We study the solvability of the system of the adjointable operator equations A₁X = 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.

1. Introduction

Many results have been made on the study of solvability of equations for operators on Hilbert spaces and Hilbert C^*-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.

Hilbert C^*-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 Hilbert C^*-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 Hilbert C^*-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-product 𝒜-module H which is complete with respect to the induced norm ∥x∥ = ∥〈x,x〉∥^1/2 is called a (right) Hilbert 𝒜-module.

Suppose that H₁ and H₂ are two Hilbert 𝒜-modules; let L_𝒜(H₁, H₂) be the set of all maps T : H₁ → H₂ for which there is a map T^* : H₂ → H₁ such that

()

It is known that any element T of L_𝒜(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 ∈ L_𝒜(H₁, H₂), the range and the null space of T are denoted by R(T) and N(T), respectively. We call L_𝒜(H₁, H₂) the set of adjointable operators from H₁ to H₂. We denote by B_𝒜(H₁, H₂) the set of all bounded 𝒜-linear maps, and therefore we have L_𝒜(H₁, H₂)⊆B_𝒜(H₁, H₂). In case H₁ = H₂, L_𝒜(H₁), to which we abbreviate L_𝒜(H₁, H₂), is a C^*-algebra. Then for A ∈ L_𝒜(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 A^1/2 the unique positive element B such that B² = A in the C^*-algebra L_𝒜(H) and then

. Let L_𝒜(H) _sa, L_𝒜(H) ₊ be the sets of Hermitian and positive elements of L_𝒜(H), respectively. For any A, B ∈ L_𝒜(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 L_𝒜(H). For an operator T ∈ L_𝒜(H), set Re(T) = T + T^*, and T is called real positive if Re(T) ≥ 0.

We say that a closed submodule H₁ 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 , called the direct sum of H₁ and H₂. Moreover, if where 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, and there exists unique orthogonal projection (i.e., idempotent and self-adjointable operator in L_𝒜(H)) onto H₁. For two submodules H₁ and H₂ of H, if H₁⊆H₂, then .

Let T ∈ L_𝒜(H₁, H₂); then (1) N(T) = R(T^*) ^⊥ and ; (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 element T⁻ of {X ∈ L_𝒜(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

()

In this case, (T⁺) ^* = (T^*) ⁺, R(T⁺) = R(T^*) and

. Thus TT⁺ and T⁺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 ∈ L_𝒜(H₁, H₂), if is orthogonally complemented, then and there exists an orthogonal decomposition . Let denote the orthogonal projection of H₁ onto and N_T the projection ; then .

Lemma 1 (see [6], Theorem 1.1.)Let T^′ ∈ L_𝒜(H₁, H₂) and T ∈ L_𝒜(H₃, H₂) with being orthogonally complemented. The following statements are equivalent:

(1)
T^′T^′* ≤ λTT^* for some λ ≥ 0;
(2)
there exists μ ≥ 0 such that ∥T^′*z∥ ≤ μ∥T^*z∥ for all z ∈ H₂;
(3)
there exists D ∈ L_𝒜(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

()

In this case,

()

and D is called the reduced solution of the equation TX = T^′.

The general solution to TX = T^′ is of the form

()

where K ∈ L_𝒜(H₃, H₁) is arbitrary.

Lemma 2 (see [6], Theorem 2.1.)Let A ∈ L_𝒜(H₁, H₂), C ∈ L_𝒜(H₃, H₂), B ∈ L_𝒜(H₄, H₃), and D ∈ L_𝒜(H₄, H₁), suppose and are orthogonally complemented submodules in H₁ and H₃, respectively, Then AX = C and XB = D have a common solution X ∈ L_𝒜(H₃, H₁) if and only if

()

In this case, the general solution is of the form:

()

where D₁ and D₂ are the reduced solutions of AX = C and B^*X = D^*, respectively, and V ∈ L_𝒜(H₃, H₁) is arbitrary.

Lemma 3 (see [11], Theorem 3.1.)Let A ∈ L_𝒜(H₁, H₂), B ∈ L_𝒜(H₃, H₄), and C ∈ L_𝒜(H₃, H₂).

(1)
If the equation AXB = C has a solution X ∈ L_𝒜(H₄, H₁), then
()
(2)
Suppose and are orthogonally complemented submodules of H₄ and H₁, respectively. If
()
or
()

then AXB = C has a unique solution D ∈ L_𝒜(H₄, H₁) such that

()

which is called the reduced solution, and the general solution to AXB = C is of the form

()

where V₁, V₂ ∈ L_𝒜(H₄, H₁).

Then, from R(D)⊆N(A) ^⊥ and R(D^*)⊆N(B^*) ^⊥, one knows that . As in [11], one can obtain that .

Lemma 4 (see [6], Theorem 1.3.)Let A, C ∈ L_𝒜(H₁, H₂) such that is orthogonally complemented. Then AX = C has a positive solution X ∈ L_𝒜(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

()

where D is the positive reduced solution and K ∈ L_𝒜(H₁) ₊ is an arbitrary positive operator.

Lemma 5 (see [6], Lemma 2.1.)Let U ∈ L_𝒜(H₁), V ∈ L_𝒜(H₂, H₁), and L ∈ L_𝒜(H₂). Then 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.)Let A₁, C₁ ∈ L_𝒜(H₁, H₂) and B₂, C₂ ∈ L_𝒜(H₃, H₁),

()

such that

is orthogonally complemented. Then A₁X = C₁ and XB₂ = C₂ have a common positive solution X ∈ L_𝒜(H₁), if and only if F ≥ 0 and R(E)⊆R(D).

In this case, the general positive solution can be expressed as X = Y₀ + N_DYN_D, where Y₀ ∈ L_𝒜(H₁) ₊ is the positive reduced solution and Y ∈ L_𝒜(H₁) ₊ is an arbitrary positive operator.

Lemma 7 (see [11], Corollary 3.3iii.)Let A ∈ L_𝒜(H₁, H₂) and C ∈ L_𝒜(H₂) such that and 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

()

is a positive solution for any V₁ ∈ L_𝒜(H₁) and V₂ ∈ L_𝒜(H₁) ₊, where D is the reduced solution.

Let A ∈ L_𝒜(H₁, H₂), B ∈ L_𝒜(H₃, H₁), and C ∈ L_𝒜(H₃, H₂). Suppose

and

are orthogonally complemented submodules of H₁; the equation AXB = C has the reduced solution D ∈ L_𝒜(H₁). Set T = N_AB and Z = N_TB^*Re(D)BN_T, and assume that

and

are orthogonally complemented submodules of H₁ and H₃, respectively. Set

()

It is clear that SS(A, B, C) is a subset of the solution space to the equation AXB = C.

Lemma 8 (see [11], Theorem 4.7.)Let A ∈ L_𝒜(H₁, H₂), B ∈ L_𝒜(H₃, H₁), and C ∈ L_𝒜(H₃, H₂). Suppose that and are orthogonally complemented submodules of H₁, AXB = C has the reduced solution D ∈ L_𝒜(H₁), T = N_AB has the closed range, and N_TB^*DBN_T is self-adjoint. Let X ∈ SS(A, B, C) with for some V ∈ L_𝒜(H₁) and W ∈ SΣ(A, B, C). Then X = X^* if and only if there exist V₁ ∈ L_𝒜(H₁) and V₂ ∈ L_𝒜(H₁) _sa such that

()

in which case,

As a consequence,

()

Lemma 9 (see [11], Theorem 4.12.)Let A ∈ L_𝒜(H₁, H₂), B ∈ L_𝒜(H₃, H₁), and C ∈ L_𝒜(H₃, H₂) such that , and let and be orthogonally complemented submodules of H₁. Suppose that AXB = C has the reduced solution D ∈ L_𝒜(H₁).

(1)
If AXB = C has a positive solution X ∈ L_𝒜(H₁), then B^*DB ≥ 0 and there exists a positive number λ such that
()
(2)
Suppose that is orthogonally complemented in H₁. If B^*DB ≥ 0 and
()

for some positive number λ, then AXB = C has a positive solution X ∈ SS(A, B, C) _sa.

3. Main Results

Theorem 10. Let H_i (i = 1,2, 3,4, 5) be Hilbert 𝒜-modules, A₁ ∈ L_𝒜(H₁, H₂), A₃ ∈ L_𝒜(H₁, H₄), B₂ ∈ L_𝒜(H₃, H₁), B₃ ∈ L_𝒜(H₅, H₁), C₁ ∈ L_𝒜(H₁, H₂), C₂ ∈ L_𝒜(H₃, H₁), and C₃ ∈ L_𝒜(H₅, H₄).

(1)
If the system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a solution X ∈ L_𝒜(H₁), then
()
(2)
Set . Suppose , , , and are orthogonally complemented submodules of H₁. If
()
or
()

and A₁C₂ = C₁B₂, then the system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a unique solution D ∈ L_𝒜(H₁) such that

()

In this case, the general solution is of the form

()

where D₁, D₂, and D₃ are the reduced solutions of A₁X = C₁,

, and

respectively. V₁, V₂ ∈ L_𝒜(H₁) are arbitrary.

Proof. (1) If the system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a solution X ∈ L_𝒜(H₁), it is easy to know that

()

(2) Since and are orthogonally complemented, , and A₁C₂ = C₁B₂. By Lemma 2, we know that A₁X = C₁ and XB₂ = C₂ have a common solution X₀ and it has the form:

()

where D₁ and D₂ are the reduced solutions of

respectively.

Take into A₃XB₃ = C₃, then we can get

()

Since R(C₃)⊆R(A₃) and (or and ), then

()

By Lemma 3(2), we know that the operator equation

has a unique reduced solution D₃ ∈ L_𝒜(H₁) and the general solution has the form

()

where V₁ and V₂ ∈ L_𝒜(H₁) are arbitrary.

Hence the system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a solution and it has the form

()

And it is easy to see that is the reduced solution to the system of operator equations A₁X = C₁, XB₂ = C₂ and A₃XB₃ = C₃; and .

If and are orthogonally complemented submodules, then and are orthogonally complemented submodules.

Remark 11. Let H_i (i = 1,2, 3,4, 5) be Hilbert 𝒜-modules, A₁ ∈ L_𝒜(H₁, H₂), A₃ ∈ L_𝒜(H₁, H₄), B₂ ∈ L_𝒜(H₃, H₁), B₃ ∈ L_𝒜(H₅, H₁), C₁ ∈ L_𝒜(H₁, H₂), C₂ ∈ L_𝒜(H₃, H₁), and C₃ ∈ L_𝒜(H₅, H₄). Set D, E, and F as Lemma 6, and suppose , , and are orthogonally complemented submodules of H₁. If R(E)⊆R(D), A₁C₂ = C₁B₂, R(C₃)⊆R(A₃), (or, R(E)⊆R(D), A₁C₂ = C₁B₂, , ), then the system of operator equations A₁X = C₁, XB₂ = C₂, A₃XB₃ = C₃ has a solution from Theorem 10. Let C ∈ L_𝒜(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 Remark 11. Set A₄ = A₃N_D, B₄ = N_DB₃, and .

(1)
If the system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a positive solution, then F ≥ 0, , and there exists a positive number λ such that .
(2)
Suppose is orthogonally complemented submodule in H₁. If F ≥ 0, , and , 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 equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a positive solution X₀ ∈ L_𝒜(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

()

where Y₀ is the positive reduced solution to the system of equations A₁X = C₁, XB₂ = C₂. Taking X₀ = Y₀ + N_DYN_D into A₃XB₃ = C₃ yields that A₄YB₄ = C₃ − A₃Y₀B₃. And it has a positive solution. From Lemma 9, we know that

, and there exists a positive number λ such that

(2) If F ≥ 0, we can get that the system of equations A₁X = C₁, XB₂ = C₂ has a positive solution X₀ ∈ L_𝒜(H₁) ₊, and it has the form

()

where Y₀ is the positive reduced solution to the system of equations A₁X = C₁, XB₂ = C₂. Take X₀ = Y₀ + N_DYN_D into A₃XB₃ = C₃; then we can obtain that A₄YB₄ = C₃ − A₃Y₀B₃.

If is orthogonally complemented submodule in H₁, and 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 equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃. First we propose a lemma as follows.

Lemma 13. Suppose that M ∈ L_𝒜(H₁, H₂), N ∈ L_𝒜(H₃, H₁), and and are orthogonally complemented submodules of H₁. Let , and suppose T has a closed range; then TX = P_N and both have positive solutions.

In this case, the general positive solutions are of the forms

()

where P and Q are the reduced positive solutions, respectively, U and V ∈ L_𝒜(H₁) ₊ are arbitrary. Furthermore, on has

and P_N(Q + P) = P_N.

Proof. It is clear that , P_N ≥ 0, so T ≥ 0, and . Consider R(P_N)⊆R(T), , , , 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

()

where U, V ∈ L_𝒜(H₁) ₊ are arbitrary.

Consider TP = P_N, , ; then PT = P_N, , and T(Q + P) = T, , , and . By Lemma 4, we know that , so . Similarly, we can obtain that (Q + P)P_N = P_N; then P_N(Q + P) = P_N.

For simplicity, put

()

Y₀ is the common reduced positive solution to A₁X = C₁, XB₂ = C₂. Consider M = A₃N_D, N = N_DB₃; L is the reduced solution to MYN = C₃ − A₃Y₀B₃. Consider

. P is the positive reduced solution to TX = P_N. Q is the positive reduced solution to

. Y₁ is the positive solution to XP = LQ. Y₂ is the positive solution to QX = PL. Take

()

By Lemma 6, we know that Y₀ uniquely exists when is orthogonally complemented, F ≥ 0, and R(E) ⊂ R(D). From Lemma 3, we know that L uniquely exists when and , are orthogonally complemented, R(C₃ − A₃Y₀B₃) ⊂ R(M) and . From Lemma 13, we know that P and Q uniquely exist when and 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. Let H_i (i = 1,2, 3,4, 5) be Hilbert 𝒜-modules, A₁ ∈ L_𝒜(H₁, H₂), A₃ ∈ L_𝒜(H₁, H₄), B₂ ∈ L_𝒜(H₃, H₁), B₃ ∈ L_𝒜(H₅, H₁), C₁ ∈ L_𝒜(H₁, H₂), C₂ ∈ L_𝒜(H₃, H₁), and C₃ ∈ L_𝒜(H₅, H₄). Suppose , , and are orthogonally complemented submodules of H₁, , (or R(C₃)⊆R(A₃), ), and T has closed range. The system of operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ has a positive solution X ∈ L_𝒜(H₁) if and only if

()

in which case the general common positive solution to A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ can be expressed as

()

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 L_𝒜(H₁).

Proof. Suppose X₀ 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

()

where Y₀ is the positive reduced solution of A₁X = C₁, XB₂ = C₂, and Y ∈ L_𝒜(H₁) ₊ is an arbitrary operator.

Take X₀ = Y₀ + N_DYN_D into A₃XB₃ = C₃; we can get that

()

has a positive solution. Consider

()

By Lemma 13, we know that P, Q ≥ 0. Hence, if Y is positive, so is S.

For all x ∈ 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).

If F ≥ 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

()

where Y₀ is the positive reduced solution of the system of the adjointable operator equations A₁X = C₁, XB₂ = C₂, and Y ∈ L_𝒜(H₁) ₊ is an arbitrary operator.

Taking X₀ = Y₀ + N_DYN_D into A₃XB₃ = C₃, we can get

()

Now, we want to show that the equation MYN = C₃ − A₃Y₀B₃ has a positive solution. By Lemma 3, we know that the equation MYN = C₃ − A₃Y₀B₃ has a solution; then L exists.

We rewrite Y₁P = LQ, QY₂ = PL as Y₁P = LQ = L₁, QY₂ = PL = L₂; then L₂Q^* = PLQ = S ≥ 0, , and . 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

()

where D₁ and D₂ are the positive reduced solutions to the equations Y₁P = L₁, QY₂ = L₂, respectively, and V, W ∈ L_𝒜(H₁) ₊ are arbitrary.

For operator equations SX = L₂, XS = L₁, we can obtain that L₂S^* = PLQL^*P ≥ 0, and R(L₂)⊆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

()

Let V = I and . By Lemma 5, R ≥ 0.

For the operator equation TXT^* = R, by Lemma 7, TXT^* = R has a positive solution U and U has the form

()

where C is the positive reduced solution to the operator equation TXT^* = R and V₁ ∈ L_𝒜(H₁) and V₂ ∈ L_𝒜(H₁) ₊ are arbitrary.

Then we claim that is the positive solution to A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃. In fact, we only need to prove that MUN = C₃ − A₃Y₀B₃. Consider

()

Suppose that is a positive solution to the system of the operator equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃. It follows from Lemma 6 that can be expressed as

()

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

()

Let , Y₂ = P_NUP_N; it follows from

()

that Y₁ and Y₂ are positive solutions to the operator equations Y₁P = LQ, QY₂ = PL, respectively. Consider

()

By Lemma 7, U can be expressed as

()

where C is the positive reduced solution to TXT^* = R and V₁ ∈ L_𝒜(H₁) and V₂ ∈ L_𝒜(H₁) ₊ are arbitrary.

Take into ; we know that has the form that is expressed as (41).

Let A ∈ L_𝒜(H₁, H₂), B ∈ L_𝒜(H₃, H₁), and C ∈ L_𝒜(H₃, H₁), and suppose that , are orthogonally complemented submodules of H₁. If and R(C^*)⊆R(B^*) (or R(C)⊆R(A), ) 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 and suppose it has a closed range, where P is the positive reduced solution to TX = P_B. Q is the positive reduced solution to . Y₁ is the positive solution to Y₁P = LQ. Y₂ is the positive solution to QY₂ = PL.

Consider

()

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 equation AXB = C has a positive solution in L_𝒜(H₁) ₊ if and only if

()

in this case the form of general positive solution to AXB = C is

()

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 ∈ L_𝒜(H₁) and V₂ ∈ L_𝒜(H₁) ₊ are arbitrary.

The results obtained in Lemma 9 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].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This paper is supported by the National Natural Science Foundation of China (11071188) and “Chen Guang” Project (supported by Shanghai Municipal Education Commission and Shanghai Education Development Foundation) (10CGB25).

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Solutions to the System of Operator Equations A₁X = C₁, XB₂ = C₂, and A₃XB₃ = C₃ on Hilbert C^*-Modules