Positive Solutions of Operator Equations AX = B, XC = D

Abstract

In this paper, using the technique of operator matrix, we consider the positive solution of the system of operator equations $AX=B,XC=D$ in the framework of the Hilbert space; here, the ranges $R\left(A\right)$ of A and $R\left(C\right)$ of C are not necessarily closed. Firstly, we provide a new necessary and sufficient condition for the existence of positive solutions of $AX=B$ and also provide a representation of positive solutions, which generalize previous conclusions. Furthermore, using the above result, a condition of equivalence for the existence of common positive solutions of $AX=B,XC=D$ is given, as well as the general forms of positive solutions.

1. Introduction

The problem of solving operator equations has important applications in physics, control theory, and many other fields. It has been paid attention to by many scholars. For example, equations

$$AX=B$$

(1)

and

$$AX=B,XC=D$$

(2)

have been widely studied in matrix algebra [1], the operator space on the Hilbert space [2,3,4,5,6], and the adjointable operator space on Hilbert $C^{*}$-modules [7,8,9,10]. There is a classical result addressing the existence of solutions for Equation (1), which is the famous Douglas range inclusion theorem [4]. Subsequent to this, many scholars have studied the existence and formulae of solutions for Equation (1) and common solutions for Equation (2). On occasion, specific requirements were imposed on these solutions, such as them being self-adjoint, positive, real positive, and so on.

Daji$\acute{\mathrm{c}}$ and Koliha [2,3] investigated positive solutions for Equation (1) and common positive solutions for Equation (2) in operator spaces where operators A and C have closed ranges. Qingxiang Xu [10] generalized the results to adjointable operators on Hilbert $C^{*}$-modules. The main tool used for closed-range operators is the Moore–Penrose generalized inverses. Unfortunately, this method does not work if the ranges of A and C are not closed. Liang and Deng [5] provided equivalent conditions for the existence of positive solutions for Equations (1) and (2), with no restrictions on the range. But determining the formulas for the common positive solution remains an open question. Recently, the authors of [9] considered the common positive solutions of system (2) for adjointable operators under certain conditions.

For the sake of the description, let us introduce some symbols. Throughout this note, $\mathcal{H}$ and $\mathcal{K}$ are complex separable Hilbert spaces. We denote the set of all bounded linear operators from $\mathcal{H}$ into $\mathcal{K}$ by $\mathcal{B}(\mathcal{H},\mathcal{K})$ and abbreviate $\mathcal{B}(\mathcal{H},\mathcal{K})$ to $\mathcal{B}\left(\mathcal{H}\right)$ if $\mathcal{K}=\mathcal{H}$. For an operator $A\in \mathcal{B}(\mathcal{H},\mathcal{K})$, we denote the adjoint operator, and the absolute values of A by $A^{*}$ and $\left|A\right|$, respectively. Let $R\left(A\right)$, $\overline{R\left(A\right)}$, and $N\left(A\right)$ be the range, the closure of the range, and the null space of A, respectively. The Moore–Penrose inverse of $A\in \mathcal{B}(\mathcal{H},\mathcal{K})$ is defined as the operator $A^{†}\in \mathcal{B}(\mathcal{K},\mathcal{H})$, satisfying the equations,

$$A{A}^{†}A=A,A^{†}A{A}^{†}=A^{†},{\left(A^{†}A\right)}^{*}=A^{†}A,{\left(A{A}^{†}\right)}^{*}=A{A}^{†}.$$

An operator A has the (unique) Moore–Penrose inverse if and only if A is closed-range [11]. An operator $A\in \mathcal{B}\left(\mathcal{H}\right)$ is termed self-adjoint if $A^{*}=A$, and positive if $\langle Ax,x\rangle \ge 0$ for all $x\in \mathcal{H}$. We write $A\ge 0$ if A is positive. By ${\mathcal{B}}_{+}\left(\mathcal{H}\right)$, we denote the set of all positive elements in $\mathcal{B}\left(\mathcal{H}\right)$. We assume that M is a closed subspace of $\mathcal{H}$, and the orthogonal projection onto M is denoted by $P_M$. $M^{\perp }$ refers to the orthogonal complementary subspace of M. For an operator $A\in \mathcal{B}\left(\mathcal{H}\right)$, we define P as the projection onto $\overline{R\left(A^{*}\right)}$ in the sequel.

In this paper, we discuss positive solutions for Equations (1) and (2) in the framework of the Hilbert space; the ranges of A and B may not be closed. Our main goal is to find the common positive solutions of system Equation (2). Firstly, we need to discuss the positive solutions for Equation (1).

In Section 2, using the operator matrix technique, we provide a new equivalent condition for the existence of positive solutions for Equation (1) by the range of reduced solutions without the condition for the range to be closed. Simultaneously, a specific representation of positive solutions is given. This representation is not as concise as in [5], which provides a method to study the common positive solutions of system (2). Moreover, we show that our result is equivalent to the related result in [2] when A and $B{A}^{*}$ have closed ranges.

In Section 3, we study common positive solutions for Equation (2). We use the result in Section 2; a necessary and sufficient condition for the existence of common positive solutions is given, and a formula for common positive solutions is obtained.

In Section 4, an example is provided to demonstrate that our results are valid. As shown by Example 1, Equation (1) has positive solutions, where $R\left(A\right)$ and $R\left(B{A}^{*}\right)$ are not closed. It also indicates that system (2) has common positive solutions, which are calculated by Theorem 2.

2. The Positive Solutions of $\mathit{AX}=\mathit{B}$

In this section, the positive solutions for Equation (1) are discussed. An equivalent condition for the existence was given by the operator inequality in [5]; simultaneously, a formula of positive solutions was provided by the reduced solution of ${\left(B{A}^{*}\right)}^{\frac{1}{2}}X=B$. But this expression does not work for studying the common positive solution of system (2), to our knowledge. Here, we offer a new equivalent condition and a new formula of positive solutions, which can help us solve the relative question about the system of operator equations.

Lemma 1

([4]). Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$. The following three conditions are equivalent:

 (i)

There exists $D\in \mathcal{B}\left(\mathcal{H}\right)$, such that $AD=B$;

 (ii)

$B{B}^{*}\le \lambda A{A}^{*}$ for some $\lambda >0$;

 (iii)

$R\left(B\right)\subseteq R\left(A\right)$.

If one of these conditions holds, then there exists a unique solution $G\in \mathcal{B}\left(\mathcal{H}\right)$ of equation $AX=B$, such that $R\left(G\right)\subseteq \overline{R\left(A^{*}\right)}$ and $N\left(G\right)=N\left(B\right)$. This solution is called the Douglas-reduced solution. If the operator A has a closed range and $R\left(B\right)\subseteq R\left(A\right)$, then the Douglas-reduced solution is $G=A^{†}B$.

Definition 1

([12]). An operator sequence $\left\{A_n\right\}$ converges to $A\in \mathcal{B}\left(\mathcal{H}\right)$ in the strong operator topology if $\left|\right|A_{n}h-Ah\left|\right|\to 0$ as $n\to \infty$ for all h in $\mathcal{H}$. We denote by

$$A=s.o.-li{m}_{n\to \infty }{A}_n.$$

Lemma 2

([6]). Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ with $R\left(B\right)\subseteq R\left(A\right)$ and $A=U_A\left|A\right|$ be the polar decomposition. Then the formula for the reduced solution G of Equation (1) is

$$G=s.o.-li{m}_{n\to \infty }\left(\right|A|+\frac{1}{n}{I)}^{-1}{U}_A^{*}B.$$

(3)

In this case, the general solution is of the form

$$X=G+(I-P)Y,$$

for all $Y\in \mathcal{B}\left(\mathcal{H}\right)$.

Lemma 3

([6]). Let $A=\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{12}^{*}& A_{22}\end{array}\right)\in \mathcal{B}(\mathcal{H}\oplus \mathcal{H})$. Then $A\ge 0$ if and only if

 (i)

$A_{11}\ge 0$;

 (ii)

$A_{11}^{\frac{1}{2}}X=A_{12}$ has a solution;

 (iii)

$A_{22}-E^{*}E\ge 0,$ where E is the reduced solution of $A_{11}^{\frac{1}{2}}X=A_{12}$.

Lemma 4

([8]). Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $AX=B$ is solvable and G is the reduced solution. Then we have

 (i)

If $R\left(B{A}^{*}\right)$ is closed, this implies that $R\left(GP\right)$ is closed;

 (ii)

$B{A}^{*}\ge 0$ if and only if $GP\ge 0$.

Theorem 1.

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $AX=B$ is solvable. Then $AX=B$ has positive solutions if and only if $B{A}^{*}\ge 0$ and $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$, where G is the reduced solution of $AX=B$. In this case, the formula of positive solutions is

$$X=G+(I-P)G^{*}+(I-P)H^{*}H(I-P)+(I-P)Y(I-P),$$

(4)

for all Y in ${\mathcal{B}}_{+}\left(\mathcal{H}\right)$, where H is the reduced solution of ${\left(GP\right)}^{\frac{1}{2}}X=G(I-P)$.

Proof. 

Suppose that X is a solution of $AX=B$ and G is the reduced solution. It follows from Lemma 2 that there is an operator $\overline{Y}\in \mathcal{B}\left(\mathcal{H}\right)$ satisfying

$$X=G+(I-P)\overline{Y}.$$

With respect to the space decomposition $\mathcal{H}=\overline{R\left(A^{*}\right)}\oplus {\overline{R\left(A^{*}\right)}}^{\perp }$, G and $\overline{Y}$ have the matrix forms as follows,

$$G=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ 0& 0\end{array}\right)$$

(5)

and

$$\overline{Y}=\left(\begin{array}{cc}{\overline{Y}}_{11}& {\overline{Y}}_{12}\\ {\overline{Y}}_{21}& {\overline{Y}}_{22}\end{array}\right),$$

(6)

respectively. Then X is of the form

$$X=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ {\overline{Y}}_{21}& {\overline{Y}}_{22}\end{array}\right).$$

(7)

‘⇒’ Assume that X is a positive operator. According to Lemma 3, $G_{11}\ge 0$ and $R\left(G_{12}\right)\subseteq R\left({\left(G_{11}\right)}^{\frac{1}{2}}\right)$. This shows $GP\ge 0$ and $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$. It is obvious that $B{A}^{*}\ge 0$, according to Lemma 4.

‘⇐’ Suppose that $B{A}^{*}\ge 0$. From Lemma 4 again, $GP\ge 0$, and so $G_{11}\ge 0$, according to Lemma 3 and form (5). Moreover, $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$ implies that $R\left(G(I-P)\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$. We infer that $R\left(G_{12}\right)\subseteq R\left({\left(G_{11}\right)}^{\frac{1}{2}}\right)$. This follows $G_{12}=G_{11}^{\frac{1}{2}}X$ is solvable, denote the reduced solution by $H_{12}$. Write

$$H=\left(\begin{array}{cc}0& H_{12}\\ 0& 0\end{array}\right)$$

(8)

and $X_0=G+(I-P)G^{*}+(I-P)H^{*}H(I-P)$. Then H is the reduced solution of ${\left(GP\right)}^{\frac{1}{2}}X=G(I-P)$ since $R\left(G\right)\subseteq \overline{R\left({\left(GP\right)}^{\frac{1}{2}}\right)}$. Combining matrix forms (5) with (8), it is clear that

$$X_0=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ G_{12}^{*}& H_{12}^{*}{H}_{12}\end{array}\right)=\left(\begin{array}{cc}{G}_{11}& {G_{11}}^{\frac{1}{2}}{H}_{12}\\ H_{12}^{*}{G_{11}}^{\frac{1}{2}}& H_{12}^{*}{H}_{12}\end{array}\right)={\left(\begin{array}{cc}{G}_{11}^{\frac{1}{2}}& H_{12}\\ 0& 0\end{array}\right)}^{*}\left(\begin{array}{cc}{G}_{11}^{\frac{1}{2}}& H_{12}\\ 0& 0\end{array}\right)\ge 0.$$

Therefore, $AX=B$ has a positive solution $X_0$.

Next, we consider the general representation of the positive solution X. Recall X has matrix form (7), ${\overline{Y}}_{21}=G_{12}^{*}$ and ${\overline{Y}}_{22}\ge {G_{12}}^{*}{G}_{12}$, according to Lemma 3. We denote $Y_{22}={\overline{Y}}_{22}-{G_{12}}^{*}{G}_{12}$ and then $Y_{22}\ge 0$. Therefore,

$$X=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ G_{12}^{*}& Y_{22}+{H_{12}}^{*}{H}_{12}\end{array}\right)=X_0+\left(\begin{array}{cc}0& 0\\ 0& Y_{22}\end{array}\right).$$

This shows that X can be represented by Formula (4). The proof is completed. □

In fact, the above equivalent condition is consistent with the relative results in [2,5]. Here, we will demonstrate this using Theorem 1.

Corollary 1

([5]). Let A and B be operators in $\mathcal{B}\left(\mathcal{H}\right)$. Then $AX=B$ has positive solutions if and only if there exists $\lambda >0$, such that $B{B}^{*}\le \lambda B{A}^{*}$.

Proof. 

If $AX=B$ has positive solutions and G is the reduced solution, then $B{A}^{*}\ge 0$ and $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$ by Theorem 1. From Lemma 1, there exists $\lambda >0$, such that $G{G}^{*}\le \lambda GP$. It is natural to obtain that

$$B{B}^{*}=AG{G}^{*}{A}^{*}\le \lambda AGP{A}^{*}=\lambda AG{A}^{*}=\lambda B{A}^{*}.$$

Conversely, if $B{B}^{*}\le \lambda B{A}^{*}$ for some $\lambda >0$, define a mapping T on $R\left(A^{*}\right)$:

$$T\left(A^{*}h\right)=B^{*}h\mathrm{for}\mathrm{any}h\in \mathcal{H}.$$

In fact,

$$\parallel B^{*}{h\parallel }^2=<B{B}^{*}h,h>\le <\lambda B{A}^{*}h,h>=\lambda <A^{*}h,C^{*}h>\le \lambda \parallel A^{*}h\parallel ·\parallel B^{*}h\parallel .$$

So, $\parallel B^{*}h\parallel \le \lambda \parallel A^{*}h\parallel$ and T is well-defined since

$$|T{A}^{*}{h|}^2=\parallel B^{*}{h\parallel }^2\le {\lambda }^2{\parallel A^{*}h\parallel }^2.$$

Hence, T can be uniquely extended to $\overline{R\left(A^{*}\right)}$. Define T on ${\overline{R\left(A^{*}\right)}}^{\perp }$ to be 0, T is a linear bounded operator an $\mathcal{H}$ and satisfies $A{T}^{*}=B$. That is to say, $AX=B$ is solvable. Suppose that G is the reduced solution, we have $G{G}^{*}\le \lambda GP$, since

$$<G{G}^{*}{A}^{*}h,A^{*}h>=<B{B}^{*}h,h>\le \lambda <B{A}^{*}h,h>=\lambda <G{A}^{*}h,A^{*}h>=\lambda <GP{A}^{*}h,A^{*}h>,$$

for any $h\in \mathcal{H}.$ So, $R\left(G\right)\subseteq R{\left(GP\right)}^{\frac{1}{2}}$. Therefore, $AX=B$ has positive solutions by Theorem 1. The proof is completed. □

In particular, if the operators A and $B{A}^{*}$ have closed ranges, Theorem 1 can be transformed into the following form, which is Theorem 5.2 provided in [2].

Corollary 2

([2]). Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A,B{A}^{*}$ have closed ranges. Then $AX=B$ has positive solutions if and only if $R\left(B\right)\subseteq R\left(B{A}^{*}\right)$ and $B{A}^{*}\ge 0$. In this case, the positive solution has the formula

$$X=B^{*}{\left(B{A}^{*}\right)}^{†}B+(I-P)Y(I-P),\forall Y\in {\mathcal{B}}_{+}\left(H\right).$$

(9)

Proof. 

Let A and $B{A}^{*}$ have closed ranges. According to Lemma 1 and Theorem 1, it is only to prove that $R\left(B\right)\subseteq R\left(B{A}^{*}\right)$ if and only if $R\left(B\right)\subseteq R\left(A\right)$ and $R\left(G\right)\subset R\left({\left(GP\right)}^{\frac{1}{2}}\right)$ when $B{A}^{*}\ge 0$. In this case, we have $\mathcal{H}=R\left(A^{*}\right)\oplus R{\left(A^{*}\right)}^{\perp }=R\left(A\right)\oplus R{\left(A\right)}^{\perp }$ since $R\left(A^{*}\right)$ is also closed. As is well-known, ${\overline{R\left(A^{*}\right)}}^{\perp }=N\left(A\right)$ and ${\overline{R\left(A\right)}}^{\perp }=N\left(A^{*}\right)$. Therefore, A and B have the matrix forms as follows,

$$A=\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right):\left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right)$$

(10)

and

$$B=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right):\left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right),$$

(11)

where $A_{11}$ is an invertible operator from $R\left(A^{*}\right)$ onto $R\left(A\right)$. Then $A^{†}$ and $B{A}^{*}$ can be represented by the following forms,

$$A^{†}=\left(\begin{array}{cc}{A}_{11}^{-1}& 0\\ 0& 0\end{array}\right):\left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right),$$

(12)

$$B{A}^{*}=\left(\begin{array}{cc}{B}_{11}{A}_{11}^{*}& 0\\ B_{21}{A}_{11}^{*}& 0\end{array}\right):\left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right).$$

(13)

$B{A}^{*}\ge 0$ implies $B{A}^{*}$ is a self-adjoint operator. From matrix form (13), we have $B_{21}{A}_{11}^{*}=0$. Hence, $B_{21}=0$.

If $R\left(B\right)\subseteq R\left(B{A}^{*}\right)$, comparing the matrix forms (11) with (13), we have $B_{22}=0$ and $R\left(B\right)\subseteq R\left(A\right)$. Recall that $G=A^{†}B$ from Lemma 2, $R\left(G\right)=R\left(A^{†}B\right)\subseteq R\left(A^{†}B{A}^{*}\right)=R\left(GP\right).$ Moreover, the closedness of $R\left(C{A}^{*}\right)$ implies that $R\left(GP\right)$ is closed, according to Lemma 4, and then $R\left(GP\right)=R\left({\left(GP\right)}^{\frac{1}{2}}\right)$. Thus, $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right).$

On the contrary, suppose that $R\left(B\right)\subseteq R\left(A\right)$ and $R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$. Because $R\left(GP\right)$ is closed, $R\left({\left(GP\right)}^{\frac{1}{2}}\right)=R\left(GP\right)$. Therefore,

$$R\left(B\right)=R\left(AG\right)\subseteq R\left(A{\left(GP\right)}^{\frac{1}{2}}\right)=R\left(AGP\right)=R\left(B{A}^{*}\right).$$

Next, we prove that the general positive solution X is of form (9). According to Theorem 1, X has form (4), where $G=A^{†}B,G={\left({\left(GP\right)}^{\frac{1}{2}}\right)}^{†}G(I-P)$, according to Lemma 2. We denote $X_0=G+(I-P)G^{*}+(I-P)H^{*}H(I-P)$, it is only needed to prove $X_0=B^{*}{\left(B{A}^{*}\right)}^{†}B$. Notice that matrix form (11) of B can be written as follows:

$$B=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ 0& 0\end{array}\right):\left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A\right)\\ R{\left(A\right)}^{\perp }\end{array}\right).$$

(14)

Combining $R\left(B{A}^{*}\right)\subseteq R\left(B\right)$ with $R\left(B\right)\subseteq R\left(B{A}^{*}\right)$, we have $R\left(B\right)=R\left(B{A}^{*}\right)=R\left(BP\right)$ is closed. Therefore, $R\left(B_{11}\right)$ which is closed, and so $B_{11}$ is Moore–Penrose invertible. From $B{A}^{*}\ge 0$ and matrix form (13), it is natural that $B_{11}{A}_{11}^{*}\ge 0$. This infers that ${\left(B_{11}{A}_{11}^{*}\right)}^{†}={A_{11}^{*}}^{-1}{B}_{11}^{†}={\left(A_{11}{B}_{11}^{*}\right)}^{†}={B_{11}^{*}}^{†}{A}_{11}^{-1}$. By matrix forms (12) and (14), G and G have the following forms:

$$G=A^{†}B=\left(\begin{array}{cc}{A}_{11}^{†}{B}_{11}& A_{11}^{†}{B}_{12}\\ 0& 0\end{array}\right):\left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right),$$

(15)

$$H={\left({\left(GP\right)}^{\frac{1}{2}}\right)}^{†}G(I-P)=\left(\begin{array}{cc}0& {\left({\left(A_{11}^{-1}{B}_{11}\right)}^{\frac{1}{2}}\right)}^{†}{A}_{11}^{-1}{B}_{12}\\ 0& 0\end{array}\right):\left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R\left(A^{*}\right)\\ R{\left(A^{*}\right)}^{\perp }\end{array}\right).$$

(16)

Further, by direct computing, we have

$$\begin{array}{ccc}\hfill X_0& =& \left(\begin{array}{cc}{A}_{11}^{-1}{B}_{11}& A_{11}^{-1}{B}_{12}\\ B_{12}^{*}{A_{11}^{*}}^{-1}& B_{12}^{*}{A_{11}^{-1}}^{*}{\left({\left(A_{11}^{-1}{B}_{11}\right)}^{\frac{1}{2}}\right)}^{†}{\left({\left(A_{11}^{-1}{B}_{11}\right)}^{\frac{1}{2}}\right)}^{†}{A}_{11}^{-1}{B}_{12}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{A}_{11}^{-1}{B}_{11}& A_{11}^{-1}{B}_{12}\\ B_{12}^{*}{A_{11}^{*}}^{-1}& B_{12}^{*}{A_{11}^{*}}^{-1}{B}_{11}^{†}{B}_{12}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{B}_{11}^{*}{B_{11}^{*}}^{†}{A}_{11}^{-1}{B}_{11}& B_{11}^{*}{B_{11}^{*}}^{†}{A}_{11}^{-1}{B}_{12}\\ B_{12}^{*}{A_{11}^{*}}^{-1}{B_{11}}^{†}{B}_{11}& B_{12}^{*}{\left(B_{11}{A_{11}}^{*}\right)}^{†}{B}_{12}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{B}_{11}^{*}& 0\\ B_{12}^{*}& 0\end{array}\right)\left(\begin{array}{cc}{B_{11}^{*}}^{†}{A}_{11}^{-1}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{B}_{11}& B_{12}\\ 0& 0\end{array}\right)\hfill \\ & =& B^{*}{\left(B{A}^{*}\right)}^{†}B.\hfill \end{array}$$

Therefore, X has the formula form (9). The proof is completed. □

3. The Common Positive Solutions of $\mathit{AX}=\mathit{B}$ and $\mathit{XC}=\mathit{D}$

The common positive solutions of system (2) were discussed under the condition that $A,C,B{A}^{*}$ are closed-range operators in [2,3,10], and under the assumption that $B{A}^{*}$ is a closed range operator in [9]. In this section, we consider the existence and the general form of the positive solutions of system (2) without the restriction on the closed range. For $A,C\in \mathcal{B}\left(\mathcal{H}\right)$, we denote $P=P_{\overline{R\left(A^{*}\right)}}$ and $Q=P_{\overline{R\left(\right(I-P\left)C\right)}}$.

Lemma 5

([6]). Let $A,B,C$, and D be operators in $\mathcal{B}\left(\mathcal{H}\right)$. Then system (2) has solutions if and only if $R\left(D^{*}\right)\subseteq R\left(C^{*}\right)$, $R\left(B\right)\subseteq R\left(A\right)$ and $AD=BC$. In this case, the common solution is

$$X=G-(I-P)K^{*}+(I-P)Y(I-P_{\overline{R\left(C\right)}}),$$

for all $Y\in \mathcal{B}\left(\mathcal{H}\right)$, where G and K are reduced solutions of $AX=B$ and $C^{*}X=D^{*}$, respectively. If operators A and C have closed ranges, the general solution is of the form

$$X=A^{†}B+D{C}^{†}-A^{†}AD{C}^{†}+(I-A^{†}A)Y(I-C{C}^{†}),\forall Y\in \mathcal{B}\left(\mathcal{H}\right).$$

Theorem 2.

Let $A,B,C$, and D be operators in $\mathcal{B}\left(\mathcal{H}\right)$. System (2) has positive solutions if and only if

 (i)

$R\left(D^{*}\right)\subseteq R\left(C^{*}\right)$, $R\left(B\right)\subseteq R\left(A\right)$ and $AD=BC$;

 (ii)

$R\left(G\right)\subseteq R\left({\left(GP\right)}^{\frac{1}{2}}\right)$ and $B{A}^{*}\ge 0$;

 (iii)

$(D^{*}-C^{*}G-C^{*}{H}^{*}H)(I-P)C\ge 0$ and $R\left((D^{*}-C^{*}G)(I-P)\right)\subseteq R\left(C^{*}(I-P)\right),$$R\left(K\right)\subseteq R\left({\left(KQ\right)}^{\frac{1}{2}}\right),$

where G, H, and K are reduced solutions of $AX=B$, ${\left(GP\right)}^{\frac{1}{2}}X=G(I-P)$ and $C^{*}(I-P)X=(D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P)$ respectively. In this case, the common positive solution is

$$\begin{array}{cc}X\hfill & =G+(I-P)G^{*}+(I-P)H^{*}H(I-P)+(I-P)K^{*}(I-P)+(I-P)K(I-P-Q)\hfill \\ & +(I-P-Q)L^{*}L(I-P-Q)+(I-P-Q)Z(I-P-Q),\hfill \end{array}$$

for any positive operator $Z\in \mathcal{B}\left(\mathcal{H}\right)$, where L is the reduced solution of ${\left(KQ\right)}^{\frac{1}{2}}X=K(I-Q)$.

Proof. 

‘⇒’: Combining Lemma 5 and Theorem 1, statements (i) and (ii) hold. Suppose that X is a common positive solution of system (2). Then there is a positive operator $Y\in \mathcal{B}\left(\mathcal{H}\right)$, such that X has the form

$$X=G+(I-P)G^{*}+(I-P)H^{*}H(I-P)+(I-P)Y(I-P),$$

(17)

where G and H are, respectively, reduced solutions of $AX=B$ and ${\left(GP\right)}^{\frac{1}{2}}X=G(I-P)$. Since $\mathcal{H}=\overline{R\left(C^{*}\right)}\oplus {\overline{R\left(C^{*}\right)}}^{\perp }$, $\mathcal{H}=\overline{R\left(A\right)}\oplus {\overline{R\left(A\right)}}^{\perp }=\overline{R\left(A^{*}\right)}\oplus {\overline{R\left(A^{*}\right)}}^{\perp }$ and $R\left(H\right)\subseteq \overline{R\left(A^{*}\right)}$, $R\left(A^{*}\right)\subseteq N\left(G(I-P)\right)=N\left(H\right)$, we know that operators A, G, H, and C are of matrix forms, as follows:

$$A=\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right):\left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A\right)}\\ {\overline{R\left(A\right)}}^{\perp }\end{array}\right),$$

(18)

$$G=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ 0& 0\end{array}\right):\left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right),$$

(19)

$$H=\left(\begin{array}{cc}0& H_{12}\\ 0& 0\end{array}\right):\left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right),$$

(20)

and

$$C=\left(\begin{array}{cc}{C}_{11}& 0\\ C_{21}& 0\end{array}\right):\left(\begin{array}{c}\overline{R\left(C^{*}\right)}\\ {\overline{R\left(C^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right),$$

(21)

respectively, where $A_{11}\in \mathcal{B}(\overline{R\left(A^{*}\right)},\overline{R\left(A\right)})$ is an injective operator with dense range. Moreover, $R\left(D^{*}\right)\subseteq R\left(B^{*}\right)$ implies D has the matrix form

$$D=\left(\begin{array}{cc}{D}_{11}& 0\\ D_{21}& 0\end{array}\right):\left(\begin{array}{c}\overline{R\left(C^{*}\right)}\\ {\overline{R\left(C^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right).$$

(22)

We denote

$$Y=\left(\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right):\left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right).$$

(23)

From Lemma 3, it is easy to obtain that $Y_{22}\ge 0$ since $Y\ge 0$. Substituting the matrix forms (19), (20), and (23) into Formula (1); we see that X has the following matrix form

$$X=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ G_{12}^{*}& H_{12}^{*}{H}_{12}+Y_{22}\end{array}\right):\left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right)\to \left(\begin{array}{c}\overline{R\left(A^{*}\right)}\\ {\overline{R\left(A^{*}\right)}}^{\perp }\end{array}\right).$$

(24)

From matrix forms (21) and (24), $C^{*}{X}^{*}$ is of the form

$$C^{*}{X}^{*}=\left(\begin{array}{cc}{C}_{11}^{*}{G}_{11}^{*}+C_{21}^{*}{G}_{12}^{*}& C_{11}^{*}{G}_{12}+C_{21}^{*}(H_{12}^{*}{H}_{12}+Y_{22}^{*})\\ 0& 0\end{array}\right).$$

(25)

Recall that X is also a positive solution of $C^{*}{X}^{*}=D^{*}$. This implies

$$C_{21}^{*}{Y}_{22}^{*}=D_{21}^{*}-C_{11}^{*}{G}_{12}-C_{21}^{*}{H}_{12}^{*}{H}_{12}$$

(26)

by matrix forms (25) and (22). We conclude that $(I-P)Y^{*}(I-P)$ is a positive solution of equation

$$C^{*}(I-P)\widehat{Y}=(D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P).$$

(27)

By Theorem 1, $(D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P)C\ge 0$ and $R\left((D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P)\right)\subseteq R\left(C^{*}(I-P)\right)$. Noticing that $R\left(\left(C^{*}(I-P)H^{*}H\right)(I-P)\right)\subseteq R\left(C^{*}(I-P)\right)$, we obtain $R\left((D^{*}-C^{*}G)(I-P)\right)\subseteq R\left(C^{*}(I-P)\right)$. Suppose that K is the reduced solution of Equation (17). Then $N\left(K\right)=N\left((D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P)\right)$ and $R\left(K\right)\subseteq \overline{R\left(\right(I-P\left)C\right)}$. This follows that $N\left(K(I-P)\right)=N\left((D^{*}-C^{*}G-C^{*}(I-P)H^{*}H)(I-P)\right)$ and $R\left(K(I-P)\right)\subseteq \overline{R\left(\right(I-P\left)C\right)}$. This shows $K(I-P)$ is also the reduced solution of (17) and then $K=K(I-P)$. Thus, $R\left(K\right)\subseteq R\left({\left(KQ\right)}^{\frac{1}{2}}\right).$

‘⇐’: By statements (i) and (ii), it is immediate that Equation (1) has a positive solution X, which can be represented by form (17). Meanwhile, operators $A,G,H,C,D,Y$, and X have matrix forms (18)–(24), respectively. Combining $AD=BC=AGC$ and matrix forms (18), (19), (21), and (22) of $A,F,B,D$, respectively, we obtain

$$AD=\left(\begin{array}{cc}{A}_{11}{D}_{11}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{A}_{11}(G_{11}{C}_{11}+G_{12}{C}_{21})& 0\\ 0& 0\end{array}\right)=AGC.$$

This shows $A_{11}(G_{11}{C}_{11}+G_{12}{C}_{21})=A_{11}{D}_{11}$. That is, $(C_{11}^{*}{G}_{11}^{*}+C_{21}^{*}{G}_{12}^{*})A_{11}^{*}=D_{11}^{*}{A}_{11}^{*}$ since $R\left(A_{11}\right)$ is dense in $\overline{R\left(A\right)}$. Hence,

$$C_{11}^{*}{G}_{11}^{*}+C_{21}^{*}{G}_{12}^{*}=D_{11}^{*}.$$

(28)

From statement (iii) and Theorem 1, Equation (26) has positive solutions. There exists an operator $Y_{22}\ge 0$, such that Equation (26) holds. Combining matrix forms (25) and (22) with equality (28), we see that X is also a positive solution of $CX=D$. This infers that system (2) has common positive solutions.

For the positive solution X of Equation (2), there exists a positive operator $Y\in \mathcal{B}\left(\mathcal{H}\right)$, such that X has form (17) by the above proof. From Theorem 1 and $K=K(I-P)$, we have the reduced solution of Equation (17); the positive solution of Equation (17) has the following formula:

$$\widehat{Y}=K+(I-Q)K^{*}+(I-Q)L^{*}L+(I-Q)Z(I-Q),$$

(29)

for all Z in ${\mathcal{B}}_{+}\left(\mathcal{H}\right)$, where L is the reduced solution of ${\left(KQ\right)}^{\frac{1}{2}}X=K(I-Q)$. Since $(I-P)Y^{*}(I-P)\ge 0$ is a solution of Equation (17), there is an operator $Z\in {\mathcal{B}}_{+}\left(\mathcal{H}\right)$, such that $(I-P)Y^{*}(I-P)$ has form (29). With multiples $I-P$ on two sides of the left of Equation (29), we obtain

$$(I-P)Y^{*}(I-P)=(I-P)(K+(I-Q)K^{*}+(I-Q)L^{*}L+(I-Q)Z(I-Q))(I-P).$$

Moreover, $I-P\ge P_{\overline{R\left(\right(I-P\left)C\right)}}=Q$, and so $(I-P)(I-Q)=(I-Q)(I-P)=(I-P-Q)$. Therefore,

$$\begin{array}{cc}X\hfill & =G+(I-P)G^{*}+(I-P)H^{*}H(I-P)+(I-P)Y(I-P)\hfill \\ & =G+(I-P)G^{*}+(I-P)H^{*}H(I-P)+(I-P)K^{*}(I-P)+(I-P)K(I-P-Q)\hfill \\ & +(I-P-Q)L^{*}L(I-P-Q)+(I-P-Q)Z(I-P-Q).\hfill \end{array}$$

It is clearly that X is a common solution of Equation (2) if X has the above form. The proof is completed. □

Consequently, we have the following result from Theorem 2 and Corollary 1.

Corollary 3.

Suppose that $A,B,C$, and D are operators in $\mathcal{B}\left(\mathcal{H}\right)$ with $A,B{A}^{*}$ and $C^{*}(I-P)$ having closed ranges. We denote $P=P_{R\left(A^{*}\right)}$, $Q=P_{R\left(\right(I-P\left)C\right)}$, and $M=(D^{*}-C^{*}{A}^{†}B-C^{*}(I-P)B^{*}{\left(B{A}^{*}\right)}^{†}B(I-P)).$ If $R\left(M\right(I-P\left)C\right)$ is closed, then the system of equations $AX=B$, $XC=D$ has positive solutions if and only if

 (1)

$R\left(D^{*}\right)\subseteq R\left(C^{*}\right)$, $AD=BC$;

 (2)

$B{A}^{*}$ is positive, $R\left(B\right)\subseteq R\left(B{A}^{*}\right)$;

 (3)

$M(I-P)C$ is positive and $R\left(M\right(I-P\left)\right)\subseteq R\left(M\right(I-P\left)C\right).$

In this case, the common positive solution is

$$X=B^{*}{\left(B{A}^{*}\right)}^{†}B+(I-P)M^{*}{\left(M(I-P)C\right)}^{†})M(I-P)+(I-P-Q)Z(I-P-Q),\mathrm{for}\mathrm{all}Z\in {\mathcal{B}}_{+}\left(\mathcal{H}\right).$$

4. Example

In this section, we present an example to demonstrate that the results in Theorems 1 and 2 are valid. It is shown that Equation (1) has positive solutions where $R\left(A\right)$ and $R\left(B{A}^{*}\right)$ are not closed. It also indicates that the system of Equation (2) has common positive solutions, which are calculated by Theorem 2. It is shown that our results generalize the relative results in [2,5], to some extent.

Example 1.

Let $\{e_1,e_2,\cdots \}$ be a basis of the separable Hilbert space $\mathcal{H}$. We define an operator T on $\mathcal{H}$, $T{e}_k=\frac{1}{k}{e}_k,$ for any integer $k\ge 1.$ As is well-known, T is a positive and compact operator with $\overline{R\left(T\right)}=\mathcal{H}$ and the range $R\left(T\right)$ is not closed. Assume that operators $A,B,C,D\in \mathcal{B}(\mathcal{H}\oplus \mathcal{H})$ have the following forms:

$$A=\left(\begin{array}{cc}T& 0\\ 0& 0\end{array}\right),B=\left(\begin{array}{cc}\frac{1}{4}T& \frac{1}{3}T\\ 0& 0\end{array}\right),C=\left(\begin{array}{cc}0& T\\ 0& 0\end{array}\right),D=\left(\begin{array}{cc}0& \frac{1}{4}T\\ 0& \frac{1}{3}T\end{array}\right).$$

The matrix forms of A and C imply

$$B{A}^{*}=\left(\begin{array}{cc}\frac{1}{4}{T}^2& 0\\ 0& 0\end{array}\right).$$

It is clear that operators A and $B{A}^{*}$ have no closed ranges. For the operator equation $AX=B$, the positive solution cannot be calculated by Corollary 1. Next, we solve equation $AX=B$ and system $AX=B,XC=D.$

Since $R\left(B\right)=R\left(A\right)$, equation $AX=B$ is solvable. By direct computation, $U_A=\left(\begin{array}{cc}{I}_{\mathcal{H}}& 0\\ 0& 0\end{array}\right)$ and

$$\left(\right|A|+\frac{1}{n}{I)}^{-1}{U}_A^{*}B=\left(\begin{array}{cc}\frac{1}{4}{(T+\frac{1}{n}{I}_{\mathcal{H}})}^{-1}T& \frac{1}{3}{(T+\frac{1}{n}{I}_{\mathcal{H}})}^{-1}T\\ 0& 0\end{array}\right).$$

According to Lemma 2,

$$G=s.o.-li{m}_{n\to \infty }\left(\right|A|+\frac{1}{n}{I)}^{-1}{U}_A^{*}B=\left(\begin{array}{cc}\frac{1}{4}{I}_{\mathcal{H}}& \frac{1}{3}{I}_{\mathcal{H}}\\ 0& 0\end{array}\right)$$

is the reduced solution of $AX=B$. Moreover, $P=P_{\overline{R\left(A^{*}\right)}}=\left(\begin{array}{cc}{I}_{\mathcal{H}}& 0\\ 0& 0\end{array}\right)$ and ${\left(GP\right)}^{\frac{1}{2}}=\left(\begin{array}{cc}\frac{1}{2}{I}_{\mathcal{H}}& 0\\ 0& 0\end{array}\right),G(I-P)=\left(\begin{array}{cc}0& \frac{1}{3}{I}_{\mathcal{H}}\\ 0& 0\end{array}\right).$ It is immediate that $H=\left(\begin{array}{cc}0& \frac{2}{3}{I}_{\mathcal{H}}\\ 0& 0\end{array}\right)$ is the reduced solution of ${\left(GP\right)}^{\frac{1}{2}}X=G(I-P)$. Combining $B{A}^{*}\ge 0$ and Theorem 1, we know that equation $AX=B$ has positive solutions and the formula is

$$X=\left(\begin{array}{cc}\frac{1}{4}{I}_{\mathcal{H}}& \frac{1}{3}{I}_{\mathcal{H}}\\ \frac{1}{3}{I}_{\mathcal{H}}& \frac{4}{9}{I}_{\mathcal{H}}+Y_{22}\end{array}\right),\mathrm{for}\mathrm{any}{Y}_{22}\in {\mathcal{B}}_{+}\left(\mathcal{H}\right).$$

Further, through calculating,

 (i)

$R\left(D^{*}\right)=R\left(C^{*}\right),R\left(B\right)=R\left(A\right),AD=BC$;

 (ii)

$Q=P_{\overline{R\left(\right(I-P\left)C\right)}}=0$ and $B{A}^{*}=\left(\begin{array}{cc}\frac{1}{4}{T}^2& 0\\ 0& 0\end{array}\right)\ge 0$;

 (iii)

$(D^{*}-C^{*}G-C^{*}{H}^{*}H)(I-P)C=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ and $C^{*}(I-P)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.

By Theorem 2, system $AX=B,XC=D$ has a common positive solution which is of the form

$$X=G+(I-P)G^{*}+(I-P-Q)Z(I-P-Q)=\left(\begin{array}{cc}\frac{1}{4}{I}_{\mathcal{H}}& \frac{1}{3}{I}_{\mathcal{H}}\\ \frac{1}{3}{I}_{\mathcal{H}}& \frac{4}{9}{I}_{\mathcal{H}}+Y_{22}\end{array}\right),\mathrm{for}\mathrm{any}{Y}_{22}\in {\mathcal{B}}_{+}\left(\mathcal{H}\right).$$

5. Conclusions

In this work, to find the common positive solutions for Equation (2) without the limit on the closeness of the range, we first characterize the positive solutions of Equation (1). Theorem 1 provides an equivalent condition for the existence of positive solutions for Equation (1) by the reduced solution. Meanwhile, the formula of positive solutions is given, which implies the matrix form of solutions with respect to the space decomposition $\mathcal{H}=\overline{R\left(A^{*}\right)}\oplus {\overline{R\left(A^{*}\right)}}^{\perp }$. This result offers us a method to consider common positive solutions for Equation (2). By using operator-blocking techniques, the general formula of common positive solutions for Equation (2) is obtained in Theorem 2. Through Example 1, it is illustrated that Theorems 1 and 2 are useful for finding positive solutions. The approaches may also be valid for other operator equations, such as $AXB=C,A{X}^{*}+X{A}^{*}=B$, and so on.