On positive definite solutions of the nonlinear matrix ... - Hikari Ltd.
Applied Mathematical Sciences, Vol. 9, 2015, no. 3, 107 - 120 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2015.410884
On Positive Definite Solutions of the Nonlinear Matrix Equation X A* X n A I Sana'a A. Zarea* Mathematical Sciences Department Faculty of Sciences Princess Nourah Bint Abdul Rahman University Riyadh, Saudi Arabia *Corresponding author Salah Mohammed El-Sayed Department of Scientific Computing Faculty of Computers and Informatics Benha University, Benha, Egypt Copyright © 2014 Sana'a A. Zarea and Salah Mohammed El-Sayed. 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.
Abstract In this paper, we have discussed some properties of a positive definite solution of the non linear matrix equation X A* X n A I , where n is a positive integer. Two effective iterative methods for computing a positive definite solution of this equation are proposed. The necessary and sufficient conditions for existence of a positive definite solution are derived. Some numerical examples will be presented for illustrative purposes. Mathematics Subject Classification: 15A24; 65F35 Keywords: Nonlinear matrix equation, Properties of solutions, Iterative algorithms
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Sana'a A. Zarea and Salah Mohammed El-Sayed
1 Introduction We consider the matrix equation: (1) X A* X n A I where n is a positive integer. The nonlinear matrix equation (1) arises in a wide variety of applications and research areas, including automatic control, ladder networks, dynamic programming, stochastic filtering and statistics [4, 7, 9]. If n is a negative real numbers, the positive definite solutions of (1) have been studied in some special cases [1-6, 8, 10-15]. In this paper, we'll mainly discuss the basic properties and some analysis of the positive definite solutions of (1). The rest of the paper is organized as follows: In Section 2, we show that (1) always has a positive definite solution and we give the necessary and sufficient conditions for the equation to have solutions. In Sections 3 and 4, we'll discuss two iterations methods for solving (1). Finally, we'll give three numerical examples in Section 5 to illustrate our theoretical results. The aim of this paper is to find the positive definite solution of (1). In this paper, we will use mathematical induction technique in the most proofs. The following notations are used throughout the rest of the paper. The notation A 0(A 0 ) means that A is positive semidefinite (positive definite), A * denotes the complex conjugate transpose of A, and I is the identity matrix. Moreover, A B(A B ) is used as a different notation for A B 0 (A B 0 ) . We denote by (A) A
the spectral radius of A. The norm used in this paper is the spectral norm of the matrix A unless otherwise noted.
2 Some properties of the solutions In this section, we'll discuss some properties of positive definite solutions of the matrix (1). Theorem 2.1 If m and M are the smallest and the largest eigenvalues of a solution X of (1), respectively, and is an eigenvalues of A, then
m 1 M 1 . n m Mn Proof. Let v be an eigenvector corresponding to an eigenvalue of the matrix A and v 1 . Since the solution X of (1) is a positive definite matrix then
0 m 1. n v ,v v ,v v ,v
On positive definite solutions of the nonlinear matrix equation
109
v ,v n v ,v v ,v v ,v n v , v 1 v ,v
2
nv ,v 1
m m n 1 M M n . 2
2
Consequently,
m 1 M 1 . n m Mn
Theorem 2.2 If (1) has a positive definite solution
X , then .
Proof. Since X be a positive definite solution of (1), X I ,
X A X n A ,
X A A . 3 The first iteration method (Fixed point iteration method): In this section, we'll establish the first iterative method for obtaining a positive definite solution of (1). Algorithm 3.1 Take 0 . For k 0 ,1 , 2 ,.... compute
k 1 * kn ,
(2)
where 0 . We'll prove that the sequence (2) converges to a positive definite solution will satisfies (1). Lemma 3.1 [13] Let and Z be positive definite matrices for which and hold. Then n n . Now, we'll give some properties of the terms of the sequence: Proposition 3.1 If the matrix A is normal as ( ), then k k , k 0 ,1, 2 ,.... , where k is the iterates in Algorithm 3.1. Proof.
Since 0 then 0 0 . Since , we have
n n , n n ,
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Sana'a A. Zarea and Salah Mohammed El-Sayed
n n , 1 1 .
Assume that for each s, s s is satisfied. We will prove
s 1 s 1 for all s.
From s s , we compute
sn sn ,
sn sn , s 1 s 1 , for all s. Proposition 3.2
If the matrix A is normal,
k 0,1, 2,..., where k is
then k 1 k k k 1 ,
the iterates in Algorithm 3.1.
Since 0 and 1 n then 0 1 1 0 .We compute
Proof.
1 2 n 1n n 1n n 1n ,
2 1 1n n n 1n n 1n , from , we obtain ,
n n , n n ,
n
n ,
1n 1n , 1n 1n , n 1n n 1n . Hence, 1 2 2 1 . Assume that for each s
s s 1 s 1 s ,
(3)
is satisfied. We 'll prove s 1 s s s 1 , for each s. Since s s we compute sn sn ,
sn sn , sn1 sn sn1 sn , sn1 sn sn1 sn .
(4)
On positive definite solutions of the nonlinear matrix equation
111
According to Proposition 3.1 and equality (3), we have sn1 sn sn sn1 . From the equality (4) we get
sn1 sn sn sn1 , sn1 sn sn sn1 . Hence
n s
n s 1
sn1 sn ,
s 1 s s s 1 . The next theorems give the necessary and sufficient conditions for the existence of a solution of (1). Theorem 3.1 Let the sequence k be determined by the Algorithm 3.1 and two real numbers and , 1 , for which
1
n
1 I
A A
1
n
1 I
and n n 1
2
1.
(5)
k
converges to , which is a positive definite solution of (1) for all numbers and , 1 , for which 1 1 1 I A A n 1 I . Moreover, if k 0 for every k , then (1) n has a positive definite solution. If (1) has a positive definite solution, then
From algorithm 3.1, we have X 0 I , 1 X 1 n A A I n ( n ( 1 ) I ), X 1 I I I I X 0 and X 2 X 1n n X 1 , i.e.,
Proof.
X 0 X 1 X 2. To prove X s 1 X s , suppose that s s 1 for all s. According to Lemma 3.1 we have that A X sn A A X sn1 A , I A X sn A I A X sn1A , X s 1 X s .
s
is decreasing. To prove it is bounded, let X 0 I I , 1 X 1 I n A A I n A A I n ( n ( 1)I ) I , i.e., X 0 I and X 1 I .
The sequence
To prove that
X s I , for all s, assume that X k I , we want to prove
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X k 1 I , X k I , X kn n I ,
X s I . The sequence
s
1
( 1)I ) I , thus X k 1 I and n is monotonic decreasing and bounded from
I A * X kn A I A * n IA I n (
below by I . Moreover its limit exists, that is, lim X k X , by taking the k
limit of 3.1, we get X I A X A and the solution X satisfies (1). Suppose X s 0, for all s, we proved that the limit of s exists. Let X s 1 I A X sn A 0 , X I A X n A 0 .
n
by taking the limit of both sided, we have Consequently, (1) has positive definite solution.
Theorem 3.2 Let k be the iterates in Algorithm 3.1 and there exists two real numbers & , 1 , for which 1 1 (i) 1 I A A n 1 I , n (ii)
q n n 1 1 . 2
k k Then k q 0 q , where solution of (1).
is a positive definite
Proof. From Theorem 3.1, it follows that the sequence (2) is convergent to a positive definite Solution X of (1). We compute the spectral norm of the matrix k , we obtain
X k X I A X kn1A I A X n A A X kn1 X A
2
n
X k 1 X X kn11X
j 1
n
A
A
2
X kn1 X
n
,
.
j 1
n
From Theorem 3.1, we have X & X k 1 , 2 n n 1 , let j 1
q n
n 1
2
A . Consequently,
X k X n n 1 A
2
X k 1 X ,
q X k 1 X , after k-steps, qk X 0 X
q k . Corollary 3.1 Assume that (1) has a solution. If q n n 1 1 , then converges to with at least the linear convergence rate. 2
X k
On positive definite solutions of the nonlinear matrix equation
Proof.
As we have k n n 1
choose a real number that satisfies such that for any k N ,
113
k 1 , q n n 1 . Then
2
2
q 1. Since k , there exists a N,
q . Hence
X k 1 X 2 X k X .
Theorem 3.3 If (1) has a positive definite solution and after k iterative steps of Algorithm 3.1 and, we have k1 k 1 , 0 , then
k kn n n , where 2
k
is the iterates in Algorithm 3.1.
Proof. Since, k kn k kn k kn1 kn1 kn , then take norm in both sides,
k kn
2
kn1 kn n n 1
2
k k1 k 1 . Since
k and k , k1 1 and X k 1 X as k , n n consequently, I k1 k 1 0 then, k k n . 2
4 The second iteration method (Two sided iteration method of the fixed point iteration method): In this section, we'll establish the second iterative method for obtaining a positive definite solution of (1). Algorithm 4.1 Take X 0 I and Y 0 I . For k 0,1, 2,..... compute X k 1 I A X kn A and Y k 1 I A Y kn A ,
(6)
where & are two positive real numbers. Our theorems give sufficient conditions for the existence of a solution of (1). Theorem 4.1 Let the two sequences
k
and Υ
Algorithm 4.1 and two real numbers 1 1 1 I A A n 1 I and n
k
are determined by the
and , 1 , for which
q n n 1 1 2
If (1) has a positive definite solution, then
(7)
k
and Υ
k
converges to the
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Sana'a A. Zarea and Salah Mohammed El-Sayed
same limit
which is a positive definite solution of (1) for all numbers
and , 1 , for which
k 0 1
n
1 I
Proof.
A A
1
n
1 I
for every
Yk 0
and
1 n
1 I ,
A A
k
and
1
1 I .
n
Moreover,
1 ,
for
if
which
then (1) has a positive definite solution.
From algorithm 4.1, we have 1 n 0
and 2 X 1n 0n 1 , that is, 2 1 0 . Assume that s s 1 is true, To prove that s 1 s , we have X s 1 I A X sn A I A X sn1A X s , thus k 1 k , k 0 ,1, 2 ,... . The
sequence
X s is increasing. For Υ s , From sequence (6), we have
Y 1 I A Y 0n A I n A A I Y 0 and Y 2 I A Y 1n A I A Y 0n A Y 1 ,
that is, Y 2 Y 1 Y 0 . Suppose that s s 1 is true. To prove that s 1 s , we have Y s 1 I A Y sn A I A Y sn1A Y s , k 1 k , k 0,1, 2,..... , then sequence
X s
is decreasing.
To prove that Y k X k , for k 0,1, 2,... . From algorithm 4.1, Y 0 X 0 I I I 0 , 0 0 ,
Y 1 X 1 I A Y 0n A I A X 0n A A Y 0n X 0n A n n A A 0 ,
1, that is Y 1 X 1 , for Y s 1 X s 1 . Suppose that Y s X s , Y sn X sn and Y s 1 I A Y sn A I A X sn A X s 1 . Hence k k , k 0 ,1 , 2 ,... . We have k k 1 k 1 k , k 0 ,1, 2 ,... and
X k , Y k , for all
k 0 ,1 , 2 ,... . To prove the two sequences X s and Y s have the same limit,
that is, Y k X k 0 as k , from algorithm 4.1, we have
k 1 k k 1 k ,
Y k X k A Y kn1 X kn1 A Y kn1 X kn1 A 2
2
Y k 1 X k 1 Y kn1 j X kj 11 j 1 2 n 1 n A Y k 1 X k 1 , let q n n 1 A
A
2
q Y0 X 0 k
q k ( ).
Y k 1 X k 1
n
Y j 1
n
2
after k-steps,
nj k 1
X kj 11
On positive definite solutions of the nonlinear matrix equation
115
k k 0 as k .
From (7), q k 1, we get
X k and Y k are monotonic,
We have proved that the two sequences
bounded and went to the same limit as k , say that X is, lim X k limY k X . By taking the limit of 4.1, we got X I A X n A and k
k
the solution X satisfies (1). Moreover, Suppose X k 0 and Y k 0 for all k, we proved that the limits of
X k
and Y k exist. Let X k 1 I A X kn A 0 and Y k 1 I A Y kn A 0 ,
by taking the limits as k for the sequences
X k and Y k , we have
X I A X n A 0. Consequently, (1) has positive definite matrix.
Theorem 4.2
Let the two sequences
and Υ
k
k
are determined by the
Algorithm 4.1 and two real numbers and , 1 , for which 1 1 2 1 I A A n 1 I and q n n 1 1 , then n
X k X q k X 0 X q k and
Y k Y q k Y 0 X q k ,
where X is a positive definite solution of (1) and k , k , k 0 ,1 , 2 ,... are defined in (6). Proof. From Theorem 4.1, it follows that the sequence (6) is convergent to a positive definite solution X of (1). We'll compute the spectral norm of the matrix k , k , then
X k X I A X kn1A I A X n A A X kn1 X A
2
X kn1 X
n
A
2
X k 1 X
n
X kn11X
q n n 1 A . Consequently, X k X n n 1 A
2
X k 1 X ,
q X k 1 X , after k-steps, qk X 0 X q k . . Similarly, we can prove that
Y k X q k Y 0 X q k .
A
j 1
.
j 1
From Theorem 4.1, we had X & X k 1 , 2
n
n
j 1
2
n n 1 , let
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Theorem 4.3 If the equation (1) has a solution and after k - iterative steps of the Algorithm 4.1, X k1X k 1 I and Y k1Y k 1 I , X k 1 X
k
X k Xn and k 1 k k kn , then k
k kn k kn n n , where k and Y k are 2
the iterates in Algorithm 4.1. Proof. Since Y k Y k 1 Y k A Y kn Y k A Y kn A Y k A Y kn 1A =A Y kn Y kn 1 A . Take norm in both sides,
Y k Y k 1 Y k A Y kn A I A Y kn Y kn1 n n 1 A Y k Y k 1 2
2
n n 1 A Y k I Y k1Y k 1 n n A . 2
For Y k1Y k 1 , Y k Y
2
as
k . Consequently, Y k1Y k 1 0 as
k , that is, Y k1Y k 1 , for 0 and from theorem 4.1, for every k , thus
Yk
Y k A Y kn A I n n A . From Theorem 4.1, we had 2
k 1 k k 1 . k Since X k 1 X k X k X kn
and
k 1 k k kn , then
k kn k kn n n . 2
5 Numerical Experiments In this section, the numerical experiments are used to display the flexibility of the algorithms. The solutions are computed for some different matrices with different sizes n . For the following examples, the practical stopping criterion is k 10 9 and the solution is 500 .
5.1. Numerical experiments for Algorithm 3.1 In table 1, we denote
q n n 1 , 1 k , 2 k kn , where 2
is the solution which is obtained by the iterative method 3.1.
On positive definite solutions of the nonlinear matrix equation
117
Example 1: Let
A aij :
1 1 m i aij 0.01
i j, i j,
see Table 1.
5.2 Numerical experiments for Algorithm 4.1: In table 2 and 3, we denote 2 q n n 1 , 1 k , 2 k , 3 k k ,
4 k kn , 5 k kn , where is the solution which is obtained by the iterative method 4.1. Example 2: Let 1.001, 1.0003 and 0 0 . 1 0 . 0 4 0 . 2 0 . 2 0 . 0 8 0 . 5 0 . 0 1 2 , A 0.283853, A is normal, see Table 2. A 0 0.5 0 0 . 1 0.012 0 . 1 0 . 0 4 0.1
Example 3:
Let 1.25 , 1.50 , A 0.0070375 and
0.8 0.1 1 A 0.1 0.8 200 0.4 0.9
0.5 0.1 , A is not normal, see Table 3. 0.3
6 Conclusion In this paper, we have studied some properties of a positive definite solution of the matrix equation (1). The two effective iterative methods were proposed for computing a positive definite solution of this equation. Some properties of the terms of the sequence (2) were given. Also, the necessary and sufficient conditions for the convergence to positive definite solutions were deduced. Finally, we gave three numerical examples to illustrate the efficiency of the proposed algorithms.
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Sana'a A. Zarea and Salah Mohammed El-Sayed
N 2
8 10
Table 1: Relationship Between N, 1 and 2 for Different Sizes of A q K M A 1 2 qk 1.1 1.1 1.1 1.1 1.1 1.01 1.01 1.1 1.01
15 25 50 15 15 25 30 15 25
0.182416 0.266331 0.503586 0.182416 0.182416 0.266331 0.312163 0.182416 0.266331
0.0798612 0.156051 0.608636 0.23 0.953857 0.608393 0.8358 0.784619 0.775777
5 5 6 6 8 6 6 7 6
3.92389E-11 2.89156E-11 3.8666E-11 2.27291E-11 2.40752E-12 4.60475E-11 9.24066E-11 1.46333E-10 1.83597E-10
3.24848E-06 0.0000925414 0.0508332 0.0000740186 0.685279 0.0507113 0.34089 0.183068 0.217982
1.18108E-11 5.75835E-12 5.39865E-12 5.11757E-12 7.90503E-13 8.99492E-12 1.64107E-11 4.65931E-11 3.56181E-11
Table 2: Relationship Between N, 1 , 2 , 3 , 4 and 5 N
K
q
1
2
3
4
5
2 4 N6 8
9 1 2 6 3 1 0
0.161306 0.323258 0.404477 0.649107
5.67168E-11 2.31813E-10 2.43184E-10 9.97108E-10
5.59390E-11 2.29249E-10 2.41266E-10 9.94297E-10
7.77865E-13 2.56378E-12 1.91800E-12 2.81071E-12
3.57664E-11 1.26091E-10 1.03201E-10 1.80797E-10
3.52763E-11 1.24697E-10 1.02387E-10 1. 80287E-10
Table 3: Relationship Between N, 1 , 2 , 3 , 4 and 5 q N K 1 2 4 3 2 4 6 8 1 0
3 0.000148579 3 0.000668607 3 0.00225655 3 0.0067696 3 0.0190396
4.07674E-14 4.17987E-13 1. 83669E-12 5.75702E-12 1.50848E-11
4.06024E-14 1.17809E-12 6. 78421E-12 2.86359E-11 1.03411E-10
4.98351E-14 7.60106E-13 4. 94752E-12 2.28789E-11 8.83266E-11
3.49625E-14 3.58531E-13 1.57553E-12 4.93814E-12 1.29389E-11
5 7.77189E-14 1.01057E-12 5.81933E-12 2. 4565E-11 8.87329E-11
Acknowledgements. This work was financially supported by Princess Nourah Bint Abdulrahman University.
References [1] B. Zhou, G.-B. Cai and J. Lam, Positive definite solutions of the nonlinear matrix equation X A H X 1A I , Applied Mathematics and Computation, 219 (2013), 7377 - 7391. http://dx.doi.org/10.1016/j.amc.2013.01.021 [2] C. H. Guo and P. Lancaster, Iterative solution of two matrix equations, Mathematics of Computations, 68 (1999), 1589 - 1603. http://dx.doi.org/10.1090/s0025-5718-99-01122-9 [3] C-H. Guo and W-W. Lin, The matrix equation X A T X 1 A Q and its application in Nano research, SIAM Journal on Scientific Computing, 32(5) (2010), 3020 - 3038. http://dx.doi.org/10.1137/090758209
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[4] D. Hua and P. Lancaster, Linear matrix equation from an inverse problem of vibration theory, Linear Algebra and its Applications, 246 (1996), 31 - 47. http://dx.doi.org/10.1016/0024-3795(94)00311-4 [5] J. Engwerda, On the existence of a positive definite solution of the matrix equation X AT X 1 A I , Linear Algebra and its Applications, 194 (1993), 91108. http://dx.doi.org/10.1016/0024-3795(93)90115-5 [6] L. Hung, The explicit solution and solvability of the linear matrix equations, Linear Algebra and its Applications, 311 (2000), 195 - 199. http://dx.doi.org/10.1016/s0024-3795(00)00049-5 [7] P. Lancaster and L. Rodman, Algebraic Riccati Equations, Oxford Science Publishers, Oxford, 1995. [8] Q. Li and P. Liu, Positive Definite Solutions of a Kind of Nonlinear Matrix Equations, Journal of Information & Computational Science, 7 (2010), 15271533. [9] R. Bhatia, Matrix Analysis, Graduate Texts in Mathematics 169, Springer -Verlag , New York, 1997. http://dx.doi.org/10.1007/978-1-4612-0653-8 [10] S. A. Zarea, S. M. El-Sayed and A. A. Al-Eidan, Iteration method to solve the equation X A X r A I , Assiut University Journal of Mathematics & Computer Science, 38 (1) (2009), 11 - 24. [11] S. A. Zarea and S. M. El-Sayed, On the Matrix Equation X + A* X-a A = I, International Journal of Computational Mathematics and Simulation, 1(1) (2008), 89 - 97. [12] S. M. El-Sayed, Two iterations processes for computing positive definite solutions of the matrix equation X A X n A Q , Computers & Mathematics with Applications, 41 (2001), 579 - 588. http://dx.doi.org/10.1016/s0898-1221(00)00301-1 [13] S. M. El-Sayed and A. M. Al-Dbiban, On positive definite solutions of nonlinear matrix equation X+A*X-nA=I, Applied Mathematics and Computation, 151 (2004), 533 - 541. http://dx.doi.org/10.1016/s0096-3003(03)00360-6 [14] X. Yin, S. Liu and T. Li, On positive definite solutions of a matrix equation X+A*X-qA=Q, (0 q 1 ) , The Taiwanese Journal of Mathematics, 16(4) (2012), 1391 - 1407. [15] X. Zhan, Computing the extremal positive definite solutions of a matrix
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Received: November 10, 2014; Published: December 22, 2014
On Positive Definite Solutions of the Nonlinear Matrix Equation X A* X n A I Sana'a A. Zarea* Mathematical Sciences Department Faculty of Sciences Princess Nourah Bint Abdul Rahman University Riyadh, Saudi Arabia *Corresponding author Salah Mohammed El-Sayed Department of Scientific Computing Faculty of Computers and Informatics Benha University, Benha, Egypt Copyright © 2014 Sana'a A. Zarea and Salah Mohammed El-Sayed. 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.
Abstract In this paper, we have discussed some properties of a positive definite solution of the non linear matrix equation X A* X n A I , where n is a positive integer. Two effective iterative methods for computing a positive definite solution of this equation are proposed. The necessary and sufficient conditions for existence of a positive definite solution are derived. Some numerical examples will be presented for illustrative purposes. Mathematics Subject Classification: 15A24; 65F35 Keywords: Nonlinear matrix equation, Properties of solutions, Iterative algorithms
108
Sana'a A. Zarea and Salah Mohammed El-Sayed
1 Introduction We consider the matrix equation: (1) X A* X n A I where n is a positive integer. The nonlinear matrix equation (1) arises in a wide variety of applications and research areas, including automatic control, ladder networks, dynamic programming, stochastic filtering and statistics [4, 7, 9]. If n is a negative real numbers, the positive definite solutions of (1) have been studied in some special cases [1-6, 8, 10-15]. In this paper, we'll mainly discuss the basic properties and some analysis of the positive definite solutions of (1). The rest of the paper is organized as follows: In Section 2, we show that (1) always has a positive definite solution and we give the necessary and sufficient conditions for the equation to have solutions. In Sections 3 and 4, we'll discuss two iterations methods for solving (1). Finally, we'll give three numerical examples in Section 5 to illustrate our theoretical results. The aim of this paper is to find the positive definite solution of (1). In this paper, we will use mathematical induction technique in the most proofs. The following notations are used throughout the rest of the paper. The notation A 0(A 0 ) means that A is positive semidefinite (positive definite), A * denotes the complex conjugate transpose of A, and I is the identity matrix. Moreover, A B(A B ) is used as a different notation for A B 0 (A B 0 ) . We denote by (A) A
the spectral radius of A. The norm used in this paper is the spectral norm of the matrix A unless otherwise noted.
2 Some properties of the solutions In this section, we'll discuss some properties of positive definite solutions of the matrix (1). Theorem 2.1 If m and M are the smallest and the largest eigenvalues of a solution X of (1), respectively, and is an eigenvalues of A, then
m 1 M 1 . n m Mn Proof. Let v be an eigenvector corresponding to an eigenvalue of the matrix A and v 1 . Since the solution X of (1) is a positive definite matrix then
0 m 1. n v ,v v ,v v ,v
On positive definite solutions of the nonlinear matrix equation
109
v ,v n v ,v v ,v v ,v n v , v 1 v ,v
2
nv ,v 1
m m n 1 M M n . 2
2
Consequently,
m 1 M 1 . n m Mn
Theorem 2.2 If (1) has a positive definite solution
X , then .
Proof. Since X be a positive definite solution of (1), X I ,
X A X n A ,
X A A . 3 The first iteration method (Fixed point iteration method): In this section, we'll establish the first iterative method for obtaining a positive definite solution of (1). Algorithm 3.1 Take 0 . For k 0 ,1 , 2 ,.... compute
k 1 * kn ,
(2)
where 0 . We'll prove that the sequence (2) converges to a positive definite solution will satisfies (1). Lemma 3.1 [13] Let and Z be positive definite matrices for which and hold. Then n n . Now, we'll give some properties of the terms of the sequence: Proposition 3.1 If the matrix A is normal as ( ), then k k , k 0 ,1, 2 ,.... , where k is the iterates in Algorithm 3.1. Proof.
Since 0 then 0 0 . Since , we have
n n , n n ,
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Sana'a A. Zarea and Salah Mohammed El-Sayed
n n , 1 1 .
Assume that for each s, s s is satisfied. We will prove
s 1 s 1 for all s.
From s s , we compute
sn sn ,
sn sn , s 1 s 1 , for all s. Proposition 3.2
If the matrix A is normal,
k 0,1, 2,..., where k is
then k 1 k k k 1 ,
the iterates in Algorithm 3.1.
Since 0 and 1 n then 0 1 1 0 .We compute
Proof.
1 2 n 1n n 1n n 1n ,
2 1 1n n n 1n n 1n , from , we obtain ,
n n , n n ,
n
n ,
1n 1n , 1n 1n , n 1n n 1n . Hence, 1 2 2 1 . Assume that for each s
s s 1 s 1 s ,
(3)
is satisfied. We 'll prove s 1 s s s 1 , for each s. Since s s we compute sn sn ,
sn sn , sn1 sn sn1 sn , sn1 sn sn1 sn .
(4)
On positive definite solutions of the nonlinear matrix equation
111
According to Proposition 3.1 and equality (3), we have sn1 sn sn sn1 . From the equality (4) we get
sn1 sn sn sn1 , sn1 sn sn sn1 . Hence
n s
n s 1
sn1 sn ,
s 1 s s s 1 . The next theorems give the necessary and sufficient conditions for the existence of a solution of (1). Theorem 3.1 Let the sequence k be determined by the Algorithm 3.1 and two real numbers and , 1 , for which
1
n
1 I
A A
1
n
1 I
and n n 1
2
1.
(5)
k
converges to , which is a positive definite solution of (1) for all numbers and , 1 , for which 1 1 1 I A A n 1 I . Moreover, if k 0 for every k , then (1) n has a positive definite solution. If (1) has a positive definite solution, then
From algorithm 3.1, we have X 0 I , 1 X 1 n A A I n ( n ( 1 ) I ), X 1 I I I I X 0 and X 2 X 1n n X 1 , i.e.,
Proof.
X 0 X 1 X 2. To prove X s 1 X s , suppose that s s 1 for all s. According to Lemma 3.1 we have that A X sn A A X sn1 A , I A X sn A I A X sn1A , X s 1 X s .
s
is decreasing. To prove it is bounded, let X 0 I I , 1 X 1 I n A A I n A A I n ( n ( 1)I ) I , i.e., X 0 I and X 1 I .
The sequence
To prove that
X s I , for all s, assume that X k I , we want to prove
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X k 1 I , X k I , X kn n I ,
X s I . The sequence
s
1
( 1)I ) I , thus X k 1 I and n is monotonic decreasing and bounded from
I A * X kn A I A * n IA I n (
below by I . Moreover its limit exists, that is, lim X k X , by taking the k
limit of 3.1, we get X I A X A and the solution X satisfies (1). Suppose X s 0, for all s, we proved that the limit of s exists. Let X s 1 I A X sn A 0 , X I A X n A 0 .
n
by taking the limit of both sided, we have Consequently, (1) has positive definite solution.
Theorem 3.2 Let k be the iterates in Algorithm 3.1 and there exists two real numbers & , 1 , for which 1 1 (i) 1 I A A n 1 I , n (ii)
q n n 1 1 . 2
k k Then k q 0 q , where solution of (1).
is a positive definite
Proof. From Theorem 3.1, it follows that the sequence (2) is convergent to a positive definite Solution X of (1). We compute the spectral norm of the matrix k , we obtain
X k X I A X kn1A I A X n A A X kn1 X A
2
n
X k 1 X X kn11X
j 1
n
A
A
2
X kn1 X
n
,
.
j 1
n
From Theorem 3.1, we have X & X k 1 , 2 n n 1 , let j 1
q n
n 1
2
A . Consequently,
X k X n n 1 A
2
X k 1 X ,
q X k 1 X , after k-steps, qk X 0 X
q k . Corollary 3.1 Assume that (1) has a solution. If q n n 1 1 , then converges to with at least the linear convergence rate. 2
X k
On positive definite solutions of the nonlinear matrix equation
Proof.
As we have k n n 1
choose a real number that satisfies such that for any k N ,
113
k 1 , q n n 1 . Then
2
2
q 1. Since k , there exists a N,
q . Hence
X k 1 X 2 X k X .
Theorem 3.3 If (1) has a positive definite solution and after k iterative steps of Algorithm 3.1 and, we have k1 k 1 , 0 , then
k kn n n , where 2
k
is the iterates in Algorithm 3.1.
Proof. Since, k kn k kn k kn1 kn1 kn , then take norm in both sides,
k kn
2
kn1 kn n n 1
2
k k1 k 1 . Since
k and k , k1 1 and X k 1 X as k , n n consequently, I k1 k 1 0 then, k k n . 2
4 The second iteration method (Two sided iteration method of the fixed point iteration method): In this section, we'll establish the second iterative method for obtaining a positive definite solution of (1). Algorithm 4.1 Take X 0 I and Y 0 I . For k 0,1, 2,..... compute X k 1 I A X kn A and Y k 1 I A Y kn A ,
(6)
where & are two positive real numbers. Our theorems give sufficient conditions for the existence of a solution of (1). Theorem 4.1 Let the two sequences
k
and Υ
Algorithm 4.1 and two real numbers 1 1 1 I A A n 1 I and n
k
are determined by the
and , 1 , for which
q n n 1 1 2
If (1) has a positive definite solution, then
(7)
k
and Υ
k
converges to the
114
Sana'a A. Zarea and Salah Mohammed El-Sayed
same limit
which is a positive definite solution of (1) for all numbers
and , 1 , for which
k 0 1
n
1 I
Proof.
A A
1
n
1 I
for every
Yk 0
and
1 n
1 I ,
A A
k
and
1
1 I .
n
Moreover,
1 ,
for
if
which
then (1) has a positive definite solution.
From algorithm 4.1, we have 1 n 0
and 2 X 1n 0n 1 , that is, 2 1 0 . Assume that s s 1 is true, To prove that s 1 s , we have X s 1 I A X sn A I A X sn1A X s , thus k 1 k , k 0 ,1, 2 ,... . The
sequence
X s is increasing. For Υ s , From sequence (6), we have
Y 1 I A Y 0n A I n A A I Y 0 and Y 2 I A Y 1n A I A Y 0n A Y 1 ,
that is, Y 2 Y 1 Y 0 . Suppose that s s 1 is true. To prove that s 1 s , we have Y s 1 I A Y sn A I A Y sn1A Y s , k 1 k , k 0,1, 2,..... , then sequence
X s
is decreasing.
To prove that Y k X k , for k 0,1, 2,... . From algorithm 4.1, Y 0 X 0 I I I 0 , 0 0 ,
Y 1 X 1 I A Y 0n A I A X 0n A A Y 0n X 0n A n n A A 0 ,
1, that is Y 1 X 1 , for Y s 1 X s 1 . Suppose that Y s X s , Y sn X sn and Y s 1 I A Y sn A I A X sn A X s 1 . Hence k k , k 0 ,1 , 2 ,... . We have k k 1 k 1 k , k 0 ,1, 2 ,... and
X k , Y k , for all
k 0 ,1 , 2 ,... . To prove the two sequences X s and Y s have the same limit,
that is, Y k X k 0 as k , from algorithm 4.1, we have
k 1 k k 1 k ,
Y k X k A Y kn1 X kn1 A Y kn1 X kn1 A 2
2
Y k 1 X k 1 Y kn1 j X kj 11 j 1 2 n 1 n A Y k 1 X k 1 , let q n n 1 A
A
2
q Y0 X 0 k
q k ( ).
Y k 1 X k 1
n
Y j 1
n
2
after k-steps,
nj k 1
X kj 11
On positive definite solutions of the nonlinear matrix equation
115
k k 0 as k .
From (7), q k 1, we get
X k and Y k are monotonic,
We have proved that the two sequences
bounded and went to the same limit as k , say that X is, lim X k limY k X . By taking the limit of 4.1, we got X I A X n A and k
k
the solution X satisfies (1). Moreover, Suppose X k 0 and Y k 0 for all k, we proved that the limits of
X k
and Y k exist. Let X k 1 I A X kn A 0 and Y k 1 I A Y kn A 0 ,
by taking the limits as k for the sequences
X k and Y k , we have
X I A X n A 0. Consequently, (1) has positive definite matrix.
Theorem 4.2
Let the two sequences
and Υ
k
k
are determined by the
Algorithm 4.1 and two real numbers and , 1 , for which 1 1 2 1 I A A n 1 I and q n n 1 1 , then n
X k X q k X 0 X q k and
Y k Y q k Y 0 X q k ,
where X is a positive definite solution of (1) and k , k , k 0 ,1 , 2 ,... are defined in (6). Proof. From Theorem 4.1, it follows that the sequence (6) is convergent to a positive definite solution X of (1). We'll compute the spectral norm of the matrix k , k , then
X k X I A X kn1A I A X n A A X kn1 X A
2
X kn1 X
n
A
2
X k 1 X
n
X kn11X
q n n 1 A . Consequently, X k X n n 1 A
2
X k 1 X ,
q X k 1 X , after k-steps, qk X 0 X q k . . Similarly, we can prove that
Y k X q k Y 0 X q k .
A
j 1
.
j 1
From Theorem 4.1, we had X & X k 1 , 2
n
n
j 1
2
n n 1 , let
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Theorem 4.3 If the equation (1) has a solution and after k - iterative steps of the Algorithm 4.1, X k1X k 1 I and Y k1Y k 1 I , X k 1 X
k
X k Xn and k 1 k k kn , then k
k kn k kn n n , where k and Y k are 2
the iterates in Algorithm 4.1. Proof. Since Y k Y k 1 Y k A Y kn Y k A Y kn A Y k A Y kn 1A =A Y kn Y kn 1 A . Take norm in both sides,
Y k Y k 1 Y k A Y kn A I A Y kn Y kn1 n n 1 A Y k Y k 1 2
2
n n 1 A Y k I Y k1Y k 1 n n A . 2
For Y k1Y k 1 , Y k Y
2
as
k . Consequently, Y k1Y k 1 0 as
k , that is, Y k1Y k 1 , for 0 and from theorem 4.1, for every k , thus
Yk
Y k A Y kn A I n n A . From Theorem 4.1, we had 2
k 1 k k 1 . k Since X k 1 X k X k X kn
and
k 1 k k kn , then
k kn k kn n n . 2
5 Numerical Experiments In this section, the numerical experiments are used to display the flexibility of the algorithms. The solutions are computed for some different matrices with different sizes n . For the following examples, the practical stopping criterion is k 10 9 and the solution is 500 .
5.1. Numerical experiments for Algorithm 3.1 In table 1, we denote
q n n 1 , 1 k , 2 k kn , where 2
is the solution which is obtained by the iterative method 3.1.
On positive definite solutions of the nonlinear matrix equation
117
Example 1: Let
A aij :
1 1 m i aij 0.01
i j, i j,
see Table 1.
5.2 Numerical experiments for Algorithm 4.1: In table 2 and 3, we denote 2 q n n 1 , 1 k , 2 k , 3 k k ,
4 k kn , 5 k kn , where is the solution which is obtained by the iterative method 4.1. Example 2: Let 1.001, 1.0003 and 0 0 . 1 0 . 0 4 0 . 2 0 . 2 0 . 0 8 0 . 5 0 . 0 1 2 , A 0.283853, A is normal, see Table 2. A 0 0.5 0 0 . 1 0.012 0 . 1 0 . 0 4 0.1
Example 3:
Let 1.25 , 1.50 , A 0.0070375 and
0.8 0.1 1 A 0.1 0.8 200 0.4 0.9
0.5 0.1 , A is not normal, see Table 3. 0.3
6 Conclusion In this paper, we have studied some properties of a positive definite solution of the matrix equation (1). The two effective iterative methods were proposed for computing a positive definite solution of this equation. Some properties of the terms of the sequence (2) were given. Also, the necessary and sufficient conditions for the convergence to positive definite solutions were deduced. Finally, we gave three numerical examples to illustrate the efficiency of the proposed algorithms.
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Sana'a A. Zarea and Salah Mohammed El-Sayed
N 2
8 10
Table 1: Relationship Between N, 1 and 2 for Different Sizes of A q K M A 1 2 qk 1.1 1.1 1.1 1.1 1.1 1.01 1.01 1.1 1.01
15 25 50 15 15 25 30 15 25
0.182416 0.266331 0.503586 0.182416 0.182416 0.266331 0.312163 0.182416 0.266331
0.0798612 0.156051 0.608636 0.23 0.953857 0.608393 0.8358 0.784619 0.775777
5 5 6 6 8 6 6 7 6
3.92389E-11 2.89156E-11 3.8666E-11 2.27291E-11 2.40752E-12 4.60475E-11 9.24066E-11 1.46333E-10 1.83597E-10
3.24848E-06 0.0000925414 0.0508332 0.0000740186 0.685279 0.0507113 0.34089 0.183068 0.217982
1.18108E-11 5.75835E-12 5.39865E-12 5.11757E-12 7.90503E-13 8.99492E-12 1.64107E-11 4.65931E-11 3.56181E-11
Table 2: Relationship Between N, 1 , 2 , 3 , 4 and 5 N
K
q
1
2
3
4
5
2 4 N6 8
9 1 2 6 3 1 0
0.161306 0.323258 0.404477 0.649107
5.67168E-11 2.31813E-10 2.43184E-10 9.97108E-10
5.59390E-11 2.29249E-10 2.41266E-10 9.94297E-10
7.77865E-13 2.56378E-12 1.91800E-12 2.81071E-12
3.57664E-11 1.26091E-10 1.03201E-10 1.80797E-10
3.52763E-11 1.24697E-10 1.02387E-10 1. 80287E-10
Table 3: Relationship Between N, 1 , 2 , 3 , 4 and 5 q N K 1 2 4 3 2 4 6 8 1 0
3 0.000148579 3 0.000668607 3 0.00225655 3 0.0067696 3 0.0190396
4.07674E-14 4.17987E-13 1. 83669E-12 5.75702E-12 1.50848E-11
4.06024E-14 1.17809E-12 6. 78421E-12 2.86359E-11 1.03411E-10
4.98351E-14 7.60106E-13 4. 94752E-12 2.28789E-11 8.83266E-11
3.49625E-14 3.58531E-13 1.57553E-12 4.93814E-12 1.29389E-11
5 7.77189E-14 1.01057E-12 5.81933E-12 2. 4565E-11 8.87329E-11
Acknowledgements. This work was financially supported by Princess Nourah Bint Abdulrahman University.
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Received: November 10, 2014; Published: December 22, 2014
