# Solving systems of nonlinear matrix equations involving Lipshitzian mappings

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## Abstract

In this study, both theoretical results and numerical methods are derived for solving different classes of systems of nonlinear matrix equations involving Lipshitzian mappings.

2000 Mathematics Subject Classifications: 15A24; 65H05.

## 1 Introduction

Fixed point theory is a very attractive subject, which has recently drawn much attention from the communities of physics, engineering, mathematics, etc. The Banach contraction principle [1] is one of the most important theorems in fixed point theory. It has applications in many diverse areas.

Definition 1.1 Let M be a nonempty set and f: MM be a given mapping. We say that x* M is a fixed point of f if fx* = x*.

Theorem 1.1 (Banach contraction principle [1]). Let (M, d) be a complete metric space and f: MM be a contractive mapping, i.e., there exists λ [0, 1) such that for all x, y M,

$d ( f x , f y ) ≤ λ d ( x , y ) .$
(1)

Then the mapping f has a unique fixed point x* M. Moreover, for every x0 M, the sequence (x k ) defined by: xk+1= fx k for all k = 0, 1, 2, ... converges to x*, and the error estimate is given by:

$d ( x k , x * ) ≤ λ k 1 - λ d ( x 0 , x 1 ) , f o r a l l k = 0 , 1 , 2 , …$

Many generalizations of Banach contraction principle exists in the literature. For more details, we refer the reader to [24].

To apply the Banach fixed point theorem, the choice of the metric plays a crucial role. In this study, we use the Thompson metric introduced by Thompson [5] for the study of solutions to systems of nonlinear matrix equations involving contractive mappings.

We first review the Thompson metric on the open convex cone P(n) (n ≥ 2), the set of all n×n Hermitian positive definite matrices. We endow P(n) with the Thompson metric defined by:

$d ( A , B ) = max log M ( A B ) , log M ( B A ) ,$

where M(A/B) = inf{λ > 0: AλB} = λ+(B-1/2AB-1/2), the maximal eigenvalue of B-1/2AB-1/2. Here, XY means that Y - X is positive semidefinite and X < Y means that Y - X is positive definite. Thompson [5] (cf. [6, 7]) has proved that P(n) is a complete metric space with respect to the Thompson metric d and d(A, B) = ||log(A-1/2BA-1/2)||, where ||·|| stands for the spectral norm. The Thompson metric exists on any open normal convex cones of real Banach spaces [5, 6]; in particular, the open convex cone of positive definite operators of a Hilbert space. It is invariant under the matrix inversion and congruence transformations, that is,

$d ( A , B ) = d ( A - 1 , B - 1 ) = d ( M A M * , M B M * )$
(2)

for any nonsingular matrix M. The other useful result is the nonpositive curvature property of the Thompson metric, that is,

$d ( X r , Y r ) ≤ r d ( X , Y ) , r ∈ [ 0 , 1 ] .$
(3)

By the invariant properties of the metric, we then have

$d ( M X r M * , M Y r M * ) ≤ | r | d ( X , Y ) , r ∈ [ - 1 , 1 ]$
(4)

for any X, Y P(n) and nonsingular matrix M.

Lemma 1.1 (see [8]). For all A, B, C, D P(n), we have

$d ( A + B , C + D ) ≤ max { d ( A , C ) , d ( B , D ) } .$

In particular,

$d ( A + B , A + C ) ≤ d ( B , C ) .$

## 2 Main result

In the last few years, there has been a constantly increasing interest in developing the theory and numerical approaches for HPD (Hermitian positive definite) solutions to different classes of nonlinear matrix equations (see [821]). In this study, we consider the following problem: Find (X1, X2, ..., X m ) (P(n)) m solution to the following system of nonlinear matrix equations:

$X i r i = Q i + ∑ j = 1 m A j * F i j ( X j ) A j α i j , i = 1 , 2 , … , m ,$
(5)

where r i ≥ 1, 0 < |α ij | ≤ 1, Q i ≥ 0, A i are nonsingular matrices, and F ij : P(n) → P (n) are Lipshitzian mappings, that is,

$sup X , Y ∈ P ( n ) , X ≠ Y d ( F i j ( X ) , F i j ( Y ) ) d ( X , Y ) = k i j < ∞ .$
(6)

If m = 1 and α11 = 1, then (5) reduces to find X P(n) solution to Xr = Q + A*F(X)A. Such problem was studied by Liao et al. [15]. Now, we introduce the following definition.

Definition 2.1 We say that Problem (5) is Banach admissible if the following hypothesis is satisfied:

$max 1 ≤ i ≤ m max 1 ≤ j ≤ m { | α i j | k i j ∕ r i } < 1 .$

Our main result is the following.

Theorem 2.1 If Problem (5) is Banach admissible, then it has one and only one solution$( X 1 * , X 2 * , … , X m * ) ∈ ( P ( n ) ) m$. Moreover, for any (X1(0), X2(0), ..., X m (0)) (P(n)) m, the sequences (X i (k))k≥0, 1 ≤ im, defined by:

$X i ( k + 1 ) = Q i + ∑ j = 1 m ( A j * F i j ( X j ( k ) ) A j ) α i j 1 ∕ r i ,$
(7)

converge respectively to$X 1 * , X 2 * ,…, X m *$, and the error estimation is

$max { d ( X 1 ( k ) , X 1 * ) , d ( X 2 ( k ) , X 2 * ) , … , d ( X m ( k ) , X m * ) } ≤ q m k 1 - q m max { d ( X 1 ( 1 ) , X 1 ( 0 ) ) , d ( X 2 ( 1 ) , X 2 ( 0 ) ) , … , d ( X m ( 1 ) , X m ( 0 ) ) } ,$
(8)

where

$q m = max 1 ≤ i ≤ m max 1 ≤ j ≤ m { | α i j | k i j ∕ r i } .$

Proof. Define the mapping G: (P(n)) m → (P(n)) m by:

$G ( X 1 , X 2 , … , X m ) = ( G 1 ( X 1 , X 2 , … , X m ) , G 2 ( X 1 , X 2 , … , X m ) , … , G m ( X 1 , X 2 , … , X m ) ) ,$

for all X = (X1, X2, ..., X m ) (P(n)) m , where

$G i ( X ) = Q i + ∑ j = 1 m ( A j * F i j ( X j ) A j ) α i j 1 ∕ r i ,$

for all i = 1, 2, ..., m. We endow (P(n)) m with the metric d m defined by:

$d m ( ( X 1 , X 2 , … , X m ) , ( Y 1 , Y 2 , … , Y m ) ) = max d ( X 1 , Y 1 ) , d ( X 2 , Y 2 ) , … , d ( X m , Y m ) ,$

for all X = (X1, X2, ..., X m ), Y = (Y1, Y2, ..., Y m ) (P (n)) m . Obviously, ((P(n)) m , d m ) is a complete metric space.

We claim that

(9)

For all X, Y (P(n)) m , We have

$d m ( G ( X ) , G ( Y ) ) = max 1 ≤ i ≤ m { d ( G i ( X ) , G i ( Y ) ) } .$
(10)

On the other hand, using the properties of the Thompson metric (see Section 1), for all i = 1, 2, ..., m, we have

$d ( G i ( X ), G i ( Y ) ) = d ( ( Q i + ∑ j = 1 m ( A j * F i j ( X j ) A j ) α i j ) 1 / r i , ( Q i + ∑ j = 1 m ( A j * F i j ( Y j ) A j ) α i j ) 1 / r i ) ≤ 1 r i d ( Q i + ∑ j = 1 m ( A j * F i j ( X j ) A j ) α i j , Q i + ∑ j = 1 m ( A j * F i j ( Y j ) A j ) α i j ) ≤ 1 r i d ( ∑ j = 1 m ( A j * F i j ( X j ) A j ) α i j , ∑ j = 1 m ( A j * F i j ( Y j ) A j ) α i j ) ≤ 1 r i d ( ( A 1 * F i 1 ( X 1 ) A 1 ) α i 1 + ∑ j = 2 m ( A j * F i j ( X j ) A j ) α i j , ( A 1 * F i 1 ( Y 1 ) A 1 ) α i 1 + ∑ j = 2 m ( A j * F i j ( Y j ) A j ) α i j ) ≤ 1 r i m a x { d ( ( A 1 * F i 1 ( X 1 ) A 1 ) α i 1 , ( A 1 * F i 1 ( Y 1 ) A 1 ) α i 1 ), d ( ∑ j = 2 m ( A j * F i j ( X j ) A j ) α i j , ∑ j = 2 m ( A j * F i j ( Y j ) A j ) α i j ) } ≤ ⋅ ⋅ ⋅ ≤ 1 r i m a x { d ( ( A 1 * F i 1 ( X 1 ) A 1 ) α i 1 , ( A 1 * F i 1 ( Y 1 ) A 1 ) α i 1 ), … , d ( ( A m * F i m ( X m ) A m ) α i m , ( A m * F i m ( Y m ) A m ) α i m ) } ≤ 1 r i m a x { | α i 1 | d ( A 1 * F i 1 ( X 1 ) A 1 , A 1 * F i 1 ( Y 1 ) A 1 ), … , | α i m | d ( A m * F i m ( X m ) A m , A m * F i m ( Y m ) A m ) } ≤ 1 r i m a x { | α i 1 | d ( F i 1 ( X 1 ), F i 1 ( Y 1 ) ) , … , | α i m | d ( F i m ( X m ), F i m ( Y m ) ) } ≤ 1 r i m a x { | α i 1 | k i 1 d ( X 1 , Y 1 ), … , | α i m | k i m d ( X m , Y m ) } ≤ max 1 ≤ j ≤ m { | α i j | k i j } r i m a x { d ( X 1 , Y 1 ), … , d ( X m , Y m ) } ≤ max 1 ≤ j ≤ m { | α i j | k i j / r i } d m ( X , Y ) .$

Thus, we proved that for all i = 1, 2, ..., m, we have

$d ( G i ( X ) , G i ( Y ) ) ≤ max 1 ≤ j ≤ m { | α i j | k i j ∕ r i } d m ( X , Y ) .$
(11)

Now, (9) holds immediately from (10) and (11). Applying the Banach contraction principle (see Theorem 1.1) to the mapping G, we get the desired result. □

## 3 Examples and numerical results

### 3.1 The matrix equation: $X = ( ( ( X 1 / 2 + B 1 ) − 1 / 2 + B 2 ) 1 / 3 + B 3 ) 1 / 2$

We consider the problem: Find X P(n) solution to

$X = ( ( ( X 1 / 2 + B 1 ) − 1 / 2 + B 2 ) ) 1 / 3 + B 3 ) 1 / 2 ,$
(12)

where B i ≥ 0 for all i = 1, 2, 3.

Problem (12) is equivalent to: Find X1 P (n) solution to

$X 1 r 1 = Q 1 + ( A 1 * F 11 ( X 1 ) A 1 ) α 11 ,$
(13)

where r1 = 2, Q1 = B3, A1 = I n (the identity matrix), α11 = 1/3 and F11 : P(n) → P (n) is given by:

$F 11 ( X ) = ( X 1 ∕ 2 + B 1 ) - 1 ∕ 2 + B 2 .$

Proposition 3.1 F11is a Lipshitzian mapping with k11 ≤ 1/4.

Proof. Using the properties of the Thompson metric, for all X, Y P(n), we have

$d ( F 11 ( X ) , F 11 ( Y ) ) = d ( ( X 1 ∕ 2 + B 1 ) - 1 ∕ 2 + B 2 , ( Y 1 ∕ 2 + B 1 ) - 1 ∕ 2 + B 2 ) ≤ d ( ( X 1 ∕ 2 + B 1 ) - 1 ∕ 2 , ( Y 1 ∕ 2 + B 1 ) - 1 ∕ 2 ) ≤ 1 2 d ( X 1 ∕ 2 + B 1 , Y 1 ∕ 2 + B 1 ) ≤ 1 2 d ( X 1 ∕ 2 , Y 1 ∕ 2 ) ≤ 1 4 d ( X , Y ) .$

Thus, we have k11 ≤ 1/4. □

Proposition 3.2 Problem (13) is Banach admissible.

Proof. We have

$| α 11 | k 11 r 1 ≤ 1 3 1 4 2 = 1 24 < 1 .$

This implies that Problem (13) is Banach admissible. □

Theorem 3.1 Problem (13) has one and only one solution$X 1 * ∈P ( n )$. Moreover, for any X1(0) P(n), the sequence (X1(k))k≥0defined by:

$X 1 ( k + 1 ) = ( X 1 ( k ) 1 ∕ 2 + B 1 ) - 1 ∕ 2 + B 2 1 ∕ 3 + B 3 1 ∕ 2 ,$
(14)

converges to$X 1 *$, and the error estimation is

$d ( X 1 ( k ) , X 1 * ) ≤ q 1 k 1 - q 1 d ( X 1 ( 1 ) , X 1 ( 0 ) ) ,$
(15)

where q1 = 1/4.

Proof. Follows from Propositions 3.1, 3.2 and Theorem 2.1. □

Now, we give a numerical example to illustrate our result given by Theorem 3.1.

We consider the 5 × 5 positive matrices B1, B2, and B3 given by:

$B 1 = 1 . 0000 0 . 5000 0 . 3333 0 . 2500 0 0 . 5000 1 . 0000 0 . 6667 0 . 5000 0 0 . 3333 0 . 6667 1 . 0000 0 . 7500 0 0 . 2500 0 . 5000 0 . 7500 1 . 0000 0 0 0 0 0 0 , B 2 = 1 . 4236 1 . 3472 1 . 1875 1 . 0000 0 1 . 3472 1 . 9444 1 . 8750 1 . 6250 0 1 . 1875 1 . 8750 2 . 1181 1 . 9167 0 1 . 0000 1 . 6250 1 . 9167 1 . 8750 0 0 0 0 0 0$

and

$B 3 = 2 . 7431 3 . 3507 3 . 3102 2 . 9201 0 3 . 3507 4 . 6806 4 . 8391 4 . 3403 0 3 . 3102 4 . 8391 5 . 2014 4 . 7396 0 2 . 9201 4 . 3403 4 . 7396 4 . 3750 0 0 0 0 0 0 .$

We use the iterative algorithm (14) to solve (12) for different values of X1(0):

$X 1 ( 0 ) = M 1 = 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0 5 , X 1 ( 0 ) = M 2 = 0 . 02 0 . 01 0 0 0 0 . 01 0 . 02 0 . 01 0 0 0 0 . 01 0 . 02 0 . 01 0 0 0 0 . 01 0 . 02 0 . 01 0 0 0 0 . 01 0 . 02$

and

$X 1 ( 0 ) = M 3 = 30 15 10 7 . 5 6 15 30 20 15 12 10 20 30 22 . 5 18 7 . 5 15 22 . 5 30 24 6 12 18 24 30 .$

For X1(0) = M1, after 9 iterations, we get the unique positive definite solution

$X 1 ( 9 ) = 1 . 6819 0 . 69442 0 . 61478 0 . 51591 0 0 . 69442 1 . 9552 0 . 96059 0 . 84385 0 0 . 61478 0 . 96059 2 . 0567 0 . 9785 0 0 . 51591 0 . 84385 0 . 9785 1 . 9227 0 0 0 0 0 1$

and its residual error

$R ( X 1 ( 9 ) ) = X 1 ( 9 ) - X 1 ( 9 ) 1 ∕ 2 + B 1 - 1 ∕ 2 + B 2 1 ∕ 3 + B 3 1 ∕ 2 = 6 . 346 × 1 0 - 13 .$

For X1(0) = M2, after 9 iterations, the residual error

$R ( X 1 ( 9 ) ) = 1 . 5884 × 1 0 - 12 .$

For X1(0) = M3, after 9 iterations, the residual error

$R ( X 1 ( 9 ) ) = 1 . 1123 × 1 0 - 12 .$

The convergence history of the algorithm for different values of X1(0) is given by Figure 1, where c1 corresponds to X1(0) = M1, c2 corresponds to X1(0) = M2, and c3 corresponds to X1(0) = M3.

### 3.2 System of three nonlinear matrix equations

We consider the problem: Find (X1, X2, X3) (P(n))3 solution to

$X 1 = I n + A 1 * ( X 1 1 ∕ 3 + B 1 ) 1 ∕ 2 A 1 + A 2 * ( X 2 1 ∕ 4 + B 2 ) 1 ∕ 3 A 2 + A 3 * ( X 3 1 ∕ 5 + B 3 ) 1 ∕ 4 A 3 , X 2 = I n + A 1 * ( X 1 1 ∕ 5 + B 1 ) 1 ∕ 4 A 1 + A 2 * ( X 2 1 ∕ 3 + B 2 ) 1 ∕ 2 A 2 + A 3 * ( X 3 1 ∕ 4 + B 3 ) 1 ∕ 3 A 3 , X 3 = I n + A 1 * ( X 1 1 ∕ 4 + B 1 ) 1 ∕ 3 A 1 + A 2 * ( X 2 1 ∕ 5 + B 2 ) 1 ∕ 4 A 2 + A 3 * ( X 3 1 ∕ 3 + B 3 ) 1 ∕ 2 A 3 ,$
(16)

where A i are n × n singular matrices.

Problem (16) is equivalent to: Find (X1, X2, X3) (P(n))3 solution to

$X i r i = Q i + ∑ j = 1 3 ( A j * F i j ( X j ) A j ) α i j , i = 1 , 2 , 3 ,$
(17)

where r1 = r2 = r3 = 1, Q1 = Q2 = Q3 = I n and for all i, j {1, 2, 3}, α ij = 1,

$F i j ( X j ) = ( X j θ i j + B j ) γ i j , θ = ( θ i j ) = 1 ∕ 3 1 ∕ 4 1 ∕ 5 1 ∕ 5 1 ∕ 3 1 ∕ 4 1 ∕ 4 1 ∕ 5 1 ∕ 3 , γ = ( γ i j ) = 1 ∕ 2 1 ∕ 3 1 ∕ 4 1 ∕ 4 1 ∕ 2 1 ∕ 3 1 ∕ 3 1 ∕ 4 1 ∕ 2 .$

Proposition 3.3 For all i, j {1, 2, 3}, F ij : P(n) → P(n) is a Lipshitzian mapping with k ij γ ij θ ij .

Proof. For all X, Y P(n), since θ ij , γ ij (0, 1), we have

$d ( F i j ( X ) , F i j ( Y ) ) = d ( ( X θ i j + B j ) γ i j , ( Y θ i j + B j ) γ i j ) ≤ γ i j d ( X θ i j + B j , Y θ i j + B j ) ≤ γ i j d ( X θ i j , Y θ i j ) ≤ γ i j θ i j d ( X , Y ) .$

Then, F ij is a Lipshitzian mapping with k ij γ ij θ ij . □

Proposition 3.4 Problem (17) is Banach admissible.

Proof. We have

$max 1 ≤ i ≤ 3 max 1 ≤ j ≤ 3 { | α i j | k i j ∕ r i } = max 1 ≤ i , j ≤ 3 k i j ≤ max 1 ≤ i , j ≤ 3 γ i j θ i j = 1 ∕ 6 < 1 .$

This implies that Problem (17) is Banach admissible. □

Theorem 3.2 Problem (16) has one and only one solution$( X 1 * , X 2 * , X 3 * ) ∈ ( P ( n ) ) 3$. Moreover, for any (X1(0), X2(0), X3(0)) (P(n))3, the sequences (X i (k))k≥0, 1 ≤ i ≤ 3, defined by:

$X i ( k + 1 ) = I n + ∑ j = 1 3 A j * F i j ( X j ( k ) ) A j ,$
(18)

converge respectively to$X 1 * , X 2 * , X 3 *$, and the error estimation is

$max { d ( X 1 ( k ) , X 1 * ) , d ( X 2 ( k ) , X 2 * ) , d ( X 3 ( k ) , X 3 * ) } ≤ q 3 k 1 - q 3 max { d ( X 1 ( 1 ) , X 1 ( 0 ) ) , d ( X 2 ( 1 ) , X 2 ( 0 ) ) , d ( X 3 ( 1 ) , X 3 ( 0 ) ) } ,$
(19)

where q3 = 1/6.

Proof. Follows from Propositions 3.3, 3.4 and Theorem 2.1. □

Now, we give a numerical example to illustrate our obtained result given by Theorem 3.2.

We consider the 3 × 3 positive matrices B1, B2 and B3 given by:

$B 1 = 1 . 0 . 5 0 0 . 5 1 0 0 0 0 , B 2 = 1 . 25 1 0 1 1 . 25 0 0 0 0 and B 3 = 1 . 75 1 . 625 0 1 . 625 1 . 75 0 0 0 0 .$

We consider the 3 × 3 nonsingular matrices A1, A2 and A3 given by:

$A 1 = 0 . 3107 - 0 . 5972 0 . 7395 0 . 9505 0 . 1952 - 0 . 2417 0 - 0 . 7780 - 0 . 6282 , A 2 = 1 . 5 - 2 0 . 5 0 . 5 0 - 0 . 5 - 0 . 5 2 - 1 . 5$

and

$A 3 = - 1 - 1 1 1 - 1 1 - 1 - 1 - 1 .$

We use the iterative algorithm (18) to solve Problem (16) for different values of (X1(0), X2(0), X3(0)):

$X 1 ( 0 ) = X 2 ( 0 ) = X 3 ( 0 ) = M 1 = 1 0 0 0 2 0 0 0 3 ,$
$X 1 ( 0 ) = X 2 ( 0 ) = X 3 ( 0 ) = M 2 = 0 . 02 0 . 01 0 0 . 01 0 . 02 0 . 01 0 0 . 01 0 . 02$

and

$X 1 ( 0 ) = X 2 ( 0 ) = X 3 ( 0 ) = M 3 = 30 15 10 15 30 20 10 20 30 .$

The error at the iteration k is given by:

$R ( X 1 ( k ) , X 2 ( k ) , X 3 ( k ) ) = max 1 ≤ i ≤ 3 X i ( k ) - I 3 - ∑ j = 1 3 A j * F i j ( X j ( k ) ) A j .$

For X1(0) = X2(0) = X3(0) = M1, after 15 iterations, we obtain

$X 1 ( 15 ) = 10 . 565 - 4 . 4081 2 . 7937 - 4 . 4081 16 . 883 - 6 . 6118 2 . 7937 - 6 . 6118 9 . 7152 , X 2 ( 15 ) = 11 . 512 - 5 . 8429 3 . 1922 - 5 . 8429 19 . 485 - 7 . 9308 3 . 1922 - 7 . 9308 10 . 68$

and

$X 3 ( 15 ) = 11 . 235 - 3 . 5241 3 . 2712 - 3 . 5241 17 . 839 - 7 . 8035 3 . 2712 - 7 . 8035 11 . 618 .$

The residual error is given by:

$R ( X 1 ( 15 ) , X 2 ( 15 ) , X 3 ( 15 ) ) = 4 . 722 × 1 0 - 15 .$

For X1(0) = X2(0) = X3(0) = M2, after 15 iterations, the residual error is given by:

$R ( X 1 ( 15 ) , X 2 ( 15 ) , X 3 ( 15 ) ) = 4 . 911 × 1 0 - 15 .$

For X1(0) = X2(0) = X3(0) = M3, after 15 iterations, the residual error is given by:

$R ( X 1 ( 15 ) , X 2 ( 15 ) , X 3 ( 15 ) ) = 8 . 869 × 1 0 - 15 .$

The convergence history of the algorithm for different values of X1(0), X2(0), and X3(0) is given by Figure 2, where c1 corresponds to X1(0) = X2(0) = X3(0) = M1, c2 corresponds to X1(0) = X2(0) = X3(0) = M2 and c3 corresponds to X1(0) = X2(0) = X3(0) = M3.

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## Author information

Correspondence to Maher Berzig.

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• #### DOI

https://doi.org/10.1186/1687-1812-2011-89

### Keywords

• nonlinear matrix equations
• Lipshitzian mappings
• Banach contraction principle
• iterative method
• fixed point
• Thompson metric