# Solutions to the System of Operator EquationsA1X=C1, XB2=C2, and A3XB3=C3 on Hilbert C*-Modules

Authors: Xiaochun Fang; Enran Hou; Ge Dong
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the solvability of the system of the adjointable operator equationsA1X=C1, XB2=C2, and A3XB3=C3 over Hilbert C*-modules. We give necessary and sufficient conditions for the existence of a solution and a positive solution of the system. We also derive representations for a general solution and a positive solution to this system. The above results generalize some recent results concerning the equations for operators with closed ranges.


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## Body

## 1. Introduction

Many results have been made on the study of solvability of equations for operators on Hilbert spaces and HilbertC*-modules. In 1966, Douglas presented the famous Douglas theorem in [1]. He gave the conditions of the existence of the solution to the equation AX=B for operators on a Hilbert space. By using the generalized inverse of operators, Dajić and Koliha [2] got the existence of the common Hermitian and positive solution to the equations AX=C, XB=D for operators on a Hilbert space.HilbertC*-module is a natural generalization both of Hilbert space and C*-algebra and it has been an important tool in the theory of C*-algebra, especially in the study of KK-groups and induced representations (see [3–9]). Therefore it is meaningful to put forward a generalized version of the previous results about operator equations in the context of Hilbert C*-modules.By using the generalized inverses of adjointable operators on a HilbertC*-module, Wang and Dong recently obtained the necessary and sufficient conditions for the existence of a positive solution to the system of adjointable operator equations A1X=C1, XB2=C2, and A3XB3=C3 for operators on Hilbert C*-modules in [10].To use the generalized inverse, the authors mentioned above have to focus their attentions on those adjointable operators whose ranges are closed. However, closed range is a very strong condition in infinite dimensional case which general bounded (adjointable) linear operators may not satisfy. In fact an operator with closed range is also called a generalized Fredholm operator. In [6, 11], Fang et al. generalize the famous Douglas theorem from the case of the Hilbert spaces to the one of the Hilbert C*-modules and get some results about solutions to some equations and some systems of equations without the assumption of the closed ranges by use of some new approaches. They put the attention to the operators whose adjoint operators’ range closures are orthogonally complemented, which is automatically satisfied in Hilbert space case.In this paper, along the same way as in [6, 11], we obtain the existence of the solution to the system of equations A1X=C1, XB2=C2, and A3XB3=C3 which was studied in [10] and then two theorems about the existence of the positive solution to this system, which extend the main results in [10] from operators with closed range to operators whose adjoint operators’ range closures are orthogonally complemented.


## 2. Preliminaries

First of all, we recall some knowledge about HilbertC*-modules.Throughout this paper,𝒜 is a C*-algebra. An inner-product 𝒜-module is a linear space H which is a right 𝒜-module, together with a map (x,y)→〈x,y〉:H×H→𝒜 such that, for any x,y,z∈H, α,β∈ℂ, and a∈𝒜, the following conditions hold:(1)
〈x,αy+βz〉=α〈x,y〉+β〈x,z〉;(2)
〈x,ya〉=〈x,y〉a;(3)
〈x,y〉=〈y,x〉*;(4)
〈x,x〉≥0, and 〈x,x〉=0 if and only if x=0.An inner-product𝒜-module H which is complete with respect to the induced norm ∥x∥=∥〈x,x〉∥1/2 is called a (right) Hilbert 𝒜-module.Suppose thatH1 and H2 are two Hilbert 𝒜-modules; let L𝒜(H1,H2) be the set of all maps T:H1→H2 for which there is a map T*:H2→H1 such that
(1)〈Tx,y〉=〈x,T*y〉,foreachx∈H1,y∈H2.
It is known that any element T of L𝒜(H1,H2) must be a bounded linear operator, which is also 𝒜-linear in the sense that T(xa)=T(x)a for x∈H1 and a∈𝒜. For any T∈L𝒜(H1,H2), the range and the null space of T are denoted by R(T) and N(T), respectively. We call L𝒜(H1,H2) the set of adjointable operators from H1 to H2. We denote by B𝒜(H1,H2) the set of all bounded 𝒜-linear maps, and therefore we have L𝒜(H1,H2)⊆B𝒜(H1,H2). In case H1=H2, L𝒜(H1), to which we abbreviate L𝒜(H1,H2), is a C*-algebra. Then for A∈L𝒜(H), A is Hermitian (self-adjointable) if and only if 〈Ax,y〉=〈x,Ay〉 for any x,y∈H, and positive if and only if 〈Ax,x〉≥0 for any x∈H, in which case, we denote by A1/2 the unique positive element B such that B2=A in the C*-algebra L𝒜(H) and then R(A)¯=R(A1/2)¯. Let L𝒜(H)sa,L𝒜(H)+ be the sets of Hermitian and positive elements of L𝒜(H), respectively. For any A,B∈L𝒜(H)sa, we say A≥B if 〈(A-B)x,x〉≥0 for any x∈H. For 𝒜+, the set of positive elements of the C*-algebra 𝒜 is a positive cone; we could easily verify that “≥” is a partial order on L𝒜(H). For an operator T∈L𝒜(H), set Re(T)=T+T*, and T is called real positive if Re(T)≥0.We say that a closed submoduleH1 of H is topologically complemented if there is a closed submodule H2 of H such that H1+H2=H and H1∩H2=0 and briefly denote the sum by H=H1⊕~H2, called the direct sum of H1 and H2. Moreover, if H2=H1⊥ where H1⊥={x∈H:〈x,y〉=0 for all y∈H1}, we say H1 is orthogonally complemented and briefly denote the sum by H=H1⊕H2, called the orthogonal sum of H1 and H2. In this case, H1=H1⊥⊥ and there exists unique orthogonal projection (i.e., idempotent and self-adjointable operator in L𝒜(H)) onto H1. For two submodules H1 and H2 of H, if H1⊆H2, then H1⊥⊇H2⊥.LetT∈L𝒜(H1,H2); then (1) N(T)=R(T*)⊥ and N(T)⊥⊇R(T*)¯; (2) if R(T) is closed, then so is R(T*), and in this case both R(T) and R(T*) are orthogonally complemented and R(T)⊥=N(T*), R(T*)⊥=N(T) (see [7], Theorem 3.2).Any elementT- of {X∈L𝒜(H1,H2):TXT=T} is called the inner inverse of T and R(TT-)=R(T). R(T) is closed if and only if T has a inner inverse. The Moore-Penrose inverse T+ of T is the unique inner inverse of T which satisfies
(2)T+TT+=T+,TT+=(TT+)*,T+T=(T+T)*.
In this case, (T+)*=(T*)+, R(T+)=R(T*) and T+|R(T)⊥=0. Thus TT+andT+T are the projections onto R(T) and R(T*), respectively.Throughout this paper,H1, H2, H3, H4, and H5 are Hilbert 𝒜-modules. For an operator T∈L𝒜(H1,H2), if R(T*)¯ is orthogonally complemented, then R(T*)¯⊥=N(T) and there exists an orthogonal decomposition E=R(T*)¯⊕N(T). Let PT* denote the orthogonal projection of H1 onto R(T*)¯ and NT the projection I-PT*; then PT*+NT=IH1.Lemma 1 (see [6, Theorem 1.1]).
LetT′∈L𝒜(H1,H2) and T∈L𝒜(H3,H2) with R(T*)¯ being orthogonally complemented. The following statements are equivalent:(1)
T′T′*≤λTT* for some λ≥0;(2)
there existsμ≥0 such that ∥T′*z∥≤μ∥T*z∥ for all z∈H2;(3)
there existsD∈L𝒜(H3,H1) such that T′=TD; that is, TX=T′ has a solution;(4)
R(T′)⊆R(T).Moreover there exists a unique operator D which satisfies the conditions
(3)T′=TD,R(D)⊆N(T)⊥.
In this case,
(4)∥D∥2=inf{λ:T′T′*≤λTT*},R(D)⊆R(T*)¯;N(D)⊆N(T′),
and D is called the reduced solution of the equation TX=T′.
The general solution toTX=T′ is of the form
(5)X=D+NTK,
where K∈L𝒜(H3,H1) is arbitrary.Lemma 2 (see [6, Theorem 2.1]).
LetA∈L𝒜(H1,H2), C∈L𝒜(H3,H2), B∈L𝒜(H4,H3), and D∈L𝒜(H4,H1), suppose R(A*)¯ and R(B)¯ are orthogonally complemented submodules in H1 and H3, respectively, Then AX=C and XB=D have a common solution X∈L𝒜(H3,H1) if and only if
(6)R(C)⊆R(A),R(D*)⊆R(B*),AD=CB.
In this case, the general solution is of the form:
(7)X=D1+NAD2*+NAVNB*,
where D1 and D2 are the reduced solutions of AX=C and B*X=D*, respectively, and V∈L𝒜(H3,H1) is arbitrary.Lemma 3 (see [11, Theorem 3.1]).
LetA∈L𝒜(H1,H2), B∈L𝒜(H3,H4), and C∈L𝒜(H3,H2).(1)
If the equationAXB=C has a solution X∈L𝒜(H4,H1), then
(8)R(C)⊆R(A),R(C*)⊆R(B*).(2)
SupposeR(B)¯ and R(A*)¯ are orthogonally complemented submodules of H4 and H1, respectively. If
(9)R(C)⊆R(A),R(C*)¯⊆R(B*)or(10)R(C)¯⊆R(A),R(C*)⊆R(B*),then AXB=C has a unique solution D∈L𝒜(H4,H1) such that
(11)R(D)⊆N(A)⊥,R(D*)⊆N(B*)⊥,
which is called the reduced solution, and the general solution to AXB=C is of the form
(12)X=D+NAV1+V2NB*,
where V1,V2∈L𝒜(H4,H1).
Then, fromR(D)⊆N(A)⊥ and R(D*)⊆N(B*)⊥, one knows that D=PA*DPB. As in [11], one can obtain that AXB=0⇔PA*XPB=0.Lemma 4 (see [6, Theorem 1.3]).
LetA,C∈L𝒜(H1,H2) such that R(A*)¯ is orthogonally complemented. Then AX=C has a positive solution X∈L𝒜(H1) if and only if R(C)⊆R(A),CA*≥0.
In this case,D≥0,R(D)⊆N(A)⊥, and the general positive solution is of the form
(13)X=D+NAKNA,
where D is the positive reduced solution and K∈L𝒜(H1)+ is an arbitrary positive operator.Lemma 5 (see [6, Lemma 2.1]).
LetU∈L𝒜(H1),V∈L𝒜(H2,H1), and L∈L𝒜(H2). Then (UVV*L)≥0 if and only if U≥0, L≥0, and φ(〈x,Vy〉)φ(〈Vy,x〉)≤φ(〈Ux,x〉)φ(〈Ly,y〉) for any x∈H1, y∈H2, and any state φ∈S(𝒜).Lemma 6 (see [6, Proposition 2.2]).
LetA1,C1∈L𝒜(H1,H2) and B2,C2∈L𝒜(H3,H1),
(14)D=(A1B2*),E=(C1C2*),F=ED*=(C1A1*C1B2C2*A1*C2*B2)
such that R(D*)¯ is orthogonally complemented. Then A1X=C1 and XB2=C2 have a common positive solution X∈L𝒜(H1), if and only if F≥0 and R(E)⊆R(D).
In this case, the general positive solution can be expressed asX=Y0+NDYND, where Y0∈L𝒜(H1)+ is the positive reduced solution and Y∈L𝒜(H1)+ is an arbitrary positive operator.Lemma 7 (see [11, Corollary 3.3(iii)]).
LetA∈L𝒜(H1,H2) and C∈L𝒜(H2) such that R(A)¯ and R(A*)¯ are orthogonally complemented, and A or C has the closed range; the equation AXA*=C has a positive solution if and only if C≥0 and R(C)⊆R(A). In this case, the operator
(15)X=D+DV1NA+NAV1*D+NAV1*DV1NA+NAV2NA
is a positive solution for any V1∈L𝒜(H1) and V2∈L𝒜(H1)+, where D is the reduced solution.LetA∈L𝒜(H1,H2),B∈L𝒜(H3,H1), and C∈L𝒜(H3,H2). Suppose R(A*)¯ and R(B)¯ are orthogonally complemented submodules of H1; the equation AXB=C has the reduced solution D∈L𝒜(H1). Set T=NAB and Z=NTB*Re(D)BNT, and assume that R(T)¯ and R(T*)¯ are orthogonally complemented submodules of H1 and H3, respectively. Set
(16)S𝒴(A,B,C)={Z+ZV1PT*+PT*V1*Z+PT*V1*ZV1PT*+PT*V2PT*:V1∈L𝒜(H3)suchthatNTB*DBPT*=NTB*DBNTV1PT*,V2∈L𝒜(H3)+},SΣ(A,B,C)1={X1∈L𝒜(H1):X1B=12PTT*-1PT*×(Y-B*Re(D)B)(NT+I)forsomeY∈S𝒴(A,B,C),X1NB*=0},SΣ(A,B,C)2={X2∈L𝒜(H1):X2B=12PTT*-1(PT*B*(D*-D)BPT*)+NT*(V3NT+iV4T)forsomeV4∈L𝒜(E),V3∈L𝒜(H3,H1)withR(V3*NT*)⊆R(B*)},SΣ(A,B,C)={X1+X2:X1∈SΣ(A,B,C)1,X2∈SΣ(A,B,C)2},SS(A,B,C)={D-D*NB*+NAWPB-PBW*NANB*+VNB*:V∈L𝒜(H1),W∈SΣ(A,B,C)}.It is clear thatSS(A,B,C) is a subset of the solution space to the equation AXB=C.Lemma 8 (see [11, Theorem 4.7]).
LetA∈L𝒜(H1,H2), B∈L𝒜(H3,H1), and C∈L𝒜(H3,H2). Suppose that R(A*)¯ and R(B)¯ are orthogonally complemented submodules of H1, AXB=C has the reduced solution D∈L𝒜(H1), T=NAB has the closed range, and NTB*DBNT is self-adjoint. Let X∈SS(A,B,C) with X=D-D*NB*+NAWPB-PBW*NANB*+VNB* for some V∈L𝒜(H1) and W∈SΣ(A,B,C). Then X=X* if and only if there exist V1∈L𝒜(H1) and V2∈L𝒜(H1)sa such that
(17)V=-12(PB(D+NAWPB)-(D*+PBW*NA)(I+3NB*))-iPBV1*PB-NB*V2,
in which case, X=D+D*NB*+NAWPB+PBW*NANB*+NB*(-V2)NB*.
As a consequence,(18)SS(A,B,C)sa={D+D*NB*+NAWPB+PBW*NANB*+NB*VNB*:V∈L𝒜(H1)sa,W∈SΣ(A,B,C)}.Lemma 9 (see [11, Theorem 4.12]).
LetA∈L𝒜(H1,H2), B∈L𝒜(H3,H1), and C∈L𝒜(H3,H2) such that R(B)⊆R(A*)¯, and let R(A*)¯ and R(B)¯ be orthogonally complemented submodules of H1. Suppose that AXB=C has the reduced solution D∈L𝒜(H1).(1)
IfAXB=C has a positive solution X∈L𝒜(H1), then B*DB≥0 and there exists a positive number λ such that
(19)B*D*NB*DB≤λB*DB.(2)
Suppose thatR(PBDPB)¯ is orthogonally complemented in H1. If B*DB≥0 and
(20)B*D*NB*DB≤λB*DB,for some positive number λ, then AXB=C has a positive solution X∈SS(A,B,C)sa.


## 3. Main Results

Theorem 10.
LetHi (i=1,2,3,4,5) be Hilbert 𝒜-modules, A1∈L𝒜(H1,H2), A3∈L𝒜(H1,H4), B2∈L𝒜(H3,H1), B3∈L𝒜(H5,H1), C1∈L𝒜(H1,H2), C2∈L𝒜(H3,H1), and C3∈L𝒜(H5,H4).(1)
If the system of operator equationsA1X=C1,XB2=C2, and A3XB3=C3 has a solution X∈L𝒜(H1), then
(21)R(C1)⊆R(A1),R(C2*)⊆R(B2*),R(C3)⊆R(A3),R(C3*)⊆R(B3*),A1C2=C1B2.(2)
SetA4=A3NA1,B4=NB2*B3. Suppose R(A1*)¯, R(B2)¯, R(A4*)¯, and R(B4)¯ are orthogonally complemented submodules of H1. If
(22)R(C1)⊆R(A1),R(C2*)⊆R(B2*),R(C3)⊆R(A3),R(C3*)¯⊆R(B3*)or(23)R(C1)⊆R(A1),R(C2*)⊆R(B2*),R(C3)¯⊆R(A3),R(C3*)⊆R(B3*),and A1C2=C1B2, then the system of operator equations A1X=C1, XB2=C2, and A3XB3=C3 has a unique solution D∈L𝒜(H1) such that
(24)R(D)⊆N(A1)⊥⋂‍N(B2*)⊥⋂‍N(A4)⊥,R(D*)⊆N(A1*)⊥⋂‍N(B2)⊥⋂‍N(B4*)⊥.
In this case, the general solution is of the form
(25)X=D1+NA1D2*+NA1(D3+NA4V1+V2NB4*)NB2*,
where D1, D2, and D3 are the reduced solutions of A1X=C1, B2*X=C2*, and A4XB4=C3-A3D1B3-A3NA1D2*B3 respectively. V1,V2∈L𝒜(H1) are arbitrary.Proof.
(1) If the system of operator equationsA1X=C1,XB2=C2,andA3XB3=C3 has a solution X∈L𝒜(H1), it is easy to know that
(26)R(C1)⊆R(A1),R(C2*)⊆R(B2*),R(C3)⊆R(A3),R(C3*)⊆R(B3*),A1C2=A1XB2=C1B2.
(2) SinceR(A1*)¯ and B2¯ are orthogonally complemented, R(C1)⊆R(A1),R(C2*)⊆R(B2*), and A1C2=C1B2. By Lemma 2, we know that A1X=C1andXB2=C2 have a common solution X0 and it has the form:
(27)X0=D1+NA1D2*+NA1YNB2*,
where D1 and D2 are the reduced solutions of A1X=C1,B2*X=C2* respectively.
TakeX0=D1+NA1D2*+NA1YNB2* into A3XB3=C3, then we can get
(28)A3NA1YNB2*B3=C3-A3D1B3-A3NA1D2*B3.(29)A4YB4=C3-A3D1B3-A3NA1D2*B3.
SinceR(C3)⊆R(A3) and R(C3*)¯⊆R(B3*) (or R(C3)¯⊆R(A3) and R(C3*)⊆R(B3*)), then
(30)R(C3-A3D1B3-A3NA1D2*B3)⊆R(A3),R(C3*-B3*D1*A3*-B3*D2NA1A3*)¯⊆R(B3*),
or
(31)R(C3-A3D1B3-A3NA1D2*B3)¯⊆R(A3),R(C3*-B3*D1*A3*-B3*D2NA1A3*)⊆R(B3*).
By Lemma 3(2), we know that the operator equation A4YB4=C3-A3D1B3-A3NA1D2*B3 has a unique reduced solution D3∈L𝒜(H1) and the general solution has the form
(32)Y=D3+NA4V1+V2NB4*,
where V1 and V2∈L𝒜(H1) are arbitrary.
Hence the system of operator equationsA1X=C1,XB2=C2, and A3XB3=C3 has a solution and it has the form
(33)X=D1+NA1D2*+NA1(D3+NA4V1+V2NB4*)NB2*.
And it is easy to see thatD=D1+NA1D2*+NA1D3NB2* is the reduced solution to the system of operator equations A1X=C1, XB2=C2 and A3XB3=C3; R(D)⊆N(A1)⊥⋂‍N(B2*)⊥⋂‍N(A4)⊥ and R(D*)⊆N(A1*)⊥⋂‍N(B2)⊥⋂‍N(B4*)⊥.IfR(A3*)¯ and R(B3)¯ are orthogonally complemented submodules, then R(A4*)¯ and R(B4)¯ are orthogonally complemented submodules.Remark 11.
LetHi (i=1,2,3,4,5) be Hilbert 𝒜-modules, A1∈L𝒜(H1,H2), A3∈L𝒜(H1,H4), B2∈L𝒜(H3,H1), B3∈L𝒜(H5,H1), C1∈L𝒜(H1,H2), C2∈L𝒜(H3,H1), and C3∈L𝒜(H5,H4). Set D, E, and F as Lemma 6, and suppose R(D*)¯, R(A3*)¯, and R(B3*)¯ are orthogonally complemented submodules of H1. If R(E)⊆R(D), A1C2=C1B2, R(C3)⊆R(A3), R(C3*)¯⊆R(B3*) (or, R(E)⊆R(D), A1C2=C1B2, R(C3)¯⊆R(A3), R(C3*)⊆R(B3*)), then the system of operator equations A1X=C1, XB2=C2, A3XB3=C3 has a solution from Theorem 10. Let C∈L𝒜(H1) be the reduced solution to the system, and then we can obtain the next theorem about the positive solution to the system.Theorem 12.
Let the notions and conditions as Remark11. Set A4=A3ND, B4=NDB3, and R(B3)⊆R(A3*)¯.(1)
If the system of operator equationsA1X=C1, XB2=C2, and A3XB3=C3 has a positive solution, then F≥0, B4*CB4≥0, and there exists a positive number λ such that B4*C*NB4*CB4≤λB4*CB4.(2)
SupposeR(PB4CPB4)¯ is orthogonally complemented submodule in H1. If F≥0, B4*CB4≥0, and B4*C*NB4*CB4≤λB4*CB4, for some λ>0, then the system of operator equations A1X=C1, XB2=C2, and A3XB3=C3 has a positive solution.Proof.
(1) If the system of operator equationsA1X=C1, XB2=C2, and A3XB3=C3 has a positive solution X0∈L𝒜(H1)+, then X0 is the positive solution to the system of equations A1X=C1, XB2=C2. by Lemma 6, we know that F≥0 and X0 can be expressed as
(34)X0=Y0+NDYND,Y∈L𝒜(H1)+,
where Y0 is the positive reduced solution to the system of equations A1X=C1, XB2=C2. Taking X0=Y0+NDYND into A3XB3=C3 yields that A4YB4=C3-A3Y0B3. And it has a positive solution. From Lemma 9, we know that B4*CB4*≥0, and there exists a positive number λ such that B4*C*NB4*CB4≤λB4*CB4.
(2) IfF≥0, we can get that the system of equations A1X=C1, XB2=C2 has a positive solution X0∈L𝒜(H1)+, and it has the form
(35)X0=Y0+NDYND,Y∈L𝒜(H1)+,
where Y0 is the positive reduced solution to the system of equations A1X=C1, XB2=C2. Take X0=Y0+NDYND into A3XB3=C3; then we can obtain that A4YB4=C3-A3Y0B3.
IfR(PB4CPB4)¯ is orthogonally complemented submodule in H1, B4*CB4*≥0 and B4*C*NB4*CB4≤λB4*CB4 for some λ>0, by Lemma 9(2), we can easily get that the equation A4YB4=C3-A3Y0B3 has a positive solution. Therefore the system of operator equations A1X=C1, XB2=C2, and A3XB3=C3 has a positive solution.Next, we give another theorem about the existence of the positive solution to the system of operator equationsA1X=C1, XB2=C2, and A3XB3=C3. First we propose a lemma as follows.Lemma 13.
Suppose thatM∈L𝒜(H1,H2), N∈L𝒜(H3,H1), and R(M*)¯ and R(N)¯ are orthogonally complemented submodules of H1. Let T=PM*+PN, and suppose T has a closed range; then TX=PN and YT=PM* both have positive solutions.
In this case, the general positive solutions are of the forms(36)X=P+NTUNT,Y=Q+NTVNT,
where P and Q are the reduced positive solutions, respectively, U and V∈L𝒜(H1)+ are arbitrary. Furthermore, on has (Q+P)PM*=PM* and PN(Q+P)=PN.Proof.
It is clear thatPM*≥0, PN≥0, so T≥0, PNPM*≥0 and PNPM*≥0. Consider R(PN)⊆R(T), PNT*=PNT=PN(PM*+PN)=PNPM*+PN≥0, R(PM*)⊆R(T), PM**T=PM*T=PM*(PM*+PN)=PM*+PM*PN≥0, then it follows from Lemma 4 that P≥0, Q≥0, R(P),R(Q)⊆N(T)⊥ and the general positive solutions are of the forms
(37)X=P+NTUNT,Y=Q+NTVNT,
where U,V∈L𝒜(H1)+ are arbitrary.
ConsiderTP=PN, QT=PM*, T,P,Q,PN,PM*≥0; then PT=PN, TQ=PM*, and T(Q+P)=T, T(Q+P)PM*=TPM*, T((Q+P)PM*-PM*)=0, and (Q+P)PM*-PM*⊆N(T). By Lemma 4, we know that (Q+P)PM*-PM*⊆N(T)⊥, so (Q+P)PM*=PM*. Similarly, we can obtain that (Q+P)PN=PN; then PN(Q+P)=PN.For simplicity, put(38)D=(A1B2*),E=(C1C2*),F=(C1A1*C1B2C2*A1*C2*B2).Y
0 is the common reduced positive solution to A1X=C1, XB2=C2. Consider M=A3ND, N=NDB3; L is the reduced solution to MYN=C3-A3Y0B3. Consider T=PM*+PN. P is the positive reduced solution to TX=PN. Q is the positive reduced solution to XT=PM*. Y1 is the positive solution to XP=LQ. Y2 is the positive solution to QX=PL. Take
(39)R=L+L*+Y1+Y2,S=PLQ.By Lemma6, we know that Y0 uniquely exists when R(D*)¯ is orthogonally complemented, F≥0, and R(E)⊂R(D). From Lemma 3, we know that L uniquely exists when R(N)¯ and R(M*)¯, are orthogonally complemented, R(C3-A3Y0B3)⊂R(M) and R((C3-A3Y0B3)*)¯⊂R(M*). From Lemma 13, we know that PandQ uniquely exist when R(M*)¯ and R(N)¯ are orthogonally complemented, T has a closed range. By Lemmas 4 and 13, we know that Y1 and Y2 exist. So S uniquely exists. In fact, if the conditions in the next theorem are satisfied, we can easily get that Y0, L, P, Q, and S uniquely exist.Theorem 14.
LetHi (i=1,2,3,4,5) be Hilbert 𝒜-modules, A1∈L𝒜(H1,H2), A3∈L𝒜(H1,H4), B2∈L𝒜(H3,H1), B3∈L𝒜(H5,H1), C1∈L𝒜(H1,H2), C2∈L𝒜(H3,H1), and C3∈L𝒜(H5,H4). Suppose R(D*)¯, R(M*)¯, and R(N)¯ are orthogonally complemented submodules of H1, R(C3)¯⊆R(A3), R(C3*)⊆R(B3*) (or R(C3)⊆R(A3), R(C3*)¯⊆R(B3*)), and T has closed range. The system of operator equations A1X=C1, XB2=C2, and A3XB3=C3 has a positive solution X∈L𝒜(H1) if and only if
(40)F≥0,S≥0,R(E)⊆R(D),R(PL)⊆R(S),R(QL*)⊆R(S),
in which case the general common positive solution to A1X=C1, XB2=C2, and A3XB3=C3 can be expressed as
(41)X=Y0+ND(C+CV1NT+NTV1*C+NTV1*CV1NT+NTV2NT)ND,
where Y0 is the common positive reduced solution to A1X=C1, XB2=C2, C is the positive reduced solution to TXT*=R, Y1 and Y2 are arbitrary positive solutions to Y1P=LQ, QY2=PL, respectively, such that R is positive, V1 is arbitrary, and V2 is arbitrary positive operator in L𝒜(H1).Proof.
SupposeX0 is a positive solution to the system of the adjointable operator equations A1X=C1, XB2=C2, and A3XB3=C3; then X0 is a positive solution to the operator equations A1X=C1, XB2=C2. It follows from Lemma 6 that F≥0, R(E)⊆R(D), and X0 has the form
(42)X0=Y0+NDYND,
where Y0 is the positive reduced solution of A1X=C1, XB2=C2, and Y∈L𝒜(H1)+ is an arbitrary operator.
TakeX0=Y0+NDYND into A3XB3=C3; we can get that
(43)MYN=C3-A3Y0B3
has a positive solution. Consider
(44)S=PLQ=PPM*LPNQ=PPM*YPNQ=PYQ.
By Lemma 13, we know that P,Q≥0. Hence, if Y is positive, so is S.
For allx∈H1, S*x=0; that is, QL*Px=0; then L*Px=0 since that Q≥0. We can get N(S*)⊆N(L*P)=N((PL)*); then R(PL)⊆R(S). Similarly, R(QL*)⊆R(S).
IfF≥0 and R(E)⊆R(D), then equations A1X=C1, XB2=C2 have a positive solution by Lemma 6 and this positive solution can be expressed as
(45)X0=Y0+NDYND,
where Y0 is the positive reduced solution of the system of the adjointable operator equations A1X=C1, XB2=C2, and Y∈L𝒜(H1)+ is an arbitrary operator.
TakingX0=Y0+NDYND into A3XB3=C3, we can get
(46)MYN=C3-A3Y0B3.
Now, we want to show that the equationMYN=C3-A3Y0B3 has a positive solution. By Lemma 3, we know that the equation MYN=C3-A3Y0B3 has a solution; then L exists.
We rewriteY1P=LQ, QY2=PL as Y1P=LQ=L1, QY2=PL=L2; then L2Q*=PLQ=S≥0, L1*P=Q*L*P=S*=S≥0, and R(S)=R(L1*)=R(L2). Consider R(P*L)⊆R(S), R(QL*)⊆R(S); the equations Y1P=L1, QY2=L2 both have positive solutions by Lemma 4 and they can be expressed as
(47)Y1=D1+NPVNP,Y2=D2+NQWNQ,
where D1 and D2 are the positive reduced solutions to the equations Y1P=L1, QY2=L2, respectively, and V,W∈L𝒜(H1)+ are arbitrary.
For operator equationsSX=L2, XS=L1, we can obtain that L2S*=PLQL*P≥0, L1*S*=QL*PLQ*≥0 and R(L2)⊆R(S), R(L1*)⊆R(S*). By Lemma 4, we can get that SX=L2 and XS=L1 both have positive solutions. Let Z1 and Z2 be the positive reduced solutions, respectively. Hence
(48)SZ1=L2,PLQZ1=PL,PL1Z1=L,P(L-L1Z1)=0,Z2S=L1,Z2PLQ=LQ,Z2L2Q=LQ,(L-Z2L2)Q=0,R=L+L*+Y1+Y2=L+L*+D1+NPVNP+D2+NQWNQ=L1Z1+Z1*L1*+D1+D2+NPVNP+NQWNQ+L-L1Z1+L*-Z1*L1*=Z2SZ1+Z1*SZ2*+D1+D2+(NPNQ)(VL-L1Z1L*-Z1*L1*W)(NPNQ)=(Z2+Z1*)S(Z1+Z2*)+D1+D2+(NPNQ)(VL-L1Z1L*-Z1*L1*W)(NPNQ).
LetV=I and W=(L*-Z1*L1*)(L-L1Z1). By Lemma 5, R≥0.
For the operator equationTXT*=R, by Lemma 7, TXT*=R has a positive solution U and U has the form
(49)U=C+CV1NT+NTV1*C+NTV1*CV1NT+NTV2NT,
where C is the positive reduced solution to the operator equation TXT*=R and V1∈L𝒜(H1) and V2∈L𝒜(H1)+ are arbitrary.
Then we claim thatX=Y0+ND(C+CV1NT+NTV1*C+NTV1*CV1NT+NTV2NT)ND is the positive solution to A1X=C1, XB2=C2, and A3XB3=C3. In fact, we only need to prove that MUN=C3-A3Y0B3. Consider
(50)MUN=MQTUT*PN=MQRPN=MQ(L+L*+Y1+Y2)PN=MQLPN+MQL*PN+MQY1PN+MQY2PN=MQLPN+MPLQN+MQLQN+MPLPN=M(Q+P)L(Q+P)N=M(Q+P)PM*LPN(Q+P)N=MPM*LPNN=MLN=C3-A3Y0B3.
Suppose thatX¯ is a positive solution to the system of the operator equations A1X=C1, XB2=C2, and A3XB3=C3. It follows from Lemma 6 that X¯ can be expressed as
(51)X¯=Y0+NDYND,Y∈L𝒜(H1)+,
where Y0 is the positive reduced solution of the system of the operator equations A1X=C1, XB2=C2. Hence there is a positive operator U such that
(52)X¯-Y0=NDUND.
LetY1=PM*UPM*, Y2=PNUPN; it follows from
(53)Y1P=PM*UPM*P=PM*U(PN-PNP)=PM*UPN-PM*UPNP=PM*LPN-PM*LPNP=PM*L(PN-PNP)=PM*LPNQ=LQ,QY2=QPNUPN=(PM*-QPM*)UPN=PM*UPN-QPM*UPN=PM*LPN-QPM*LPN=(PM*-QPM*)LPN=PPM*LPN=PL
that Y1 and Y2 are positive solutions to the operator equations Y1P=LQ, QY2=PL, respectively. Consider
(54)R=L+L*+Y1+Y2=PM*LPN+PNLPM*+PM*UPM*+PNUPN=PM*UPN+PNUPM*+PM*UPM*+PNUPN=(PM*+PN)U(PM*+PN)=TUT=TUT*.
By Lemma 7, U can be expressed as
(55)U=C+CV1NT+NTV1*C+NTV1*CV1NT+NTV2NT,
where C is the positive reduced solution to TXT*=R and V1∈L𝒜(H1) and V2∈L𝒜(H1)+ are arbitrary.
TakeU=C+CV1NT+NTV1*C+NTV1*CV1NT+NTV2NT into X¯-Y0=NDUND; we know that X¯ has the form that is expressed as (41).LetA∈L𝒜(H1,H2), B∈L𝒜(H3,H1), and C∈L𝒜(H3,H1), and suppose that R(A*)¯, R(B)¯ are orthogonally complemented submodules of H1. If R(C)¯⊆R(A) and R(C*)⊆R(B*) (or R(C)⊆R(A), R(C*)¯⊆R(B*)) we can obtain that the operator equation AXB=C has a solution by Lemma 3. Let L be the reduced solution to AXB=C and T=PA*+PB and suppose it has a closed range, where P is the positive reduced solution to TX=PB. Q is the positive reduced solution to XT=PA*. Y1 is the positive solution to Y1P=LQ. Y2 is the positive solution to QY2=PL.Consider(56)R=L+L*+Y1+Y2,S=PLQ.
By Theorem 14, we can give necessary and sufficient conditions for the existence of a positive solution to operator equation AXB=C.Corollary 15.
Let the notions and conditions be as described above. Then the operator equationAXB=C has a positive solution in L𝒜(H1)+ if and only if
(57)S≥0,R(PL)⊆R(S),R(QL*)⊆R(S),
in this case the form of general positive solution to AXB=C is
(58)X=D+DV1NT+NTV1*D+NTV1*DV1NT+NTV2NT,
where D is the positive reduced solution to TXT*=R, Y1 and Y2 are arbitrary positive solutions to Y1P=LQ, QY2=PL, respectively, such that R is positive, and V1∈L𝒜(H1) and V2∈L𝒜(H1)+ are arbitrary.The results obtained in Lemma9 give us the condition of the existence of the positive solution to AXB=C, but they are restricted by the assumption that R(B)⊆R(A*). The result in Corollary 15 does not have the constraint mentioned above. Clearly Theorem 14 extends Theorem 3.6 in [10], and Corollary 15 extends Corollary 4.1 in [10].



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*Source: 826564-2013-11-25.xml*