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On the Hermitian positive definite solutions of nonlinear matrix equation Xs+∑i=1mAi∗X-tiAi=Q
Applied Mathematics and Computation 243 (2014) 950–959
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On the Hermitian positive definite solutions of nonlinear matrix P t i equation X s þ m Ai ¼ Q q i¼1 Ai X Aijing Liu a,b, Guoliang Chen b,⇑ a b
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, PR China Department of Mathematics, East China Normal University, Shanghai 200062, PR China
a r t i c l e
i n f o
Keywords: Nonlinear matrix equations Hermitian positive definite solutions Iterative method Maximal solution Error estimation
a b s t r a c t In this paper, the Hermitian positive definite solutions of nonlinear matrix equation P ti Ai ¼ Q are considered, where Q is a Hermitian positive definite matrix, Ai Xs þ m i¼1 Ai X are nonsingular complex matrices, s; m are positive numbers, and 0 < t i 1; i ¼ 1; . . . ; m. 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 some numerical examples are given to show the efficiency of the proposed iterative method. Ó 2014 Published by Elsevier Inc.
1. Introduction We consider the nonlinear matrix equation
Xs þ
m X Ai X ti Ai ¼ Q ;
ð1:1Þ
i¼1
where Q is an n n Hermitian positive definite matrix, Ai are n n nonsingular complex matrices, s; m are positive numbers, and 0 < t i 1; i ¼ 1; . . . ; m. Here Ai stands for the conjugate transpose of the matrix Ai ; i ¼ 1; . . . ; m. Nonlinear matrix equations with the form of (1.1) have many applications in: control theory; dynamic programming; ladder networks; stochastic filtering; statistics and etc. The solutions of practical interest are their Hermitian positive definite (HPD) solutions. The existence of HPD solutions of Eq. (1.1) has been investigated in some special cases. The authors [12,13,3,2] studied the matrix equation when m ¼ 1; s ¼ 1; t ¼ 1. In Hasanov [10,11], the authors investigated the matrix equation when m ¼ 1; s ¼ 1; t 2 ð0; 1. Then Peng et al. [17] proposed iterative methods for the extremal positive definite solutions of Eq. (1.1) for m ¼ 1; s ¼ 1 with two cases: 0 < t 1 and t P 1. Cai and Chen [4,5] studied the matrix equation for m ¼ 1 with two cases: s and t are positive integers, and s P 1; 0 < t 1 or 0 < s 6 1; t P 1 respectively. He and Long [16] considered the matrix equation when s ¼ 1; t ¼ 1; Q ¼ I. And Sarhan et al. [1] investigated the matrix equation when Q ¼ I. q Research supported by Natural Science Foundation of China (No. 11071079), Shanghai Natural Science Foundation (No. 09ZR1408700) and Doctoral Scientific Research Starting Foundation of Qufu Normal University (No. bsqd20130141). ⇑ Corresponding author. E-mail address: [email protected] (G. Chen).
http://dx.doi.org/10.1016/j.amc.2014.05.090 0096-3003/Ó 2014 Published by Elsevier Inc.
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
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In this paper, we study the HPD solutions of Eq. (1.1). The paper is organized as follows. In Section 2, we derive necessary and sufficient conditions for the existence of HPD solutions of Eq. (1.1), and give a sufficient condition for the existence of a unique HPD solution of Eq. (1.1). We also present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1). Then in Section 3, we propose an iterative method for obtaining the maximal HPD solution of Eq. (1.1). We give some numerical examples in Section 4 to show the efficiency of the proposed iterative method. Conclusions will be put in Section 5. We start with some notations which we use throughout this paper. The symbol Cmn denotes the set of m n complex matrices. We write B > 0ðB P 0Þ if the matrix B is positive definite (semidefinite). If B C is positive definite (semidefinite), then we write B > CðB P CÞ. We use k1 ðBÞ and kn ðBÞ to denote the maximal and minimal eigenvalues of a matrix B. We use kBk and kBkF to denote the spectral and Frobenius norm of a matrix B, and we also use kbk to denote l2 -norms of a vector b. We use X S and X L to denote the minimal and maximal HPD solution of Eq. (1.1), that is, for any HPD solution X of Eq. (1.1), then X S 6 X 6 X L . The symbol I denotes the n n identity matrix. The symbol qðBÞ denotes the spectral radius of B. Let ½B; C ¼ fXjB 6 X 6 Cg and ðB; CÞ ¼ fXjB < X < Cg. For matrices B ¼ ðb1 ; b2 ; . . . ; bn Þ ¼ ðbij Þ and C; B C ¼ ðbij CÞ is a Kronecker T T T T product and v ecðBÞ is a vector defined by v ecðBÞ ¼ ðb1 ; b2 ; . . . ; bn Þ . 2. Solvability conditions and properties of the HPD solutions In this section, we shall derive the necessary and sufficient conditions for Eq. (1.1) to have a HPD solution, and give a sufficient condition for the existence of a unique HPD solution of Eq. (1.1). We also shall present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1) Lemma 2.1. [7]. If A P B > 0(or A > B > 0), then Aa P Ba > 0(or Aa > Ba > 0) for all a 2 ð0; 1, and Ba P Aa > 0(or Ba > Aa > 0) for all a 2 ½1; 0Þ.
Lemma 2.2 [9]. Let A and B be positive operators on a Hilbert space H such that 0 < m1 I 6 A 6 M 1 I; 0 < m2 I 6 B 6 M 2 I, and 0 < A 6 B. Then
At 6
t1 M1 Bt ; m1
At 6
M2 m2
t1
Bt
hold for any t P 1. Lemma 2.3 [14]. Let f ðxÞ ¼ xt ðg xs Þ; g > 0; x P 0. Then 1s 1s t t (1) f is increasing on 0; sþt g g ; þ1 ; and decreasing on sþt 1s st t t s t g gsþ1 . ¼ sþt (2) fmax ¼ f sþt sþt
Lemma 2.4 [15]. If C and P are Hermitian matrices of the same order with P > 0, then CPC þ P1 P 2C. Lemma 2.5 [8]. If 0 < h 6 1, and P and Q are positive definite matrices of the same order with P; Q P bI > 0, then h1 ðhþ1Þ kP h Q h k 6 hb kP Q k and kP h Q h k 6 hb kP Q k. Here k k stands for one kind of matrix norm. 1
Theorem 2.1. Eq. (1.1) has a HPD solution if and only if Ai Q 2 can factor as ti
1
Ai Q 2 ¼ ðL LÞ2s N i ; i ¼ 1; 2; . . . ; 0 m;
1 1 LQ 2 B N C B 1 C where L is a nonsingular matrix and B . C is column-orthonormal. @ .. A
ð2:1Þ
Nm Proof. If Eq. (1.1) has a HPD solution, then X s > 0. Let X s ¼ L L be the Cholesky factorization, where L is a nonsingular matrix. Then Eq. (1.1) can be rewritten as 1
1
Q 2 L LQ 2 þ
m t t X 1 1 i i Q 2 Ai ðL LÞ 2s ðL LÞ 2s Ai Q 2 ¼ I:
ð2:2Þ
i¼1
Let N i ¼ ðL LÞ
t
2si
1
1
ti
Ai Q 2 ; i ¼ 1; 2; . . . ; m, then Ai Q 2 ¼ ðL LÞ2s N i ; i ¼ 1; 2; . . . ; m. Moreover, (2.2) turns into
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959 1
1
Q 2 L LQ 2 þ N1 N1 þ þ Nm Nm ¼ I:
ð2:3Þ
i.e.,
0
1
LQ 2
B B N1 B B . B . @ . Nm
1 0 C C C C C A
1
LQ 2
B B N1 B B . B . @ . Nm
1 C C C C ¼ I; C A
0
1 1 LQ 2 B N C B 1 C which means that B . C is column-orthonormal. @ .. A
Nm 1
1
Conversely, if Ai Q 2 has a decomposition as (2.1), i ¼ 1; . . . ; m, let X ¼ ðL LÞ s , then X is a HPD matrix, and it follows from (2.1) and (2.3) that
Xs þ
m X 1 1 1 1 1 1 1 1 Ai X ti Ai ¼ L L þ Q 2 N1 N 1 Q 2 þ þ Q 2 Nm Nm Q 2 ¼ Q 2 ðQ 2 L LQ 2 þ N1 N1 þ þ N m Nm ÞQ 2 ¼ Q: i¼1
Hence Eq. (1.1) has a HPD solution. h As for m ¼ 2, Eq. (1.1) turns into
X s þ A1 X t1 A1 þ A2 X t2 A2 ¼ Q :
ð2:4Þ
And we have the following theorem. Theorem 2.2. Eq. (2.4) has a HPD solution if and only if there exist a unitary matrix V 2 Cnn , a column-orthonormal matrix U1 U¼ 2 C2nn , in which U 1 2 Cnn and U 2 2 Cnn , and diagonal matrices C > 0 and S P 0 with C 2 þ S2 ¼ I such that U2 1
t1 1 2s
1
1
t2 1 2s
1
A1 ¼ ðQ 2 V C 2 VQ 2 Þ U 1 SVQ 2 ;
ð2:5Þ
A2 ¼ ðQ 2 V C 2 VQ 2 Þ U 2 SVQ 2 :
ð2:6Þ
Proof. If Eq. (2.4) has a HPD solution, we have by Theorem 2.1 that, 1
t1
1
t2
A1 Q 2 ¼ ðL LÞ2s N 1 ;
ð2:7Þ
A2 Q 2 ¼ ðL LÞ2s N 2 ; 0
12
ð2:8Þ
1
LQ and the matrix @ N 1 A is column-orthonormal. According to the CS decomposition theorem (Theorem 3.8 in [6]), there N2 P1 0 exist unitary matrices P ¼ 2 C3n3n ; V 2 Cnn , in which P 1 2 Cnn ; P2 2 C2n2n , such that 0 P2
P1 0
0 0 1 1 1 C LQ 2 0 B C B C @ N 1 AV ¼ @ S A; P2 0 N2
ð2:9Þ
where C ¼ diagðcos h1 ; . . . ; cos hn Þ; S ¼ diagðsin h1 ; . . . ; sin hn Þ and 0 6 h1 6 . . . 6 hn 6 p2 . Thus the diagonal matrices C; S P 0 and C 2 þ S2 ¼ I. Furthermore, noting that L is nonsingular, by (2.9), we have 1
C ¼ P 1 LQ 2 V > 0; P2
N1 S V ¼ : 0 N2
Eq. (2.11) is equivalent to Then we have
ð2:10Þ ð2:11Þ
N1 N2
S U1 ¼ P 2 V, let P2 be partitioned as P2 ¼ 0 U2
U3 , in which U i 2 Cnn ; i ¼ 1; 2; 3; 4. U4
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
N1 N2
¼
U 1 SV S V¼ ; U 2 SV 0 U4
U1
U3
U2
1
from which it follows that N 1 ¼ U 1 SV; N 2 ¼ U 2 SV. By (2.10), we have L ¼ P 1 CVQ 2 . Then by (2.7) and (2.8), we have t1
1
1
t1 1 2s
1
t2
1
1
t2 1 2s
1
A1 ¼ ðL LÞ2s N1 Q 2 ¼ ðQ 2 V C 2 VQ 2 Þ U 1 SVQ 2 ; A2 ¼ ðL LÞ2s N2 Q 2 ¼ ðQ 2 V C 2 VQ 2 Þ U 2 SVQ 2 : 1
1
1 s
Conversely, assume that A1 ; A2 have the decomposition (2.5) and (2.6). Let X ¼ ðQ 2 V C 2 VQ 2 Þ , which is a HPD matrix. Then it is easy to verify that X is a HPD solution of Eq. (2.4). h Theorem 2.3. If Eq. (1.1) has a HPD solution X, then X 2 ðM; NÞ,where
M¼ in which
1t m t i 1 1X li i ðAi Q 1 Ai Þti ; m i¼1 mi
N¼
Q
m X ti Ai Q s Ai
!1s ð2:12Þ
;
i¼1
li and mi are the minimal and maximal eigenvalues of Ai Q 1 Ai respectively, i ¼ 1; . . . ; m. ti
Proof. Let X be a HPD solution of Eq. (1.1), then it follows from 0 < X s < Q , and Lemma 2.1 that X ti > Q s ; i ¼ 1; . . . ; m. Hence
Xs ¼ Q
m m X X ti Ai X ti Ai < Q Ai Q s Ai : i¼1
i¼1
Thus we have
X<
Q
m X
t
i Ai Q s Ai
!1s ¼ N:
i¼1
On the other hand, from Ai X ti Ai < Q ; i ¼ 1; . . . ; m, it follows that ti
1
ti
1
Q 2 Ai X 2 X 2 Ai Q 2 < I; i ¼ 1; . . . ; m; ti
ti
X 2 Ai Q 1 Ai X 2 < I; i ¼ 1; . . . ; m; Ai Q 1 Ai < X ti ; i ¼ 1; . . . ; m: Let
li and mi are the minimal and maximal eigenvalues of Ai Q 1 Ai respectively, i ¼ 1; . . . ; m. Since
li I 6 Ai Q 1 Ai 6 mi I; i ¼ 1; . . . ; m, by Lemma 2.2, we get 1t Pm li ti i 1
X>m
i¼1
mi
Theorem 2.4. If
p¼
Pm
1 ti
ðAi Q 1 Ai Þ ¼ M:
t i Ai i¼1 Ai X
i 1t t
li mi
i
1 ti
ðAi Q 1 Ai Þ < X; i ¼ 1; . . . ; m, Hence we have
h 1
6 Q M s for all X 2 ½M; Q s , and
m 1X t i kðsþti Þ ðMÞkAi k2F < 1; s i¼1 n
where M is defined by (2.12), then Eq. (1.1) has a unique HPD solution. Proof. By the definition of M, we have M > 0. Hence kn ðMÞ > 0. n o 1 P 1 t i We consider the map FðXÞ ¼ ðQ m Ai Þ s and let X 2 X ¼ XjM 6 X 6 Q s . i¼1 Ai X Obviously, X is a convex, closed and bounded set and FðXÞ is continuous on X. By the hypothesis of the theorem, we have 1 1 P 1 t i Ai Þ s P ðQ Q þ M s Þ s ¼ M, Q s P ðQ m i¼1 Ai X i.e., 1 M 6 FðXÞ 6 Q s . Hence FðXÞ # X.
1 ti
P 1, and
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
For arbitrary X; Y 2 X, we have Pm ti P t i Ai 6 Q M s and m Ai 6 Q M s ; i¼1 Ai X i¼1 Ai Y Hence
FðXÞ ¼
Q
m X Ai X ti Ai
!1s
1
P ðQ Q þ M s Þ s ¼ M P kn ðMÞI;
ð2:13Þ
i¼1
FðYÞ ¼
Q
m X Ai Y ti Ai
!1s
1
P ðQ Q þ M s Þ s ¼ M P kn ðMÞI;
ð2:14Þ
i¼1
From (2.13) and (2.14), it follows that
" # X s1 s1 X kFðXÞs FðYÞs kF ¼ FðXÞi ðFðXÞ FðYÞÞFðYÞs1i ¼ vec FðXÞi ðFðXÞ FðYÞÞFðYÞs1i i¼0 i¼0 F X X s1 s1 i s1i s1i i ¼ vec½FðXÞ ðFðXÞ FðYÞÞFðYÞ ¼ ðFðYÞ FðXÞ ÞvecðFðXÞ FðYÞÞ i¼0 i¼0 P
s1 X kns1 ðMÞkvecðFðXÞ FðYÞÞk ¼ sks1 n ðMÞkðFðXÞ FðYÞÞkF :
ð2:15Þ
i¼0
According to the definition of the map F, we have s
m X Q Ai X ti Ai
s
FðXÞ FðYÞ ¼
!
m X Q Ai Y ti Ai
i¼1
! ¼
i¼1
m X Ai ðY ti X ti ÞAi
ð2:16Þ
i¼1
Combining (2.15) and (2.16), we by Lemma 2.5 have
X m m X 1 2 ti t i t i t kFðXÞ FðYÞ kF ¼ s1 kFðXÞ FðYÞkF 6 s1 Ai ðY X ÞAi 6 s1 Ai kF kY X i F skn ðMÞ skn ðMÞ i¼1 skn ðMÞ i¼1 F 1
6
s
1
m X
sks1 n ðMÞ
i¼1
s
1
i þ1Þ ti kðt ðMÞkAi k2F kY XkF ¼ n
m 1X t i kðsþti Þ ðMÞkAi k2F kX YkF ¼ pkX YkF : s i¼1 n
Since p < 1, we know that the map FðXÞ is a contraction map in X. By Banach fixed point theorem, the map FðXÞ has a 1 unique fixed point in X and this shows that Eq. (1.1) has a unique HPD solution in ½M; Q s . h Theorem 2.5. If Eq. (1.1) has a HPD solution X, then
kn
m X
Q
12
t
1 i Ai Q s Ai Q 2
!
t sþt
6
i¼1
st
1 s ^Q s ; ; and X 6 a sþt
^ is a solution of equation yt ð1 ys Þ ¼ kn where t ¼ min ti , and a
P m
i¼1 Q
16i6m
12
ti
1
Ai Q s Ai Q 2
in
t sþt
1s
;1
Proof. Consider the sequence defined as follows: P 11s 0 t 1 1
a0 ¼ 1; akþ1 ¼ @1
kn
m
i¼1
Q
i 2 A Q s Ai Q 2 i
atk
A ; k ¼ 0; 1; 2; . . .
Let X be a HPD solution of Eq. (1.1), then 1 P 1 1 t i s X¼ Q m Ai < Q s ¼ a0 Q s . i¼1 Ai X 1
Assuming that X < ak Q s , then by Lemma 2.1, we have
! Pm tsi Pm 1 tsi 1 m m X X 2 1 t i 1 1 Ai Ai Q 2 ti i¼1 Ai Q i¼1 Q Ai Q s 2 Q2 X ¼Q Ai X Ai < Q Ai ðak Q Þ Ai < Q ¼Q I t t s
0
i¼1
kn 1B 6 Q 2 @1
P m
i¼1 Q
i¼1 12
Ai Q
atk 1
t si
1 1
ak
ak
Ai Q 2 C 1 AQ 2 ¼ askþ1 Q: 1
Therefore X < akþ1 Q s . Then by the principle of induction, we get X < ak Q s ; k ¼ 0; 1; 2; . . ..
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
^ , then Noting that the sequence ak is monotonically decreasing and positive, hence ak is convergent. Let limk!1 ak ¼ a P 11s 0 t i m 1 1 P kn Q 2 Ai Q s Ai Q 2 t 1 i¼1 m 12 si A , i.e., a ^ is a solution of the equation yt ð1 ys Þ ¼ kn a^ ¼ @1 Ai Q 2 . i¼1 Q Ai Q a^ t 1 t s s t s t t s since maxy2½0;1 ¼ f ¼ sþt thet function f ðyÞ t ¼ y ð1 y Þ, Consider sþt sþt , from which it follows that Pm 1 i 1 s t s s A Q 2 2A Q 6 kn Q . i i¼1 i sþt sþt 1s 1s t t ^ 2 sþt ^ 6 1. On the other hand, for the sequence ak , since a0 ¼ 1 > sþt Next we shall prove that a ; 1 . Obviously, a , 1s t without loss of generality. Then we may assume that ak > sþt
0
akþ1 ¼ B @1
kn
P m
i¼1 Q
12
11s ti 1 Ai Q s Ai Q 2 C A P t
1
ak
1
atk
t sþt
ts
s sþt
!1s
0
1 t B > @1 t s sþt t
ts
11s
1s s C t : A ¼ sþt sþt
sþt
t sþt
1s
^ ¼ limk!1 ak P Hence ak > ; k ¼ 0; 1; 2; . . .. So a 1 s t ^ 2 sþt Consequently, we have a ;1 . This completes the proof.
t sþt
1s
.
h 1
1
Theorem 2.6. Suppose that Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m. If Eq. (1.1) has a HPD solution, then
tsj t j s tj þ1 ðqðAj ÞÞ 6 k1s ðQÞ; j ¼ 1; . . . ; m: s þ tj s þ tj 2
1
1
1
1
Proof. If Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m, then Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m. So Eq. (1.1) can be written as 1
1
Q 2 X s Q 2 þ
m X 1 1 Ai Q 2 X ti Q 2 Ai ¼ I:
ð2:17Þ
i¼1
For any eigenvalue k of Aj ; j ¼ 1; . . . ; m, let x be a corresponding eigenvector. Multiplying left side of Eq. (2.17) by x and right side by x, we have 1
1
1
1
x Q 2 X s Q 2 x þ x Aj Q 2 X tj Q 2 Aj x þ
m X 1 1 x Ai Q 2 X ti Q 2 Ai x ¼ x x; i¼1 i–j
which yields 1
1
1
1
x Q 2 X s Q 2 x þ jkj2 x Q 2 X tj Q 2 x 6 x x:
ð2:18Þ
Since X > 0, there exists an unitary matrix U such that X ¼ U KU, where K ¼ diagðg1 ; . . . gn Þ > 0. Then (2.18) turns into the following form 1
1
1
1
x Q 2 U Ks UQ 2 x þ jkj2 x Q 2 U Ktj UQ 2 x 6 x x:
ð2:19Þ
12
T
Let y ¼ ðy1 ; y2 ; . . . ; yn Þ ¼ UQ x, then (2.19) reduces to
y Ks y þ jkj2 y Ktj y 6 y UQU y; from which we obtain
jkj2 6
y ðUQU Ks Þy y ðk1 ðQ ÞI Ks Þy 6 ¼ y Ktj y y Ktj y
Pm
2 i¼1 yi ðk1 ðQ Þ Pm 2 tj i¼1 yi i
Form Lemma 2.3, we know that tsj tj þ1 tj gti j ðk1 ðQ Þ gsi Þ 6 sþts j sþt k1s ðQ Þ, j i.e., s ðk1 ðQ Þ gsi Þ 6 sþt j
tj sþt j
tsj
tj
þ1
k1s
t
ðQ Þgi j .
Noting that y – 0, we get tsj tj Pm 2 P þ1 tj s s 2 t j y ðk ðQ Þ g Þ 6 k1s ðQ Þ m 1 i¼1 i i¼1 yi gi . i sþt j sþt j
g
gsi Þ
:
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
Consequently, 2
Pm
jkj 6
2 i¼1 yi ðk1 ðQÞ Pm 2 tj i¼1 yi i t 2
Then ðqðAj ÞÞ 6
s sþt j
g
tj sþt j
sj
gsi Þ tj
tsj t j s tj þ1 6 k1s ðQ Þ: s þ tj s þ tj
þ1
k1s ðQ Þ; j ¼ 1; . . . ; m.
h
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 Eq. (1.1). We propose the following algorithm which avoids calculating matrix inversion in the process of iteration. Algorithm 3.1. step1. Input initial matrices: 1
X 0 ¼ cQ s ; Y0 ¼
c þ 1 1s Q ; 2c
^ ; 1Þ, and a ^ is defined in Theorem 2.5. where c 2 ða step2. For k ¼ 0; 1; 2; . . ., compute:
Y kþ1 ¼ Y k ð2I X k Y k Þ; 1
X kþ1 ¼ ðQ
m s X ti Ai Y kþ1 Ai Þ : i¼1
Theorem 3.1. If Eq. (1.1) has a HPD solution, then it has the maximal one X L . Moreover, to the sequences X k and Y k generated by Algorithm 3.1, we have
Y 0 < Y 1 < Y 2 < ; lim Y k ¼ X 1 L :
X 0 > X 1 > X 2 > ; lim X k ¼ X L ; k!1
k!1
1
^ Q s , thus Proof. Since X L is a HPD solution of Eq. (1.1), by Theorem 2.5, we have X L 6 a 1
1
^Q s P XL; X 0 ¼ cQ s > a
Y0 ¼
c þ 1 1s 1 1s 1 1s Q < Q < Q 6 X 1 L : 2c c a^
By Lemma 2.1 and Lemma 2.4, we have 1 Y 1 ¼ Y 0 ð2I X 0 Y 0 Þ ¼ 2Y 0 Y 0 X 0 Y 0 6 X 1 0 < XL ;
and
Y 1 Y 0 ¼ Y 0 Y 0 X 0 Y 0 ¼ Y 0 ðY 1 0 X 0 ÞY 0 ¼
1 c2 1s Q > 0: 4c
According to Lemma 2.1 and Y 1 < X 1 L , we have
X1 ¼
m X t Q Ai Y 1i Ai i¼1
!1s
1
> ðQ
m s X t Ai X L i Ai Þ ¼ X L ; i¼1
and m X t t X s1 X s0 ¼ Ai ðY 1i Y 0i ÞAi < 0; i¼1
i.e., X s1 < X s0 , by Lemma 2.1 again, it follows that X 1 < X 0 . Hence X 0 > X 1 > X L , and Y 0 < Y 1 < X 1 L . 1 Assume that X k1 > X k > X L , and Y k1 < Y k < X 1 L , we will prove the inequalities X k > X kþ1 > X L , and Y k < Y kþ1 < X L .
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
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By Lemmas 2.1 and 2.4, we have 1 Y kþ1 ¼ 2Y k Y k X k Y k 6 X 1 k < XL ;
and
X kþ1 ¼
Q
m X
!1s ti Ai Y kþ1 Ai
Q
>
i¼1
m X
!1s t Ai X L i Ai
¼ XL:
i¼1
1 1 Since Y k 6 X 1 k1 < X k , we have Y k > X k , thus we have by Lemma 2.1 that 1 Y kþ1 Y k ¼ Y k ðY k X k ÞY k > 0, and
m X ti t X skþ1 X sk ¼ Ai ðY kþ1 Y ki ÞAi < 0; i¼1
i.e., X skþ1 < X sk , by Lemma 2.1 again, it follows that X kþ1 < X k . Hence we have by induction that
X 0 > X 1 > X 2 > > X k > X L and Y 0 < Y 1 < Y 2 < < Y k < X 1 L b ; limk!1 Y k ¼ Y b , taking the limit in are true for all k ¼ 0; 1; 2; . . ., and so limk!1 X k and limk!1 Y k exist. Suppose limk!1 X k ¼ X 1 Pm b t s ^1 b b b b 6 X L . Morethe Algorithm 3.1 leads to Y ¼ X and X ¼ ðQ i¼1 Ai X i Ai Þ . Therefore X is a HPD solution of Eq. (1.1), thus X b P X L , then X b ¼ X L . The theorem is proved. h over, as each X k > X L , so X Theorem 3.2. If Eq. (1.1) has a HPD solution and after n iterative steps of Algorithm 3.1, we have kI X n Y n k < e, then
m m 1ti X s X ti Ai X n Ai Q 6 ek1 t i k1 s ðQ ÞkAi k2 ; X n þ n ðMÞ i¼1 i¼1
where M is defined by (2.12). 1
1
s 1 Proof. From the proof of Theorem 3.1, we have Q s < c2þ1 < Y n < X 1 for all n ¼ 1; 2; . . .. Thus we have by Theon < XL c Q 1s 1 1 rem 2.3 that Q < Y n < X n < M . And this implies
1
1 k1 s ðQ ÞI < Y n < X 1 n < kn ðMÞI:
Since
X sn þ
m X i Ai X t n Ai Q ¼
Q
i¼1
m X Ai Y tni Ai
!
i¼1
þ
m m X X i
ti i Ai X t Ai X t n Ai Q ¼ n Y n Ai ; i¼1
i¼1
we have by Lemma 2.5 that
X X m m m m t 1 X i s X ti ti ti i i Ai X n Ai Q ¼ Ai ðX t Y ÞA kAi k2 kX t t i k1 s ðQ ÞkAi k2 kX 1 X n þ i 6 n n n Yn k 6 n Y nk i¼1 i¼1 i¼1 i¼1 6
m m 1ti 1ti X X 1 ti k1 s ðQ ÞkAi k2 kX 1 t i k1 s ðQ ÞkAi k2 : n kkI X n Y n k 6 ekn ðMÞ i¼1
i¼1
4. Numerical examples In this section, we give some numerical examples to illustrate the efficiency of the proposed algorithm. All the tests are performed by MATLAB 7.0 with machine precision around 1016 . We stop the practical iteration when the residual P t i kX sk þ m i¼1 Ai X k Ai Q kF 1:0e 010. Example 4.1. Let m ¼ 1; s ¼ 2; t1 ¼ 0:2, and
0
2 B 1 B B B 0 A1 ¼ B B 1 B B @ 1 0
0 2
0 0
0
3
0
0
0
2
0
1
0
1
0
0
1 0 0 C C C 1 0 C C; 0 1 C C C 3 0 A
1 0 0 1
1
2
1 310 12 48 60 135 62 B 12 168 38 54 116 10 C C B C B B 48 38 289 22 168 76 C C: Q ¼B B 60 54 22 202 16 220 C C B C B @ 135 116 168 16 374 50 A 0
62
10
76
220
50
278
958
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
^ 0:9984984, so we choose c ¼ 0:999. By using Algorithm 3.1 and iterating nine steps, we obtain the By calculating, a maximal HPD solution X L of Eq. (1.1) as follows:
0
16:6193
0:0450
2:3234
1:9639
4:2784
1:4450
1
C B B 0:0450 12:0544 0:6912 2:4099 3:6401 0:2898 C C B C B 5:3575 2:3218 C B 2:3234 0:6912 15:5761 0:4933 C: XL X9 ¼ B C B 2:4099 0:4933 9:8559 1:1515 9:4797 C B 1:9639 C B B 4:2784 3:6401 5:3575 1:1515 17:4170 1:3839 C A @ 1:4450 with the residual
kX 29
þ
0:2898
A1 X 0:2 A1 9
2:3218
9:4797
1:3839
13:2603
Q kF ¼ 8:1270e 011.
Example 4.2. Let m ¼ 2; s ¼ 5; t 1 ¼ 0:2; t 2 ¼ 0:5; A1 is defined in Example 4.1, and
0
1
0
C 4 7 1 3 0C C C 9 2 4 7 8C C; C 5 3 0 0 1C C 5 0 2 1 7C A
B B B B B Q ¼B B B B B @
2 1 6 0 5 7
B B3 B B B0 A2 ¼ B B B8 B B2 @
4 0
105
62
62
154
56
67
11
50
41
82
71
121
0 1 4 9
56
11
41
71
1
C 121 C C C 109 15 59 61 C C: C 15 28 37 53 C C 59 37 113 132 C A 67
50
82
61
53 132 250
^ 0:8412949, so we choose c ¼ 0:87. By using Algorithm 3.1 and iterating 27 steps, we obtain the maxBy calculating, a imal HPD solution X L of Eq. (1.1) as follows:
0
X L X 27
1:6740 B 0:1069 B B B 0:2218 ¼B B 0:0033 B B @ 0:0775 0:2502
with the residual
kX 527
þ
1 0:0033 0:0775 0:2502 0:2854 0:2327 0:2553 C C C 0:2356 1:7428 0:0088 0:2549 0:0884 C C; 0:2854 0:0088 1:1526 0:1433 0:1666 C C C 0:2327 0:2549 0:1433 1:6075 0:4349 A
0:1069 1:8446
0:2218 0:2356
0:2553
0:0884
A1 X 0:2 27 A1
þ
A2 X 0:5 27 A2
0:1666
0:4349 2:1978
Q kF ¼ 4:9831e 011.
5. Conclusions In this paper, we discuss the HPD solutions of Eq. (1.1). We derive necessary and sufficient conditions for the existence of a HPD solution of Eq. (1.1), give a sufficient condition for the existence of a unique HPD solutio of Eq. (1.1), and also present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1). We propose an iterative method for obtaining the maximal HPD solution of Eq. (1.1). Numerical results show that the proposed algorithm is quite efficient. References [1] A.M. PSarhan, N.M. El-Shazly, E.M. Shehata, On the existence of extremal positive definite solutions of the nonlinear matrix equation di Xr þ m Ai ¼ I, Math. Comput. Model. 51 (2010) 1107–1117. i¼1 Ai X [2] J.C. Engwerda, On the existence of a positive definite solution of the matrix equation X þ A X 1 A ¼ I, Linear Algebra Appl. 194 (1993) 91–108. [3] J.C. Engwerda, A.C.M. Ran, A.L. Rijkeboer, Necessary and sufficient conditions for the existence of a positive definite solution of the matrix equation X þ A X 1 A ¼ Q , Linear Algebra Appl. 186 (1993) 255–275. [4] J. Cai, G. Chen, Some investigation on Hermitian positive definite solutions the matrix equation X s þ A X t A ¼ Q , Linear Algebra Appl. 430 (2009) 2448– 2456. [5] J. Cai, G. Chen, On the Hemitian positive definite solutions of nonlinear matrix equation X s þ A X t A ¼ Q , Appl. Math. Comput. 217 (2010) 117–123. [6] J. Sun, Matrix Perturbation Analysis, Science Press, Beijing, China, 2001. [7] M. Parodi, La localisation des valeurs caracterisiques desmatrices etses applications, Gauthiervillars, Paris, 1959. [8] R. Bhatia, Matrix analysis, Graduate Texts in Mathematics, Spring, Berlin, 1997. [9] T. Furuta, Operator inequalities associated with Holder-McCarthy and Kantorovich inequalities, J. Inequal. Appl. 6 (1998) 137–148. [10] V.I. Hasanov, Positive definite solutions of the matrix equations X A X q A ¼ Q , Linear Algebra Appl. 404 (2005) 166–182. [11] V.I. Hasanov, S.M. El-sayed, On the positive definite solutions of nonlinear matrix equation X þ A X d A ¼ Q , Linear Algebra Appl. 412 (2006) 154–160. [12] W.N. Anderson, T.D. Morley, G.E. Trapp, Positive solutions to X ¼ A BX 1 B , Linear Algebra Appl. 134 (1990) 53–62. [13] X. Chen, W. Li, On the matrix equation X þ A X 1 A ¼ P: solution and perturbation analysis, Math. Numer. Sin. 27 (2005) 303–310 (in Chinese). [14] X. Duan, A. Liao, On the existence of Hermitian positive definite solutions of the matrix equation X s þ A X t A ¼ Q , Linear Algebra Appl. 429 (2008) 673–687. [15] X. Zhan, Computing the extremal positive definite solutions of a matrix equation, SIAMJ. Sci. Comput. 17 (1996) 1167–1174.
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P 1 [16] Y. He, J. Long, On the Hemitian positive definite solution of the nonlinear matrix equation X þ m Ai ¼ I, Appl. Math. Comput. 216 (2010) 3480– i¼1 Ai X 3485. [17] Z. Peng, S.M. El-Sayed, X. Zhang, Iterative methods for the extremal positive solution of the matrix equation X þ A X a A ¼ Q , J. Comput. Appl. Math. 200 (2007) 520–527.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On the Hermitian positive definite solutions of nonlinear matrix P t i equation X s þ m Ai ¼ Q q i¼1 Ai X Aijing Liu a,b, Guoliang Chen b,⇑ a b
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, PR China Department of Mathematics, East China Normal University, Shanghai 200062, PR China
a r t i c l e
i n f o
Keywords: Nonlinear matrix equations Hermitian positive definite solutions Iterative method Maximal solution Error estimation
a b s t r a c t In this paper, the Hermitian positive definite solutions of nonlinear matrix equation P ti Ai ¼ Q are considered, where Q is a Hermitian positive definite matrix, Ai Xs þ m i¼1 Ai X are nonsingular complex matrices, s; m are positive numbers, and 0 < t i 1; i ¼ 1; . . . ; m. 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 some numerical examples are given to show the efficiency of the proposed iterative method. Ó 2014 Published by Elsevier Inc.
1. Introduction We consider the nonlinear matrix equation
Xs þ
m X Ai X ti Ai ¼ Q ;
ð1:1Þ
i¼1
where Q is an n n Hermitian positive definite matrix, Ai are n n nonsingular complex matrices, s; m are positive numbers, and 0 < t i 1; i ¼ 1; . . . ; m. Here Ai stands for the conjugate transpose of the matrix Ai ; i ¼ 1; . . . ; m. Nonlinear matrix equations with the form of (1.1) have many applications in: control theory; dynamic programming; ladder networks; stochastic filtering; statistics and etc. The solutions of practical interest are their Hermitian positive definite (HPD) solutions. The existence of HPD solutions of Eq. (1.1) has been investigated in some special cases. The authors [12,13,3,2] studied the matrix equation when m ¼ 1; s ¼ 1; t ¼ 1. In Hasanov [10,11], the authors investigated the matrix equation when m ¼ 1; s ¼ 1; t 2 ð0; 1. Then Peng et al. [17] proposed iterative methods for the extremal positive definite solutions of Eq. (1.1) for m ¼ 1; s ¼ 1 with two cases: 0 < t 1 and t P 1. Cai and Chen [4,5] studied the matrix equation for m ¼ 1 with two cases: s and t are positive integers, and s P 1; 0 < t 1 or 0 < s 6 1; t P 1 respectively. He and Long [16] considered the matrix equation when s ¼ 1; t ¼ 1; Q ¼ I. And Sarhan et al. [1] investigated the matrix equation when Q ¼ I. q Research supported by Natural Science Foundation of China (No. 11071079), Shanghai Natural Science Foundation (No. 09ZR1408700) and Doctoral Scientific Research Starting Foundation of Qufu Normal University (No. bsqd20130141). ⇑ Corresponding author. E-mail address: [email protected] (G. Chen).
http://dx.doi.org/10.1016/j.amc.2014.05.090 0096-3003/Ó 2014 Published by Elsevier Inc.
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
951
In this paper, we study the HPD solutions of Eq. (1.1). The paper is organized as follows. In Section 2, we derive necessary and sufficient conditions for the existence of HPD solutions of Eq. (1.1), and give a sufficient condition for the existence of a unique HPD solution of Eq. (1.1). We also present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1). Then in Section 3, we propose an iterative method for obtaining the maximal HPD solution of Eq. (1.1). We give some numerical examples in Section 4 to show the efficiency of the proposed iterative method. Conclusions will be put in Section 5. We start with some notations which we use throughout this paper. The symbol Cmn denotes the set of m n complex matrices. We write B > 0ðB P 0Þ if the matrix B is positive definite (semidefinite). If B C is positive definite (semidefinite), then we write B > CðB P CÞ. We use k1 ðBÞ and kn ðBÞ to denote the maximal and minimal eigenvalues of a matrix B. We use kBk and kBkF to denote the spectral and Frobenius norm of a matrix B, and we also use kbk to denote l2 -norms of a vector b. We use X S and X L to denote the minimal and maximal HPD solution of Eq. (1.1), that is, for any HPD solution X of Eq. (1.1), then X S 6 X 6 X L . The symbol I denotes the n n identity matrix. The symbol qðBÞ denotes the spectral radius of B. Let ½B; C ¼ fXjB 6 X 6 Cg and ðB; CÞ ¼ fXjB < X < Cg. For matrices B ¼ ðb1 ; b2 ; . . . ; bn Þ ¼ ðbij Þ and C; B C ¼ ðbij CÞ is a Kronecker T T T T product and v ecðBÞ is a vector defined by v ecðBÞ ¼ ðb1 ; b2 ; . . . ; bn Þ . 2. Solvability conditions and properties of the HPD solutions In this section, we shall derive the necessary and sufficient conditions for Eq. (1.1) to have a HPD solution, and give a sufficient condition for the existence of a unique HPD solution of Eq. (1.1). We also shall present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1) Lemma 2.1. [7]. If A P B > 0(or A > B > 0), then Aa P Ba > 0(or Aa > Ba > 0) for all a 2 ð0; 1, and Ba P Aa > 0(or Ba > Aa > 0) for all a 2 ½1; 0Þ.
Lemma 2.2 [9]. Let A and B be positive operators on a Hilbert space H such that 0 < m1 I 6 A 6 M 1 I; 0 < m2 I 6 B 6 M 2 I, and 0 < A 6 B. Then
At 6
t1 M1 Bt ; m1
At 6
M2 m2
t1
Bt
hold for any t P 1. Lemma 2.3 [14]. Let f ðxÞ ¼ xt ðg xs Þ; g > 0; x P 0. Then 1s 1s t t (1) f is increasing on 0; sþt g g ; þ1 ; and decreasing on sþt 1s st t t s t g gsþ1 . ¼ sþt (2) fmax ¼ f sþt sþt
Lemma 2.4 [15]. If C and P are Hermitian matrices of the same order with P > 0, then CPC þ P1 P 2C. Lemma 2.5 [8]. If 0 < h 6 1, and P and Q are positive definite matrices of the same order with P; Q P bI > 0, then h1 ðhþ1Þ kP h Q h k 6 hb kP Q k and kP h Q h k 6 hb kP Q k. Here k k stands for one kind of matrix norm. 1
Theorem 2.1. Eq. (1.1) has a HPD solution if and only if Ai Q 2 can factor as ti
1
Ai Q 2 ¼ ðL LÞ2s N i ; i ¼ 1; 2; . . . ; 0 m;
1 1 LQ 2 B N C B 1 C where L is a nonsingular matrix and B . C is column-orthonormal. @ .. A
ð2:1Þ
Nm Proof. If Eq. (1.1) has a HPD solution, then X s > 0. Let X s ¼ L L be the Cholesky factorization, where L is a nonsingular matrix. Then Eq. (1.1) can be rewritten as 1
1
Q 2 L LQ 2 þ
m t t X 1 1 i i Q 2 Ai ðL LÞ 2s ðL LÞ 2s Ai Q 2 ¼ I:
ð2:2Þ
i¼1
Let N i ¼ ðL LÞ
t
2si
1
1
ti
Ai Q 2 ; i ¼ 1; 2; . . . ; m, then Ai Q 2 ¼ ðL LÞ2s N i ; i ¼ 1; 2; . . . ; m. Moreover, (2.2) turns into
952
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959 1
1
Q 2 L LQ 2 þ N1 N1 þ þ Nm Nm ¼ I:
ð2:3Þ
i.e.,
0
1
LQ 2
B B N1 B B . B . @ . Nm
1 0 C C C C C A
1
LQ 2
B B N1 B B . B . @ . Nm
1 C C C C ¼ I; C A
0
1 1 LQ 2 B N C B 1 C which means that B . C is column-orthonormal. @ .. A
Nm 1
1
Conversely, if Ai Q 2 has a decomposition as (2.1), i ¼ 1; . . . ; m, let X ¼ ðL LÞ s , then X is a HPD matrix, and it follows from (2.1) and (2.3) that
Xs þ
m X 1 1 1 1 1 1 1 1 Ai X ti Ai ¼ L L þ Q 2 N1 N 1 Q 2 þ þ Q 2 Nm Nm Q 2 ¼ Q 2 ðQ 2 L LQ 2 þ N1 N1 þ þ N m Nm ÞQ 2 ¼ Q: i¼1
Hence Eq. (1.1) has a HPD solution. h As for m ¼ 2, Eq. (1.1) turns into
X s þ A1 X t1 A1 þ A2 X t2 A2 ¼ Q :
ð2:4Þ
And we have the following theorem. Theorem 2.2. Eq. (2.4) has a HPD solution if and only if there exist a unitary matrix V 2 Cnn , a column-orthonormal matrix U1 U¼ 2 C2nn , in which U 1 2 Cnn and U 2 2 Cnn , and diagonal matrices C > 0 and S P 0 with C 2 þ S2 ¼ I such that U2 1
t1 1 2s
1
1
t2 1 2s
1
A1 ¼ ðQ 2 V C 2 VQ 2 Þ U 1 SVQ 2 ;
ð2:5Þ
A2 ¼ ðQ 2 V C 2 VQ 2 Þ U 2 SVQ 2 :
ð2:6Þ
Proof. If Eq. (2.4) has a HPD solution, we have by Theorem 2.1 that, 1
t1
1
t2
A1 Q 2 ¼ ðL LÞ2s N 1 ;
ð2:7Þ
A2 Q 2 ¼ ðL LÞ2s N 2 ; 0
12
ð2:8Þ
1
LQ and the matrix @ N 1 A is column-orthonormal. According to the CS decomposition theorem (Theorem 3.8 in [6]), there N2 P1 0 exist unitary matrices P ¼ 2 C3n3n ; V 2 Cnn , in which P 1 2 Cnn ; P2 2 C2n2n , such that 0 P2
P1 0
0 0 1 1 1 C LQ 2 0 B C B C @ N 1 AV ¼ @ S A; P2 0 N2
ð2:9Þ
where C ¼ diagðcos h1 ; . . . ; cos hn Þ; S ¼ diagðsin h1 ; . . . ; sin hn Þ and 0 6 h1 6 . . . 6 hn 6 p2 . Thus the diagonal matrices C; S P 0 and C 2 þ S2 ¼ I. Furthermore, noting that L is nonsingular, by (2.9), we have 1
C ¼ P 1 LQ 2 V > 0; P2
N1 S V ¼ : 0 N2
Eq. (2.11) is equivalent to Then we have
ð2:10Þ ð2:11Þ
N1 N2
S U1 ¼ P 2 V, let P2 be partitioned as P2 ¼ 0 U2
U3 , in which U i 2 Cnn ; i ¼ 1; 2; 3; 4. U4
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
N1 N2
¼
U 1 SV S V¼ ; U 2 SV 0 U4
U1
U3
U2
1
from which it follows that N 1 ¼ U 1 SV; N 2 ¼ U 2 SV. By (2.10), we have L ¼ P 1 CVQ 2 . Then by (2.7) and (2.8), we have t1
1
1
t1 1 2s
1
t2
1
1
t2 1 2s
1
A1 ¼ ðL LÞ2s N1 Q 2 ¼ ðQ 2 V C 2 VQ 2 Þ U 1 SVQ 2 ; A2 ¼ ðL LÞ2s N2 Q 2 ¼ ðQ 2 V C 2 VQ 2 Þ U 2 SVQ 2 : 1
1
1 s
Conversely, assume that A1 ; A2 have the decomposition (2.5) and (2.6). Let X ¼ ðQ 2 V C 2 VQ 2 Þ , which is a HPD matrix. Then it is easy to verify that X is a HPD solution of Eq. (2.4). h Theorem 2.3. If Eq. (1.1) has a HPD solution X, then X 2 ðM; NÞ,where
M¼ in which
1t m t i 1 1X li i ðAi Q 1 Ai Þti ; m i¼1 mi
N¼
Q
m X ti Ai Q s Ai
!1s ð2:12Þ
;
i¼1
li and mi are the minimal and maximal eigenvalues of Ai Q 1 Ai respectively, i ¼ 1; . . . ; m. ti
Proof. Let X be a HPD solution of Eq. (1.1), then it follows from 0 < X s < Q , and Lemma 2.1 that X ti > Q s ; i ¼ 1; . . . ; m. Hence
Xs ¼ Q
m m X X ti Ai X ti Ai < Q Ai Q s Ai : i¼1
i¼1
Thus we have
X<
Q
m X
t
i Ai Q s Ai
!1s ¼ N:
i¼1
On the other hand, from Ai X ti Ai < Q ; i ¼ 1; . . . ; m, it follows that ti
1
ti
1
Q 2 Ai X 2 X 2 Ai Q 2 < I; i ¼ 1; . . . ; m; ti
ti
X 2 Ai Q 1 Ai X 2 < I; i ¼ 1; . . . ; m; Ai Q 1 Ai < X ti ; i ¼ 1; . . . ; m: Let
li and mi are the minimal and maximal eigenvalues of Ai Q 1 Ai respectively, i ¼ 1; . . . ; m. Since
li I 6 Ai Q 1 Ai 6 mi I; i ¼ 1; . . . ; m, by Lemma 2.2, we get 1t Pm li ti i 1
X>m
i¼1
mi
Theorem 2.4. If
p¼
Pm
1 ti
ðAi Q 1 Ai Þ ¼ M:
t i Ai i¼1 Ai X
i 1t t
li mi
i
1 ti
ðAi Q 1 Ai Þ < X; i ¼ 1; . . . ; m, Hence we have
h 1
6 Q M s for all X 2 ½M; Q s , and
m 1X t i kðsþti Þ ðMÞkAi k2F < 1; s i¼1 n
where M is defined by (2.12), then Eq. (1.1) has a unique HPD solution. Proof. By the definition of M, we have M > 0. Hence kn ðMÞ > 0. n o 1 P 1 t i We consider the map FðXÞ ¼ ðQ m Ai Þ s and let X 2 X ¼ XjM 6 X 6 Q s . i¼1 Ai X Obviously, X is a convex, closed and bounded set and FðXÞ is continuous on X. By the hypothesis of the theorem, we have 1 1 P 1 t i Ai Þ s P ðQ Q þ M s Þ s ¼ M, Q s P ðQ m i¼1 Ai X i.e., 1 M 6 FðXÞ 6 Q s . Hence FðXÞ # X.
1 ti
P 1, and
954
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
For arbitrary X; Y 2 X, we have Pm ti P t i Ai 6 Q M s and m Ai 6 Q M s ; i¼1 Ai X i¼1 Ai Y Hence
FðXÞ ¼
Q
m X Ai X ti Ai
!1s
1
P ðQ Q þ M s Þ s ¼ M P kn ðMÞI;
ð2:13Þ
i¼1
FðYÞ ¼
Q
m X Ai Y ti Ai
!1s
1
P ðQ Q þ M s Þ s ¼ M P kn ðMÞI;
ð2:14Þ
i¼1
From (2.13) and (2.14), it follows that
" # X s1 s1 X kFðXÞs FðYÞs kF ¼ FðXÞi ðFðXÞ FðYÞÞFðYÞs1i ¼ vec FðXÞi ðFðXÞ FðYÞÞFðYÞs1i i¼0 i¼0 F X X s1 s1 i s1i s1i i ¼ vec½FðXÞ ðFðXÞ FðYÞÞFðYÞ ¼ ðFðYÞ FðXÞ ÞvecðFðXÞ FðYÞÞ i¼0 i¼0 P
s1 X kns1 ðMÞkvecðFðXÞ FðYÞÞk ¼ sks1 n ðMÞkðFðXÞ FðYÞÞkF :
ð2:15Þ
i¼0
According to the definition of the map F, we have s
m X Q Ai X ti Ai
s
FðXÞ FðYÞ ¼
!
m X Q Ai Y ti Ai
i¼1
! ¼
i¼1
m X Ai ðY ti X ti ÞAi
ð2:16Þ
i¼1
Combining (2.15) and (2.16), we by Lemma 2.5 have
X m m X 1 2 ti t i t i t kFðXÞ FðYÞ kF ¼ s1 kFðXÞ FðYÞkF 6 s1 Ai ðY X ÞAi 6 s1 Ai kF kY X i F skn ðMÞ skn ðMÞ i¼1 skn ðMÞ i¼1 F 1
6
s
1
m X
sks1 n ðMÞ
i¼1
s
1
i þ1Þ ti kðt ðMÞkAi k2F kY XkF ¼ n
m 1X t i kðsþti Þ ðMÞkAi k2F kX YkF ¼ pkX YkF : s i¼1 n
Since p < 1, we know that the map FðXÞ is a contraction map in X. By Banach fixed point theorem, the map FðXÞ has a 1 unique fixed point in X and this shows that Eq. (1.1) has a unique HPD solution in ½M; Q s . h Theorem 2.5. If Eq. (1.1) has a HPD solution X, then
kn
m X
Q
12
t
1 i Ai Q s Ai Q 2
!
t sþt
6
i¼1
st
1 s ^Q s ; ; and X 6 a sþt
^ is a solution of equation yt ð1 ys Þ ¼ kn where t ¼ min ti , and a
P m
i¼1 Q
16i6m
12
ti
1
Ai Q s Ai Q 2
in
t sþt
1s
;1
Proof. Consider the sequence defined as follows: P 11s 0 t 1 1
a0 ¼ 1; akþ1 ¼ @1
kn
m
i¼1
Q
i 2 A Q s Ai Q 2 i
atk
A ; k ¼ 0; 1; 2; . . .
Let X be a HPD solution of Eq. (1.1), then 1 P 1 1 t i s X¼ Q m Ai < Q s ¼ a0 Q s . i¼1 Ai X 1
Assuming that X < ak Q s , then by Lemma 2.1, we have
! Pm tsi Pm 1 tsi 1 m m X X 2 1 t i 1 1 Ai Ai Q 2 ti i¼1 Ai Q i¼1 Q Ai Q s 2 Q2 X ¼Q Ai X Ai < Q Ai ðak Q Þ Ai < Q ¼Q I t t s
0
i¼1
kn 1B 6 Q 2 @1
P m
i¼1 Q
i¼1 12
Ai Q
atk 1
t si
1 1
ak
ak
Ai Q 2 C 1 AQ 2 ¼ askþ1 Q: 1
Therefore X < akþ1 Q s . Then by the principle of induction, we get X < ak Q s ; k ¼ 0; 1; 2; . . ..
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
^ , then Noting that the sequence ak is monotonically decreasing and positive, hence ak is convergent. Let limk!1 ak ¼ a P 11s 0 t i m 1 1 P kn Q 2 Ai Q s Ai Q 2 t 1 i¼1 m 12 si A , i.e., a ^ is a solution of the equation yt ð1 ys Þ ¼ kn a^ ¼ @1 Ai Q 2 . i¼1 Q Ai Q a^ t 1 t s s t s t t s since maxy2½0;1 ¼ f ¼ sþt thet function f ðyÞ t ¼ y ð1 y Þ, Consider sþt sþt , from which it follows that Pm 1 i 1 s t s s A Q 2 2A Q 6 kn Q . i i¼1 i sþt sþt 1s 1s t t ^ 2 sþt ^ 6 1. On the other hand, for the sequence ak , since a0 ¼ 1 > sþt Next we shall prove that a ; 1 . Obviously, a , 1s t without loss of generality. Then we may assume that ak > sþt
0
akþ1 ¼ B @1
kn
P m
i¼1 Q
12
11s ti 1 Ai Q s Ai Q 2 C A P t
1
ak
1
atk
t sþt
ts
s sþt
!1s
0
1 t B > @1 t s sþt t
ts
11s
1s s C t : A ¼ sþt sþt
sþt
t sþt
1s
^ ¼ limk!1 ak P Hence ak > ; k ¼ 0; 1; 2; . . .. So a 1 s t ^ 2 sþt Consequently, we have a ;1 . This completes the proof.
t sþt
1s
.
h 1
1
Theorem 2.6. Suppose that Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m. If Eq. (1.1) has a HPD solution, then
tsj t j s tj þ1 ðqðAj ÞÞ 6 k1s ðQÞ; j ¼ 1; . . . ; m: s þ tj s þ tj 2
1
1
1
1
Proof. If Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m, then Aj Q 2 ¼ Q 2 Aj ; j ¼ 1; . . . ; m. So Eq. (1.1) can be written as 1
1
Q 2 X s Q 2 þ
m X 1 1 Ai Q 2 X ti Q 2 Ai ¼ I:
ð2:17Þ
i¼1
For any eigenvalue k of Aj ; j ¼ 1; . . . ; m, let x be a corresponding eigenvector. Multiplying left side of Eq. (2.17) by x and right side by x, we have 1
1
1
1
x Q 2 X s Q 2 x þ x Aj Q 2 X tj Q 2 Aj x þ
m X 1 1 x Ai Q 2 X ti Q 2 Ai x ¼ x x; i¼1 i–j
which yields 1
1
1
1
x Q 2 X s Q 2 x þ jkj2 x Q 2 X tj Q 2 x 6 x x:
ð2:18Þ
Since X > 0, there exists an unitary matrix U such that X ¼ U KU, where K ¼ diagðg1 ; . . . gn Þ > 0. Then (2.18) turns into the following form 1
1
1
1
x Q 2 U Ks UQ 2 x þ jkj2 x Q 2 U Ktj UQ 2 x 6 x x:
ð2:19Þ
12
T
Let y ¼ ðy1 ; y2 ; . . . ; yn Þ ¼ UQ x, then (2.19) reduces to
y Ks y þ jkj2 y Ktj y 6 y UQU y; from which we obtain
jkj2 6
y ðUQU Ks Þy y ðk1 ðQ ÞI Ks Þy 6 ¼ y Ktj y y Ktj y
Pm
2 i¼1 yi ðk1 ðQ Þ Pm 2 tj i¼1 yi i
Form Lemma 2.3, we know that tsj tj þ1 tj gti j ðk1 ðQ Þ gsi Þ 6 sþts j sþt k1s ðQ Þ, j i.e., s ðk1 ðQ Þ gsi Þ 6 sþt j
tj sþt j
tsj
tj
þ1
k1s
t
ðQ Þgi j .
Noting that y – 0, we get tsj tj Pm 2 P þ1 tj s s 2 t j y ðk ðQ Þ g Þ 6 k1s ðQ Þ m 1 i¼1 i i¼1 yi gi . i sþt j sþt j
g
gsi Þ
:
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
Consequently, 2
Pm
jkj 6
2 i¼1 yi ðk1 ðQÞ Pm 2 tj i¼1 yi i t 2
Then ðqðAj ÞÞ 6
s sþt j
g
tj sþt j
sj
gsi Þ tj
tsj t j s tj þ1 6 k1s ðQ Þ: s þ tj s þ tj
þ1
k1s ðQ Þ; j ¼ 1; . . . ; m.
h
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 Eq. (1.1). We propose the following algorithm which avoids calculating matrix inversion in the process of iteration. Algorithm 3.1. step1. Input initial matrices: 1
X 0 ¼ cQ s ; Y0 ¼
c þ 1 1s Q ; 2c
^ ; 1Þ, and a ^ is defined in Theorem 2.5. where c 2 ða step2. For k ¼ 0; 1; 2; . . ., compute:
Y kþ1 ¼ Y k ð2I X k Y k Þ; 1
X kþ1 ¼ ðQ
m s X ti Ai Y kþ1 Ai Þ : i¼1
Theorem 3.1. If Eq. (1.1) has a HPD solution, then it has the maximal one X L . Moreover, to the sequences X k and Y k generated by Algorithm 3.1, we have
Y 0 < Y 1 < Y 2 < ; lim Y k ¼ X 1 L :
X 0 > X 1 > X 2 > ; lim X k ¼ X L ; k!1
k!1
1
^ Q s , thus Proof. Since X L is a HPD solution of Eq. (1.1), by Theorem 2.5, we have X L 6 a 1
1
^Q s P XL; X 0 ¼ cQ s > a
Y0 ¼
c þ 1 1s 1 1s 1 1s Q < Q < Q 6 X 1 L : 2c c a^
By Lemma 2.1 and Lemma 2.4, we have 1 Y 1 ¼ Y 0 ð2I X 0 Y 0 Þ ¼ 2Y 0 Y 0 X 0 Y 0 6 X 1 0 < XL ;
and
Y 1 Y 0 ¼ Y 0 Y 0 X 0 Y 0 ¼ Y 0 ðY 1 0 X 0 ÞY 0 ¼
1 c2 1s Q > 0: 4c
According to Lemma 2.1 and Y 1 < X 1 L , we have
X1 ¼
m X t Q Ai Y 1i Ai i¼1
!1s
1
> ðQ
m s X t Ai X L i Ai Þ ¼ X L ; i¼1
and m X t t X s1 X s0 ¼ Ai ðY 1i Y 0i ÞAi < 0; i¼1
i.e., X s1 < X s0 , by Lemma 2.1 again, it follows that X 1 < X 0 . Hence X 0 > X 1 > X L , and Y 0 < Y 1 < X 1 L . 1 Assume that X k1 > X k > X L , and Y k1 < Y k < X 1 L , we will prove the inequalities X k > X kþ1 > X L , and Y k < Y kþ1 < X L .
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
957
By Lemmas 2.1 and 2.4, we have 1 Y kþ1 ¼ 2Y k Y k X k Y k 6 X 1 k < XL ;
and
X kþ1 ¼
Q
m X
!1s ti Ai Y kþ1 Ai
Q
>
i¼1
m X
!1s t Ai X L i Ai
¼ XL:
i¼1
1 1 Since Y k 6 X 1 k1 < X k , we have Y k > X k , thus we have by Lemma 2.1 that 1 Y kþ1 Y k ¼ Y k ðY k X k ÞY k > 0, and
m X ti t X skþ1 X sk ¼ Ai ðY kþ1 Y ki ÞAi < 0; i¼1
i.e., X skþ1 < X sk , by Lemma 2.1 again, it follows that X kþ1 < X k . Hence we have by induction that
X 0 > X 1 > X 2 > > X k > X L and Y 0 < Y 1 < Y 2 < < Y k < X 1 L b ; limk!1 Y k ¼ Y b , taking the limit in are true for all k ¼ 0; 1; 2; . . ., and so limk!1 X k and limk!1 Y k exist. Suppose limk!1 X k ¼ X 1 Pm b t s ^1 b b b b 6 X L . Morethe Algorithm 3.1 leads to Y ¼ X and X ¼ ðQ i¼1 Ai X i Ai Þ . Therefore X is a HPD solution of Eq. (1.1), thus X b P X L , then X b ¼ X L . The theorem is proved. h over, as each X k > X L , so X Theorem 3.2. If Eq. (1.1) has a HPD solution and after n iterative steps of Algorithm 3.1, we have kI X n Y n k < e, then
m m 1ti X s X ti Ai X n Ai Q 6 ek1 t i k1 s ðQ ÞkAi k2 ; X n þ n ðMÞ i¼1 i¼1
where M is defined by (2.12). 1
1
s 1 Proof. From the proof of Theorem 3.1, we have Q s < c2þ1 < Y n < X 1 for all n ¼ 1; 2; . . .. Thus we have by Theon < XL c Q 1s 1 1 rem 2.3 that Q < Y n < X n < M . And this implies
1
1 k1 s ðQ ÞI < Y n < X 1 n < kn ðMÞI:
Since
X sn þ
m X i Ai X t n Ai Q ¼
Q
i¼1
m X Ai Y tni Ai
!
i¼1
þ
m m X X i
ti i Ai X t Ai X t n Ai Q ¼ n Y n Ai ; i¼1
i¼1
we have by Lemma 2.5 that
X X m m m m t 1 X i s X ti ti ti i i Ai X n Ai Q ¼ Ai ðX t Y ÞA kAi k2 kX t t i k1 s ðQ ÞkAi k2 kX 1 X n þ i 6 n n n Yn k 6 n Y nk i¼1 i¼1 i¼1 i¼1 6
m m 1ti 1ti X X 1 ti k1 s ðQ ÞkAi k2 kX 1 t i k1 s ðQ ÞkAi k2 : n kkI X n Y n k 6 ekn ðMÞ i¼1
i¼1
4. Numerical examples In this section, we give some numerical examples to illustrate the efficiency of the proposed algorithm. All the tests are performed by MATLAB 7.0 with machine precision around 1016 . We stop the practical iteration when the residual P t i kX sk þ m i¼1 Ai X k Ai Q kF 1:0e 010. Example 4.1. Let m ¼ 1; s ¼ 2; t1 ¼ 0:2, and
0
2 B 1 B B B 0 A1 ¼ B B 1 B B @ 1 0
0 2
0 0
0
3
0
0
0
2
0
1
0
1
0
0
1 0 0 C C C 1 0 C C; 0 1 C C C 3 0 A
1 0 0 1
1
2
1 310 12 48 60 135 62 B 12 168 38 54 116 10 C C B C B B 48 38 289 22 168 76 C C: Q ¼B B 60 54 22 202 16 220 C C B C B @ 135 116 168 16 374 50 A 0
62
10
76
220
50
278
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A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
^ 0:9984984, so we choose c ¼ 0:999. By using Algorithm 3.1 and iterating nine steps, we obtain the By calculating, a maximal HPD solution X L of Eq. (1.1) as follows:
0
16:6193
0:0450
2:3234
1:9639
4:2784
1:4450
1
C B B 0:0450 12:0544 0:6912 2:4099 3:6401 0:2898 C C B C B 5:3575 2:3218 C B 2:3234 0:6912 15:5761 0:4933 C: XL X9 ¼ B C B 2:4099 0:4933 9:8559 1:1515 9:4797 C B 1:9639 C B B 4:2784 3:6401 5:3575 1:1515 17:4170 1:3839 C A @ 1:4450 with the residual
kX 29
þ
0:2898
A1 X 0:2 A1 9
2:3218
9:4797
1:3839
13:2603
Q kF ¼ 8:1270e 011.
Example 4.2. Let m ¼ 2; s ¼ 5; t 1 ¼ 0:2; t 2 ¼ 0:5; A1 is defined in Example 4.1, and
0
1
0
C 4 7 1 3 0C C C 9 2 4 7 8C C; C 5 3 0 0 1C C 5 0 2 1 7C A
B B B B B Q ¼B B B B B @
2 1 6 0 5 7
B B3 B B B0 A2 ¼ B B B8 B B2 @
4 0
105
62
62
154
56
67
11
50
41
82
71
121
0 1 4 9
56
11
41
71
1
C 121 C C C 109 15 59 61 C C: C 15 28 37 53 C C 59 37 113 132 C A 67
50
82
61
53 132 250
^ 0:8412949, so we choose c ¼ 0:87. By using Algorithm 3.1 and iterating 27 steps, we obtain the maxBy calculating, a imal HPD solution X L of Eq. (1.1) as follows:
0
X L X 27
1:6740 B 0:1069 B B B 0:2218 ¼B B 0:0033 B B @ 0:0775 0:2502
with the residual
kX 527
þ
1 0:0033 0:0775 0:2502 0:2854 0:2327 0:2553 C C C 0:2356 1:7428 0:0088 0:2549 0:0884 C C; 0:2854 0:0088 1:1526 0:1433 0:1666 C C C 0:2327 0:2549 0:1433 1:6075 0:4349 A
0:1069 1:8446
0:2218 0:2356
0:2553
0:0884
A1 X 0:2 27 A1
þ
A2 X 0:5 27 A2
0:1666
0:4349 2:1978
Q kF ¼ 4:9831e 011.
5. Conclusions In this paper, we discuss the HPD solutions of Eq. (1.1). We derive necessary and sufficient conditions for the existence of a HPD solution of Eq. (1.1), give a sufficient condition for the existence of a unique HPD solutio of Eq. (1.1), and also present some necessary conditions and sufficient conditions for the existence of Hermitian positive definite solutions of Eq. (1.1). We propose an iterative method for obtaining the maximal HPD solution of Eq. (1.1). Numerical results show that the proposed algorithm is quite efficient. References [1] A.M. PSarhan, N.M. El-Shazly, E.M. Shehata, On the existence of extremal positive definite solutions of the nonlinear matrix equation di Xr þ m Ai ¼ I, Math. Comput. Model. 51 (2010) 1107–1117. i¼1 Ai X [2] J.C. Engwerda, On the existence of a positive definite solution of the matrix equation X þ A X 1 A ¼ I, Linear Algebra Appl. 194 (1993) 91–108. [3] J.C. Engwerda, A.C.M. Ran, A.L. Rijkeboer, Necessary and sufficient conditions for the existence of a positive definite solution of the matrix equation X þ A X 1 A ¼ Q , Linear Algebra Appl. 186 (1993) 255–275. [4] J. Cai, G. Chen, Some investigation on Hermitian positive definite solutions the matrix equation X s þ A X t A ¼ Q , Linear Algebra Appl. 430 (2009) 2448– 2456. [5] J. Cai, G. Chen, On the Hemitian positive definite solutions of nonlinear matrix equation X s þ A X t A ¼ Q , Appl. Math. Comput. 217 (2010) 117–123. [6] J. Sun, Matrix Perturbation Analysis, Science Press, Beijing, China, 2001. [7] M. Parodi, La localisation des valeurs caracterisiques desmatrices etses applications, Gauthiervillars, Paris, 1959. [8] R. Bhatia, Matrix analysis, Graduate Texts in Mathematics, Spring, Berlin, 1997. [9] T. Furuta, Operator inequalities associated with Holder-McCarthy and Kantorovich inequalities, J. Inequal. Appl. 6 (1998) 137–148. [10] V.I. Hasanov, Positive definite solutions of the matrix equations X A X q A ¼ Q , Linear Algebra Appl. 404 (2005) 166–182. [11] V.I. Hasanov, S.M. El-sayed, On the positive definite solutions of nonlinear matrix equation X þ A X d A ¼ Q , Linear Algebra Appl. 412 (2006) 154–160. [12] W.N. Anderson, T.D. Morley, G.E. Trapp, Positive solutions to X ¼ A BX 1 B , Linear Algebra Appl. 134 (1990) 53–62. [13] X. Chen, W. Li, On the matrix equation X þ A X 1 A ¼ P: solution and perturbation analysis, Math. Numer. Sin. 27 (2005) 303–310 (in Chinese). [14] X. Duan, A. Liao, On the existence of Hermitian positive definite solutions of the matrix equation X s þ A X t A ¼ Q , Linear Algebra Appl. 429 (2008) 673–687. [15] X. Zhan, Computing the extremal positive definite solutions of a matrix equation, SIAMJ. Sci. Comput. 17 (1996) 1167–1174.
A. Liu, G. Chen / Applied Mathematics and Computation 243 (2014) 950–959
959
P 1 [16] Y. He, J. Long, On the Hemitian positive definite solution of the nonlinear matrix equation X þ m Ai ¼ I, Appl. Math. Comput. 216 (2010) 3480– i¼1 Ai X 3485. [17] Z. Peng, S.M. El-Sayed, X. Zhang, Iterative methods for the extremal positive solution of the matrix equation X þ A X a A ¼ Q , J. Comput. Appl. Math. 200 (2007) 520–527.