Finite Iterative Algorithm for Solving a Class of Complex Matrix Equation with Two Unknowns of General Form

(1)

Published online November 20, 2014 (http://www.sciencepublishinggroup.com/j/acm) doi: 10.11648/j.acm.20140305.23

ISSN: 2328-5605 (Print); ISSN: 2328-5613 (Online)

Finite iterative algorithm for solving a class of complex matrix

equation with two unknowns of general form

Mohamed A. Ramadan

1

, Mokhtar A. Abdel Naby

2

, Talaat S. El-Danaf

1

, Ahmed M. E. Bayoumi

2

1

Department of Mathematics, Faculty of Science, Menoufia University, Shebeen El- Koom, Egypt

2

Department of Mathematics, Faculty of Education, Ain Shams University, Cairo, Egypt

Email address:

[email protected] (M. A. Ramadan), [email protected] (M. A. Ramadan), Abdalnaby @hotmail.com (M. A. Abdel Naby), [email protected](Talaat S. El-Danaf), [email protected] (A. M. E. Bayoumi)

To cite this article:

Mohamed A. Ramadan, Mokhtar A. Abdel Naby, Talaat S. El-Danaf, Ahmed M.E. Bayoumi. Finite Iterative Algorithm for Solving a Class of Complex Matrix Equation with Two Unknowns of General Form.Applied and Computational Mathematics.

Vol. 3, No. 5, 2014, pp. 273-284. doi: 10.11648/j.acm.20140305.23

Abstract:

This paper is concerned with an efficient iterative algorithm to solve general the Sylvester-conjugate matrix equation of the form AVB CWD mE VF C

1 l l l j

t 1 j j s

1

i i i+ = +

∑ ∑

∑

= =

= The proposed algorithm is an extension to our proposed general Sylvester-conjugate equation of the form AV BW mE VF C

1 l l l t

1 j j s

1

i i + = +

∑ ∑

∑

= =

= When a solution exists for this matrix equation, for any initial matrices, the solutions can be obtained within finite iterative steps in the absence of round off errors. Some lemmas and theorems are stated and proved where the iterative solutions are obtained. Finally, a numerical example is given to verify the effectiveness of the proposed algorithm.

Keywords:

General Sylvester-Conjugate matrix Equations, Finite Iterative Algorithm, Orthogonality, Inner Product Space, Frobenius norm

1. Introduction

We know that matrix equation is one of the topics of very active research in computational mathematics, and a large number of papers have presented several methods for solving several matrix equations [1–5]. Ding and Chen presented the hierarchical gradient iterative algorithms for general matrix equations [6, 12] and hierarchical least squares iterative algorithms for generalized coupled Sylvester matrix equations and general coupled matrix equations [13, 14]. The hierarchical gradient iterative algorithms [6, 12] and hierarchical least squares iterative algorithms [6,14,15] for solving general (coupled) matrix equations are innovational and computationally efficient numerical ones and were proposed based on the hierarchical identification principle [13,16] which regards the unknown matrix as the system parameter matrix to be identified.

In [7], the necessary and sufficient conditions for the solvability of the matrix equationAHXB₌C

, over reflexive and anti-reflexive matrices are given, and the general expression of the reflexive and anti-reflexive solutions for a

solvable case is obtained. Ramadan et al. [8] introduced a complete, general and explicit solution to the Yakubovich matrix equationV−AVF=BW, with

_F

in an arbitrary form. Also with the help of the concept of Kronecker map, an explicit solution for the matrix equation XF−AX=C was established in [9]. Zhou et al. [10] proposed gradient based iterative algorithms for solving the general coupled Sylvester matrix equations with weighted least squares solutions. In [11], a general parametric solution to a family of generalized Sylvester matrix equations arising in linear system theory is presented by using the so-called generalized Sylvester mapping which has some elegant properties.

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equation of the form AV BW mEVF C 1

l l l t

1 j j s

1 i i

+

∑

=

∑

+

∑

= =

= in [20].

This paper is organized as follows: First, in section 2, we introduce some notations, lemmas and theorems that will be needed to develop this work. In section 3, we propose iterative method to obtain numerical solution to the matrix equations AV BW mEVF C

l l l

t

j j

s

i i

+

∑

=

∑

+

∑

= =

=1 1 1

using iterative method. In section 4, a numerical example is given to explore the simplicity and the neatness of the presented methods.

2. Preliminaries

The following notations, definitions, lemmas and theorems will be used to develop the proposed work. We use

H T

A A

A , , and tr(A) to denote the transpose, conjugate,

conjugate transpose and the trace of a matrixA respectively. We denote the set of all

m

×

n

complex

matrices by ℂ m×n ,

Re(

a

)

denote the real part of number

a

.

Definition 1. Inner product [38]

A real inner product space is a vector space V over the real field ℝ together with an inner product. i.e. with a map

→ ×V V :

.,. _ℝ Satisfying the following three axioms for all vectors

V

z

y

x

,

∈

and all scalars

a

∈

ℝ 1. Symmetry:

x

,

y

=

y

,

x

. 2. Linearity in the first argument:

y

x

a

y

ax

,

=

,

z

y

z

x

z

y

x

+

,

=

,

+

,

3. Positive definiteness:

x

,

x

>

0

for all

x

≠

0

. Two vectors u,v∈V are said to be orthogonal if u,v =0. The following theorem defines a real inner product on space ℂm×n_{over the field}_ℝ

Theorem 1. [29]

In the space ℂm×n_{over the field}_ℝ_{, an inner product can}

be defined as

)]

(

Re[

,

B

tr

A

B

A

=

H . (2) Proof.

(1) For

A

,

B

∈

ℂm×n_{, according to the properties of}

trace of a matrix one has

A B A B tr

A B tr A

B tr B

A tr B

A

H

T T

H

, )] ( Re[

] ) ( Re[ )] ( Re[ )] ( Re[ ,

= =

=

(2) For a real number a, and

A

,

B

,

C

∈

ℂm×n_{, one has}

B A a B A tr a

B A tr a B aA tr B aA tr B aA

H

H H

H

, )] ( Re[

)] ( Re[ )] ( Re[ )] ) (( Re[ ,

= =

=

C B C A C B tr C A tr

C B A tr C B A tr C B A

H H

H H H

, , )] ( Re[ )] ( Re[

] ) (

Re[ )] ) (( Re[ ,

+ = +

=

+ =

+

(3) It is well-known that tr(AHA)≻0

for all

_x

≠

₀

. Thus, _A,_A =Re[_tr(_AH_A)]≻0 for all

x

≠

0

.

According to definition 1, all the above argument reveals that the space ℂm×n_{over field}_ℝ_{with the inner product}

defined by (2) is an inner product space. Definition2. Frobenius Norm

The matrix norm of A induced by the inner product is Frobenius norm and denoted by A A2= A,A =trace(AHA).

3. Main Results

In this section, we propose an iterative solution to the complex matrix equation

G F V E D W C VB

A m

1 l l l j

t 1 j j s

1 i i i

+

∑

=

∑

+

∑

= =

= , (1)

where

A

_i

,

C

_j

,

E

_l

∈

ℂ n×n _, ∈

l j i D F

B, , _ℂ p×p _and

∈

G

ℂ n×p _{are given matrices, while}_V_,_W∈_ℂ n×p _are

matrices to be determined.

The solution the matrix equation (1) is based on the following algorithm.

Algorithm I (Finite Iterative Algorithm for (1)) 1 Input A_i,B_i,C_j,D_jE_l,F_l,G

2 Chosen arbitrary matrices

V

₁

,

W

₁

∈

ℂn×p_;

3 Set

j t

j j s

i

i i m

l

V

F

A

V

B

C

W

D

E

G

R

∑

= =

=

−

+

==

1 1 1

1 1

1

1 ;

H

l m

l H

i s

i H

i

R

B

E

R

F

A

P

₁

1 1

1

∑

= =

−

=

;

H j t

j H j

R

D

C

Q

₁

1

∑

=

;

1 :

=

k

1 If

R

_k

=

0 ,

then stop; and Vk,Wk are the solution; else

let

k

:

=

k

+

1

go to STEP 5. 2 Compute

k k k

k k

k

P

Q

P

R

V

₂ ₂

2 1

+

=

(3)

k k k

k k

k Q

Q P

R W

W ₂ ₂

2 1

+ + =

+ ;

];

[

1 1

1 2 2

2

1 1 1

1 1

j t

j k j m

l l k l i s

i k i k k

k k

j t

j k j s

i

i k i m

l

l k l k

D

Q

C

F

P

E

B

P

A

Q

P

R

D

W

C

B

V

A

F

V

E

G

R

∑

= =

=

= +

+

−

+

−

=

−

+

=

k k k H l k m

l H l H

i s

i

k H i

k

P

R

F

R

E

B

R

A

P

₂

2 1 1

1 1

+ +

=

= +

+

=

∑

−

∑

+

;

k k k H j k t

j H j

k

Q

R

D

R

C

Q

₂

2 1 1

1 1

+ +

=

+

=

∑

+

;

3 If

R

_k₊₁

=

0

;then stop ;else let

k

:

=

k

+

1

; go to STEP 5.

To prove the convergence property of Algorithm I, we first establish the following basic properties

Lemma 1

Suppose the matrix equation (1) is consistent and let

* *

,

W

V

be its solution. Then, for any initial matrices

1 1

,

W

V

, we have

2 *

* *

*

2 )] ( ) ( [ )] ( ) (

[P_iH V V_i Q_iHW W_i trP_iHV V_i Q_iHW W_i R_i

tr − + − + − + − = (3)

Or, equivalently

2 *

* ₎ ₍ _)]}

( [

Re{tr P_iH V −V_i +Q_iH W −W_i = R_i

Where the sequences

{ } { } { } { }

Vi , Pi , Wi , Qi and { }Ri are

generated by Algorithm I for _i=₁_,₂_,…

Proof

We apply mathematical induction to prove the conclusion For

i

=

1

, from Algorithm I we have

* * *

1 1 1 1 1 1 1

1 1

*

1 1

1

[ (

)

(

)]

[(

) (

)

(

) (

)]

s m _H _H

H H H H H

i i l l

i l

t

H H H

j j

j

tr P V V Q W W

tr

A RB

E RF

V V

C RD

W W

= =

=

− +

− =

−

+

−

∑

* *

1 1 1 1

1 1 1

*

1 1 1 1 1

1 1 1

[

]

s _H m t

H H

i i l l j j

i l j

s _H m t

H H

i i l l j j

i l j

tr R

AV B

R

E V F

R

C W D

R

AV B

R

E V F

R

C W D

= = =

=

+

−

∑

In view that

_V

*

_,

_W

* are solutions of the matrix equation (1), it is easy that one can obtain from above relation

]

[

]

[

)]

(

)

(

[

)]

(

)

(

[

1 1 1 1

* 1

1 1 1

*

1 1 1

1 1

1 * 1

1 1 1

1

*

1 1 1

1 *

1 1 1

1 1 1 *

1

1 * 1 1 * 1 1

* 1 1 * 1

j t

j j H l m

l l H

i s

i i H j t

j j H m

l

l l H i s

i i H j

t

j j H

s

i

l m

l l H i i H j t

j j H s

i

l m

l l H i i H

H H

D

W

C

R

F

V

E

R

B

V

A

R

D

W

C

R

F

V

E

R

B

V

A

R

tr

D

W

C

R

F

V

E

R

B

V

A

R

D

W

C

R

F

V

E

R

B

V

A

R

tr

W

Q

V

P

tr

W

Q

V

P

tr

∑

= =

=

= =

=

= =

−

+

−

+

=

−

+

−

+

−

+

−

)]

(

)

(

)

(

)

(

[

1 1 1

1 1

* 1 *

1 1

* 1

1 1 1

1 1

1 * 1

* 1

l m

l l j

t

j j i

s

i i H

l m

l l j

t

j j i

s

i i H

m

l

l l j

t

j j s

i

i i H m

l

l l j

t

j j s

i

i i H

F

V

E

D

W

C

B

V

A

R

F

V

E

D

W

C

B

V

A

R

F

V

E

D

W

C

B

V

A

R

F

V

E

D

W

C

B

V

A

R

tr

∑

= =

−

+

−

+

−

+

−

+

=

1

1 1 1 1 1 1 1

1 1 1 1 1 1

[

(

)

(

m s t _H m s t

H

l l i i j j l l i i j j

l i j l i j

tr R

G

E V F

AV B

C W D

R

G

E V F

A V B

C W D

= = = = = =

=

+

∑

−

∑

−

∑

+

∑

−

∑

−

∑

2 1 1

1 1 1

(

H

)

(

H

)

2 tr R R

tr R R

R

=

+

=

This implies that (3) holds for

i

=

1

. Assume that (3) holds for

i

=

k

. That is,

Then we have to prove that the conclusion holds

for

i

=

k

+

1

. It follows from Algorithm I that

2 *

* *

*

2 )]

(

)

(

[

)]

(

)

(

[

P

_kH

V

_k

Q

_kH

W

_k

tr

P

_kH

V

_k

Q

_kH

W

_k

R

_k

(4)

)]

(

)

(

)

(

)

[(

)]

(

)

(

[

1 * 2

2 1 1

1

1 * 2

2 1 1

1 1

* 1 1

* 1

+ +

+ =

+ +

+ =

= +

+ +

−

+

−

+

−

=

−

+

−

∑

k H

k k k H j k t

j H j

k H

k

k k H

l k m

l H

i s

i

k H i k

H k k

H k

W

Q

R

D

R

C

V

P

R

F

R

E

B

R

A

tr

W

Q

V

P

tr

)]

(

)

(

[

]

)

(

)

(

)

(

[

2 2

2 *

2 2

2 *

2 2 1

1 * 1

1 1

1 * 1

1 * 1 1

k

k k

H k k

k k

H k

k k

j k t

j j H k l m

l

k l

H k i k s

i i H k

Q

P

R

W

Q

P

Q

P

R

V

P

tr

R

D

W

C

R

F

V

E

R

B

V

A

R

tr

+

−

+

−

+

−

+

−

=

+

+ =

+

= +

+ +

=

+

∑

* * *

1

1 1 1 1 1

1 1 1

2 2 2

2 2

1 * * 1

2 2 2 2

[

(

)

(

)

(

)

]

[ (

(

)

(

))]

[

(

)]

s _H m t

H H

k

k i k i l k l k j k j

i l j

k H H k k

k k k k k k

k k k k

tr R

A V

V

B

R

E V

V

F

R

C W

W

D

R

tr P

V

Q

W

P

Q

R

P

Q

+

+ + + + +

= = =

+ +

=

−

+

−

+

−

+

−

+

∑

* * *

1

1 1 1 1 1

1 1 1

2

1 * *

1 2

[

(

)

(

)

(

)

]

[ (

(

)

(

))]

s m t

H

H H

k

k i k i l k l k j k j

i l j

k H H

k k k k k

k

tr R

A V

V

B

R

E V

V

F

R

C W

W

D

R

tr P

V

Q

W

R

+

+ + + + +

= = =

+

=

−

+

−

+

−

+

−

∑

(4)

In view that

V

*

,

W

* is a solution of matrix equation (1), with this relation and (4) one has

))]

(

)

(

[

]

)

(

)

(

)

(

[

2 ))]

(

)

(

[

]

)

(

)

(

)

(

[

)]

(

)

(

[

)]

(

)

(

[

* *

2 2 1

1 * 1

1 1

1 * 1

1 * 1 1

2 1 *

* 2

2 1

1 * 1

1 1

1 * 1

1 * 1 1

* 1 1

* 1

k H

k k H

k

k k

j k t

j j H k l m

l

k l

H k i k s

i i H k

k k

H k k H

k

k k

j k t

j j H k l m

l

k l

H k i k s

i i H k

k H

k k

H k k

H k

W

Q

V

P

tr

R

D

W

C

R

F

V

E

R

B

V

A

R

tr

R

W

Q

V

P

tr

R

D

W

C

R

F

V

E

R

B

V

A

R

tr

W

Q

V

P

tr

W

Q

V

P

tr

−

+

−

+

−

+

−

+

−

+

−

+

−

+

−

=

−

+

−

+

−

+

−

+

+ =

+

= +

+ +

= +

+ +

+ =

+

= +

+ +

= +

+ +

∑

]

))

(

)

(

]

)

(

[

2 ))

(

)

(

[

]

)

(

[

* *

1 1 1

1 1

* *

1 *

1 1

2 1 *

* 2

2 1

1 1 1

1 1

* *

1 *

1 1

k H

k k H

k

l m

l k l j

k t

j j i

k s

i i l

m

l l j

t

j j i

s

i i H k

k k

H k k H

k k

k

l m

l k l j

k t

j j i

k s

i i l

m

l l j

t

j j i

s

i i H k

W

Q

V

P

tr

F

V

E

D

W

C

B

V

A

F

V

E

D

W

C

B

V

A

R

tr

R

W

Q

V

P

tr

R

F

V

E

D

W

C

B

V

A

F

V

E

D

W

C

B

V

A

R

tr

−

+

−

+

−

+

−

+

−

+

−

+

=

∑

= +

+ =

+ = =

= =

+

+ +

= +

+ =

+ = =

= =

(5)

)

2 (

)

(

[

2 )]

(

[

2 2

2 1 1

1 1

2 1 1

1 1

k

k k j t

j

k j s

i

i k i m

l

l k l H

k

k j

t

j

k j s

i

i k i m

l

l k l H

k

R

D

W

C

B

V

A

F

V

E

G

R

tr

R

D

W

C

B

V

A

F

V

E

G

R

tr

+

= +

+

= +

+

−

+

−

+

=

∑

2 1

2 1 2

1 1

1

2

2 )

(

)

(

+

+ +

+ + +

+

=

−

+

=

k

k k

k H k k

H k

R

tr

R

tr

Hence relation (3) holds by principle of induction. Lemma 2

Suppose that the matrix equation (1) is consistent and the sequences

{ }

R

_i ,

{ }

P

_i and

{ }

Q

_i are generated by Algorithm I with any initial matrices

V

₁

,

W

₁ , such that

R

_i

≠

0

for all

k

i

=

1 ,

2 ,

…

,

_then,

{

(

)

}

0 Re

i

=

H j

R

trace

(5)

and

Re

{

(

+

i

)

}

=

0

H j i H

j

P

Q

P

trace

for

j

i

k

j

i

,

=

1 ,

2 ,

…

,

≠

_{. (6)}

Proof

We apply mathematical induction

Step 1: We prove that

0 )

(

R

_iH₊₁

R

_i

=

tr

(7) and

(

+1

+

+1 i

)

=

0

H i i H

i

P

Q

P

tr

for

i

=

1 ,

2 ,

…

,

k

_{. (8)}

First from Algorithm I we have

j t

j k j s

i

i k i m

l

l k l

k

G

E

V

F

A

V

B

C

W

D

R

∑

= +

+

=

+

−

1 1 1

1 1

j t

j

k

k k

j

s

i

i k

k k

i m

l

l k

k k

l

D

Q

P

R

W

C

B

P

Q

P

R

V

A

F

P

Q

P

R

V

E

G

∑

=

= =

+

−

+

−

+

=

1

2 2

2

1

2 2

2

1

2 2

2

)

(

)

(

)

(

)

(

1 1

1 2 2

2

1 1

1

j t

j k j i

s

i k i m

l

l k l k

k k j

t

j

k j s

i

i k i m

l

l k

l

E

P

F

A

P

B

C

Q

D

Q

P

R

D

W

C

B

V

A

F

V

E

G

∑

= =

−

+

−

+

=

];

[

1 1

1 2 2

2

j t

j

k j m

l

l k l i

s

i k i k

k k

k

A

P

B

E

P

F

C

Q

D

Q

P

R

∑

= =

=

+

−

+

−

=

(9)

For

i

=

1

, it follows from (9) that

)

])

[

((

)

(

₁

1 1 1

1 1

1 2

1 2 1

2 1 1

1

2

A

P

B

E

P

F

C

Q

D

R

Q

P

R

tr

R

tr

_j H

t

j j m

l

l l i

s

i i

H

∑

= =

=

+

−

+

−

=

2 1

1 1 2 2 1 1 1 1 1 1

1 1 1

1 1

(

[

])

s _H m t

H H H H H H H H H

i i l l j j

i l j

R

tr R R

P

A R B

P

E R F

Q

C R D

P

Q

= = =

=

−

+

∑

.

(6)

]

[

]

[

2 )

(

)

(

1 1 1

1 1 1 1

1 1 2 1 2 1

2 1

1 1 1

2 1 2 1

2 1 2

1 1

2 1

2

H l m

l H l H H j t

j H j H H i s

i H i H

t

j

H j H j H m

l

H l H l H H i s

i H i H H

H

F

R

E

P

D

R

C

Q

B

R

A

P

tr

Q

P

R

D

R

C

Q

F

R

E

P

B

R

A

P

tr

Q

P

R

tr

R

tr

∑

= =

=

= =

=

−

+

−

+

−

+

−

=

+

]

)

(

)

(

[

2

1 1 1 1

1 1

1 1 1 1

1 1

2 1 2 1

2 1 2

1

H j t

j H j H m

l

H l H l H

i s

i H i H

H j t

j H j H H

l m

l H

i s

i H i H

D

R

C

Q

F

R

E

B

R

A

P

D

R

C

Q

F

R

E

B

R

A

P

tr

Q

P

R

∑

= =

=

= =

=

+

−

+

−

+

−

=

]

[

2

₂ ₁ ₁ ₁ ₁ ₁ ₁ ₁ ₁

1 2 1

2 1 2

1

tr

P

Q

P

Q

P

R

H

+

H

+

H

+

H

+

−

=

0 ]

2

2 [

2

₂ ₁ 2 ₁ 2

1 2 1

2 1 2

1

+

=

+

−

=

P

Q

P

R

This implies that (7) is satisfied for

i

=

1

.

From Algorithm I we have

]

)

(

)

[(

)

(

₂ ₁ ₁

1 2 2 2

1 1 1 2 1

2 2 2

1 1

2 1

2

Q

R

D

R

C

P

R

F

R

E

B

R

A

tr

Q

P

tr

H_j H

t

j H j H

H

l m

l H

i s

i H i H

H

+

=

∑

−

∑

+

∑

+

= =

=

]

[

₂ ₁ ₁

1 2 2 1

1 2 1 1 2 1

2 2 1

1 2 1

1

2

Q

R

D

Q

C

R

P

R

F

P

E

R

B

P

A

R

tr

_j H

t

j j H H

l m

l l H i s

i i

H

−

+

=

∑

= =

=

From this last relation one has

]

[

)

(

]

[

)

(

)

(

1 1 2 1

1 2 1 1 2

1 1 1 1 1 1 1 1 2 1

2 2

1 1 2 1

1 2 1 1 2 1

2 1 2 1

2 1 2

j t

j j H l m

l l H i s

i i H

H H

j t

j j H l m

l l H i s

i i H H

H H

H

D

Q

C

R

F

P

E

R

B

P

A

R

tr

Q

P

tr

R

D

Q

C

R

F

P

E

R

B

P

A

R

tr

Q

P

tr

Q

P

tr

∑

= =

=

= =

=

+

−

+

−

=

+

)]

(

[

)

(

2 )]

(

[

1 1 1

1 1

1 2

2 1 2 1 2 1

2 2 1

1 1

1 2

l m

l l j

t

j j i

s

i i H

l m

l l j

t

j j i

s

i i H

F

P

E

D

Q

C

B

P

A

R

tr

Q

P

R

F

P

E

D

Q

C

B

P

A

R

tr

∑

= =

=

= =

=

−

+

−

+

(7)

0 )

(

2 ]

2 [

)

(

2 ]

))

(

))

(

[

2 1 2 1 2 1

2 2 2

2 2

1 2 1 2 1

2 1 2 1 2 1

2 2 2

1 2 2

1 2 1 2 1

=

+

−

=

+

−

+

−

+

=

Q

P

R

Q

P

Q

P

R

tr

R

tr

R

Q

P

_H _H

This implies that (8) is satisfied for

i

=

1

.

Assume (7) and (8) hold for

i

=

k

−

1

. From (9) and applying mathematical induction assumption, from Algorithm I we have