Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 676978, 6 pages http://dx.doi.org/10.1155/2013/676978 Research Article On the Positive Definite Solutions of a Nonlinear Matrix Equation Panpan Liu,1 Shugong Zhang,1 and Qingchun Li2 1 School of Mathematics, Key Laboratory of Symbolic Computation and Knowledge Engineering (Ministry of Education), Jilin University, Changchun 130012, China 2 Department of Mathematics, Beihua University, Jilin 132013, China Correspondence should be addressed to Shugong Zhang; sgzh@mail.jlu.edu.cn Received 28 November 2012; Revised 19 April 2013; Accepted 21 April 2013 Academic Editor: J. Biazar Copyright © 2013 Panpan Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The positive definite solutions of the nonlinear matrix equation ππ + π΄∗ π(π)π΄ = π 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. 1. Introduction Denote the set of all π × π positive definite matrices by π(π). In this paper, we consider the matrix equation ππ + π΄∗ π (π) π΄ = π, (1) where π΄ is nonsingular, π is a Hermitian positive definite matrix, π is a positive real number, π is a continuous map from π(π) into π(π), and π is either monotone (meaning that 0 ≤ π ≤ π implies that π(π) ≤ π(π)) or antimonotone (meaning that 0 ≤ π ≤ π implies that π(π) ≥ π(π)). 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 π = 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 π(π) = π−π‘ is proposed by Yueting [10]. Liu and Gao [9] proved the existence of the symmetric positive definite solutions of (1) with π(π) = ±π−π‘ and π = πΌ. Many other authors investigated (1) for particular choices of the map π [2–4, 6–8]. In Section 2, 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 matrix π΄, π max (π΄) and π min (π΄) stand for the maximal and minimal eigenvalues of matrix π΄, respectively. π΄∗ is the conjugate transpose of the matrix π΄, and π΄−∗ is the inversion of π΄∗ . βπ΄β denotes the spectral norm of π΄. π΄ > 0 (π΄ ≥ 0) denotes that π΄ is a positive definite (semidefinite) matrix, and π΄ > π΅ (π΄ ≥ π΅) means that π΄ − π΅ > 0 (π΄ − π΅ ≥ 0). The notation πΏ π΄,π΅ denotes the line segment joining π΄ and π΅; that is, πΏ π΄,π΅ = {π‘π΄ + (1 − π‘) π΅ | π‘ ∈ [0, 1]} . (2) 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 that π΄ ≥ π΅ ≥ 0 (π΄ > π΅ ≥ 0); if π ∈ (0, 1], then π΄π ≥ π΅π (π΄π > π΅π ), and if π ∈ [−1, 0), then π΄π ≤ π΅π (π΄π < π΅π ). 2 Journal of Applied Mathematics If there is unique Hermitian positive definite matrix π, such that π(π) = π2 , π ∈ π(π), then we denote that π = π1/2 (π). Theorem 2. Assume that π is a continuous map from π(π) into π»π(π), where π»π(π) denotes the set of all π × π Hermitian positive definite matrices. Then, (1) has a Hermitian positive definite solution if and only if there is a nonsingular matrix π, such that π∗ π = ππ∗ , and π΄ = (π1/2 (π∗ π))−1 ππ1/2 , where −1/2 ∗ π ∗ π −1/2 ) (π ) (π ) π (π ∗ + π π = πΌ. (3) In this case, (1) has a Hermitian positive definite solution π = π∗ π. Proof. If π is a Hermitian positive definite solution of (1), then there is unique Hermitian positive definite matrix π, such that π = π2 (see [11]). So, π(π) = π(π∗ π) ∈ π»π(π), and therefore there is unique Hermitian positive definite matrix π, such that π(π∗ π) = π2 . Substituting π = π2 = π∗ π into (1) gives π (π∗ π) + π΄∗ π (π∗ π) π΄ = π. (4) ∗ (ππ ) (ππ ) + π΄∗ π∗ ππ΄ = π. (5) ∗ (π−1/2 ) (ππ ) (ππ ) π−1/2 + (π−1/2 ) π΄∗ π∗ ππ΄π−1/2 = πΌ; (6) that is ∗ ππ π−1/2 ππ π−1/2 ] = πΌ. −1/2 ] [ ππ΄π ππ΄π−1/2 [ (7) Let π = ππ΄π−1/2 . Then, π΄ = π−1 ππ1/2 = (π1/2 (π∗ π))−1 ππ1/2 , and by (7) we know that (π−1/2 )∗ (ππ )∗ (ππ )π−1/2 + π∗ π = πΌ. Conversely, if π΄ = (π1/2 (π∗ π))−1 ππ1/2 , let π = π∗ π. Then, ∗ π ππ + π΄∗ π (π) π΄ = (π∗ π) + (π1/2 ) π∗ (π1/2 (π∗ π)) −∗ ∗ = (ππ ) ππ + (π1/2 ) π∗ ππ1/2 = (π1/2 ) (9) In this section, we assume that π΄, π, π in (1) satisfy π−1 (π΄−∗ ππ΄−1 ) ≤ π1/π . Theorem 3. Suppose that π−1 exists and that π is antimonotone. Let π = 1. Equation (1) has a positive definite solution in the interval (0, π max (π)πΌ) if and only if there is a number π ∈ (0, π max (π)), such that ππ ≤ ππΌ for all π. Moreover, in this case, the iteration (9) with π0 = 0 converges to the smallest positive definite solution of (1). Proof. Since π−1 exists and π is anti-monotone, then π−1 is also anti-monotone. Assume that there is a number π as in the theorem. Since π0 = 0, one has (10) Furthermore, we get ≥ π−1 (π΄−∗ (π − π0 ) π΄−1 ) = π1 . (11) Now, if ππ ≥ ππ−1 , we have ππ+1 = π−1 (π΄−∗ (π − ππ ) π΄−1 ) ≥ π−1 (π΄−∗ (π − ππ−1 ) π΄−1 ) = ππ . (12) Then, the sequence {ππ } is a monotonically nondecreasing sequence and bounded above by some positive definite matrix ππΌ. Consequently, the sequence {ππ } converges to a positive definite matrix π, which is a solution of (1); that is, π = π−1 (π΄−∗ (π − π) π΄−1 ) . (13) π0 = 0 < π, ∗ π π = 0, 1, 2, . . . Conversely, let (1) have a positive definite solution π ∈ (0, π max (π)πΌ), and let π ∈ (0, π max (π)) be the largest eigenvalue of π. In order to prove that ππ ≤ ππΌ for every ππ generated from (9), we will prove that ππ ≤ π. Consider −1 × π (π∗ π) (π1/2 (π∗ π)) ππ1/2 ∗ ππ+1 = π−1 (π΄−∗ (π − πππ ) π΄−1 ) , π2 = π−1 (π΄−∗ (π − π1 ) π΄−1 ) π is Hermitian positive definite, and so ∗ In order to discuss an iterative method for solving (1), we assume that for a given matrix π΅, the equation π(π) = π΅ 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: π1 = π−1 (π΄−∗ ππ΄−1 ) ≥ π0 = 0. Then, we have ∗ 3. Iterative Method −1/2 ∗ × ((π π π −1/2 ) (π π ∗ 1/2 ) + π π) π = π. (8) So, π = π∗ π is a Hermitian positive definite solution of (1). If π = 1, then the restriction π∗ π = ππ∗ in Theorem 2 can be omitted. π1 = π−1 (π΄−∗ (π − π0 ) π΄−1 ) (14) ≤ π−1 (π΄−∗ (π − π) π΄−1 ) = π. Assume that ππ ≤ π, for some fixed π, and then we have ππ+1 = π−1 (π΄−∗ (π − ππ ) π΄−1 ) ≤ π−1 (π΄−∗ (π − π) π΄−1 ) = π. So, ππ+1 ∈ (0, π). Then, the theorem is proved. (15) Journal of Applied Mathematics 3 Lemma 4 (see [1]). Let π : π → π(π)(π ⊂ π(π) ππππ) be differentiable at any point of π. Then, From Lemma 4, σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π (π) − π (π)σ΅©σ΅©σ΅© ≤ sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© βπ − πβ , σ΅©σ΅© σ΅© σ΅©σ΅©π − πσ΅©σ΅©σ΅© = σ΅© σ΅© π∈πΏ π,π (16) π σ΅© −π−1 (π΄−∗ (π − π ) π΄−1 )σ΅©σ΅©σ΅©σ΅© for all π, π ∈ π. σ΅©σ΅© σ΅© ≤ sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−∗ (π − ππ ) π΄−1 π∈πΏ −1 Lemma 5. Suppose that π exists and that π is antimonotone. If (1) with π ≥ 1 has a Hermitian positive definite solution π, then π ∈ [π−1 (π΄−∗ ππ΄−1 ), π1/π ]. π1 ,π2 π σ΅© −π΄−∗ (π − π ) π΄−1 σ΅©σ΅©σ΅©σ΅© Proof. Since π−1 exists and π is anti-monotone, then π−1 is also anti-monotone. Assume that (1) has a Hermitian positive definite solution π. Since π maps into π(π), we have π(π) > 0 and π΄∗ π(π)π΄ ≥ 0. Therefore, ππ = π − π΄∗ π(π)π΄ ≤ π. By Lemma 1 we know that π ≤ π1/π . On the other hand, we have π΄∗ π(π)π΄ ≤ π, π(π) ≤ π΄−∗ ππ΄−1 . Then, we obtain π ≥ π−1 (π΄−∗ ππ΄−1 ) because π−1 is anti-monotone. Theorem 6. Suppose that π−1 exists and that π is antimonotone, and suppose that π, π−1 are differentiable at any point of Ω1 = [π−1 (π΄−∗ ππ΄−1 ), π1/π ] and Ω2 = [π(π1/π ), π΄−∗ ππ΄−1 ], respectively. Let σ΅© σ΅© π2 = sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© , π∈Ω σ΅© σ΅© π1 = sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© , π∈Ω1 σ΅©σ΅© −1 −∗ σ΅©σ΅©π (π΄ (π − ππ ) π΄−1 ) σ΅© 2 = sup π∈πΏ π(π),π(π) π σ΅©σ΅© σ΅©σ΅© σ΅© σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π΄−∗ (π − ππ ) π΄−1 σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© π σ΅©σ΅© σ΅© σ΅© ≤ sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−∗ (π − ππ ) π΄−1 σ΅©σ΅©σ΅©σ΅© π∈Ω 2 σ΅© σ΅©2 σ΅© π σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π − ππ σ΅©σ΅©σ΅©σ΅© . (20) Let σ΅© σ΅© π1 = sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© . (17) π∈Ω1 (21) 2 and π = π1 π2 βπ΄−1 β βπ΄β2 . Then, (i) If (1) with π > 1 has a Hermitian positive definite solution π and π < 1, then π is the unique solution of (1). (ii) Assume that there is a closed set Ω ⊆ Ω1 satisfying that π : Ω → Ω and π(π) = π−1 (π΄−∗ (π − ππ )π΄−1 ); if π < 1, then (1) with π > 1 has a unique solution π in Ω. Furthermore, one considers the iterative method (9) with π0 ∈ Ω. The sequence {ππ } in (9) converges to the unique solution π; moreover, ππ σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©σ΅©ππ − πσ΅©σ΅©σ΅© ≤ σ΅©π − π0 σ΅©σ΅©σ΅© , 1−πσ΅© 1 π σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©σ΅©ππ − πσ΅©σ΅©σ΅© ≤ σ΅©π − ππ−1 σ΅©σ΅©σ΅© . 1−πσ΅© π (18) σ΅©σ΅© π σ΅© σ΅©σ΅©π − ππ σ΅©σ΅©σ΅© = σ΅© σ΅© σ΅©σ΅© σ΅© σ΅©σ΅©π − π΄∗ π (π) π΄ − π + π΄∗ π (π) π΄σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© 2σ΅© ≤ βπ΄β σ΅©σ΅©σ΅©σ΅©π (π) − π (π)σ΅©σ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© ≤ βπ΄β2 sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅© π∈πΏ π,π (22) σ΅© σ΅©σ΅© σ΅© ≤ βπ΄β2 sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅© π∈Ω1 σ΅© σ΅© ≤ βπ΄β2 π1 σ΅©σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅© . By π < 1 we have Proof. (i) Assume that π and π are two different Hermitian positive definite solutions of (1), and by Lemma 5, π, π ∈ [π−1 (π΄−∗ ππ΄−1 ), π1/π ] = Ω1 ; then, π(π), π(π) ∈ [π(π1/π ), π΄−∗ ππ΄−1 ] = Ω2 . Let π1 = π΄−∗ (π − ππ )π΄−1 , π2 = π π΄−∗ (π − π )π΄−1 , and σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅©σ΅©π − πσ΅©σ΅©σ΅© ≤ π2 σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅© βπ΄β2 π1 σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© = π σ΅©σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅© < σ΅©σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅© , σ΅© σ΅© π2 = sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© . π∈Ω which is a contradiction; so, π = π. That is, π is the unique solution of (1). 2 (19) (23) 4 Journal of Applied Mathematics (ii) Let π1 , π2 ∈ Ω ⊆ Ω1 . Then, π(π(π1 )), π(π(π2 )) ∈ Ω2 ; from Lemma 4, σ΅© σ΅©σ΅© σ΅©σ΅©π (π1 ) − π (π2 )σ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©σ΅©π−1 (π΄−∗ (π − π1π ) π΄−1 ) −1 −∗ −π (π΄ ≤ sup π∈πΏ π(π(π1 )),π(π(π2 )) (π − Μ be the positive definite solutions of (1) Theorem 8. Let π, π and its perturbation equation (26), respectively. Map π−1 is differentiable at any point of Ω4 with Ω4 = {πΌπ + (1 − πΌ) π | πΌ ∈ [0, 1] , π ∈ π ([π−1 (π΄−∗ ππ΄−1 ) , π1/π ]) , σ΅© π2π ) π΄−1 )σ΅©σ΅©σ΅©σ΅© Μπ΄ Μ−1 ) , π Μ−∗ π Μ1/π ])} . π ∈ π ([π−1 (π΄ σ΅©σ΅© σ΅©σ΅© σ΅© σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π΄−∗ (π2π − π1π ) π΄−1 σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© Let σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© ≤ sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π2π − π1π σ΅©σ΅©σ΅© π∈Ω σ΅© σ΅© π4 = sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© . π∈Ω 2 4 σ΅© σ΅©2 σ΅© σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π − π΄∗ π (π2 ) π΄ − π + π΄∗ π (π1 ) π΄σ΅©σ΅©σ΅© σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©π − π σ΅© σ΅© βπβ σ΅© σ΅©2 σ΅©σ΅© σ΅© σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© βπ΄β2 sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π1 − π2 σ΅©σ΅©σ΅© π∈πΏ π1 ,π2 ≤ σ΅© σ΅©2 σ΅©σ΅© σ΅© σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© βπ΄β2 sup σ΅©σ΅©σ΅©Dπ (π)σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π1 − π2 σ΅©σ΅©σ΅© π π min (π−1 (π΄−∗ ππ΄−1 )) (π4−1 π∈Ω1 σ΅© σ΅©2 σ΅© σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© βπ΄β2 π1 σ΅©σ΅©σ΅©π1 − π2 σ΅©σ΅©σ΅© (24) The interval Ω is a complete metric space because it is a closed subset of π(π). And π < 1; so, π is a contraction on Ω. Then, it follows from contractive mapping principle that the map π has a unique fixed point π in Ω. Furthermore, the sequence {ππ } in (9) converges to the unique solution of (1); moreover, σ΅© σ΅© Μ−1 σ΅©σ΅©σ΅© Μ π −1 , σ΅© − σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅©σ΅© π π ) (30) Μ π πΌββπ΄−1 β + βπ Μ π πΌββπ΄ Μ−1 β(βπ − π Μ−π Μ−1 β) + where π = βπ΄−1 − π΄ −1 Μ−1 −1 −∗ −1 Μ Μ βπ΄ ββπ΄ ββπ−πβ, π = min{π min (π (π΄ ππ΄ )), π min (π−1 Μπ΄ Μ−1 ))}. Μ−∗ π (π΄ Μ are the positive definite solutions of (1) and (26), Proof. π, π Μ−∗ (π− Μ π Μπ )π΄ Μ−1 . respectively. Let π1 = π΄−∗ (π−ππ )π΄−1 , π2 = π΄ From Lemma 4, σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© = σ΅©σ΅©π − π σ΅© σ΅© π π σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© σ΅©σ΅©ππ − πσ΅©σ΅©σ΅© ≤ σ΅©π − π0 σ΅©σ΅©σ΅© , 1−πσ΅© 1 π σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©σ΅©ππ − πσ΅©σ΅©σ΅© ≤ σ΅©π − ππ−1 σ΅©σ΅©σ΅© . 1−πσ΅© π (29) Μ π −1 < 1, one has Μ−1 βπ π If π4 βπ΄−1 ββπ΄ σ΅© σ΅©2 σ΅© σ΅© ≤ π2 σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© βπ΄β2 σ΅©σ΅©σ΅©π (π1 ) − π (π2 )σ΅©σ΅©σ΅© σ΅© σ΅© = π σ΅©σ΅©σ΅©π1 − π2 σ΅©σ΅©σ΅© . (28) σ΅©σ΅© −1 −∗ σ΅©σ΅©π (π΄ (π − ππ ) π΄−1 ) σ΅© Μ−∗ (π Μ−π Μπ ) π΄ Μ−1 )σ΅©σ΅©σ΅©σ΅© −π−1 (π΄ σ΅© σ΅©σ΅© σ΅© ≤ sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−∗ (π − ππ ) π΄−1 π∈πΏ (25) π1 ,π2 Μ−∗ (π Μ−π Μπ ) π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© . −π΄ σ΅© (31) By Lemma 5, 4. Perturbation Analysis π ∈ [π−1 (π΄−∗ ππ΄−1 ) , π1/π ] , Let Μπ + π΄ Μ∗ π (π) Μ π΄ Μ= π Μ π Μ ∈ [π−1 (π΄ Μπ΄ Μ−1 ) , π Μ−∗ π Μ1/π ] . π (32) (26) Let Μ be nonsingular be the perturbation equation of (1). Let π΄, π΄ Μ be positive definite, and 0 < π < 1. Suppose matrices and π, π that π−1 : π(π) → π(π) exists and π is anti-monotone. Lemma 7 (see [12]). If 0 < π < 1, the operators π, π satisfy π ≥ ππΌ and π ≥ ππΌ for some positive number π; then, σ΅©σ΅© π πσ΅© π−1 σ΅©σ΅©π − π σ΅©σ΅©σ΅© ≤ ππ βπ − πβ . (27) Ω4 = {πΌπ + (1 − πΌ) π | πΌ ∈ [0, 1] , π ∈ π ([π−1 (π΄−∗ ππ΄−1 ) , π1/π ]) , Μπ΄ Μ−1 ) , π Μ−∗ π Μ1/π ])} , π ∈ π ([π−1 (π΄ σ΅© σ΅© π4 = sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© . π∈Ω 4 (33) Journal of Applied Mathematics Then, we have σ΅© σ΅© sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© = π∈πΏ π1 ,π2 5 sup π∈πΏ π(π),π(π) Μ σ΅© σ΅©σ΅© σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© ≤ sup σ΅©σ΅©σ΅©σ΅©Dπ−1 (π)σ΅©σ΅©σ΅©σ΅© ≤ π4 . π∈Ω (34) 4 ≤ Notice that π ≥ π−1 (π΄−∗ ππ΄−1 ) ≥ π min (π−1 (π΄−∗ ππ΄−1 )) πΌ, Μ ≥ π−1 (π΄ Μπ΄ Μ−1 ) ≥ π min (π−1 (π΄ Μπ΄ Μ−1 )) πΌ. Μ−∗ π Μ−∗ π π (35) Μ = min{π (π−1 (π΄−∗ ππ΄−1 )), π (π−1 (π΄ Μπ΄ Μ−1 ))}. Μ−∗ π Let π min min π π π Μ Μ Μ ≥ ππΌ. By Lemma 1, π , π Μ ≥ π πΌ since 0 < π < 1. Then, π, π And from Lemma 7, σ΅©σ΅©σ΅©π΄−∗ (π − ππ ) π΄−1 − π΄ Μ−∗ (π Μ−π Μπ ) π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅©σ΅© −∗ Μ−1 ) = σ΅©σ΅©σ΅©π΄ (π − ππ ) (π΄−1 − π΄ −∗ +π΄ Μ+ π )π΄ (π − π − π π Μπ Μ−1 π π min (π−1 (π΄−∗ ππ΄−1 )) (π4−1 σ΅© σ΅© Μ−1 σ΅©σ΅©σ΅© Μ π −1 , σ΅© − σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅©σ΅© π π ) (39) where σ΅© Μ π πΌσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π Μ Μ π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© (σ΅©σ΅©σ΅©σ΅©π − π π = σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅© σ΅© σ΅© σ΅© − π πΌσ΅©σ΅© σ΅©σ΅©π΄ σ΅©σ΅©) σ΅©σ΅© σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© . σ΅©σ΅© σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅©σ΅© (40) Acknowledgments The authors wish to thank Professor J. Biazar and the reviewers for providing valuable comments and suggestions which improve this paper. This work is supported by NSFC (Grant no. 11171133). Μ−∗ ) (π Μ−π Μπ ) π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© + (π΄−∗ − π΄ σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© ≤ σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π − ππ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅© σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© π σ΅© σ΅©σ΅© Μπ Μσ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π σ΅©σ΅© (σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅© − π σ΅©σ΅©) σ΅© σ΅© References σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π Μ Μπ σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ + σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅©σ΅© σ΅© σ΅© − π σ΅©σ΅© σ΅© Μ Μπ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© (σ΅©σ΅©σ΅©π − ππ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π = σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© − π σ΅©σ΅© σ΅©σ΅©π΄ σ΅©σ΅©) σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© π σ΅© σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π Μπ σ΅©σ΅© σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© − π σ΅©σ΅© σ΅© Μ π πΌσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π Μ Μ π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© (σ΅©σ΅©σ΅©σ΅©π − π ≤ σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅© σ΅© σ΅© σ΅© − π πΌσ΅©σ΅© σ΅©σ΅©π΄ σ΅©σ΅©) σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© Μ π −1 σ΅©σ΅©σ΅©σ΅©π Μσ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© π π Μ σ΅© σ΅©σ΅© σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© − πσ΅©σ΅© . (36) Therefore, σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© ≤ π4 (σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ Μ−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©π − π σ΅© σ΅© σ΅© σ΅© σ΅©σ΅© Μ π πΌσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π Μ Μ π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅© × (σ΅©σ΅©σ΅©π − π σ΅© σ΅© − π πΌσ΅©σ΅© σ΅©σ΅©π΄ σ΅©σ΅©) σ΅©σ΅© σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅©σ΅© σ΅©σ΅© −1 σ΅©σ΅© σ΅©σ΅© Μ−1 σ΅©σ΅© Μ π −1 σ΅©σ΅© Μ σ΅© + σ΅©σ΅©σ΅©π΄ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π΄ σ΅©σ΅©σ΅© π π σ΅©σ΅©σ΅©π − πσ΅©σ΅©σ΅©σ΅©) . (37) That is, σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© Μ π −1 σ΅©σ΅© Μ σ΅© σ΅©σ΅© π π ) σ΅©σ΅©π − πσ΅©σ΅©σ΅© (π4−1 − σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅© σ΅© σ΅© Μ π πΌσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©π Μ Μ π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅© Μ−1 σ΅©σ΅©σ΅©σ΅© (σ΅©σ΅©σ΅©σ΅©π − π ≤ σ΅©σ΅©σ΅©σ΅©π΄−1 − π΄ σ΅© σ΅© σ΅© σ΅© − π πΌσ΅©σ΅© σ΅©σ΅©π΄ σ΅©σ΅©) σ΅©σ΅© σ΅© σ΅© σ΅© Μ−1 σ΅©σ΅© σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© . σ΅©σ΅© σ΅©σ΅©π − π + σ΅©σ΅©σ΅©σ΅©π΄−1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©σ΅©π΄ σ΅© σ΅©σ΅© Μ π −1 < 1, by π ≥ π (π−1 (π΄−∗ ππ΄−1 ))πΌ we Μ−1 βπ π If π4 βπ΄−1 ββπ΄ min have σ΅©σ΅© Μσ΅©σ΅©σ΅©σ΅© σ΅©σ΅©π − π σ΅© σ΅© βπβ (38) [1] A. C. M. Ran and M. C. B. Reurings, “On the nonlinear matrix equation π + π΄∗ F(π)π΄ = π: solutions and perturbation theory,” Linear Algebra and Its Applications, vol. 346, pp. 15–26, 2002. [2] I. G. 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Hasanov, “Positive definite solutions of the matrix equations π ± π΄∗ π−π π΄ = π(0 < π ≤ 1),” Linear Algebra and Its Applications, vol. 404, pp. 166–182, 2005. [8] X. F. Duan and A. P. Liao, “The Hermitian positive definite solution of the matrix equation π + π΄∗ π−π π΄ = π(π ≥ 1) and its perturbation analysis,” Numerical Mathematics, vol. 30, no. 3, pp. 280–288, 2008 (Chinese). [9] X.-G. Liu and H. Gao, “On the positive definite solutions of the matrix equations ππ ± π΄π π−π‘ π΄ = πΌπ ,” Linear Algebra and Its Applications, vol. 368, pp. 83–97, 2003. 6 [10] Y. Yueting, “The iterative method for solving nonlinear matrix equation ππ + π΄∗ π−π‘ π΄ = π,” Applied Mathematics and Computation, vol. 188, no. 1, pp. 46–53, 2007. [11] G. W. Stewart and J. G. Sun, Matrix Perturbation Theory, Computer Science and Scientific Computing, Academic Press, Boston, Mass, USA, 1990. [12] R. Bhatia, Matrix Analysis, vol. 169 of Graduate Texts in Mathematics, Springer, New York, NY, USA, 1997. 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