On the Hermitian Positive Definite Solutions of Nonlinear Matrix Equation 𝑋^𝑠+𝐴^∗𝑋^−𝑡𝟏𝐴+𝐵^∗𝑋^−𝑡𝟐𝐵=𝑄

Abstract

Nonlinear matrix equation 𝑋^𝑠+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 𝑋^𝑠+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<𝑡_𝑖≤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.

1. Introduction

We consider the nonlinear matrix equation𝑋^𝑠+𝐴^∗𝑋^−𝑡1𝐴+𝐵^∗𝑋^−𝑡2𝐵=𝑄,(1.1) 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<𝑡_𝑖≤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 of𝑋^𝑠+𝐴^∗𝑋^−𝑡𝐴=𝑄.(1.2) 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 symbol C^𝑚×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 𝜆_𝑛(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 𝑋_𝑆 and 𝑋_𝐿 to denote the minimal and maximal HPD solution of (1.1), that is, for any HPD solution 𝑋 of (1.1), then 𝑋_𝑆≤𝑋≤𝑋_𝐿. 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,…,𝑑_𝑛)=(𝑑_𝑖𝑗) and 𝐸, 𝐷⊗𝐸=(𝑑_𝑖𝑗𝐸) is a Kronecker product and vec(𝐷) is a vector defined by vec(𝐷)=(𝑑^𝑇1,𝑑^𝑇2,…,𝑑^𝑇𝑛)^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]). If 𝐷≥𝐸>0(or 𝐷>𝐸>0), then 𝐷^𝑝≥𝐸^𝑝>0(or 𝐷^𝑝>𝐸^𝑝>0) for all 𝑝∈(0,1], and 𝐸^𝑝≥𝐷^𝑝>0(or 𝐸^𝑝>𝐷^𝑝>0) for all 𝑝∈[−1,0).

Lemma 2.2 (see [12]). Let 𝐷 and 𝐸 be positive operators on a Hilbert space such that 0<𝑚₁𝐼≤𝐷≤𝑀₁𝐼, 0<𝑚₂𝐼≤𝐸≤𝑀₂𝐼, and 0<𝐷≤𝐸. Then 𝐷^𝑞≤𝑀¹𝑚₁^𝑞−1𝐸^𝑞,𝐷^𝑞≤𝑀²𝑚₂^𝑞−1𝐸^𝑞(2.1) hold for any 𝑞≥1.

Lemma 2.3 (see [13]). Let 𝑓(𝑥)=𝑥^𝑡(𝜁−𝑥^𝑠), 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)𝑓_max=𝑓(((𝑡/(𝑠+𝑡))𝜁)^1/𝑠)=(𝑠/(𝑠+𝑡))(𝑡/(𝑠+𝑡))^𝑡/𝑠𝜁^{(𝑡/𝑠)+1}.

Lemma 2.4 (see [14]). If 𝐷 and 𝐸 are Hermitian matrices of the same order with 𝐸>0, then 𝐷𝐸𝐷+𝐸⁻¹≥2𝐷.

Lemma 2.5 (see [15]). If 0<𝜃≤1, and 𝐷 and 𝐸 are positive definite matrices of the same order with 𝐷,𝐸≥𝑏𝐼>0, then ‖𝐷^𝜃−𝐸^𝜃‖≤𝜃𝑏^𝜃−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]). Let 𝐷 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 𝐴=𝐿^∗𝐿^𝑡1/2𝑠𝑁₁,𝐵=𝐿^∗𝐿^𝑡2/2𝑠𝑁₂,(2.2) where 𝐿 is a nonsingular matrix and ^{𝐿𝑄−1/2𝑁1𝑄−1/2}_𝑁2Q−1/2 is column orthonormal.

Proof. If (1.1) has an HPD solution, then 𝑋^𝑠>0. Let 𝑋^𝑠=𝐿^∗𝐿 be the Cholesky factorization, where 𝐿 is a nonsingular matrix. Then (1.1) can be rewritten as 𝑄^−1/2𝐿^∗𝐿𝑄^−1/2+𝑄^−1/2𝐴^∗𝐿^∗𝐿^{−𝑡1/2𝑠}𝐿^∗𝐿^{−𝑡1/2𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝐿^∗𝐿^{−𝑡2/2𝑠}𝐿^∗𝐿^{−𝑡2/2𝑠}𝐵𝑄^−1/2=𝐼.(2.3) Let 𝑁₁=(𝐿^∗𝐿)^{−𝑡1/2𝑠}𝐴,𝑁₂=(𝐿^∗𝐿)^{−𝑡2/2𝑠}𝐵, then 𝐴=(𝐿^∗𝐿)^𝑡1/2𝑠𝑁₁, 𝐵=(𝐿^∗𝐿)^𝑡2/2𝑠𝑁₂. Moreover, (2.3) turns into 𝑄^−1/2𝐿^∗𝐿𝑄^−1/2+𝑄^−1/2𝑁^∗1𝑁₁𝑄^−1/2+𝑄^−1/2𝑁^∗2𝑁₂𝑄^−1/2=𝐼,(2.4) that is, ⎛⎜⎜⎜⎜⎝𝐿𝑄^−1/2𝑁₁𝑄^−1/2𝑁₂𝑄^−1/2⎞⎟⎟⎟⎟⎠^∗⎛⎜⎜⎜⎜⎝𝐿𝑄^−1/2𝑁₁𝑄^−1/2𝑁₂𝑄^−1/2⎞⎟⎟⎟⎟⎠=𝐼,(2.5) which means that ^{𝐿𝑄−1/2𝑁1𝑄−1/2}_{𝑁2𝑄−1/2} is column orthonormal.
Conversely, if 𝐴,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 𝑋^𝑠+𝐴^∗𝑋^−𝑡1𝐴+𝐵^∗𝑋^−𝑡2𝐵=𝐿^∗𝐿+𝑁^∗1𝑁₁+𝑁^∗2𝑁₂=𝑄^1/2𝑄^−1/2𝐿^∗𝐿𝑄^−1/2+𝑄^−1/2𝑁^∗1𝑁₁𝑄^−1/2+𝑄^−1/2𝑁^∗2𝑁₂𝑄^−1/2𝑄^1/2=𝑄.(2.6) 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∈C^𝑛×𝑛, a column-orthonormal matrix 𝑈=^𝑈1_𝑈2∈C^2𝑛×𝑛(in which 𝑈₁,𝑈₂∈C^𝑛×𝑛), and diagonal matrices 𝐶>0 and 𝑆≥0 with 𝐶²+𝑆²=𝐼 such that 𝐴𝐵==𝑄𝑄^1/2_1/2𝑉^∗𝐶²𝑉𝑄^1/2^𝑡1/2𝑠𝑈₁𝑆𝑉𝑄^1/2,𝑉^∗𝐶²𝑉𝑄^1/2^𝑡2/2𝑠𝑈₂𝑆𝑉𝑄^1/2.(2.7)

Proof. If (1.1) has an HPD solution, we have by Theorem 2.7 that the matrix ^{𝐿𝑄−1/2𝑁1𝑄−1/2}_{𝑁2𝑄−1/2} is column orthonormal. According to the CS decomposition theorem (Theorem  3.8 in [16]), there exist unitary matrices 𝑃=^𝑃100𝑃₂∈C^3𝑛×3𝑛 (in which 𝑃₁∈C^𝑛×𝑛,𝑃₂∈C^2𝑛×2𝑛), 𝑉∈C^𝑛×𝑛, such that ⎛⎜⎜⎝𝑃0₁𝑃0₂⎞⎟⎟⎠⎛⎜⎜⎜⎝𝐿𝑄^−1/2𝑁₁𝑄^−1/2𝑁₂𝑄^−1/2⎞⎟⎟⎟⎠𝑉_∗=⎛⎜⎜⎝𝐶𝑆0⎞⎟⎟⎠,(2.8) where 𝐶=diag(cos𝜃₁,…,cos𝜃_𝑛),𝑆=diag(sin𝜃₁,…,sin𝜃_𝑛), and 0≤𝜃₁≤⋯≤𝜃_𝑛≤𝜋/2. Thus the diagonal matrices 𝐶,𝑆≥0 and 𝐶²+𝑆²=𝐼. Furthermore, noting that 𝐿 is nonsingular, by (2.8), we have 𝐶=𝑃₁𝐿𝑄^−1/2𝑉^∗>0,(2.9)𝑃₂⎛⎜⎜⎝𝑁₁𝑁₂⎞⎟⎟⎠𝑄_−1/2𝑉^∗=⎛⎜⎜⎝𝑆0⎞⎟⎟⎠.(2.10)
Equation (2.10) is equivalent to ^𝑁1𝑁₂=𝑃^∗2^𝑆0𝑉𝑄^1/2. Let 𝑃^∗2 be partitioned as 𝑃^∗2=^𝑈1𝑈3_𝑈2𝑈4, in which 𝑈_𝑖∈C^𝑛×𝑛, 𝑖=1,2,3,4, then we have⎛⎜⎜⎝𝑁₁𝑁₂⎞⎟⎟⎠=⎛⎜⎜⎝𝑈₁𝑈₃𝑈₂𝑈₄⎞⎟⎟⎠⎛⎜⎜⎝𝑆0⎞⎟⎟⎠𝑉𝑄_1/2=⎛⎜⎜⎝𝑈¹𝑆𝑉𝑄^1/2𝑈₂𝑆𝑉𝑄^1/2⎞⎟⎟⎠,(2.11) from which it follows that 𝑁₁=𝑈₁𝑆𝑉𝑄^1/2, 𝑁₂=𝑈₂𝑆𝑉𝑄^1/2. By (2.9), we have 𝐿=𝑃^∗1𝐶𝑉𝑄^1/2. Then by (2.2), we have 𝐴𝐵==𝐿𝐿^∗_∗𝐿^𝑡1/2𝑠𝑁₁=𝑄^1/2𝑉^∗𝐶²𝑉𝑄^1/2^𝑡1/2𝑠𝑈₁𝑆𝑉𝑄^1/2,𝐿^𝑡2/2𝑠𝑁₂=𝑄^1/2𝑉^∗𝐶²𝑉𝑄^1/2^𝑡2/2𝑠𝑈₂𝑆𝑉𝑄^1/2.(2.12)
Conversely, assume that 𝐴,𝐵 have the decomposition (2.7). Let 𝑋=(𝑄^1/2𝑉^∗𝐶²𝑉𝑄^1/2)^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 𝑀=12𝜇𝜈¹₁^{(1−𝑡1)/𝑡1}𝐴𝑄⁻¹𝐴^∗^1/𝑡1+𝜇²𝜈₂^{(1−𝑡2)/𝑡2}𝐵𝑄⁻¹𝐵^∗^1/𝑡2,𝑁=𝑄−𝐴^∗𝑄^{−𝑡1/𝑠}𝐴−𝐵^∗𝑄^{−𝑡2/𝑠}𝐵^1/𝑠,(2.13) 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. Let 𝑋 be an HPD solution of (1.1), then it follows from 0<𝑋^𝑠<𝑄 and Lemma 2.1 that 𝑋^−𝑡𝑖>𝑄^{−𝑡𝑖/𝑠}, 𝑖=1,2. Hence 𝑋^𝑠=𝑄−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵<𝑄−𝐴^∗𝑄^{−𝑡1/𝑠}𝐴−𝐵^∗𝑄^{−𝑡2/𝑠}𝐵.(2.14) Thus we have 𝑋<𝑄−𝐴^∗𝑄^{−𝑡1/𝑠}𝐴−𝐵^∗𝑄^{−𝑡2/𝑠}𝐵^1/𝑠=𝑁.(2.15) On the other hand, from 𝐴^∗𝑋^−𝑡1𝐴<𝑄, it follows that 𝑄𝑋^−1/2𝐴^∗𝑋^−𝑡1/2𝑋^−𝑡1/2𝐴𝑄^−1/2<𝐼,_−𝑡1/2𝐴𝐴𝑄𝑄⁻¹₋₁𝐴^∗𝑋^−𝑡1/2<𝐼,𝐴^∗<𝑋^𝑡1.(2.16) 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 have 𝑋>1/2((𝜇₁/𝜈₁)^{(1−𝑡1)/𝑡1}(𝐴𝑄⁻¹𝐴^∗)^1/𝑡1+(𝜇₂/𝜈₂)^{(1−𝑡2)/𝑡2}(𝐵𝑄⁻¹𝐵^∗)^1/𝑡2)=𝑀.

Theorem 2.10. If 𝐴^∗𝑋^−𝑡1𝐴+𝐵^∗𝑋^−𝑡2𝐵≤𝑄−𝑀^𝑠 for all 𝑋∈[𝑀,𝑄^1/𝑠], and 𝑝=1𝑠𝑡¹𝜆^{−(𝑠+𝑡1)𝑛}𝑀‖𝐴‖^2𝐹+𝑡₂𝜆^{−(𝑠+𝑡2)𝑛}𝑀‖𝐵‖^2𝐹<1,(2.17) where 𝑀 is defined by (2.13), then (1.1) has a unique HPD solution.

Proof. By the definition of 𝑀, we have 𝑀>0. Hence 𝜆_𝑛(𝑀)>0.
We consider the map 𝐹(𝑋)=(𝑄−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵)^1/𝑠 and let 𝑋∈Ω={𝑋∣𝑀≤𝑋≤𝑄^1/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 𝑄^1/𝑠≥𝑄−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵^1/𝑠≥𝑄−𝑄+𝑀^𝑠^1/𝑠=𝑀,(2.18) that is, 𝑀≤𝐹(𝑋)≤𝑄^1/𝑠. Hence 𝐹(Ω)⊆Ω.
For arbitrary 𝑋,𝑌∈Ω, we have 𝐴^∗𝑋^−𝑡1𝐴+𝐵^∗𝑋^−𝑡2𝐵≤𝑄−𝑀^𝑠,𝐴^∗𝑌^−𝑡1𝐴+𝐵^∗𝑌^−𝑡2𝐵≤𝑄−𝑀^𝑠.(2.19) Hence 𝐹𝐹((𝑋𝑌))==𝑄𝑄−−𝐴𝐴^∗_∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵^1/𝑠≥𝑄−𝑄+𝑀^𝑠^1/𝑠=𝑀≥𝜆_𝑛𝑀𝐼,𝑌^−𝑡1𝐴−𝐵^∗𝑌^−𝑡2𝐵^1/𝑠≥𝑄−𝑄+𝑀^𝑠^1/𝑠=𝑀≥𝜆_𝑛𝑀𝐼.(2.20) From (2.20), it follows that ‖𝐹(𝑋)^𝑠−𝐹(𝑌)^𝑠‖_𝐹=‖‖‖‖^s−1_𝑖=0𝐹(𝑋)^𝑖(𝐹(𝑋)−𝐹(𝑌))𝐹(𝑌)^{𝑠−1−𝑖}‖‖‖‖_𝐹=‖‖‖‖vec^𝑠−1_𝑖=0𝐹(𝑋)^𝑖(𝐹(𝑋)−𝐹(𝑌))𝐹(𝑌)^{𝑠−1−𝑖}=‖‖‖‖_𝑠−1‖‖‖‖_𝑖=0=‖‖‖‖_𝑠−1vec𝐹(𝑋)^𝑖(𝐹(𝑋)−𝐹(𝑌))𝐹(𝑌)^{𝑠−1−𝑖}‖‖‖‖_𝑖=0𝐹(𝑌)^{𝑠−1−𝑖}≥⊗𝐹(𝑋)^𝑖v𝑒𝑐(𝐹(𝑋)−𝐹(𝑌))_𝑠−1‖‖‖‖_𝑖=0=𝑠𝜆𝜆^𝑛𝑠−1_𝑛𝑠−1𝑀‖vec(𝐹(𝑋)−𝐹(𝑌))‖𝑀‖𝐹(𝑋)−𝐹(𝑌)‖_𝐹.(2.21)
According to the definition of the map 𝐹, we have 𝐹(𝑋)^𝑠−𝐹(𝑌)^𝑠==𝐴𝑄_∗−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵−𝑄−𝐴^∗𝑌^−𝑡1𝐴−𝐵^∗𝑌^−𝑡2𝐵𝑌^−𝑡1−𝑋^−𝑡1𝐴+𝐵^∗𝑌^−𝑡2−𝑋^−𝑡2𝐵.(2.22) Combining (2.21) and (2.22), we have by Lemma 2.5 that ‖𝐹(𝑋)−𝐹(𝑌)‖_𝐹≤1𝑠𝜆^𝑛𝑠−1𝑀‖𝐹(𝑋)^𝑠−𝐹(𝑌)^𝑠‖_𝐹=1𝑠𝜆^𝑛𝑠−1𝑀‖‖𝐴^∗𝑌^−𝑡1−𝑋^−𝑡1𝐴+𝐵^∗𝑌^−𝑡2−𝑋^−𝑡2𝐵‖‖_𝐹≤1𝑠𝜆^𝑛𝑠−1𝑀‖𝐴‖^2𝐹≤1‖‖𝑌^−𝑡1−𝑋^−𝑡1‖‖_𝐹+‖𝐵‖^2𝐹‖‖𝑌^−𝑡2−𝑋^−𝑡2‖‖_𝐹𝑠𝜆^𝑛𝑠−1𝑀𝑡¹𝜆^{−(𝑡1𝑛+1)}𝑀‖𝐴‖^2𝐹+𝑡₂𝜆^{−(𝑡2𝑛+1)}𝑀‖𝐵‖^2𝐹‖𝑌−𝑋‖^𝐹=1𝑠𝑡¹𝜆^{−(𝑠+𝑡1)𝑛}𝑀‖𝐴‖^2𝐹+𝑡₂𝜆^{−(𝑠+𝑡2)𝑛}𝑀‖𝐵‖^2𝐹‖𝑋−𝑌‖^𝐹=𝑝‖𝑋−𝑌‖_𝐹.(2.23)
Since 𝑝<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,𝑄^1/𝑠].

Theorem 2.11. If (1.1) has an HPD solution 𝑋, then 𝜆_𝑛𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2≤𝑡𝑠+𝑡^t/𝑠𝑠𝑠+𝑡,𝑋≤𝛼𝑄^1/𝑠,(2.24) where 𝑡=min{𝑡₁,𝑡₂}, and 𝛼 is a solution of the equation 𝑦^𝑡(1−𝑦^𝑠)=𝜆_𝑛𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2(2.25) in [(𝑡/(𝑠+𝑡))^1/𝑠,1].

Proof. Consider the sequence defined as follows: 𝛼₀=1,𝛼_𝑘+1=1−𝜆^𝑛𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2𝛼^𝑡𝑘^1/𝑠𝑘=0,1,2,….,(2.26) Let 𝑋 be an HPD solution of (1.1), then 𝑋=𝑄−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵^1/𝑠<𝑄^1/𝑠=𝛼₀𝑄^1/𝑠.(2.27) Assuming that 𝑋<𝛼_𝑘𝑄^1/𝑠, then by Lemma 2.1, we have 𝑋^𝑠<=𝑄𝑄−−𝐴𝐴^∗_∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵<𝑄−𝐴_∗𝛼_𝑘𝑄^1/𝑠^−𝑡1𝐴−𝐵^∗𝛼_𝑘𝑄^1/𝑠^−𝑡2𝐵𝑄^{−𝑡1/𝑠}𝐴+𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝛼^𝑡𝑘=𝑄^1/2𝐼−𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2𝛼^𝑡𝑘𝑄^1/2≤𝑄^1/21−𝜆^𝑛𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2𝛼^𝑡𝑘𝑄^1/2=𝛼^𝑠𝑘+1𝑄.(2.28) Therefore 𝑋<𝛼_𝑘+1𝑄^1/𝑠. Then by the principle of induction, we get 𝑋<𝛼_𝑘𝑄^1/𝑠,𝑘=0,1,2,….
Noting that the sequence 𝛼_𝑘 is monotonically decreasing and positive, hence 𝛼_𝑘 is convergent. Let lim_𝑘→∞𝛼_𝑘=𝛼, then 𝛼=(1−𝜆_𝑛(𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2)/𝛼^𝑡)^1/𝑠, that is, 𝛼 is a solution of the equation 𝑦^𝑡(1−𝑦^𝑠)=𝜆_𝑛(𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2).
Consider the function 𝑓(𝑦)=𝑦^𝑡(1−𝑦^𝑠), since max_𝑦∈[0,1]=𝑓𝑡𝑠+𝑡^1/𝑠=𝑠+𝑡𝑡^𝑡/𝑠𝑠𝑠+𝑡,(2.29) from which it follows that 𝜆_𝑛(𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2)≤(𝑡/(𝑠+𝑡))^𝑡/𝑠(𝑠/(𝑠+𝑡)).
Next we will prove that 𝛼∈[(𝑡/(𝑠+𝑡))^1/𝑠,1]. Obviously, 𝛼≤1. On the other hand, for the sequence 𝛼_𝑘, since 𝛼₀=1>(𝑡/(𝑠+𝑡))^1/𝑠, we may assume that 𝛼_𝑘>(𝑡/(𝑠+𝑡))^1/𝑠 without loss of generality. Then 𝛼_𝑘+1=1−𝜆^𝑛𝑄^−1/2𝐴^∗𝑄^{−𝑡1/𝑠}𝐴𝑄^−1/2+𝑄^−1/2𝐵^∗𝑄^{−𝑡2/𝑠}𝐵𝑄^−1/2𝛼^𝑡𝑘^1/𝑠≥1−1𝛼^𝑡𝑘𝑠+𝑡𝑡^𝑡/𝑠𝑠𝑠+𝑡^1/𝑠>1−1(𝑡/(𝑠+𝑡))^𝑡/𝑠𝑠+𝑡𝑡^𝑡/𝑠𝑠𝑠+𝑡^1/𝑠=𝑡𝑠+𝑡^1/𝑠.(2.30) Hence 𝛼_𝑘>(𝑡/(𝑠+𝑡))^1/𝑠, 𝑘=0,1,2,…. So 𝛼=lim_𝑘→∞𝛼_𝑘≥(𝑡/(𝑠+𝑡))^1/𝑠.
Consequently, we have 𝛼∈[(𝑡/(𝑠+𝑡))^1/𝑠,1].
This completes the proof.

Theorem 2.12. If (1.1) has an HPD solution, then (𝜌(𝐴))²≤𝑠𝑠+𝑡₁𝑠+𝑡¹𝑡₁^𝑡1/𝑠(𝜌(𝑄))^{(𝑡1/𝑠)+1},(2.31)(𝜌(𝐵))²≤𝑠𝑠+𝑡₂𝑠+𝑡²𝑡₂^𝑡2/𝑠(𝜌(𝑄))^{(𝑡2/𝑠)+1}.(2.32)

Proof. For any eigenvalue 𝜆(𝐴) of 𝐴, let 𝑥 be a corresponding eigenvector. Multiplying left side of (1.1) by 𝑥^∗ and right side by 𝑥, we have 𝑥^∗𝑋^𝑠𝑥+𝑥^∗𝐴^∗𝑋^−𝑡1𝐴𝑥+𝑥^∗𝐵^∗𝑋^−𝑡2𝐵𝑥=𝑥^∗𝑄𝑥,(2.33) which yields 𝑥^∗𝑋^𝑠𝑥+||𝜆(𝐴)||²𝑥^∗𝑋^−𝑡1𝑥+𝑥^∗𝐵^∗𝑋^−𝑡2𝐵𝑥=𝑥^∗𝑄𝑥.(2.34) Since 𝑋>0, there exists an unitary matrix 𝑈 such that 𝑋=𝑈^∗Λ𝑈, where Λ=diag(𝜂₁,…,𝜂_𝑛)>0. Then (2.34) turns into the following form: 𝑥^∗𝑈^∗Λ^𝑠𝑈𝑥+||𝜆(𝐴)||²𝑥^∗U^∗Λ^−𝑡1𝑈𝑥≤𝑥^∗𝑄𝑥.(2.35) Let 𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑛)^𝑇=𝑈𝑥, then (2.35) reduces to 𝑦^∗Λ^𝑠𝑦+||𝜆(𝐴)||²𝑦^∗Λ^−𝑡1𝑦≤𝑦^∗𝑈𝑄𝑈^∗𝑦,(2.36) from which we obtain ||𝜆(𝐴)||²≤𝑦^∗𝑈𝑄𝑈^∗−Λ^𝑠𝑦𝑦^∗Λ^−𝑡1𝑦≤𝑦^∗𝜆₁(𝑄)𝐼−Λ^𝑠𝑦𝑦^∗Λ^−𝑡1𝑦=∑^𝑛𝑖=1𝑦^2𝑖𝜆₁(𝑄)−𝜂^𝑠𝑖∑^𝑛𝑖=1𝑦^2𝑖𝜂^−𝑡1_𝑖.(2.37) Form Lemma 2.3, we know that 𝜂^𝑡1_𝑖𝜆₁(𝑄)−𝜂^𝑠𝑖≤𝑠𝑠+𝑡₁𝑠+𝑡¹𝑡₁^𝑡1/𝑠𝜆^{(𝑡11/𝑠)+1}(𝑄),(2.38) that is, 𝜆₁(𝑄)−𝜂^𝑠𝑖≤𝑠+𝑠𝑡₁𝑠+𝑡¹𝑡₁^𝑡1/𝑠𝜆^{(𝑡11/𝑠)+1}(𝑄)𝜂^−𝑡1_𝑖.(2.39) Noting that 𝑦≠0, we get ^𝑛_𝑖=1𝑦^2𝑖𝜆₁(𝑄)−𝜂^𝑠𝑖≤𝑠𝑠+𝑡₁𝑠+𝑡¹𝑡₁^𝑡1/𝑠𝜆^{(𝑡11/𝑠)+1}(𝑄)^𝑛_𝑖=1𝑦^2𝑖𝜂^−𝑡1_𝑖.(2.40) Consequently, ||𝜆(𝐴)||²≤∑^𝑛𝑖=1𝑦^2𝑖𝜆₁(𝑄)−𝜂^𝑠𝑖∑^𝑛𝑖=1𝑦^2𝑖𝜂^−𝑡1_𝑖≤𝑠𝑠+𝑡₁𝑠+𝑡¹𝑡₁^𝑡1/𝑠𝜆^{(𝑡11/𝑠)+1}(𝑄).(2.41)
Then (𝜌(𝐴))²≤(𝑠/(𝑠+𝑡₁))(𝑡₁/(𝑠+𝑡₁))^𝑡1/𝑠𝜆^{(𝑡11/𝑠)+1}(𝑄).
Since 𝑄>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. If 𝑄≤𝐼 and (1.1) has an HPD solution, then 𝜌𝐴^𝑠/𝑡1+𝐴^∗^𝑠/𝑡1≤𝜌(𝑄),𝜌𝐴^𝑠/𝑡1−𝐴^∗^𝑠/𝑡1≤𝜌(𝑄),(2.42)𝜌𝐵^𝑠/𝑡2+𝐵^∗^𝑠/𝑡2≤𝜌(𝑄),𝜌𝐵^𝑠/𝑡2−𝐵^∗^𝑠/𝑡2≤𝜌(𝑄).(2.43)

Proof. If (1.1) has an HPD solution, we have by Theorem 2.7 that 𝐴=𝐿^∗𝐿^𝑡1/2𝑠𝑁₁,𝐵=𝐿^∗𝐿^𝑡2/2𝑠𝑁₂,(2.44) and the matrix ^{𝐿𝑄−1/2𝑁1𝑄−1/2}_{𝑁2𝑄−1/2} is column orthonormal. From which we have 𝐿^∗𝐿+𝑁^∗1𝑁₁+𝑁^∗2𝑁₂=𝑄.(2.45) Hence, 𝑄−𝐴^𝑠/𝑡1+𝐴^∗^𝑠/𝑡1=𝐿^∗𝐿+𝑁^∗1𝑁₁+𝑁^∗2𝑁₂−𝐿^∗𝐿^1/2𝑁^𝑠/𝑡1₁−𝑁^∗1^𝑠/𝑡1𝐿^∗𝐿^1/2=+𝑁𝐿^∗_∗2𝐿^1/2−𝑁^𝑠/𝑡1₁^∗𝐿^∗𝐿^1/2−𝑁^𝑠/𝑡1₁≥0.𝑁₂+𝑁^∗1𝑁₁−𝑁^∗1^𝑠/𝑡1𝑁^𝑠/𝑡1₁(2.46)
Similarly, we have 𝑄+(𝐴^𝑠/𝑡1+(𝐴^∗)^𝑠/𝑡1)≥0.
Thus, −𝑄≤(𝐴^𝑠/𝑡1+(𝐴^∗)^𝑠/𝑡1)≤𝑄. Hence 𝜌(𝐴^𝑠/𝑡1+(𝐴^∗)^𝑠/𝑡1)≤𝜌(𝑄).
On the other hand, by Lemma 2.6 and (2.2), we get 𝜌𝐴^𝑠/𝑡1−𝐴^∗^𝑠/𝑡1≤=𝜌𝜌𝐿𝐿^∗_∗𝐿^1/2𝑁^𝑠/𝑡1₁−𝑁^∗1^𝑠/𝑡1𝐿^∗𝐿^1/2≤𝜌𝐿_∗𝐿+𝑁^∗1^𝑠/𝑡1𝑁^𝑠/𝑡1₁≤𝜌(𝑄)𝐿.+𝑁^∗1𝑁₁(2.47) The proof of (2.43) is similar to that of (2.42).

If 𝑡₁=𝑡₂, we denote 𝑡=𝑡₁=𝑡₂. Then (1.1) turns into𝑋^𝑠+𝐴^∗𝑋^−𝑡𝐴+𝐵^∗𝑋^−𝑡𝐵=𝑄.(2.48) Consider the following equations:𝑥^𝑠+𝑡−𝜆_𝑛(𝑄)𝑥^𝑡+𝜆₁𝐴^∗𝐴+𝜆₁𝐵^∗𝐵=0,(2.49)𝑥^𝑠+𝑡−𝜆₁(𝑄)𝑥^𝑡+𝜆_𝑛𝐴^∗𝐴+𝜆_𝑛𝐵^∗𝐵=0.(2.50)

We assume that 𝐴,𝐵, and 𝑄 satisfy𝜆₁𝐴^∗𝐴+𝜆₁𝐵^∗𝐵<𝑠𝑠+𝑡𝜉^𝑡∗𝜆_𝑛(𝑄),(2.51) where 𝜉_∗=((𝑡/(𝑠+𝑡))𝜆_𝑛(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 0<𝛼₁≤𝛼₂<𝜉_∗<𝛽₁≤𝛽₂<𝜆^11/𝑠(𝑄).(2.52)

We define matrix sets as follows: 𝜑𝜑₁₂=𝑋=𝑋^∗∣0<𝑋<𝛼₁𝐼,𝜑₃=𝑋=𝑋^∗∣𝛼₁𝐼≤𝑋≤𝛼₂𝐼,𝜑₄=𝑋=𝑋^∗∣𝛼₂𝐼<𝑋<𝛽₁𝐼,𝜑₅=𝑋=𝑋^∗∣𝛽₁𝐼≤𝑋≤𝛽₂𝐼,=𝑋=𝑋^∗∣𝛽₂𝐼<𝑋<𝜆^11/𝑠(𝑄)𝐼.(2.53)

Theorem 2.14. Suppose that 𝐴,𝐵, and 𝑄 satisfy (2.51), that is, 𝜆₁𝐴^∗𝐴+𝜆₁𝐵^∗𝐵<𝑠𝑠+𝑡𝑠+𝑡𝑡^𝑡/𝑠𝜆^{𝑛(𝑡/𝑠)+1}(𝑄).(2.54) 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 map 𝐺(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, 𝜆^𝑠1(𝐺(𝑋))=𝜆₁(𝐺(𝑋)^𝑠)=≤𝜆𝜆₁₁𝑄−𝐴^∗𝑋^−𝑡𝐴−𝐵^∗𝑋^−𝑡𝐵(𝑄)−𝜆^𝑛𝐴^∗𝐴+𝜆_𝑛𝐵^∗𝐵≤𝜆₁𝜆^𝑡1(𝑋)(𝑄)−𝜆^𝑛𝐴^∗𝐴+𝜆_𝑛𝐵^∗𝐵𝛽^𝑡2=𝛽^𝑠2.(2.55) Hence, we have 𝜆₁(𝐺(𝑋))<𝛽₂. One has 𝜆^𝑠𝑛(𝐺(𝑋))=𝜆_𝑛(𝐺(𝑋)^𝑠)=≥𝜆𝜆_𝑛𝑛𝑄−𝐴^∗𝑋^−𝑡𝐴−𝐵^∗𝑋^−𝑡𝐵(𝑄)−𝜆¹𝐴^∗𝐴+𝜆₁𝐵^∗𝐵≥𝜆_𝑛𝜆^𝑡𝑛(𝑋)(𝑄)−𝜆¹𝐴^∗𝐴+𝜆₁𝐵^∗𝐵𝛽^𝑡1=𝛽^𝑠1.(2.56) Hence, we have 𝜆_𝑛(𝐺(𝑋))<𝛽₁.
Thus, 𝐺(𝑋) maps 𝜑₄ into itself.
For arbitrary 𝑋,𝑌∈𝜑₄, similar to (2.21) and (2.22), we have 𝐺(𝑋)‖_𝑠𝐺(𝑋)^𝑠−𝐺(𝑌)^𝑠‖_𝐹≥𝑠𝛽^1𝑠−1‖𝐺(𝑋)−𝐺(𝑌)‖_𝐹,−𝐺(𝑌)^𝑠=𝐴^∗𝑌^−𝑡−𝑋^−𝑡𝐴+𝐵^∗𝑌^−𝑡−𝑋^−𝑡𝐵.(2.57) Combining (2.57), we have by Lemma 2.5 and (2.49) ‖𝐺(𝑋)−𝐺(𝑌)‖_𝐹≤1𝑠𝛽^1𝑠−1‖𝐺(𝑋)^𝑠−𝐺(𝑌)^𝑠‖_𝐹=1𝑠𝛽^1𝑠−1‖‖𝐴^∗𝑌^−𝑡−𝑋^−𝑡𝐴+𝐵^∗𝑌^−𝑡−𝑋^−𝑡𝐵‖‖_𝐹≤1𝑠𝛽^1𝑠−1‖𝐴‖²²+‖𝐵‖²²‖‖𝑌^−𝑡−𝑋^−𝑡‖‖_𝐹≤1𝑠𝛽^1𝑠−1𝜆₁𝐴^∗𝐴+𝜆₁𝐵^∗𝐵𝑡𝛽^1−(𝑡+1)‖𝑌−𝑋‖_𝐹=𝑡𝑠𝜆¹𝐴^∗𝐴+𝜆₁𝐵^∗𝐵𝛽^1𝑠+𝑡‖𝑋−𝑌‖_𝐹=𝑡𝑠𝜆^𝑛(𝑄)𝛽^𝑠1−1‖𝑋−𝑌‖_𝐹<‖𝑋−𝑌‖_𝐹.(2.58)
Thus, we know that the map 𝐺(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.
Assume 𝑋 is the HPD solution of (2.48), then 𝜆^𝑠1(𝑋)=𝜆₁(𝑋^𝑠)=≤𝜆𝜆₁₁𝑄−𝐴^∗𝑋^−𝑡𝐴−𝐵^∗𝑋^−𝑡𝐵(𝑄)−𝜆^𝑛𝐴^∗𝐴+𝜆_𝑛𝐵^∗𝐵𝜆^𝑡1(𝑋),(2.59) that is, 𝜆^1𝑠+𝑡(𝑋)−𝜆₁(𝑄)𝜆^𝑡1(𝑋)+𝜆_𝑛(𝐴^∗𝐴)+𝜆_𝑛(𝐵^∗𝐵)≤0. So, 𝛼₁≤𝜆₁(𝑋)≤𝛽₂, thus (2.48) has no HPD solution in 𝜑₁,𝜑₅. 𝜆^𝑠𝑛(𝑋)=𝜆_𝑛(𝑋^𝑠)=≥𝜆𝜆_𝑛𝑛𝑄−𝐴^∗𝑋^−𝑡𝐴−𝐵^∗𝑋^−𝑡𝐵(𝑄)−𝜆¹𝐴^∗𝐴+𝜆₁𝐵^∗𝐵𝜆^𝑡𝑛(𝑋),(2.60) that is, 𝜆^{𝑛𝑠+𝑡}(𝑋)−𝜆_𝑛(𝑄)𝜆^𝑡𝑛(𝑋)+𝜆₁(𝐴^∗𝐴)+𝜆₁(𝐵^∗𝐵)≥0. So, 𝜆_𝑛(𝑋)≤𝛼₂ or 𝜆_𝑛(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 solution 𝑋_𝐿 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: 𝑋𝑌₀₀=𝛾𝑄^1/𝑠,=𝛾+12𝛾𝑄^−1/𝑠,(3.1) where 𝛾∈(𝛼,1), and 𝛼 is defined in Theorem 2.11.Step 2. For 𝑘=0,1,2,…, compute 𝑋𝑌_𝑘+1𝑘+1=𝑌_𝑘2𝐼−𝑋_𝑘𝑌_𝑘,=𝑄−𝐴^∗𝑌^𝑡1_𝑘+1𝐴−𝐵^∗𝑌^𝑡2_𝑘+1𝐵^1/𝑠.(3.2)

Theorem 3.1. If (1.1) has an HPD solution, then it has the maximal one 𝑋_𝐿. Moreover, to the sequences 𝑋_𝑘 and 𝑌_𝑘 generated by Algorithm 1, one has 𝑋₀>𝑋₁>𝑋₂>…,lim_𝑘→∞𝑋_𝑘=𝑋_𝐿;𝑌₀<𝑌₁<𝑌₂<⋯,lim_𝑘→∞𝑌_𝑘=𝑋^𝐿−1.(3.3)

Proof. Since 𝑋_𝐿 is an HPD solution of (1.1), by Theorem 2.11, we have 𝑋_𝐿≤𝛼𝑄^1/𝑠, thus 𝑋₀=𝛾𝑄^1/𝑠>𝛼𝑄^1/𝑠≥𝑋_𝐿,𝑌₀=𝛾+12𝛾𝑄^−1/𝑠<1𝛾𝑄^−1/𝑠<1𝛼𝑄^−1/𝑠≤𝑋^𝐿−1.(3.4) By Lemmas 2.1 and 2.4, we have 𝑌₁𝑌₁=𝑌₀2𝐼−𝑋₀𝑌₀=2𝑌₀−𝑌₀𝑋₀𝑌₀≤𝑋⁰⁻¹<𝑋^𝐿−1,−𝑌₀=𝑌₀−𝑌₀𝑋₀𝑌₀=𝑌₀𝑌⁰⁻¹−𝑋₀𝑌₀=1−𝛾²4𝛾𝑄^−1/𝑠>0.(3.5) According to Lemma 2.1 and 𝑌₁<𝑋^𝐿−1, we have 𝑋₁=𝑄−𝑋_𝑠1𝐴^∗𝑌^𝑡1₁𝐴−𝐵^∗𝑌^𝑡2₁𝐵^1/𝑠>𝑄−𝐴^∗𝑋^−𝑡1_𝐿𝐴−𝐵^∗𝑋^−𝑡2_𝐿𝐵^1/𝑠=𝑋_𝐿,−𝑋^𝑠0=−𝐴^∗𝑌^𝑡1₁−𝑌^𝑡1₀𝐴−𝐵^∗𝑌^𝑡2₁−𝑌^𝑡2₀𝐵<0,(3.6) that is, 𝑋^𝑠1<𝑋^𝑠0, by Lemma 2.1 again, it follows that 𝑋₁<𝑋₀.
Hence 𝑋₀>𝑋₁>𝑋_𝐿, and 𝑌₀<𝑌₁<𝑋^𝐿−1.
Assume that 𝑋_𝑘−1>𝑋_𝑘>𝑋_𝐿, and 𝑌_𝑘−1<𝑌_𝑘<𝑋^𝐿−1, we will prove the inequalities 𝑋_𝑘>𝑋_𝑘+1>𝑋_𝐿, and 𝑌_𝑘<𝑌_𝑘+1<𝑋^𝐿−1.
By Lemmas 2.1 and 2.4, we have 𝑋_𝑘+1𝑌_𝑘+1=2𝑌_𝑘−𝑌_𝑘𝑋_𝑘𝑌_𝑘≤𝑋^𝑘−1<𝑋^𝐿−1,=𝑄−𝐴^∗𝑌^𝑡1_𝑘+1𝐴−𝐵^∗𝑌^𝑡2_𝑘+1𝐵^1/𝑠>𝑄−𝐴^∗𝑋^−𝑡1_𝐿𝐴−𝐵^∗𝑋^−𝑡2_𝐿𝐵^1/𝑠=𝑋_𝐿.(3.7) Since 𝑌_𝑘≤𝑋^−1𝑘−1<𝑋^𝑘−1, we have 𝑌^𝑘−1>𝑋_𝑘, thus we have by Lemma 2.1 that 𝑋𝑌_𝑘+1−𝑌_𝑘=𝑌_𝑘𝑌^𝑘−1−𝑋_𝑘𝑌_𝑘>0,_𝑠𝑘+1−𝑋^𝑠𝑘=−𝐴^∗𝑌^𝑡1_𝑘+1−𝑌^𝑡1_𝑘𝐴−𝐵^∗𝑌^𝑡2_𝑘+1−𝑌^𝑡2_𝑘𝐵<0,(3.8) that is, 𝑋^𝑠𝑘+1<𝑋^𝑠𝑘, by Lemma 2.1 again, it follows that 𝑋_𝑘+1<𝑋_𝑘.
Hence we have by induction that 𝑋₀>𝑋₁>𝑋₂>⋯>𝑋_𝑘>𝑋_𝐿,𝑌₀<𝑌₁<𝑌₂<⋯<𝑌_𝑘<𝑋^𝐿−1(3.9) are true for all 𝑘=0,1,2,…, and so lim_𝑘→∞𝑋_𝑘 and lim_𝑘→∞𝑌_𝑘 exist. Suppose lim_𝑘→∞𝑋_𝑘=𝑋, lim_𝑘→∞𝑌_𝑘=𝑌, taking the limit in the Algorithm 1 leads to 𝑌=𝑋⁻¹ and 𝑋=(𝑄−𝐴^∗𝑋^−𝑡1𝐴−𝐵^∗𝑋^−𝑡2𝐵)^1/𝑠. Therefore 𝑋 is an HPD solution of (1.1), thus 𝑋≤𝑋_𝐿. Moreover, as each 𝑋_𝑘>𝑋_𝐿, so 𝑋≥𝑋_𝐿, then 𝑋=𝑋_𝐿. 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−𝑋_𝑘𝑌_𝑘‖<𝜀, then ‖‖𝑋^𝑠𝑘+𝐴^∗𝑋^−𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘𝐵−𝑄‖‖≤𝜀𝜆^𝑛−1𝑀𝑡₁𝜆^{(1−𝑡11)/𝑠}(𝑄)‖𝐴‖²+𝑡₂𝜆^{(1−𝑡21)/𝑠}(𝑄)‖𝐵‖²,(3.10) where 𝑀 is defined by (2.13).

Proof. From the proof of Theorem 3.1, we have 𝑄^−1/𝑠<((𝛾+1)/2𝛾)𝑄^−1/𝑠<𝑌_𝑘<𝑋^𝑘−1<𝑋^𝐿−1 for all 𝑘=1,2,…. Thus we have by Theorem 2.9 that 𝑄^−1/𝑠<𝑌_𝑘<𝑋^𝑘−1<𝑀⁻¹. And this implies 𝜆^1−1/𝑠(𝑄)𝐼<𝑌_𝑘<𝑋^𝑘−1<𝜆^𝑛−1𝑀𝐼.(3.11) Since 𝑋^𝑠𝑘+𝐴^∗𝑋^−𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘𝐵−𝑄==𝐴𝑄_∗−𝐴^∗𝑌^𝑡1_𝑘𝐴−𝐵^∗𝑌^𝑡2_𝑘𝐵+𝐴^∗𝑋^−𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘𝐵−𝑄𝑋^−𝑡1_𝑘−𝑌^𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘−𝑌^𝑡2_𝑘𝐵,(3.12) we have by Lemma 2.5 that ‖‖𝑋^𝑠𝑘+𝐴^∗𝑋^−𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡1_𝑘𝐵−𝑄‖‖≤=‖‖‖𝐴𝐴‖^∗₂𝑋^−𝑡1_𝑘−𝑌^𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘−𝑌^𝑡2_𝑘𝐵‖‖≤𝑡₁‖‖𝑋^−𝑡1_𝑘−𝑌^𝑡1_𝑘‖‖+‖𝐵‖²‖‖𝑋^−𝑡2_𝑘−𝑌^𝑡2_𝑘‖‖≤𝑡₁𝜆^{−(𝑡11−1)/𝑠}(𝑄)‖𝐴‖²+𝑡₂𝜆^{−(𝑡21−1)/𝑠}(𝑄)‖𝐵‖²‖‖𝑋^𝑘−1−𝑌_𝑘‖‖≤𝜀𝜆𝜆^{(1−𝑡11)/𝑠}_𝑛−1(𝑄)‖𝐴‖²+𝑡₂𝜆^{(1−𝑡21)/𝑠}(𝑄)‖𝐵‖²‖‖𝑋^𝑘−1‖‖‖‖𝐼−𝑋_𝑘𝑌_𝑘‖‖𝑀𝑡₁𝜆^{(1−𝑡11)/𝑠}(𝑄)‖𝐴‖²+𝑡₂𝜆^{(1−𝑡21)/𝑠}(𝑄)‖𝐵‖².(3.13)

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 around 10⁻¹⁶. We stop the practical iteration when the residual ‖𝑋^s_𝑘+𝐴^∗𝑋^−𝑡1_𝑘𝐴+𝐵^∗𝑋^−𝑡2_𝑘𝐵−𝑄‖_𝐹≤1.0𝑒−010.

Example 4.1. Let 𝑠=5, 𝑡₁=0.2, 𝑡₂=0.5, and 𝐴=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝210110𝑄020001=003010⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝110020065147068513501103111000102658652⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠708641,1617650751189𝐵512351058775=11⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝487313230824181736149550112765235672300317160⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠014021.537014708179⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,(4.1) By calculating, 𝛼≈0.8397136, so we choose 𝛾=0.84. By using Algorithm 1 and iterating 29 steps, we obtain the maximal HPD solution 𝑋_𝐿 of (1.1) as follows: 𝑋_𝐿≈𝑋₂₉=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝100000......612002613156519062701564010000......182222132864137758070063−001000.....22730.0322060973563108827−001000.....02111.0185270070233501797000010......023154560218655243668701000002......220141646787286335432911⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,(4.2) with the residual ‖𝑋⁵²⁹+𝐴^∗𝑋^−0.229𝐴+𝐵^∗𝑋^−0.529𝐵−𝑄‖_𝐹=9.9360𝑒−011.

Acknowledgments

The authors express their thanks to the editor Professor Mohammad I. Younis and the anonymous referees who made much useful and detailed suggestions that helped them to correct some minor errors and improve the quality of the paper. This project was granted financial support from Natural Science Foundation of China (11071079), Natural Science Foundation of China (11001167), Natural Science Foundation of China (10901056), Natural Science Foundation of Zhejiang Province (Y6110043), and Shanghai Natural Science Foundation (09ZR1408700).