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R E S E A R C H

Open Access

Least-squares Hermitian problem of

complex matrix equation

(AXB,

CXD) = (E

,

F

)

Peng Wang

1,2*

, Shifang Yuan

1

and Xiangyun Xie

1

*Correspondence:

p_wong@126.com

1School of Mathematics and

Computational Science, Wuyi University, Jiangmen, 529020, P.R. China

2School of Computer Science and

Technology, Wuhan University of Technology, Wuhan, 430070, P.R. China

Abstract

In this paper, we present a direct method to solve the least-squares Hermitian problem of the complex matrix equation (AXB,CXD) = (E,F) with complex arbitrary coefficient matricesA,B,C,Dand the right-hand sideE,F. This method determines the least-squares Hermitian solution with the minimum norm. It relies on a matrix-vector product and the Moore-Penrose generalized inverse. Numerical experiments are presented which demonstrate the efficiency of the proposed method.

Keywords: matrix equation; least-squares solution; Hermitian matrices; Moore-Penrose generalized inverse

1 Introduction

LetA,B, C, D,E, andF are given matrices of suitable sizes defined over the complex number field. We are interested in the analysis of the linear matrix equation

(AXB,CXD) = (E,F) ()

to be solved forXCn×n. Here and in the following Cm×ndenotes the set of allm×n complex matrices, while the set of allm×nreal matrices is denoted by Rm×n. In particular,

we focus on the least-squares Hermitian solution with the least norm of (), which can be described as follows.

Problem  GivenACm×n,BCn×s,CCm×n,DCn×t,ECm×s, andFCm×t, let

HL=

X|XHCn×n,

AXBE+CXDF= min

X∈HCn×n

AXBE+CXDF

.

FindXHHLsuch that

XH=min XHL

X, ()

where HCn×ndenotes the set of alln×nHermitian matrices.

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Correspondingly, the set of alln×nreal symmetric matrices and the set of alln×nreal anti-symmetric matrices are denoted by SRn×nand ASRn×n, respectively.

Nowadays, matrix equations are very useful in numerous applications such as control theory [, ], vibration theory [], image processing [, ] and so on. Therefore it is an active area of research to solve different matrix equations [, –]. For the real matrix equation (), in [], least-squares solutions with the minimum-norm were obtained by using generalized singular value decomposition and canonical correlation decomposition of matrices (CCD). In [], the quaternion matrix equation () was considered and the least-squares solution with the least norm was given through the use of the Kronecker product and the Moore-Penrose generalized inverse. Though the matrix equations of the form () are studied in the literature, less or even no attention was paid to the least-squares Hermitian solution of () over the complex field which is studied in this paper. A special vectorization, as we defined in [], of the matrix equation () is carried out and Problem  is turned into the least-squares unconstrained problem of a system of real linear equations. The notations used in this paper are summarized as follows: ForACm×n, the sym-bolsAT,AHandA+denote the transpose matrix, the conjugate transpose matrix and the Moore-Penrose generalized inverse of matrixA, respectively. The identity matrix is de-noted by I. ForA= (aij)∈Cm×n,B= (bij)∈Cp×q, the Kronecker product ofAandBis

defined byAB= (aijB)∈Cmp×nq. For matrixACm×n, the stretching operatorvec(A)

is defined byvec(A) = (a,a, . . . ,an)T, whereaiis theith column ofA.

For allA,BCm×n, we define the inner productA,B=tr(AHB). Then Cm×nis a Hilbert

inner product space and the norm of a matrix generated by this inner product is the matrix Frobenius norm · . Further, denote the linear space Cm×n×Cm×t={[A,B]|ACm×n,

BCm×t}, and for the matrix pairs [Ai,Bi]∈Cm×n×Cm×t (i= , ), we can define their

inner product as follows:[A,B], [A,B]=tr(AHA) +tr(BHB). Then Cm×n×Cm×t is also a Hilbert inner space. The Frobenius norm of the matrix pair [A,B]∈Cm×n×Cm×t

can be derived:

[A,B]=[A,B], [A,B]

=

trAHA+trBHB

=A+B.

The structure of the paper is the following. In Section , we deduce some results in Hilbert inner product Cm×n×Cm×t which are important for our main results. In Sec-tion , we introduce a matrix-vector product for the matrices and vectors of Cm×nbased

on which we consider the structure of (AXB,CXD) overXHCn×n. In Section , we de-rive the explicit expression for the solution of Problem . Finally we express the algorithm for Problem  and perform some numerical experiments.

2 Some preliminaries

In this section, we will prove some theorems which are important for the proof of our main result. GivenARm×nandbRn, the solution of the linear equationsAx=binvolves two

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solving the corresponding normal equations:

ATAx=ATb, ()

moreover, we have

x|xRn,Axb=min=x|xRn,ATAx=ATb.

As for the latter, () still holds, further, the solution set ofAx=band that of () is the same, that is,

x|xRn,Ax=b=x|xRn,ATAx=ATb.

It is should be noticed that () is always consistent. Therefore, solvingAx=bis usually translated into solving the corresponding consistent equations (). In the following, we will extend the conclusion to a more general case. To do this, we first give the following problem.

Problem  Given complex matrices A,A, . . . ,AlCm×n, B,B, . . . ,BlCm×n, C

Cm×n, andDCm×n, findk= (k,k, . . . ,k

l)TRlsuch that l

i=

ki[Ai,Bi] =

l

i= kiAi,

l

i= kiBi

= [C,D]. ()

Theorem  Assume the matrix equation()in Problemis consistent.Let

Ei=

⎡ ⎢ ⎢ ⎢ ⎣

Re(Ai)

Im(Ai)

Re(Bi)

Im(Bi)

⎤ ⎥ ⎥ ⎥

⎦, F=

⎡ ⎢ ⎢ ⎢ ⎣

Re(C)

Im(C)

Re(D)

Im(D)

⎤ ⎥ ⎥ ⎥ ⎦.

Then the set of vectors k that satisfies()is exactly the set that solves the following consistent system:

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

E,EE,E · · · E,El

E,EE,E · · · E,El

..

. ... ...

El,EEl,E · · · El,El

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

k

k .. .

kl

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

E,F

E,F .. .

El,F

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

. ()

Proof By (), we have

l

i= kiAi,

l

i= kiBi

= [C,D]

⇐⇒

l

i=

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⇐⇒ l i= ki ⎡ ⎢ ⎢ ⎢ ⎣

Re(Ai)

Im(Ai)

Re(Bi)

Im(Bi)

⎤ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎣

Re(C)

Im(C)

Re(D)

Im(D)

⎤ ⎥ ⎥ ⎥ ⎦ ⇐⇒ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vec(ReA)T vec(ImA)T vec(ReB)T vec(ImB)T

vec(ReA)T vec(ImA)T vec(ReB)T vec(ImB)T

..

. ... ... ...

vec(ReAl)T vec(ImAl)T vec(ReBl)T vec(ImBl)T

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ × ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vec(ReA) vec(ReA) · · · vec(ReAl)

vec(ImA) vec(ImA) · · · vec(ImAl)

vec(ReB) vec(ReB) · · · vec(ReBl)

vec(ImB) vec(ImB) · · · vec(ImBl)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ k = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vec(ReA)T vec(ImA)T vec(ReB)T vec(ImB

)T

vec(ReA)T vec(ImA)T vec(ReB)T vec(ImB)T ..

. ... ... ...

vec(ReAl)T vec(ImAl)T vec(ReBl)T vec(ImBl)T

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ × ⎡ ⎢ ⎢ ⎢ ⎣

vec(ReC)

vec(ImC)

vec(ReD)

vec(ImD)

⎤ ⎥ ⎥ ⎥ ⎦ ⇐⇒ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

E,EE,E · · · E,El

E,EE,E · · · E,El

..

. ... ...

El,EEl,E · · · El,El

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ kk .. . kl ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

E,F

E,F .. .

El,F

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .

Thus we have ().

We now turn to the case that matrix equation () in Problem  is inconsistent, just as (), the related least-squares problem should be considered.

Problem  Given matricesA,A, . . . ,AlCm×n,B,B, . . . ,BlCs×t,CCm×n, andD

Cs×t, find

k= (k,k, . . . ,kl)TRl

such that

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Based on the results above, we list the following theorem which concludes that solving Problem  is equivalent to solving the consistent matrix equation system ().

Theorem  Suppose that the notations and conditions are the same as in Theorem.Then the solution set of Problemis the solution set of system().

Proof By (), we have

l

i=

kiAiC, l

i=

kiBiD

=

l

i=

ki(ReAi) – (ReC)

+√–

l

i=

ki(ImAi) – (ImC)

+

l

i=

ki(ReBi) – (ReD)

+√–

l

i=

ki(ImBi) – (ImD)

=

l

i= ki

⎡ ⎢ ⎢ ⎢ ⎣

Re(Ai)

Im(Ai)

Re(Bi)

Im(Bi)

⎤ ⎥ ⎥ ⎥ ⎦–

⎡ ⎢ ⎢ ⎢ ⎣

Re(C)

Im(C)

Re(D)

Im(D)

⎤ ⎥ ⎥ ⎥ ⎦

=

l

i=

kiEiF

.

It follows that () holds. This implies the conclusion.

3 The structure of(AXB,CXD)overX∈HCn×n

Based on the discussion in [], we first recall a matrix-vector product for vectors and matrices in Cm×n.

Definition  Letx= (x,x, . . . ,xk)TCk,y= (y,y, . . . ,yk)TCkandA= (A,A, . . . ,Ak),

AiCm×n(i= , , . . . ,k). Define

(i) Ax=xA+xA+· · ·+xkAkCm×n;

(ii) A◦(x,y) = (Ax,Ay).

From Definition ., we list the following facts which are useful for solving Problem . Lety= (y,y, . . . ,yk)TCk,P= (P,P, . . . ,Pk),PiCm×n(i= , , . . . ,k), anda,bC.

(i) xHy=xHy= (x,y);

(ii) Ax+Px= (A+P)◦x; (iii) A◦(ax+by) =a(Ax) +b(Ay); (iv) (aA+bP)◦x=a(Ax) +b(Px);

(v) (A,P)◦[yx] =Ax+Py; (vi) [AP]◦x= [APxx];

(vii) vec(Ax) = (vec(A),vec(A), . . . ,vec(Ak))x;

(viii) vec((aA+bP)◦x) =avec(Ax) +bvec(Px).

SupposeB= (B,B, . . . ,Bs)∈Ck×s,BiCk(i= , , . . . ,s),C= (C,C, . . . ,Ct)∈Ck×t,Ci

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(ix) D(Ax) = (DA)◦x;

(x) (Ax)H= (AH,AH, . . . ,AkH)◦x.

ForX=ReX+√–ImXHCn×n, byXH=X, we have (ReX+–ImX)H=ReX+

–ImX. Thus we getReXT=ReX,ImXT= –ImX.

Definition  For matrixARn×n, leta

 = (a,a, . . . ,an), a= (a,a, . . . ,an), . . . ,

an–= (a(n–)(n–),an(n–)),an=ann, the operatorvecS(A) is denoted

vecS(A) = (a,a, . . . ,an–,an)TR

n(n+)

. ()

Definition  For matrixBRn×n, let b

 = (b,b, . . . ,bn),b= (b,b, . . . ,bn), . . . , bn–= (b(n–)(n–),bn(n–)),bn–=bn(n–), the operatorvecA(B) is denoted

vecA(B) = (b,b, . . . ,bn–,bn–)TR n(n–)

. ()

Let

Eij= (est)∈Rn×n, ()

where

est=

⎧ ⎨ ⎩

 (s,t) = (i,j),

 otherwise.

Let

KS= (E,E+E, . . . ,En+En,

E,E+E, . . . ,En+En, . . . ,E(n–)(n–),En(n–)+E(n–)n,Enn). ()

Note thatKSRn×

n(n+)

 .

Let

KA= (E–E, . . . ,En–En,E–E, . . . ,En–En, . . . ,En(n–)–E(n–)n). ()

Note thatKARn×

n(n–)

.

Based on Definition , Definition , () and () we get the following lemmas which are necessary for our main results.

Lemma  Suppose XRn×n,then

(i)

XSRn×n ⇐⇒ X=KS◦vecS(X), ()

(ii)

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Lemma  Suppose X=ReX+√–ImXCn×n,then

XHCn×n ⇐⇒ X=KS◦vecS(ReX) +

–KA◦vecA(ImX). ()

Lemma  Given ACm×n,BCn×s,CCp×n,and DCn×q,and X=ReX+–ImX

HCn×n.The complex matrices FijCm×s, GijCm×s, HijCp×q and KijCp×q (i,j=

, , . . . ,n;ij)are defined by

Fij=

⎧ ⎨ ⎩

AiBj, i=j,

AiBj+AjBi, i>j,

Gij=

⎧ ⎨ ⎩

, i=j,

–(AiBjAjBi), i>j,

Hij=

⎧ ⎨ ⎩

CiDj, i=j,

CiDj+CjDi, i>j,

Kij=

⎧ ⎨ ⎩

, i=j,

–(CiDjCjDi), i>j,

where AiCm,CiCp is the ith column vector of matrix A and C,meanwhile,BjCs,

DjCqis the jth row vector of matrix B and D,respectively.Then

[AXB,CXD] =

⎣(F,F, . . . ,Fn,F,F, . . . ,Fn, . . . ,F(n–)(n–),Fn(n–),Fnn,

G,G, . . . ,Gn,G, . . . ,Gn, . . . ,Gn(n–))◦

⎣vecS(ReX)

vecA(ImX)

⎤ ⎦,

(H,H, . . . ,Hn,H,H, . . . ,Hn, . . . ,H(n–)(n–),Hn(n–),Hnn,

K,K, . . . ,Kn,K, . . . ,Kn, . . . ,Kn(n–))◦

⎣vecS(ReX)

vecA(ImX)

⎤ ⎦ ⎤ ⎦.

Proof By Lemma , we can get

AXB=AKS◦vecS(ReX) +

–KA◦vecA(ImX)

B

=(AKS)◦vecS(ReX) +

–(AKA)◦vecA(ImX)

B

=(AKS)◦vecS(ReX)

B+√–(AKA)◦vecA(ImX)

B

=A(E,E+E, . . . ,En(n–)+E(n–)n,Enn)

◦vecS(ReX)

B

+√–A(E–E, . . . ,En–En, . . . ,En(n–)–E(n–)n)◦vecA(ImX)

B

=AEB,A(E+E)B, . . . ,A(En(n–)+E(n–)n)B,AEnnB

◦vecS(ReX)

+√–A(E–E)B, . . . ,A(En–En)B, . . . ,A(En(n–)–E(n–)n)B

◦vecA(ImX)

= (AB,AB+AB, . . . ,AnBn–+An–Bn,AnBn)◦vecS(ReX)

+√–(AB–AB, . . . ,AnB–ABn, . . . ,AnBn––An–Bn)◦vecA(ImX)

= (F,F, . . . ,Fn(n–),Fnn)◦vecS(ReX) + (G,G, . . . ,Gn(n–))◦vecA(ImX)

= (F,F, . . . ,F(n–)(n–),Fn(n–),Fnn,G,G, . . . ,Gn(n–))◦

⎣vecS(ReX)

vecA(ImX)

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Similarly, we have

CXD= (H,H, . . . ,H(n–)(n–),Hn(n–),Hnn,K,K, . . . ,Kn(n–))◦

⎣vecS(ReX)

vecA(ImX)

⎤ ⎦.

Thus we can get the structure of [AXB,CXD] and complete the proof.

4 The solution of Problem 1

Based on the above results, in this section, we will deduce the solution of Problem . From [], the least-squares problem

AXBE+CXDF=min

with respect to the Hermitian matrixXis equivalent to

⎡ ⎣P

P

⎤ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vecS(ReX)

vecA(ImX)

vecS(ReX)

vecA(ImX)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vec(ReE)

vec(ImE)

vec(ReF)

vec(ImF)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

=min, ()

where

P=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

(BTA)L S

–(BTA)L A

(DTC)L S

–(DTC)L A

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, P=Re(P), P=Im(P).

It should be noticed that () is an unconstrained problem over the real field and can easily be solved by existing methods, therefore, the original complex constrained Problem  is translated into an equivalent real unconstrained problem (). Since the process has been expressed in [], we omit it here. Based on Theorems , , and Lemma , we now turn to the least-squares Hermitian problem for the matrix equation (). The following lemmas are necessary for our main results.

Lemma ([]) Given ARm×nand bRn,the solution of equation Ax=b involves two cases:

(i) The equation has a solutionxRnand the general solution can be formulated as

x=A+b+IA+Ay ()

if and only ifAA+b=b,whereyRnis an arbitrary vector.

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For convenience, we introduce the following notations and lemmas. GivenACm×n,BCn×s,CCp×n,DCn×q,ECm×s, andFCp×q, let

ij= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Re(Fij)

Im(Fij)

Re(Hij)

Im(Hij)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, ϒij=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Re(Gij)

Im(Gij)

Re(Kij)

Im(Kij)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Re(E)

Im(E)

Re(F)

Im(F)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , W= ⎡

P U

UT V

⎦, e=

⎡ ⎣e

e

⎦, ()

wherenij≥,

P= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

, , · · · ,nn

, , · · · ,nn

..

. ... ...

nn, nn, · · · nn,nn

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , U= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

,ϒ ,ϒ · · · ,ϒn(n–)

,ϒ ,ϒ · · · ,ϒn(n–))

..

. ... ...

nn,ϒ nn,ϒ · · · nn,ϒn(n–)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , V= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

ϒ,ϒ ϒ,ϒ · · · ϒ,ϒn(n–)

ϒ,ϒ ϒ,ϒ · · · ϒ,ϒn(n–) ..

. ... ...

ϒn(n–),ϒ ϒn(n–),ϒ · · · ϒn(n–),ϒn(n–)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ,

e=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ,, .. .

nn,

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, e=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

ϒ,

ϒ, .. .

ϒn(n–),

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .

Theorem  Given ACm×n,BCn×s,CCm×n,DCn×t,ECm×s,and FCm×t,let

W,e be as in().Then

HL=

X|X= (KS,

–KA)◦

W+e+IW+Wy, ()

where yRnis an arbitrary vector.Problemhas a unique solution X

HHL.This solution

satisfies

XH= (KS,

–KA)◦

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Proof By Lemma  and Theorem , the least-squares problem

AXBE+CXDF=min

with respect to the Hermitian matrixXcan be translated into an equivalent consistent linear equations over the real field

W

vecS(ReX)

vecA(ImX)

=e.

It follows by Lemma  that

vecS(ReX)

vecA(ImX)

=W+e+IW+Wy.

Thus

X= (KS,

–KA)◦

W+e+IW+Wy.

The proof is completed.

We now turn to the consistency of the matrix equation (). Denote

N=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

N

N

N

N

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, ˆe=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

vec(ReE)

vec(ImE)

vec(ReF)

vec(ImF)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

, ()

where

N=

vecRe(F),vecRe(F), . . . ,vecRe(Fn)

,vecRe(F), . . . ,

vecRe(Fn)

, . . . ,vecRe(F(n–)(n–))

,vecRe(Fn(n–))

,vecRe(Fnn)

,

vecRe(G), . . . ,vecRe(Gn)

,vecRe(G), . . . ,

vecRe(Gn)

, . . . ,vecRe(Gn(n–))

,

N=

vecIm(F),vecIm(F), . . . ,vecIm(Fn)

,vecIm(F), . . . ,

vecIm(Fn)

, . . . ,vecIm(F(n–)(n–))

,vecIm(Fn(n–))

,vecIm(Fnn)

,

vecIm(G), . . . ,vecIm(Gn)

,vecIm(G)

, . . . ,

vecIm(Gn)

, . . . ,vecIm(Gn(n–))

,

N=

vecRe(H)

,vecRe(H)

, . . . ,vecRe(Hn)

,vecRe(H)

, . . . ,

vecRe(Hn)

, . . . ,vecRe(H(n–)(n–))

,vecRe(Hn(n–))

,vecRe(Hnn)

,

vecRe(K), . . . ,vecRe(Kn)

,vecRe(K)

, . . . ,

vecRe(Kn)

, . . . ,vecRe(Kn(n–))

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N=

vecIm(H),vecIm(H), . . . ,vecIm(Hn)

,vecIm(H), . . . ,

vecIm(Hn)

, . . . ,vecIm(H(n–)(n–))

,vecIm(Hn(n–))

,vecIm(Hnn)

,

vecIm(K), . . . ,vecIm(Kn)

,vecIm(K), . . . ,

vecIm(Kn)

, . . . ,vecIm(Kn(n–))

.

By Lemma , we have

(AXB,CXD) = (E,F) ⇐⇒ N

⎣vecS(ReX)

vecA(ImX)

⎦=eˆ. ()

Thus we can get the following conclusions by Lemma  and Theorem .

Corollary  The matrix equation()has a solution XHCn×nif and only if

NNee. ()

In this case,denote by HEthe solution set of ().Then

HE=

X|X= (KS,

–KA)◦

M+e+IW+Wy, ()

where yRnis an arbitrary vector.

Furthermore,if ()holds,then the matrix equation()has a unique solution XHEif

and only if

rank(N) =n. ()

In this case,

HE=

X|X= (KS,

–KA)◦

W+e. ()

The least norm problem

XH=min XHE

X

has a unique solution XHHEand XHcan be expressed as().

5 Numerical experiments

In this section, based on the results in the above sections, we first give the numerical al-gorithm to find the solution of Problem . Then numerical experiments are proposed to demonstrate the efficiency of the algorithm. The following algorithm provides the main steps to find the solutions of Problem .

Algorithm 

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() ComputeP,U,Vandeaccording to ().

() If () and () hold then calculateXH(XHHE) according to ().

() If () holds then calculateXH(XHHE) according to (), otherwise go to the

next step.

() CalculateXH(XHHL) according to ().

For convenience, in the following examples, the random matrix, the Hilbert matrix, the Toeplitz matrix, the matrix whose all elements are one and the magic matrix are all de-noted as by the corresponding Matlab function.

Example  LetAr= rand(, ),Br=rand(, ),Cr= rand(, ),Dr=rand(, ),

Ai = rand(, ), Bi = rand(, ), Ci = rand(, ), Di = rand(, ). Let X˜ =

rand(, ),Xr=X˜ +X˜T;Xˆ =hilb(),Xi=Xˆ–XˆT;A=Ar+

–Ai,B=Br+

–Bi,C=

Cr+

–Ci,D=Dr+

–Di,X=Xr+

–Xi,E=AXB,F=CXD. By using Matlab  and

Algorithm , we obtainrank(W) = ,W= .e+,WW+ee= .e–.

rank(N) = ,N= .e+,NN+eˆ–ˆe= .e–. From Algorithm (), it can be concluded that the complex matrix equation [AXB,CXD] = [E,F] is consistent, and it has a unique solutionXHHE; further,XHX= .e– can easily be tested.

Example  Letm= ,n= ,s= . TakeAr= [toeplitz( : ), m×],Br=ones(, ),Cr=

[hilb(), ×],Dr=ones(, ),Ai= ×,Bi=ones(, ),Ci= ×,Di= ×. Let

˜

X=rand(, ),Xr=X˜+X˜T;Xˆ =hilb(n),Xi=Xˆ–XˆT;A=Ar+

–Ai,B=Br+

–Bi,C=

Cr+

–Ci,D=Dr+

–Di,X=Xr+

–Xi,E=AXB,F=CXD. From Algorithm (),

we can obtainrank(W) = ,W= .e+,WW+ee= .e–.rank(N) = ,N= .,NN+eˆ–eˆ= .e–. From Algorithm , it can be concluded that the complex matrix equation [AXB,CXD] = [E,F] is consistent, and it has a unique solutionXHHE, further,XHX= . can easily be tested.

Example  Letm= ,n= ,s= . TakeAr= ×,Br=rand(, ),Cr= ×,Dr=

rand(, ),Ai=rand(, ),Bi= ×,Ci=rand(, ),Di= ×. LetX˜ =rand(, ),

Xr=X˜+X˜T;Xˆ =rand(, ),Xi=Xˆ–XˆT;A=Ar+

–Ai,B=Br+

–Bi,C=Cr+

–Ci,

D=Dr+

–Di,X=Xr+

–Xi,E=AXB+ ones(m,s),F=CXD+ ones(m,s). By using

Matlab  and Algorithm , we obtainrank(W) = ,W= .e+,WW+ee= .e–.rank(N) = ,N= .,NNe–ˆe= .. According to Algo-rithm () we can see that the complex matrix equation [AXB,CXD] = [E,F] is inconsis-tent, and it has a unique solutionXHHEand we can getXHX= ..

6 Conclusions

In this paper, we derive the explicit expressions of least-squares Hermitian solution with the minimum norm for the complex matrix equations (AXB,CXD) = (E,F). A numerical algorithm and examples show the effectiveness of our method.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

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Acknowledgements

This work is supported by the National Natural Science Foundation (No.11271040, 11361027), the Key Project of Department of Education of Guangdong Province (No.2014KZDXM055), the Training plan for the Outstanding Young Teachers in Higher Education of Guangdong (No.SYq2014002), Guangdong Natural Science Fund of China

(No.2014A030313625, 2015A030313646, 2016A030313005), Opening Project of Guangdong Province Key Laboratory of Computational Science at the Sun Yat-sen University (No.2016007), Science Foundation for Youth Teachers of Wuyi University (No.2015zk09) and Science and technology project of Jiangmen City, China.

Received: 20 May 2016 Accepted: 8 November 2016

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