# On the Iterative Method for the System of Nonlinear Matrix Equations

Authors: Asmaa M. Al-Dubiban
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The positive definite solutions for the system of nonlinear matrix equationsX+A∗Y−nA=I,Y+B∗X−mB=I are considered, where n,m are two positive integers and A, B are nonsingular complex matrices. Some sufficient conditions for the existence of positive definite solutions for the system are derived. Under some conditions, an iterative algorithm for computing the positive definite solutions for the system is proposed. Also, the estimation of the error is obtained. Finally, some numerical examples are given to show the efficiency of the proposed iterative algorithm.


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## Body

## 1. Introduction

Linear and nonlinear matrix equations have been widely used for solving many problems in several areas such as control theory, optimal control, optimization control, stability theory, communication system, dynamic programming, signal processing, and stochastic filtering and statistics, [1–3]. Many authors studied the existence of solutions for several classes of the matrix equations (see, e.g., [4–14]), in particular, Lyapunov matrix equation [15], Sylvester matrix equations [11, 14], algebraic Riccati equations [3], some special case of linear and nonlinear matrix equations [16–21], and coupled matrix equations [22–24].In recent years, many types of algebraic Riccati equations have been the subject of great activity, the aim being to achieve a fast and reliable algorithm that generates numerical positive definite solutions.In this paper, we will consider the system (Sys.) of nonlinear matrix equations that can be expressed in the form:(1)X+A*Y-nA=I,Y+B*X-mB=I,
where n,m are two positive integers,  X,Y are r×r unknown matrices,  I is the r×r identity matrix, and  A,B are nonsingular matrices. All matrices are defined over the complex field. The system of nonlinear matrix equations with the form of (1) is a special case of the system of algebraic discrete-type Riccati equations of the form:
(2)Xi=Vi*XiVi+QiDl-(Vi*XiBi+Ai)(Ri+Bi*XiBi)-1(Bi*XiVi+Ai*),
where i=1,2,…,k, [2, 3]. The efficient numerical solutions for some special case of the system (2) have been extensively studied by several authors [4–10, 22–26]. For example, Mukaidani [22] proposed a new algorithm for solving cross-coupled sign-indefinite algebraic Riccati equations for weakly coupled large-scale systems, while in [4, 5] Al-Dubiban has studied special cases of Sys. (2) by obtained sufficient conditions for the existence of positive definite solutions for the systems and proposed iterative algorithms to calculate the solutions. In [10], Davies proposed upper bounds for the sum of the maximal eigenvalues of the solutions of the continuous and discrete coupled algebraic Riccati equations. In [25], Ivanov has studied a set of discrete-time coupled algebraic Riccati equations which arise in quadratic optimal control and proposed two iterations for computing a symmetric solution of this system.In this paper, we derive the sufficient conditions of the existence of solutions for the Sys. (1). We introduce an iterative algorithm to obtain the positive definite solutions of Sys. (1). We discuss the convergence of this iterative algorithm. Finally, some numerical examples are given to illustrate the efficiency for suggested algorithm.The following notations are used throughout the rest of the paper. The notationA≥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 ρ(A) the spectral radius of  A;  λr(X),μr(Y) represent the eigenvalues of X and Y, respectively. The norm used in this paper is the spectral norm of the matrix A; that is,  ∥A∥=ρ(AA⋆) unless otherwise noted.


## 2. Main Theorems

In this section, we will introduce an iterative algorithm which is applicable for computing the positive definite solutions of the Sys. (1). We start with some results which will be used throughout this paper.Lemma 1 (see [27, 28]).
IfP>Q>0 (or P≥Q>0), then Pα>Qα (or Pα≥Qα>0) for all α∈(0,1], and Pα<Qα (or 0<Pα≤Qα) for all α∈[-1,0).Theorem 2 (see [29]).
Let the matricesP,Q, and R be positive definite r×r matrices, such that the integral
(3)∫0∞‍ePtReQtdt
exists and
(4)limt→∞ePtReQt=0;
then the matrix
(5)Z=-∫0∞‍ePtReQtdt
is the solution of the matrix equation:
(6)PZ+ZQ=R.The solution of Sys. (1) can be found by the following iterative algorithm.Algorithm 3.
(7)
X
0
=
Y
0
=
β
I
,
X
s
+
1
=
[
B
(
I
-
Y
s
)
-
1
B
*
]
1
/
m
,
Y
s
+
1
=
[
A
(
I
-
X
s
)
-
1
A
*
]
1
/
n
,
s
=
0,1
,
2
,
…
.Theorem 4.
If there exist numbersα,β satisfying 0<α<β≤min{n/(n+1),m/(m+1)}, and the following conditions hold:(i)
αn(1-α)I<AA*<βn(1-β)I,(ii)
αm(1-α)I<BB*<βm(1-β)I, then the sequences {Xs},{Ys} defined by Algorithm 3 converge to a positive definite solution (X,Y) of Sys. (1).Proof.
From Algorithm3, we get
(8)X1=[B(I-βI)-1B*]1/m=[1(1-β)BB*]1/m<[βm(1-β)(1-β)I]1/m=βI=X0.
Also, we have
(9)X1=[B(I-βI)-1B*]1/m=[1(1-β)BB*]1/m>[1(1-α)BB*]1/m>[αm(1-α)(1-α)I]1/m=αI.
That is,  X0>X1>αI, similarly we get
(10)Y1=[A(I-βI)-1A*]1/n=[1(1-β)AA*]1/n<[βn(1-β)(1-β)I]1/n=βI=Y0.
Also, we have
(11)Y1=[A(I-βI)-1A*]1/n=[1(1-β)AA*]1/n>[1(1-α)AA*]1/n>[αn(1-α)(1-α)I]1/n=αI.
That is,  Y0>Y1>αI.
Suppose that(12)Xs-1>Xs>αI,Ys-1>Ys>αI.
Now, we will prove that Xs>Xs+1>αI and Ys>Ys+1>αI.
By using the inequalities (12), we have
(13)Xs+1=[B(I-Ys)-1B*]1/m<[B(I-Ys-1)-1B*]1/m=Xs.
Also, we have
(14)Xs+1=[B(I-Ys)-1B*]1/m>[B(I-αI)-1B*]1/m=[1(1-α)BB*]1/m>[αm(1-α)(1-α)I]1/m=αI.
Similarly, we get
(15)Ys+1=[A(I-Xs)-1A*]1/n<[A(I-Xs-1)-1A*]1/n=Ys.
Also, we have
(16)Ys+1=[A(I-Xs)-1A*]1/n>[A(I-αI)-1A*]1/n=[1(1-α)AA*]1/n>[αn(1-α)(1-α)I]1/n=αI.
Therefore, the inequalities (12) are true for all s=1,2,3,….
Hence, the sequences{Xs},{Ys} are monotonically decreasing and bounded from below by the matrix αI. Consequently, the sequences converge to a positive definite limit (X,Y) which is a solution of Sys. (1).Theorem 5.
If there exist numbersα,β satisfying 0<α<β≤min{n/(n+1),m/(m+1)}, and the following conditions hold: (i)
αn(1-α)I<AA*<βn(1-β)I,(ii)
αm(1-α)I<BB*<βm(1-β)I,(iii)
q=β(2k+2t)/(2α)(k+t)(1-β)2<1, where m=2k,n=2t, then Sys. (1) has a positive definite solution (X,Y) which satisfies
(17)∥Xs-X∥≤q∥Xs-2-X∥≤⋯≤(q)s/2(β-α),∥Ys-Y∥≤q∥Ys-2-Y∥≤⋯≤(q)s/2(β-α).Proof.
From Theorem4, the two sequences {Xs},{Ys} defined by Algorithm 3 are convergent to a positive definite solution (X,Y) of Sys. (1). We compute the spectral norm of the matrices Xs-X,Ys-Y. For that, we have
(18)∥Xs-X∥=∥B(I-Ys-1)-1B*m-B(I-Y)-1B*m∥.
We denote
(19)P1=B(I-Ys-1)-1B*,Q1=B(I-Y)-1B*.
We use the following equality:
(20)P1m(P1m-Q1m)+(P1m-Q1m)Q1m=P1m/2-Q1m/2.Since Ys>Y>0 for each s=0,1,2,…, then by using Lemma 1 we have the matrix D1=P1m-Q1m being a positive definite solution of the matrix equation:
(21)P1mD1+D1Q1m=P1m/2-Q1m/2.
According to Theorem 2, we have
(22)D1=∫0∞‍e-P1mt(P1m/2-Q1m/2)e-Q1mtdt.
Since P1m,Q1m are positive definite matrices, then the integral (22) exists, and
(23)e-P1mt(P1m/2-Q1m/2)e-Q1mt→0,ast→∞.
By using (18) and (22), we have
(24)∥Xs-X∥≤∫0∞‍∥(P1m/2-Q1m/2)∥∥e-P1mt∥∥e-Q1mt∥dt.
However, Ys>Y≥αI; hence,
(25)P1m>αI,Q1m≥αI.
Then, we have
(26)∥Xs-X∥≤∥P1m/2-Q1m/2∥∫0∞∥e-P1mt∥∥e-Q1mt∥‍dt≤∥P1m/2-Q1m/2∥∫0∞∥e-αIt∥2dt‍=∥P1m/2-Q1m/2∥∫0∞e-2αtdt‍=12α∥P1m/2-Q1m/2∥≤(12α)2∥P1m/22-Q1m/22∥.
After k times as above, we get
(27)∥Xs-X∥≤(12α)k∥P1m/2k-Q1m/2k∥.
Let m=2k be special case, then we have
(28)∥Xs-X∥≤(12α)k∥P1-Q1∥=(12α)k∥B(I-Ys-1)-1B*-B(I-Y)-1B*∥=(12α)k∥B(I-Ys-1)-1(Ys-1-Y)(I-Y)-1B*∥≤(12α)k∥B∥2∥(I-Ys-1)-1∥∥(I-Y)-1∥∥Ys-1-Y∥.
Since Y<Ys-1≤βI, then (I-Y)-1<(I-Ys-1)-1≤(1/(1-β))I; that is,
(29)∥(I-Y)-1∥<∥(I-Ys-1)-1∥≤1(1-β).
Therefore, we get
(30)∥Xs-X∥≤(12α)kβ2k(1-β)∥Ys-1-Y∥.
Also, we have
(31)∥Ys-Y∥=∥A(I-Xs-1)-1A*n-A(I-X)-1A*n∥.
We denote
(32)P2=A(I-Xs-1)-1A*,Q2=A(I-X)-1A*.
We use the following equality:
(33)P2n(P2n-Q2n)+(P2n-Q2n)Q2n=P2n/2-Q2n/2.
SinceXs>X>0 for each s=0,1,2,…, then by using Lemma 1 we have the matrix D2=P2n-Q2n being a positive definite solution of the matrix equation:
(34)P2nD2+D2Q2n=P2n/2-Q2n/2.
According to Theorem 2, we have
(35)D2=∫0∞‍e-P2nt(P2n/2-Q2n/2)e-Q2ntdt.
Since P2n,Q2n are positive definite matrices, then the integral (35) exists, and
(36)e-P2nt(P2n/2-Q2n/2)e-Q2nt→0,ast→∞.
By using (31) and (35), we have
(37)∥Ys-Y∥≤∫0∞‍∥(P2n/2-Q2n/2)∥∥e-P2nt∥∥e-Q2nt∥dt.
However, Xs>X≥αI; hence,
(38)P2n>αI,Q2n≥αI.
Then, we have
(39)∥Ys-Y∥≤∥P2n/2-Q2n/2∥∫0∞∥e-P2nt∥∥e-Q2nt∥dt‍≤∥P2n/2-Q2n/2∥∫0∞∥e-αIt∥2‍dt=∥P2n/2-Q2n/2∥∫0∞e-2αtdt‍=12α∥P2n/2-Q2n/2∥≤(12α)2∥P2n/22-Q2n/22∥.
After t times as above, we get
(40)∥Ys-Y∥≤(12α)t∥P2n/2t-Q2n/2t∥.Let n=2t be special case, then we have
(41)∥Ys-Y∥≤(12α)t∥P2-Q2∥=(12α)t∥A(I-Xs-1)-1A*-A(I-X)-1A*∥=(12α)t∥A(I-Xs-1)-1(Xs-1-X)(I-X)-1A*∥≤(12α)t∥A∥2∥(I-Xs-1)-1∥∥(I-X)-1∥∥Xs-1-X∥.
SinceX<Xs-1≤βI, then (I-X)-1<(I-Xs-1)-1≤(1/(1-β))I; that is,
(42)∥(I-X)-1∥<∥(I-Xs-1)-1∥≤1(1-β).
Therefore, we get
(43)∥Ys-Y∥≤(12α)tβ2t(1-β)∥Xs-1-X∥.
By using (43) in (30) and (30) in (43), we have
(44)∥Xs-X∥≤q∥Xs-2-X∥≤⋯≤(q)s/2(β-α),∥Ys-Y∥≤q∥Ys-2-Y∥≤⋯≤(q)s/2(β-α),
which completes the proof.


## 3. Numerical Examples

We will give some numerical examples for computing the positive definite solution of the Sys. (1). The solution is computed for some different matrices A,B with different orders. Denote by X,Y the solutions which are obtained by Algorithm 3 and ϵ1(X)=∥X-Xs∥,ϵ1(Y)=∥Y-Ys∥.For computingZ1/n for all 1/n∈(0,1], we use the iterative algorithm.Algorithm 6.
(45)
W
0
Z
=
Z
W
0
,
W
s
+
1
=
1
n
[
(
n
-
1
)
W
s
+
W
s
1
-
n
Z
]
,
s
=
0
,

1
,

2
,
…
.
See [30].Example 7.
Consider Sys. (1) with n=3,m=2,β=0.5, and normal matrices
(46)A=diag[0.2,0.2,0.5],B=diag[0.4,0.3,0.1].
By using Algorithms 3 and 6, we have
(47)X=diag[0.535247,0.388225,0.174485],Y=diag[0.441516,0.402862,0.67154].
The results are given in Table1.Table 1
Error analysis for Example7.
s
ϵ
1
(
X
)
ϵ
1
(
Y
)
0
3.25515
E
-
01
1.71540
E
-
01
2
1.90010
E
-
01
1.28914
E
-
01
4
4.65789
E
-
02
3.11069
E
-
02
6
8.05431
E
-
03
5.57630
E
-
03
8
1.25249
E
-
03
8.71099
E
-
04
10
1.90796
E
-
04
1.32785
E
-
04
12
2.89695
E
-
05
2.01634
E
-
05
14
4.39639
E
-
06
3.06002
E
-
06
16
6.67142
E
-
07
4.64353
E
-
07
18
1.01236
E
-
07
7.04637
E
-
08
20
1.53621
E
-
08
1.06926
E
-
08Example 8.
Consider Sys. (1) with n=5,m=4,β=0.4 and normal matrices
(48)A=diag[-72r2,-62r2,…,r-82r2],B=diag[18+r,28+r,…,r8+r].
By using Algorithms 3 and 6, we have;
Whenr=4,
(49)X=diag[0.362018,0.510342,0.620206,0.707524],Y=diag[0.595686,0.590503,0.577588,0.556604].
Whenr=7,
(50)X=diag[0.290121,0.408797,0.497892,l0.570656,0.631828,0.68313,0.723835](51)Y=diag[0.372659,0.363432,0.349091,l0.329437,0.302795,0.265305,0.206669].
The results are given in Table2.Table 2
Error analysis for Example8.
r
s
ϵ
1
(
X
)
ϵ
1
(
Y
)
4
0
3.07524
E
-
01
1.95686
E
-
01
2
2.24755
E
-
01
1.44449
E
-
01
4
5.31418
E
-
02
5.07784
E
-
02
6
1.08011
E
-
02
1.03261
E
-
02
8
1.75364
E
-
03
1.66592
E
-
03
10
2.69238
E
-
04
2.55471
E
-
04
12
4.09499
E
-
05
3.88490
E
-
05
14
6.21930
E
-
06
5.90006
E
-
06
16
9.44353
E
-
07
8.95875
E
-
07
18
1.43388
E
-
07
1.36027
E
-
07
20
2.17715
E
-
08
2.06538
E
-
08
7
0
3.23835
E
-
01
1.93331
E
-
01
2
3.64875
E
-
01
4.33678
E
-
01
4
1.13231
E
-
01
2.06657
E
-
01
6
3.42800
E
-
02
6.90864
E
-
02
8
6.95159
E
-
03
7.88070
E
-
03
10
3.42099
E
-
04
2.93641
E
-
04
12
1.18241
E
-
05
1.00690
E
-
05
14
4.03846
E
-
07
3.43808
E
-
07
16
1.37875
E
-
08
1.17377
E
-
08Example 9.
Consider Sys. (1) with n=2,m=2,β=0.4, and matrices
(52)A=(0.2-0.10.10.2),B=(-0.40.10.20.3).
By using Algorithms 3 and 6, we have
(53)X=(0.497814-0.0839355-0.08393550.420533),Y=(0.333326-0.00556924-0.005569240.281592),λr(X)={0.551576,0.36677},μr(Y)={0.333919,0.280999}.
The results are given in Table3.Table 3
Error analysis for Example9.
s
ϵ
1
(
X
)
ϵ
1
(
Y
)
0
9.78137
E
-
02
1.18408
E
-
01
2
1.11953
E
-
01
9.96781
E
-
02
4
2.16876
E
-
02
1.76104
E
-
02
6
3.22178
E
-
03
2.84069
E
-
03
8
4.83451
E
-
04
4.38916
E
-
04
10
7.50050
E
-
05
6.69635
E
-
05
12
1.08812
E
-
05
1.00510
E
-
05
14
1.70240
E
-
06
1.53219
E
-
06
16
2.59464
E
-
07
2.32406
E
-
07
18
3.81681
E
-
08
3.49836
E
-
08Example 10.
Consider Sys. (1) with n=2,m=1,β=0.65, and matrices
(54)A=(-0.2-0.4-0.4-0.6-0.20.60.2-0.40.1),B=(-0.03-0.080000.0200.10).
By using Algorithms 3 and 6, we have
(55)X=(0.0573078-0.000371142-0.00702446-0.0003711420.000752950.0000206791-0.007024460.00002067910.000896794),Y=(0.596078-0.02336790.0748523-0.02336790.8870690.01045920.07485230.01045920.454549),λr(X)={0.0581718,0.000751403,0.0000343988},μr(Y)={0.888982,0.627173,0.42154}.
The results are given in Table4.Table 4
Error analysis for Example10.
s
ϵ
1
(
X
)
ϵ
1
(
Y
)
0
6.49247
E
-
01
2.37069
E
-
01
2
8.46092
E
-
02
1.48180
E
-
01
4
3.51586
E
-
03
5.04374
E
-
02
6
8.98887
E
-
04
1.30368
E
-
02
8
3.32057
E
-
04
4.71441
E
-
03
10
1.12763
E
-
04
1.57391
E
-
03
12
3.78219
E
-
05
5.11273
E
-
04
14
1.24881
E
-
05
1.64362
E
-
04
16
4.06906
E
-
06
5.26213
E
-
05
18
1.31421
E
-
06
1.68193
E
-
05
20
4.22232
E
-
07
5.37246
E
-
06
22
1.35257
E
-
07
1.71563
E
-
06
24
4.32590
E
-
08
5.47811
E
-
07
26
1.38239
E
-
08
1.74912
E
-
07
28
4.41568
E
-
09
5.58474
E
-
08


## 4. Conclusion

In this paper, the positive definite solutions for Sys. (1) have been tackled. We presented sufficient conditions for the existence of positive definite solutions for Sys. (1). Moreover, we discussed an iterative algorithm from which solutions can always be calculated numerically whenever the system is solvable. Finally, we gave numerical examples that illustrated the behavior of the proposed algorithm.



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*Source: 685753-2013-03-24.xml*