SOLUTIONS TO THE SYSTEM OF OPERATOR EQUATIONS AXB=C=BXA?

Xiao ZHANG(張肖)Guoxing JI(吉國興)

School of Mathematics and Information Science,Shaanxi Normal University,Xi’an 710119,China

E-mail:1964263480@qq.com;gxji@snnu.edu.cn

Abstract In this paper,we present some necessary and sufficient conditions for the existence of solutions,hermitian solutions and positive solutions to the system of operator equations AXB=C=BXA in the setting of bounded linear operators on a Hilbert space.Moreover,we obtain the general forms of solutions,hermitian solutions and positive solutions to the system above.

Key words operator equation;Moore-Penrose inverse;solution;hermitian solution;positive solution

1 Introduction

It is known that matrix equations and operator equations have many applications in control theory,information theory,linear system theory,sampling and other areas(cf.[1]).In 1966,Douglas[2]established the well-known “Douglas Range Inclusion Theorem”,which gave some conditions for the existence of a solution to the operator equation AX=B.Later,many scholars concerned the related problems of solutions to some deformations of AX=B(cf.[3–9]).In particular,Daji? and Koliha[10]considered the existence and concrete forms of hermitian solutions and positive solutions to the system of operator equations AX=C,XB=D.Based on that,Arias and Gonzalez[1]studied the existence and expression of positive solutions to operator equation AXB=C with arbitrary operators A,B and C.Afterwards,Vosough and Moslehian[11]restricted the case to the system of operator equations BXA=B=AXB and gave some characterizations of the existence and representations of the solutions to the system.In this paper,we extend the situation,and consider the relevant problems of the solutions to the system of operator equations AXB=C=BXA.

Throughout this paper,let H be an infinite dimensional complex Hilbert space,we denote the space of all bounded linear operators from H into K by B(H,K),and B(H)when H=K.Let PMdenote the projection onto the closed subspace M of H.For T ∈ B(H),let T?,R(T)and N(T)be the adjoint operator,the range and null space of T,respectively.An operator T∈B(H)is hermitian if T=T?,and positive if hTx,xi≥0 for all x∈H.We call T is regular if there exists some S∈B(H)such that TST=T.At this time,S is said to be an inner inverse of T and denoted by T?.

We say T is Moore-Penrose invertible if the system

has a common solution.If the system is solvable,then its solution is unique and denoted by T?.Indeed,the Moore-Penrose inverse of T exists if and only if R(T)is closed(cf.[12]).It is easy to obtain the following properties

The paper is organized as follows.In Section 2,we consider the existence of solutions to the system of operator equations AXB=C=BXA,and give some necessary and sufficient conditions for the existence of solutions,hermitian solutions and positive solutions to the system AXB=C=BXA.In Section 3,the formulaes for the solutions,hermitian solutions and positive solutions to the system above are obtained.

2 Properties and Conclusions

We give some conditions for the existence of solutions,hermitian solutions and positive solutions to the system of operator equations AXB=C=BXA.

Proposition 2.1 Let A,B,C ∈ B(H).If A and B have closed range,A?CB?=B?CA?,then the following statements are equivalent

(1)The system of operator equations AXB=C=BXA is solvable;

(2)AA?CA?A=BB?CB?B=C;

(3)R(C)? R(A),R(C)? R(B),R(C?)? R(A?)and R(C?)? R(B?).

Proof (1)?(2)If there exists ansuch that AB=C=BA,then R(C)?R(A),R(C)? R(B),R(C?)? R(A?)and R(C?)? R(B?).Since R(C)? R(A)and R(C?) ? R(A?),it follows that C=AA?C and C?=A?(A?)?C?from “Douglas Range Inclusion Theorem”(cf.[2]).We then obtain C=AA?C=CA?A,which yields C=AA?CA?A.Similarly,we have C=BB?C=CB?B and C=BB?CB?B.

(2)? (3)As C=AA?CA?A,we can see R(C)? R(A)and R(C?)? R(A?).Analogously,we have R(C)? R(B)and R(C?)? R(B?).

(3)? (1) It induces that C=BB?C and C=CB?B from the assumptions that R(C)?R(B)and R(C?) ? R(B?).Moreover,we have C=AA?C and C=CA?A from the fact that R(C?)? R(A?)and R(C?)? R(B?).This implies that C=BB?C=BB?CA?A and C=AA?C=AA?CB?B.Since A?CB?=B?CA?,it means that X=A?CB?is a solution of the system AXB=C=BXA. ?

Remark 2.2 Notice that condition(2)in Proposition 2.1 is obvious,but not superfluous.For instance,let K be a Hilbert space.Put in terms of H=K⊕K⊕K.It is evident that A?=I⊕0⊕0,this follows that A?CB?=B?CA?=0 and AA?CA?A 6=C.Then,by an elementary calculation,we know the system of operator equations AXB=C=BXA is unsolvable.

In the next proposition,we establish a relationship between solutions to the system AXB=C=BXA and solutions to the system BX=CA?,XB=A?C,which shall be useful to give the general hermitian solutions and positive solutions to the system AXB=C=BXA.

Proposition 2.3 Let A,B,C ∈ B(H).If A has closed range,R(B)? R(A)and R(B?)?R(A?),then the following conditions are equivalent

(1)The system of operator equations AXB=C=BXA is solvable;

(2)R(C)? R(A),R(C?)? R(A?),and the system of operator equations BX=CA?,XB=A?C is solvable.

Proof (1)?(2)If the system AXB=C=BXA is solvable,then we have R(C)?R(A)and R(C?) ? R(A?).Let∈ B(H)such that AB=C=BA.Then A?C=A?AB.It induces that B=AA?B from inclusion R(B)? R(A),which implies that A?C=A?AAA?B.On the other hand,we have BA?AAA?=BAA?=CA?,this means that A?AAA?is a solution of the system BX=CA?,XB=A?C.

(2)? (1) It follows from the assumptions that R(C)? R(A)and R(C?)? R(A?)that C=AA?C=CA?A.For∈ B(H)such that B=CA?andB=A?C,we easily know that BA=CA?A=C and AB=AA?C=C,which tell us thatis a solution of the system AXB=C=BXA. ?

Remark 2.4 If one of R(C)? R(A)and R(C?)? R(A?)in Proposition 2.3 vanishes,then(2)?(1)is untenable.For example,let K be a Hilbert space.Set A=B=I⊕I⊕0 and C=I with respect to H=K⊕K⊕K.Clearly,R(C)?R(A)does not hold.Moreover,A?C=CA?=I⊕I⊕0.It is easy to check that X=I⊕I⊕0 is a solution of the system BX=CA?,XB=A?C.However,a simple calculation shows that the system AXB=C=BXA is unsolvable.

We recall that,for A,B ∈ B(H),A ≤?B if and only if A=B==(cf.[13,Lemma 2.1]).This will lead to the following result.

Proposition 2.5 Let A,B,C∈B(H).If C≤?A,C≤?B and C has closed range,then the system AXB=C=BXA is solvable.

Proof Since C ≤?A,it follows that C=PR(C)A=APR(C?).So C=CC?A=AC?C.Similarly,if C ≤?B,there would be C=CC?B=BC?C.Consequently,C=BC?C=BC?CC?A=BC?A and C=AC?C=AC?CC?B=AC?B.This implies that X=C?is a solution of the system AXB=C=BXA. ?

In the next proposition,we obtain another sufficient condition for the existence of solutions to the system of operator equations AXB=C=BXA.

Proposition 2.6 Let A,B,C∈B(H).If C≤?A,C≤?B and there exists a T∈B(H)such that C ≤?ATB,C ≤?BTA,then the system of operator equations AXB=C=BXA is solvable.

Proof From the assumption that C ≤?ATB,we have C=ATB=It follows that C=A=from the fact C ≤?A.It is immediate that C=from C ≤?BTA and C ≤?B.We easily see that C=CTA=BTC and C=CTB=ATC,this yields C=BTC=BTCTA and C=ATC=ATCTB.Therefore X=TCT is a solution of the system AXB=C=BXA.?

In the remainder of this section,we shall consider the extra condition AA?=A?A.Under this hypothesis,we get a relationship between hermitian solutions to the system AXB=C=BXA and hermitian solutions to the system BX=CA?,XB=A?C.

Corollary 2.7 Let A,B,C ∈ B(H).If A has closed range with A?A=AA?and C ≤?B ≤?A,then the following statements are equivalent

(1)There is a hermitian operator∈B(H)such that AB=C=BA;

(2)There exists a hermitian operator∈ B(H)such that B=CA?andB=A?C.

Proof Suppose A has closed range and C ≤?B ≤?A,This implies the equation AXB=C=BXA is solvable by Proposition 2.5.

In the following,we present a relationship between positive solutions to the system AXB=C=BXA and positive solutions to the system BX=CA?,XB=A?C.

Corollary 2.8 Let A,B,C ∈B(H).If A has closed range,A?A=AA?and C ≤?B ≤?A,then the followings are equivalent

(1)There exists a positive operator∈B(H)satisfying AB=C=BA;

(2)There is a positive operator∈ B(H)such that BeY=CA?,B=A?C.

3 Main Theorems

In the following,we characterize the concrete forms of solutions,hermitian solutions and positive solutions to the system of operator equations AXB=C=BXA.

Theorem 3.1 Let A,B,C∈B(H).If A has closed range and C≤?B≤?A,then the general solution of the system AXB=C=BXA is

where S∈B(H)is arbitrary.

Proof Suppose A has closed range and C ≤?B ≤?A.Then we know that B and C have closed range.It follows from Proposition 2.5 that the system AXB=C=BXA is solvable.Combining[14,Theorem 2.1],one can easily obtain that the general solution of the equation AXB=C is

where U∈B(H)is an arbitrary operator.If X satisfies the equation BXA=C,then

this implies that B=BB?A=BA?A from the assumption B ≤?A.Hence

It is immediate that U is a solution of the equation

Using[14,Theorem 2.1]again,we get

where S∈B(H)is arbitrary.By putting U in equation(3.2),we obtain that the solution of the system AXB=C=BXA is

where S∈B(H)is arbitrary.?

Remark 3.2 Here are some special cases.

(i)A=B.In this way,the general solution of the system in Theorem 3.1 is

where S∈B(H)is arbitrary.

(ii)C=B.It is clear that the general solution of the system in Theorem 3.1 is

where S∈B(H)is arbitrary.

(iii)In fact,if we take A=B and S=0 in(3.1),then the inner inverse of operator A as in(3.1)cannot be replaced by its Moore-Penrose inverse.For example,we let K be a Hilbert space.Put with respect to H=K⊕K⊕K.It is evident that C≤?A.By an elementary calculation,one obtains

which implies that

It is not hard to see that the specific solution A?CA?changes along with the change of a13or a31.However,A?CA?=C.

In the next theorem,we characterize the concrete form of hermitian solutions to the system of equations AXB=C=BXA.

Theorem 3.3 Let A,B,C ∈ B(H)such that A has closed range with A?A=AA?and C ≤?B ≤?A.If the system AXB=C=BXA has a hermitian solution,then the general hermitian solutions have the matrix representation

in terms of H=R(A?)⊕ N(A),where x22is hermitian and x11=PR(A?)PR(A?)satisfying thatis a hermitian solution of BX=CA?,XB=A?C.

Proof Suppose X ∈B(H)has the matrix decomposition(3.3).Since B≤?A and A?A=AA?,we get A and B have the matrix representations

with respect to H=R(A?)⊕N(A).Thus

On the contrary,assume that X∈B(H)is a hermitian solution of AXB=C=BXA.Set

in terms of H=R(A?)⊕ N(A).From Corollary 2.7 and AA?=A?A,it induces that=PR(A?)XPR(A?)is a hermitian solution of BX=CA?,XB=A?C.Obviously,x11=PR(A?)eY PR(A?).The fact that X is hermitian implies x21=and x22is hermitian.Therefore X has the form of(3.3). ?

In the following,we obtain the representation of positive solutions to the system of equations AXB=C=BXA.

Theorem 3.4 Let A,B,C ∈ B(H)such that A has closed range with A?A=AA?and C ≤?B ≤?A.If the system AXB=C=BXA has a positive solution,then the general positive solutions have the matrix representation

with respect to H=R(A?)⊕ N(A),where f is positive,R(x12) ? Rand x11=PR(A?)PR(A?)satisfying∈ B(H)is a positive solution of BX=CA?,XB=A?C.

Proof Suppose X ∈B(H)has the matrix representation(3.4).Since B≤?A and A?A=AA?,it follows that A and B have the matrix representations

in terms of H=R(A?)⊕N(A).Thus

Conversely,assume that X∈B(H)is a positive solution of AXB=C=BXA.Set

with respect to H=R(A?)⊕ N(A).Combining Corollary 2.8 with AA?=A?A,one obtains=PR(A?)XPR(A?)is a positive solution of BX=CA?,XB=A?C.Clearly,x11=PR(A?)eY PR(A?).Since X is positive,it follows from[1,Theorem 4.2]that x21=with f is positive.Hence X has the form of(3.4).?