1.Proof.
SinceXL is an HPD solution of (1.1), by Theorem 2.11, we have XL≤α̂Q1/s, thus
(3.4)X0=γQ1/s>α̂Q1/s≥XL,Y0=γ+12γQ-1/s<1γQ-1/s<1α̂Q-1/s≤XL-1.
By Lemmas 2.1 and 2.4, we have
(3.5)Y1=Y0(2I-X0Y0)=2Y0-Y0X0Y0≤X0-1<XL-1,Y1-Y0=Y0-Y0X0Y0=Y0(Y0-1-X0)Y0=1-γ24γQ-1/s>0.
According to Lemma 2.1 and Y1<XL-1, we have
(3.6)X1=(Q-A*Y1t1A-B*Y1t2B)1/s>(Q-A*XL-t1A-B*XL-t2B)1/s=XL,X1s-X0s=-A*(Y1t1-Y0t1)A-B*(Y1t2-Y0t2)B<0,
that is, X1s<X0s, by Lemma 2.1 again, it follows that X1<X0.
HenceX0>X1>XL, and Y0<Y1<XL-1.
Assume thatXk-1>Xk>XL, and Yk-1<Yk<XL-1, we will prove the inequalities Xk>Xk+1>XL, and Yk<Yk+1<XL-1.
By Lemmas2.1 and 2.4, we have
(3.7)Yk+1=2Yk-YkXkYk≤Xk-1<XL-1,Xk+1=(Q-A*Yk+1t1A-B*Yk+1t2B)1/s>(Q-A*XL-t1A-B*XL-t2B)1/s=XL.
Since Yk≤Xk-1-1<Xk-1, we have Yk-1>Xk, thus we have by Lemma 2.1 that
(3.8)Yk+1-Yk=Yk(Yk-1-Xk)Yk>0,Xk+1s-Xks=-A*(Yk+1t1-Ykt1)A-B*(Yk+1t2-Ykt2)B<0,
that is, Xk+1s<Xks, by Lemma 2.1 again, it follows that Xk+1<Xk.
Hence we have by induction that(3.9)X0>X1>X2>⋯>Xk>XL,Y0<Y1<Y2<⋯<Yk<XL-1
are true for all k=0,1,2,…, and so limk→∞Xk and limk→∞Yk exist. Suppose limk→∞Xk=X̂, limk→∞Yk=Ŷ, taking the limit in the Algorithm 1 leads to Ŷ=X̂-1 and X̂=(Q-A*X̂-t1A-B*X̂-t2B)1/s. Therefore X̂ is an HPD solution of (1.1), thus X̂≤XL. Moreover, as each Xk>XL, so X̂≥XL, then X̂=XL. The theorem is proved.Theorem 3.2.
If (1.1) has an HPD solution and after k iterative steps of Algorithm 1, one has ∥I-XkYk∥<ε, then
(3.10)‖Xks+A*Xk-t1A+B*Xk-t2B-Q‖≤ελn-1(M¯)(t1λ1(1-t1)/s(Q)‖A‖2+t2λ1(1-t2)/s(Q)‖B‖2),
where M¯ is defined by (2.13).Proof.
From the proof of Theorem3.1, we have Q-1/s<((γ+1)/2γ)Q-1/s<Yk<Xk-1<XL-1 for all k=1,2,…. Thus we have by Theorem 2.9 that Q-1/s<Yk<Xk-1<M¯-1. And this implies
(3.11)λ1-1/s(Q)I<Yk<Xk-1<λn-1(M¯)I.
Since
(3.12)Xks+A*Xk-t1A+B*Xk-t2B-Q=(Q-A*Ykt1A-B*Ykt2B)+A*Xk-t1A+B*Xk-t2B-Q=A*(Xk-t1-Ykt1)A+B*(Xk-t2-Ykt2)B,
we have by Lemma 2.5 that
(3.13)‖Xks+A*Xk-t1A+B*Xk-t1B-Q‖=‖A*(Xk-t1-Ykt1)A+B*(Xk-t2-Ykt2)B‖≤‖A‖2‖Xk-t1-Ykt1‖+‖B‖2‖Xk-t2-Ykt2‖≤(t1λ1-(t1-1)/s(Q)‖A‖2+t2λ1-(t2-1)/s(Q)‖B‖2)‖Xk-1-Yk‖≤(t1λ1(1-t1)/s(Q)‖A‖2+t2λ1(1-t2)/s(Q)‖B‖2)‖Xk-1‖‖I-XkYk‖≤ελn-1(M¯)(t1λ1(1-t1)/s(Q)‖A‖2+t2λ1(1-t2)/s(Q)‖B‖2).


## 4. Numerical Example

In this section, we give a numerical example to illustrate the efficiency of the proposed algorithm. All the tests are performed by MATLAB 7.0 with machine precision around10-16. We stop the practical iteration when the residual ∥Xks+A*Xk-t1A+B*Xk-t2B-Q∥F≤1.0e-010.Example 4.1.
Lets=5, t1=0.2, t2=0.5, and
(4.1)A=(200100120010003010100201101030010012),B=(216057347130092478853001250217400149),Q=(105665815417366154675088121586710915716115501528375741887137113136731216157136250).
By calculating, α̂≈0.8397136, so we choose γ=0.84. By using Algorithm 1 and iterating 29 steps, we obtain the maximal HPD solution XL of (1.1) as follows:
(4.2)XL≈X29=(1.66570.11100.23910.01050.05660.26240.11101.83370.22700.28700.26560.24830.23910.22701.7238-0.00370.30580.06620.01050.2870-0.00371.15010.12270.17390.05660.26560.30580.12271.51400.48310.26240.24830.06620.17390.48312.1751),
with the residual ∥X295+A*X29-0.2A+B*X29-0.5B-Q∥F=9.9360e-011.



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*Source: 163585-2011-10-05.xml*