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

Abstract

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 𝑇𝑥,𝑦=𝑥,𝑇^∗𝑦,foreach𝑥∈𝐻₁,𝑦∈𝐻₂. (1) 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 𝑅(𝐴)=𝑅(𝐴^1/2). 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 𝐻=𝐻₁⊕𝐻₂, called the direct sum of 𝐻₁ and 𝐻₂. Moreover, if 𝐻₂=𝐻^⟂₁ where 𝐻^⟂₁={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 𝐿_𝒜(𝐻)) onto 𝐻₁. For two submodules 𝐻₁ and 𝐻₂ of 𝐻, if 𝐻₁⊆𝐻₂, then 𝐻^⟂₁⊇𝐻^⟂₂.

Let 𝑇∈𝐿_𝒜(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 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 𝑇⁺𝑇𝑇⁺=𝑇⁺,𝑇𝑇⁺=𝑇𝑇^+∗,𝑇⁺𝑇=𝑇⁺𝑇^∗. (2) 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∈𝐿_𝒜(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 𝑃_𝑇^∗ denote the orthogonal projection of 𝐻₁ onto 𝑅(𝑇^∗) and 𝑁_𝑇 the projection 𝐼−𝑃_𝑇^∗; then 𝑃_𝑇^∗+𝑁_𝑇=𝐼_𝐻1.

Lemma 1 (see [6, Theorem 1.1]). Let 𝑇^′∈𝐿_𝒜(𝐻₁,𝐻₂) and 𝑇∈𝐿_𝒜(𝐻₃,𝐻₂) with 𝑅(𝑇^∗) 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 𝑇^′=𝑇𝐷,𝑅(𝐷)⊆𝑁(𝑇)^⟂. (3) In this case, ‖𝐷‖²=inf𝜆∶𝑇^′𝑇^′∗≤𝜆𝑇𝑇^∗,𝑅(𝐷)⊆𝑅𝑇^∗;𝑁(𝐷)⊆𝑁𝑇^′, (4) and 𝐷 is called the reduced solution of the equation 𝑇𝑋=𝑇^′.
The general solution to 𝑇𝑋=𝑇^′ is of the form 𝑋=𝐷+𝑁_𝑇𝐾, (5) 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 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∈𝐿_𝒜(𝐻₃,𝐻₁) if and only if 𝑅(𝐶)⊆𝑅(𝐴),𝑅𝐷^∗⊆𝑅𝐵^∗,𝐴𝐷=𝐶𝐵. (6) In this case, the general solution is of the form: 𝑋=𝐷₁+𝑁_𝐴𝐷^∗₂+𝑁_𝐴𝑉𝑁_𝐵^∗, (7) 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 𝑅(𝐶)⊆𝑅(𝐴),𝑅𝐶^∗⊆𝑅𝐵^∗. (8)(2) Suppose 𝑅(𝐵) and 𝑅(𝐴^∗) are orthogonally complemented submodules of 𝐻₄ and 𝐻₁, respectively. If 𝑅(𝐶)⊆𝑅(𝐴),𝑅(𝐶^∗)⊆𝑅𝐵^∗ (9) or 𝑅(𝐶)⊆𝑅(𝐴),𝑅𝐶^∗⊆𝑅𝐵^∗, (10)then 𝐴𝑋𝐵=𝐶 has a unique solution 𝐷∈𝐿_𝒜(𝐻₄,𝐻₁) such that 𝑅(𝐷)⊆𝑁(𝐴)^⟂,𝑅𝐷^∗⊆𝑁𝐵^∗⟂, (11) which is called the reduced solution, and the general solution to 𝐴𝑋𝐵=𝐶 is of the form 𝑋=𝐷+𝑁_𝐴𝑉₁+𝑉₂𝑁_𝐵^∗, (12) where 𝑉₁,𝑉₂∈𝐿_𝒜(𝐻₄,𝐻₁).
Then, from 𝑅(𝐷)⊆𝑁(𝐴)^⟂ and 𝑅(𝐷^∗)⊆𝑁(𝐵^∗)^⟂, one knows that 𝐷=𝑃_𝐴^∗𝐷𝑃_𝐵. As in [11], one can obtain that 𝐴𝑋𝐵=0⇔𝑃_𝐴^∗𝑋𝑃_𝐵=0.

Lemma 4 (see [6, Theorem 1.3]). Let 𝐴,𝐶∈𝐿_𝒜(𝐻₁,𝐻₂) such that 𝑅(𝐴^∗) 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 𝑋=𝐷+𝑁_𝐴𝐾𝑁_𝐴, (13) 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 _𝑉^𝑈𝑉∗_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]). Let 𝐴₁,𝐶₁∈𝐿_𝒜(𝐻₁,𝐻₂) and 𝐵₂,𝐶₂∈𝐿_𝒜(𝐻₃,𝐻₁), 𝐴𝐷=¹𝐵^∗₂𝐶,𝐸=¹𝐶^∗₂,𝐹=𝐸𝐷^∗=𝐶₁𝐴^∗₁𝐶₁𝐵₂𝐶^∗₂𝐴^∗₁𝐶^∗₂𝐵₂ (14) such that 𝑅(𝐷^∗) 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 3.3(iii)]). Let 𝐴∈𝐿_𝒜(𝐻₁,𝐻₂) and 𝐶∈𝐿_𝒜(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 𝑋=𝐷+𝐷𝑉₁𝑁_𝐴+𝑁_𝐴𝑉^∗₁𝐷+𝑁_𝐴𝑉^∗₁𝐷𝑉₁𝑁_𝐴+𝑁_𝐴𝑉₂𝑁_𝐴 (15) 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 𝑅(𝐴^∗) and 𝑅(𝐵) 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 𝑅(𝑇) and 𝑅(𝑇^∗) are orthogonally complemented submodules of 𝐻₁ and 𝐻₃, respectively. Set 𝑆𝒴(𝐴,𝐵,𝐶)=𝑍+𝑍𝑉₁𝑃_𝑇^∗+𝑃_𝑇^∗𝑉^∗₁𝑍+𝑃_𝑇^∗𝑉^∗₁𝑍𝑉₁𝑃_𝑇^∗+𝑃_𝑇^∗𝑉₂𝑃_𝑇^∗∶𝑉₁∈𝐿_𝒜𝐻₃suchthat𝑁_𝑇𝐵^∗𝐷𝐵𝑃_𝑇^∗=𝑁_𝑇𝐵^∗𝐷𝐵𝑁_𝑇𝑉₁𝑃_𝑇^∗,𝑉₂∈𝐿_𝒜𝐻₃₊,𝑆Σ(𝐴,𝐵,𝐶)₁=𝑋₁∈𝐿_𝒜𝐻₁∶𝑋₁𝐵=12𝑃_𝑇𝑇^∗−1𝑃_𝑇^∗×𝑌−𝐵^∗Re(𝐷)𝐵𝑁_𝑇+𝐼forsome𝑌∈𝑆𝒴(𝐴,𝐵,𝐶),𝑋₁𝑁_𝐵^∗,()=0𝑆Σ𝐴,𝐵,𝐶₂=𝑋₂∈𝐿_𝒜𝐻₁∶𝑋₂𝐵=12𝑃_𝑇𝑇^∗−1𝑃_𝑇^∗𝐵^∗𝐷^∗−𝐷𝐵𝑃_𝑇^∗+𝑁_𝑇^∗𝑉₃𝑁_𝑇+𝑖𝑉₄𝑇forsome𝑉₄∈𝐿_𝒜(𝐸),𝑉₃∈𝐿_𝒜𝐻₃,𝐻₁with𝑅𝑉₃^∗𝑁_𝑇^∗⊆𝑅𝐵^∗,𝑆Σ(𝐴,𝐵,𝐶)=𝑋₁+𝑋₂∶𝑋₁∈𝑆Σ(𝐴,𝐵,𝐶)₁,𝑋₂∈𝑆Σ(𝐴,𝐵,𝐶)₂,𝑆𝑆(𝐴,𝐵,𝐶)=𝐷−𝐷^∗𝑁_𝐵^∗+𝑁_𝐴𝑊𝑃_𝐵−𝑃_𝐵𝑊^∗𝑁_𝐴𝑁_𝐵^∗+𝑉𝑁_𝐵^∗∶𝑉∈𝐿_𝒜𝐻₁,𝑊∈𝑆Σ(𝐴,𝐵,𝐶)}. (16)

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 𝑅(𝐴^∗) and 𝑅(𝐵) 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 𝑋=𝐷−𝐷^∗𝑁_𝐵^∗+𝑁_𝐴𝑊𝑃_𝐵−𝑃_𝐵𝑊^∗𝑁_𝐴𝑁_𝐵^∗+𝑉𝑁_𝐵^∗ 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 1𝑉=−2𝑃_𝐵𝐷+𝑁_𝐴𝑊𝑃_𝐵−𝐷^∗+𝑃_𝐵𝑊^∗𝑁_𝐴𝐼+3𝑁_𝐵^∗−𝑖𝑃_𝐵𝑉₁^∗𝑃_𝐵−𝑁_𝐵^∗𝑉₂, (17) in which case, 𝑋=𝐷+𝐷^∗𝑁_𝐵^∗+𝑁_𝐴𝑊𝑃_𝐵+𝑃_𝐵𝑊^∗𝑁_𝐴𝑁_𝐵^∗+𝑁_𝐵^∗(−𝑉₂)𝑁_𝐵^∗.
As a consequence, 𝑆𝑆(𝐴,𝐵,𝐶)_𝑠𝑎=𝐷+𝐷^∗𝑁_𝐵^∗+𝑁_𝐴𝑊𝑃_𝐵+𝑃_𝐵𝑊^∗𝑁_𝐴𝑁_𝐵^∗+𝑁_𝐵^∗𝑉𝑁_𝐵^∗∶𝑉∈𝐿_𝒜𝐻_1𝑠𝑎,𝑊∈𝑆Σ(𝐴,𝐵,𝐶)}. (18)

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 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∈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 𝐵^∗𝐷^∗𝑁_𝐵^∗𝐷𝐵≤𝜆𝐵^∗𝐷𝐵. (19)(2)Suppose that 𝑅(𝑃_𝐵𝐷𝑃_𝐵) is orthogonally complemented in 𝐻₁. If 𝐵^∗𝐷𝐵≥0 and 𝐵^∗𝐷^∗𝑁_𝐵^∗𝐷𝐵≤𝜆𝐵^∗𝐷𝐵, (20)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 𝑅𝐶₁⊆𝑅𝐴₁,𝑅𝐶^∗₂⊆𝑅𝐵^∗₂,𝑅𝐶₃⊆𝑅𝐴₃,𝑅𝐶^∗₃⊆𝑅𝐵^∗₃,𝐴₁𝐶₂=𝐶₁𝐵₂. (21)(2) Set 𝐴₄=𝐴₃𝑁_𝐴1,𝐵₄=𝑁_𝐵^∗₂𝐵₃. Suppose 𝑅(𝐴^∗₁), 𝑅(𝐵₂), 𝑅(𝐴^∗₄), and 𝑅(𝐵₄) are orthogonally complemented submodules of 𝐻₁. If 𝑅𝐶₁⊆𝑅𝐴₁,𝑅𝐶^∗₂⊆𝑅𝐵^∗₂,𝑅𝐶₃⊆𝑅𝐴₃,𝑅(𝐶^∗₃)⊆𝑅𝐵^∗₃ (22) or 𝑅𝐶₁⊆𝑅𝐴₁,𝑅𝐶^∗₂⊆𝑅𝐵^∗₂,𝑅𝐶₃⊆𝑅𝐴₃,𝑅𝐶^∗₃⊆𝑅𝐵^∗₃, (23)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 𝑅(𝐷)⊆𝑁𝐴₁^⟂𝑁𝐵^∗₂^⟂𝑁𝐴₄^⟂,𝑅𝐷^∗⊆𝑁𝐴^∗₁^⟂𝑁𝐵₂^⟂𝑁𝐵^∗₄^⟂. (24) In this case, the general solution is of the form 𝑋=𝐷₁+𝑁_𝐴1𝐷^∗₂+𝑁_𝐴1𝐷₃+𝑁_𝐴4𝑉₁+𝑉₂𝑁_𝐵^∗₄𝑁_𝐵^∗₂, (25) where 𝐷₁, 𝐷₂, and 𝐷₃ are the reduced solutions of 𝐴₁𝑋=𝐶₁, 𝐵^∗₂𝑋=𝐶^∗₂, and 𝐴₄𝑋𝐵₄=𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃ respectively. 𝑉₁,𝑉₂∈𝐿_𝒜(𝐻₁) are arbitrary.

Proof. (1) If the system of operator equations 𝐴₁𝑋=𝐶₁,𝑋𝐵₂=𝐶₂,and𝐴₃𝑋𝐵₃=𝐶₃ has a solution 𝑋∈𝐿_𝒜(𝐻₁), it is easy to know that 𝑅𝐶₁⊆𝑅𝐴₁,𝑅𝐶^∗₂⊆𝑅𝐵^∗₂,𝑅𝐶₃⊆𝑅𝐴₃,𝑅𝐶^∗₃⊆𝑅𝐵^∗₃,𝐴₁𝐶₂=𝐴₁𝑋𝐵₂=𝐶₁𝐵₂. (26)
(2) Since 𝑅(𝐴^∗₁) and 𝐵₂ are orthogonally complemented, 𝑅(𝐶₁)⊆𝑅(𝐴₁),𝑅(𝐶^∗₂)⊆𝑅(𝐵^∗₂), and 𝐴₁𝐶₂=𝐶₁𝐵₂. By Lemma 2, we know that 𝐴₁𝑋=𝐶₁and𝑋𝐵₂=𝐶₂ have a common solution 𝑋₀ and it has the form: 𝑋₀=𝐷₁+𝑁_𝐴1𝐷^∗₂+𝑁_𝐴1𝑌𝑁_𝐵^∗₂, (27) where 𝐷₁ and 𝐷₂ are the reduced solutions of 𝐴₁𝑋=𝐶₁,𝐵^∗₂𝑋=𝐶^∗₂ respectively.
Take 𝑋₀=𝐷₁+𝑁_𝐴1𝐷^∗₂+𝑁_𝐴1𝑌𝑁_𝐵^∗₂ into 𝐴₃𝑋𝐵₃=𝐶₃, then we can get 𝐴₃𝑁_𝐴1𝑌𝑁_𝐵^∗₂𝐵₃=𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃.𝐴₄𝑌𝐵₄=𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃. (28)(29)
Since 𝑅(𝐶₃)⊆𝑅(𝐴₃) and 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃) (or 𝑅(𝐶₃)⊆𝑅(𝐴₃) and 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃)), then 𝑅𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃⊆𝑅𝐴₃,𝑅(𝐶^∗₃−𝐵^∗₃𝐷^∗₁𝐴^∗₃−𝐵^∗₃𝐷₂𝑁_𝐴1𝐴^∗₃)⊆𝑅𝐵^∗₃, (30) or 𝑅𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃⊆𝑅𝐴₃,𝑅𝐶^∗₃−𝐵^∗₃𝐷^∗₁𝐴^∗₃−𝐵^∗₃𝐷₂𝑁_𝐴1𝐴^∗₃⊆𝑅𝐵^∗₃. (31) By Lemma 3(2), we know that the operator equation 𝐴₄𝑌𝐵₄=𝐶₃−𝐴₃𝐷₁𝐵₃−𝐴₃𝑁_𝐴1𝐷^∗₂𝐵₃ has a unique reduced solution 𝐷₃∈𝐿_𝒜(𝐻₁) and the general solution has the form 𝑌=𝐷₃+𝑁_𝐴4𝑉₁+𝑉₂𝑁_𝐵^∗₄, (32) 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 𝑋=𝐷₁+𝑁_𝐴1𝐷^∗₂+𝑁_𝐴1𝐷₃+𝑁_𝐴4𝑉₁+𝑉₂𝑁_𝐵^∗₄𝑁_𝐵^∗₂. (33)
And it is easy to see that 𝐷=𝐷₁+𝑁_𝐴1𝐷^∗₂+𝑁_𝐴1𝐷₃𝑁_𝐵^∗₂ is the reduced solution to the system of operator equations 𝐴₁𝑋=𝐶₁, 𝑋𝐵₂=𝐶₂ and 𝐴₃𝑋𝐵₃=𝐶₃; 𝑅(𝐷)⊆𝑁(𝐴₁)^⟂⋂𝑁(𝐵^∗₂)^⟂⋂𝑁(𝐴₄)^⟂ and 𝑅(𝐷^∗)⊆𝑁(𝐴^∗₁)^⟂⋂𝑁(𝐵₂)^⟂⋂𝑁(𝐵^∗₄)^⟂.

If 𝑅(𝐴^∗₃) and 𝑅(𝐵₃) are orthogonally complemented submodules, then 𝑅(𝐴^∗₄) and 𝑅(𝐵₄) 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 𝑅(𝐷^∗), 𝑅(𝐴^∗₃), and 𝑅(𝐵^∗₃) are orthogonally complemented submodules of 𝐻₁. If 𝑅(𝐸)⊆𝑅(𝐷), 𝐴₁𝐶₂=𝐶₁𝐵₂, 𝑅(𝐶₃)⊆𝑅(𝐴₃), 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃) (or, 𝑅(𝐸)⊆𝑅(𝐷), 𝐴₁𝐶₂=𝐶₁𝐵₂, 𝑅(𝐶₃)⊆𝑅(𝐴₃), 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃)), 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 𝑅(𝐵₃)⊆𝑅(𝐴^∗₃).(1)If the system of operator equations 𝐴₁𝑋=𝐶₁, 𝑋𝐵₂=𝐶₂, and 𝐴₃𝑋𝐵₃=𝐶₃ has a positive solution, then 𝐹≥0, 𝐵^∗₄𝐶𝐵₄≥0, and there exists a positive number 𝜆 such that 𝐵^∗₄𝐶^∗𝑁_𝐵^∗₄𝐶𝐵₄≤𝜆𝐵^∗₄𝐶𝐵₄.(2)Suppose 𝑅(𝑃_𝐵4𝐶𝑃_𝐵4) is orthogonally complemented submodule in 𝐻₁. If 𝐹≥0, 𝐵^∗₄𝐶𝐵₄≥0, and 𝐵^∗₄𝐶^∗𝑁_𝐵^∗₄𝐶𝐵₄≤𝜆𝐵^∗₄𝐶𝐵₄, 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 𝑋₀=𝑌₀+𝑁_𝐷𝑌𝑁_𝐷,𝑌∈𝐿_𝒜𝐻₁₊, (34) 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 𝐵^∗₄𝐶𝐵^∗₄≥0, and there exists a positive number 𝜆 such that 𝐵^∗₄𝐶^∗𝑁_𝐵^∗₄𝐶𝐵₄≤𝜆𝐵^∗₄𝐶𝐵₄.
(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 𝑋₀=𝑌₀+𝑁_𝐷𝑌𝑁_𝐷,𝑌∈𝐿_𝒜𝐻₁₊, (35) 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 𝑅(𝑃_𝐵4𝐶𝑃_𝐵4) is orthogonally complemented submodule in 𝐻₁, 𝐵^∗₄𝐶𝐵^∗₄≥0 and 𝐵^∗₄𝐶^∗𝑁_𝐵^∗₄𝐶𝐵₄≤𝜆𝐵^∗₄𝐶𝐵₄ 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 𝑅(𝑀^∗) and 𝑅(𝑁) are orthogonally complemented submodules of 𝐻₁. Let 𝑇=𝑃_𝑀^∗+𝑃_𝑁, and suppose 𝑇 has a closed range; then 𝑇𝑋=𝑃_𝑁 and 𝑌𝑇=𝑃_𝑀^∗ both have positive solutions.
In this case, the general positive solutions are of the forms 𝑋=𝑃+𝑁_𝑇𝑈𝑁_𝑇,𝑌=𝑄+𝑁_𝑇𝑉𝑁_𝑇, (36) where 𝑃 and 𝑄 are the reduced positive solutions, respectively, 𝑈 and 𝑉∈𝐿_𝒜(𝐻₁)₊ are arbitrary. Furthermore, on has (𝑄+𝑃)𝑃_𝑀^∗=𝑃_𝑀^∗ and 𝑃_𝑁(𝑄+𝑃)=𝑃_𝑁.

Proof. It is clear that 𝑃_𝑀^∗≥0, 𝑃_𝑁≥0, so 𝑇≥0, 𝑃_𝑁𝑃_𝑀^∗≥0 and 𝑃_𝑁𝑃_𝑀^∗≥0. Consider 𝑅(𝑃_𝑁)⊆𝑅(𝑇), 𝑃_𝑁𝑇^∗=𝑃_𝑁𝑇=𝑃_𝑁(𝑃_𝑀^∗+𝑃_𝑁)=𝑃_𝑁𝑃_𝑀^∗+𝑃_𝑁≥0, 𝑅(𝑃_𝑀^∗)⊆𝑅(𝑇), 𝑃^∗_𝑀^∗𝑇=𝑃_𝑀^∗𝑇=𝑃_𝑀^∗(𝑃_𝑀^∗+𝑃_𝑁)=𝑃_𝑀^∗+𝑃_𝑀^∗𝑃_𝑁≥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 𝑋=𝑃+𝑁_𝑇𝑈𝑁_𝑇,𝑌=𝑄+𝑁_𝑇𝑉𝑁_𝑇, (37) where 𝑈,𝑉∈𝐿_𝒜(𝐻₁)₊ are arbitrary.
Consider 𝑇𝑃=𝑃_𝑁, 𝑄𝑇=𝑃_𝑀^∗, 𝑇,𝑃,𝑄,𝑃_𝑁,𝑃_𝑀^∗≥0; then 𝑃𝑇=𝑃_𝑁, 𝑇𝑄=𝑃_𝑀^∗, and 𝑇(𝑄+𝑃)=𝑇, 𝑇(𝑄+𝑃)𝑃_𝑀^∗=𝑇𝑃_𝑀^∗, 𝑇((𝑄+𝑃)𝑃_𝑀^∗−𝑃_𝑀^∗)=0, and (𝑄+𝑃)𝑃_𝑀^∗−𝑃_𝑀^∗⊆𝑁(𝑇). By Lemma 4, we know that (𝑄+𝑃)𝑃_𝑀^∗−𝑃_𝑀^∗⊆𝑁(𝑇)^⟂, so (𝑄+𝑃)𝑃_𝑀^∗=𝑃_𝑀^∗. Similarly, we can obtain that (𝑄+𝑃)𝑃_𝑁=𝑃_𝑁; then 𝑃_𝑁(𝑄+𝑃)=𝑃_𝑁.

For simplicity, put 𝐴𝐷=¹𝐵^∗₂𝐶,𝐸=¹𝐶^∗₂,𝐶𝐹=¹𝐴^∗₁𝐶₁𝐵₂𝐶^∗₂𝐴^∗₁𝐶^∗₂𝐵₂. (38)

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 𝑇=𝑃_𝑀^∗+𝑃_𝑁. 𝑃 is the positive reduced solution to 𝑇𝑋=𝑃_𝑁. 𝑄 is the positive reduced solution to 𝑋𝑇=𝑃_𝑀^∗. 𝑌₁ is the positive solution to 𝑋𝑃=𝐿𝑄. 𝑌₂ is the positive solution to 𝑄𝑋=𝑃𝐿. Take 𝑅=𝐿+𝐿^∗+𝑌₁+𝑌₂,𝑆=𝑃𝐿𝑄. (39)

By Lemma 6, 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. 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 𝑅(𝐷^∗), 𝑅(𝑀^∗), and 𝑅(𝑁) are orthogonally complemented submodules of 𝐻₁, 𝑅(𝐶₃)⊆𝑅(𝐴₃), 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃) (or 𝑅(𝐶₃)⊆𝑅(𝐴₃), 𝑅(𝐶^∗₃)⊆𝑅(𝐵^∗₃)), 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 (𝐸)(𝐷),𝑅()(𝑆)𝐹≥0,𝑆≥0,𝑅⊆𝑅𝑃𝐿⊆𝑅,𝑅𝑄𝐿^∗⊆𝑅(𝑆), (40) in which case the general common positive solution to 𝐴₁𝑋=𝐶₁, 𝑋𝐵₂=𝐶₂, and 𝐴₃𝑋𝐵₃=𝐶₃ can be expressed as 𝑋=𝑌₀+𝑁_𝐷𝐶+𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐶+𝑁_𝑇𝑉^∗₁𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇𝑁_𝐷, (41) 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 𝑋₀=𝑌₀+𝑁_𝐷𝑌𝑁_𝐷, (42) 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 𝑀𝑌𝑁=𝐶₃−𝐴₃𝑌₀𝐵₃ (43) has a positive solution. Consider 𝑆=𝑃𝐿𝑄=𝑃𝑃_𝑀^∗𝐿𝑃_𝑁𝑄=𝑃𝑃_𝑀^∗𝑌𝑃_𝑁𝑄=𝑃𝑌𝑄. (44) 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 𝑋₀=𝑌₀+𝑁_𝐷𝑌𝑁_𝐷, (45) 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 𝑀𝑌𝑁=𝐶₃−𝐴₃𝑌₀𝐵₃. (46)
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, 𝐿^∗₁𝑃=𝑄^∗𝐿^∗𝑃=𝑆^∗=𝑆≥0, and 𝑅(𝑆)=𝑅(𝐿^∗₁)=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 𝑌₁=𝐷₁+𝑁_𝑃𝑉𝑁_𝑃,𝑌₂=𝐷₂+𝑁_𝑄𝑊𝑁_𝑄, (47) 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, 𝐿^∗₁𝑆^∗=𝑄𝐿^∗𝑃𝐿𝑄^∗≥0 and 𝑅(𝐿₂)⊆𝑅(𝑆), 𝑅(𝐿^∗₁)⊆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 𝑆𝑍₁=𝐿₂,𝑃𝐿𝑄𝑍₁=𝑃𝐿,𝑃𝐿₁𝑍₁=𝐿,𝑃𝐿−𝐿₁𝑍₁𝑍=0,₂𝑆=𝐿₁,𝑍₂𝑍𝑃𝐿𝑄=𝐿𝑄,₂𝐿₂𝑄=𝐿𝑄,𝐿−𝑍₂𝐿₂𝑄=0,𝑅=𝐿+𝐿^∗+𝑌₁+𝑌₂=𝐿+𝐿^∗+𝐷₁+𝑁_𝑃𝑉𝑁_𝑃+𝐷₂+𝑁_𝑄𝑊𝑁_𝑄=𝐿₁𝑍₁+𝑍^∗₁𝐿^∗₁+𝐷₁+𝐷₂+𝑁_𝑃𝑉𝑁_𝑃+𝑁_𝑄𝑊𝑁_𝑄+𝐿−𝐿₁𝑍₁+𝐿^∗−𝑍^∗₁𝐿^∗₁=𝑍₂𝑆𝑍₁+𝑍^∗₁𝑆𝑍^∗₂+𝐷₁+𝐷₂+𝑁_𝑃𝑁_𝑄𝑉𝐿−𝐿₁𝑍₁𝐿^∗−𝑍^∗₁𝐿^∗₁𝑊𝑁_𝑃𝑁_𝑄=𝑍₂+𝑍^∗₁𝑆𝑍₁+𝑍^∗₂+𝐷₁+𝐷₂+𝑁_𝑃𝑁_𝑄𝑉𝐿−𝐿₁𝑍₁𝐿^∗−𝑍^∗₁𝐿^∗₁𝑊𝑁_𝑃𝑁_𝑄. (48)
Let 𝑉=𝐼 and 𝑊=(𝐿^∗−𝑍^∗₁𝐿^∗₁)(𝐿−𝐿₁𝑍₁). 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 𝑈=𝐶+𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐶+𝑁_𝑇𝑉^∗₁𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇, (49) 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 𝑋=𝑌₀+𝑁_𝐷(𝐶+𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐶+𝑁_𝑇𝑉^∗₁𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇)𝑁_𝐷 is the positive solution to 𝐴₁𝑋=𝐶₁, 𝑋𝐵₂=𝐶₂, and 𝐴₃𝑋𝐵₃=𝐶₃. In fact, we only need to prove that 𝑀𝑈𝑁=𝐶₃−𝐴₃𝑌₀𝐵₃. Consider 𝑀𝑈𝑁=𝑀𝑄𝑇𝑈𝑇^∗𝑃𝑁=𝑀𝑄𝑅𝑃𝑁=𝑀𝑄𝐿+𝐿^∗+𝑌₁+𝑌₂𝑃𝑁=𝑀𝑄𝐿𝑃𝑁+𝑀𝑄𝐿^∗𝑃𝑁+𝑀𝑄𝑌₁𝑃𝑁+𝑀𝑄𝑌₂()𝐿()𝑁()𝑃𝑃𝑁=𝑀𝑄𝐿𝑃𝑁+𝑀𝑃𝐿𝑄𝑁+𝑀𝑄𝐿𝑄𝑁+𝑀𝑃𝐿𝑃𝑁=𝑀𝑄+𝑃𝑄+𝑃=𝑀𝑄+𝑃_𝑀^∗𝐿𝑃_𝑁(𝑄+𝑃)𝑁=𝑀𝑃_𝑀^∗𝐿𝑃_𝑁𝑁=𝑀𝐿𝑁=𝐶₃−𝐴₃𝑌₀𝐵₃. (50)
Suppose that 𝑋 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 𝑋=𝑌₀+𝑁_𝐷𝑌𝑁_𝐷,𝑌∈𝐿_𝒜𝐻₁₊, (51) 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 𝑋−𝑌₀=𝑁_𝐷𝑈𝑁_𝐷. (52)
Let 𝑌₁=𝑃_𝑀^∗𝑈𝑃_𝑀^∗, 𝑌₂=𝑃_𝑁𝑈𝑃_𝑁; it follows from 𝑌₁𝑃=𝑃_𝑀^∗𝑈𝑃_𝑀^∗𝑃=𝑃_𝑀^∗𝑈𝑃_𝑁−𝑃_𝑁𝑃=𝑃_𝑀^∗𝑈𝑃_𝑁−𝑃_𝑀^∗𝑈𝑃_𝑁𝑃=𝑃_𝑀^∗𝐿𝑃_𝑁−𝑃_𝑀^∗𝐿𝑃_𝑁𝑃=𝑃_𝑀^∗𝐿𝑃_𝑁−𝑃_𝑁𝑃=𝑃_𝑀^∗𝐿𝑃_𝑁𝑄=𝐿𝑄,𝑄𝑌₂=𝑄𝑃_𝑁𝑈𝑃_𝑁=𝑃_𝑀^∗−𝑄𝑃_𝑀^∗𝑈𝑃_𝑁=𝑃_𝑀^∗𝑈𝑃_𝑁−𝑄𝑃_𝑀^∗𝑈𝑃_𝑁=𝑃_𝑀^∗𝐿𝑃_𝑁−𝑄𝑃_𝑀^∗𝐿𝑃_𝑁=𝑃_𝑀^∗−𝑄𝑃_𝑀^∗𝐿𝑃_𝑁=𝑃𝑃_𝑀^∗𝐿𝑃_𝑁=𝑃𝐿 (53) that 𝑌₁ and 𝑌₂ are positive solutions to the operator equations 𝑌₁𝑃=𝐿𝑄, 𝑄𝑌₂=𝑃𝐿, respectively. Consider 𝑅=𝐿+𝐿^∗+𝑌₁+𝑌₂=𝑃_𝑀^∗𝐿𝑃_𝑁+𝑃_𝑁𝐿𝑃_𝑀^∗+𝑃_𝑀^∗𝑈𝑃_𝑀^∗+𝑃_𝑁𝑈𝑃_𝑁=𝑃_𝑀^∗𝑈𝑃_𝑁+𝑃_𝑁𝑈𝑃_𝑀^∗+𝑃_𝑀^∗𝑈𝑃_𝑀^∗+𝑃_𝑁𝑈𝑃_𝑁=𝑃_𝑀^∗+𝑃_𝑁𝑈𝑃_𝑀^∗+𝑃_𝑁=𝑇𝑈𝑇=𝑇𝑈𝑇^∗. (54) By Lemma 7, 𝑈 can be expressed as 𝑈=𝐶+𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐶+𝑁_𝑇𝑉^∗₁𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇, (55) where 𝐶 is the positive reduced solution to 𝑇𝑋𝑇^∗=𝑅 and 𝑉₁∈𝐿_𝒜(𝐻₁) and 𝑉₂∈𝐿_𝒜(𝐻₁)₊ are arbitrary.
Take 𝑈=𝐶+𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐶+𝑁_𝑇𝑉^∗₁𝐶𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇 into 𝑋−𝑌₀=𝑁_𝐷𝑈𝑁_𝐷; we know that 𝑋 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 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=𝑃_𝐴^∗+𝑃_𝐵 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 𝑋𝑇=𝑃_𝐴^∗. 𝑌₁ is the positive solution to 𝑌₁𝑃=𝐿𝑄. 𝑌₂ is the positive solution to 𝑄𝑌₂=𝑃𝐿.

Consider 𝑅=𝐿+𝐿^∗+𝑌₁+𝑌₂,𝑆=𝑃𝐿𝑄. (56) 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 𝑆≥0,𝑅(𝑃𝐿)⊆𝑅(𝑆),𝑅𝑄𝐿^∗⊆𝑅(𝑆), (57) in this case the form of general positive solution to 𝐴𝑋𝐵=𝐶 is 𝑋=𝐷+𝐷𝑉₁𝑁_𝑇+𝑁_𝑇𝑉^∗₁𝐷+𝑁_𝑇𝑉^∗₁𝐷𝑉₁𝑁_𝑇+𝑁_𝑇𝑉₂𝑁_𝑇, (58) 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).