Solutions of the system of operator equations $BXA=B=AXB$ via $*$-order

TL;DR

This paper investigates conditions for solutions to a specific operator equation system involving bounded linear operators on a Hilbert space, introducing the concept of inverse along an operator and characterizing solutions via the *-order.

Contribution

It establishes necessary and sufficient conditions for solutions, characterizes solutions using the *-order, and provides the general solution and characterizations of the *-order in this context.

Findings

Solutions exist under specific conditions related to the *-order.

An operator is a solution iff it satisfies a *-order inequality.

The paper provides the general form of solutions.

Abstract

In this paper, we establish some necessary and sufficient conditions for the existence of solutions to the system of operator equations $ BXA=B=AXB $ in the setting of bounded linear operators on a Hilbert space, where the unknown operator $X$ is called the inverse of $A$ along $B$. After that, under some mild conditions we prove that an operator $X$ is a solution of $ BXA=B=AXB $ if and only if $B \stackrel{*}{ \leq} AXA$, where the $*$-order $C\stackrel{*}{ \leq} D$ means $CC^*=DC^*, C^*C=C^*D$. Moreover we present the general solution of the equation above. Finally, we present some characterizations of $C \stackrel{*}{ \leq} D$ via other operator equations.

Equations

$$A{A}^{†}A=A,A^{†}A{A}^{†}=A^{†},(A{A}^{†}{)}^{\ast }=A{A}^{†},(A^{†}A{)}^{\ast }=A^{†}A.$$

$$\mathcal{R}(A^{†})=\mathcal{R}(A^{\ast })=\mathcal{R}(A^{†}A)=\mathcal{R}(A^{\ast }A),\mathcal{N}(A^{†})=\mathcal{N}(A^{\ast })=\mathcal{N}(A{A}^{†}),$$

$$\mathcal{R}(A)=\mathcal{R}(A{A}^{†})=\mathcal{R}(A{A}^{\ast }),P_{\mathcal{R}(A)}=A{A}^{†}\mathrm{and}{P}_{\overline{\mathcal{R}(A^{\ast })}}=A^{†}A.$$

$$A{A}^{\ast }=B{A}^{\ast },A^{\ast }A=A^{\ast }B.$$

$$\mathcal{H}=\mathcal{R}(A^{\ast })\oplus \mathcal{N}(A)\mathrm{and}\mathcal{K}=\mathcal{R}(A)\oplus \mathcal{N}(A^{\ast })$$

$$\displaystyle A=\left[\begin{array}[]{cc}A_{1}&0\\ 0&0\end{array}\right]:\left[\begin{array}[]{cc}\mathscr{R}(A^{*})\\ \mathscr{N}(A)\end{array}\right]\rightarrow\left[\begin{array}[]{cc}\mathscr{R}(A)\\ \mathscr{N}(A^{*})\end{array}\right],$$

$$\displaystyle A^{\dagger}=\left[\begin{array}[]{cc}A_{1}^{-1}&0\\ 0&0\end{array}\right]:\left[\begin{array}[]{cc}\mathscr{R}(A)\\ \mathscr{N}(A^{*})\end{array}\right]\rightarrow\left[\begin{array}[]{cc}\mathscr{R}(A^{*})\\ \mathscr{N}(A)\end{array}\right].$$

$$A{A}^{†}B{A}^{†}A=A{A}^{†}(A{A}^{†}\tilde{C})A^{†}A=A{A}^{†}\tilde{C}{A}^{†}A=B{A}^{†}A=(C{A}^{†}A)A^{†}A=C{A}^{†}A=B.$$

$$\displaystyle A=\left[\begin{array}[]{cc}A_{1}&0\\ 0&0\end{array}\right]{\rm~{}and~{}}A^{\dagger}=\left[\begin{array}[]{cc}A_{1}^{-1}&0\\ 0&0\end{array}\right].$$

$$\mathcal{R}(A)\subseteq \mathcal{R}(B),\mathcal{N}(B)\subseteq \mathcal{N}(A)\mathrm{and}BXA=B=AXB$$

$$\mathcal{N}(B)=\mathcal{N}(A),\mathcal{R}(B)=\mathcal{R}(A)\mathrm{and}AXA=A.$$

$$(I-AX)A=0⟹\mathcal{R}(A)\subseteq \mathcal{N}(I-AX)⟹\mathcal{R}(B)\subseteq \mathcal{N}(I-AX)⟹B=AXB,$$

$$A(I-XA)=0⟹\mathcal{R}(I-XA)\subseteq \mathcal{N}(A)⟹\mathcal{R}(I-XA)\subseteq \mathcal{N}(B)⟹B=BXA.$$

$$\mathcal{N}(B)=\mathcal{N}(A),\mathcal{R}(B)=\mathcal{R}(A)\mathrm{and}AXA=A.$$

$$P_{\overline{\mathcal{R}(B)}}(B-AXA)=B-P_{\overline{\mathcal{R}(B)}}AXA=B-BXA=0$$

$$(B-AXA)P_{\overline{\mathcal{R}(B^{\ast })}}=B-AXA{P}_{\overline{\mathcal{R}(B^{\ast })}}=B-AXB=0.$$

$$B-BXA=B-P_{\overline{\mathcal{R}(B)}}AXA=P_{\overline{\mathcal{R}(B)}}(B-AXA)=0$$

$$B-AXB=B-AXA{P}_{\overline{\mathcal{R}(B^{\ast })}}=(B-AXA)P_{\overline{\mathcal{R}(B^{\ast })}}=0.$$

$$BXA=B=AXB;$$

$$BAX=B=XAB.$$

$$X$$

$$X=A^{†}B{B}^{†}+W-A^{†}AWB{B}^{†},$$

$$B(A^{†}B{B}^{†}+W-A^{†}AWB{B}^{†})A=B.$$

$$B{A}^{†}B+BWA-B{A}^{†}AWB=B.$$

$$A^{†}B+WA-A^{†}AWB=B^{†}B+(I-B^{†}B)S,$$

$$A{A}^{†}B+AWA-A{A}^{†}AWB=A{B}^{†}B+A(I-B^{†}B)S$$

$$A{A}^{†}B+AWA-AWB=B+(A-B)S.$$

$$W=A^{†}\left[B(I-A{A}^{†})+(A-B)S\right](A-B{)}^{†}+T-A^{†}AT(A-B{)}^{†}(A-B),$$

$$X$$

$$P_{\overline{\mathcal{R}(A^{\ast })}}\tilde{X}A{A}^{†}B=(A^{†}A)\tilde{X}(A{A}^{†})B=(A^{†}A)\tilde{X}(A{A}^{†}B)=A^{†}(A\tilde{X}B)=A^{†}B.$$

$$B{P}_{\overline{\mathcal{R}(A^{\ast })}}\tilde{X}A{A}^{†}$$

Full text

1. Introduction and preliminaries

Throughout the paper, $\mathscr{H}$ and $\mathscr{K}$ are complex Hilbert spaces. We denote the space of all bounded linear operators from $\mathscr{H}$ into $\mathscr{K}$ by $\mathbb{B}(\mathscr{H},\mathscr{K})$, and write $\mathbb{B}(\mathscr{H})$ when $\mathscr{H}=\mathscr{K}.$ Recall that an operator $A\in\mathbb{B}(\mathscr{H})$ is positive if $\langle Ax,x\rangle\geq 0$ for all $x\in\mathscr{H}$ and then we write $A\geq 0$. We shall write $A>0$ if $A$ is positive and invertible. An operator $A\in\mathbb{B}(\mathscr{H})$ is a generalized projection if $A^{2}=A^{*}$. Let $\mathscr{S}(\mathscr{H}),\mathscr{Q}(\mathscr{H}),\mathscr{OP}(\mathscr{H}),\mathscr{GP}(\mathscr{H})$ be the set of all self-adjoint operators on $\mathscr{H},$ the set of all idempotents, the set of orthogonal projections and the set of all generalized projections on $\mathscr{H},$ respectively. For $A\in\mathbb{B}(\mathscr{H},\mathscr{K})$, let $\mathscr{R}(A)$ and $\mathscr{N}(A)$ be the range and the null space of $A,$ respectively. The projection corresponding to a closed subspace $\mathscr{M}$ of $\mathscr{H}$ is denoted by $P_{\mathscr{M}}$. The symbol $A^{-}$ stands for an arbitrary generalized inner inverse of A, that is, an operator $A^{-}$ satisfying $AA^{-}A=A$. The Moore–Penrose inverse of a closed range operator $A$ is the unique operator $A^{\dagger}\in\mathbb{B}(\mathscr{H})$ satisfying the following equations

[TABLE]

Then, $A^{*}AA^{\dagger}=A^{*}=A^{\dagger}AA^{*}$ and we have the following properties

[TABLE]

For $A,B\in\mathscr{S}(\mathscr{H})$, $A\leq B$ means $B-A\geq 0$. The order $\leq$ is said to be the Löwner order on $\mathscr{S}(\mathscr{H})$. If there exists $C\in\mathscr{S}(\mathscr{H})$ such that $AC=0$ and $A+C=B$, then we write $A\preceq B$. The order $\preceq$ is said to be the logic order on $\mathscr{S}(\mathscr{H})$. For $A,B\in\mathbb{B}(\mathscr{H}),$ let $A\stackrel{{\scriptstyle*}}{{\leq}}B$ mean

[TABLE]

It is known that, for $A,B\in\mathscr{S}(\mathscr{H}),A\preceq B$ if and only if $A\stackrel{{\scriptstyle*}}{{\leq}}B$; see [6]. We denote by $A\stackrel{{\scriptstyle*}}{{\wedge}}B$ the infimum (or the greatest lower bound) of $A$ and $B$ over the $*-$ order and $A\stackrel{{\scriptstyle*}}{{\vee}}B$ the supremum (or the least upper bound) of $A$ and $B$ over the $*-$ order, if they exist; cf. [12].

It is known that if $A\in\mathbb{B}(\mathscr{H},\mathscr{K})$ has closed range, then by considering

[TABLE]

we can write

[TABLE]

where $A_{1}:\mathscr{R}(A^{*})\rightarrow\mathscr{R}(A)$ is invertible; see [8, Lemma 2.1].

Therefore, the Moore–Penrose generalized inverse of $A$ can be represented as

[TABLE]

Many results have been obtained on the solvability of equations for matrices and operators on Hilbert spaces and Hilbert $C^{*}-$ modules. In 1976, Mitra [11] considered the matrix equations $AX=B,AXB=C$ and the system of linear equations $AX=C,XB=D$. He got the necessary and sufficient conditions for existence and expressions of general Hermitian solutions. In 1966, the celebrated Douglas Lemma was established in [9]. It gives some conditions for the existence of a solution to the equation $AX=B$ for operators on a Hilbert space. Using the generalized inverses of operators, in 2007, Dajić and Koliha [4] got the existence of the common Hermitian and positive solutions to the system $AX=C,XB=D$ for operators acting on a Hilbert space. In 2008, Xu [17] extended these results to the adjointable operators. Several general operator equations and systems in some general settings such as Hilbert $C^{*}$-modules have been studied by some mathematicians; see, e.g., [7, 10, 13, 16].

The matrix equation $AXB=C$ is consistent if and only if $AA^{-}CB^{-}B=C$ for some $A^{-},B^{-}$, and the general solution is $X=A^{-}CB^{-}+Y-A^{-}AYBB^{-}$, where $Y$ is an arbitrary matrix; see [11]. In 2010, Gonzalez [1] got some necessary and sufficient conditions for existence of a solution to the equation $AXB=C$ for operators on a Hilbert space.

Let $A,B$ or $C$ have closed range. Then, the operator equation $AXB=C$ is solvable if and only if $\mathscr{R}(C)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(C^{*})\subseteq\mathscr{R}(B^{*})$; see [1, Theorem 3.1]. Therefore, if $A$ or $C$ has closed range, then the equation $AXC=C$ is solvable if and only if $\mathscr{R}(C)\subseteq\mathscr{R}(A)$, and $CXA=C$ is solvable if and only if $\mathscr{R}(C^{*})\subseteq\mathscr{R}(A^{*})$. Deng [5] investigated the equation $CAX=C=XAC$, which is essentially different from ours. In this paper, we first characterize the existence of solutions of the system of operator equations $BXA=B=AXB$ by means of $*-$ order. After that, we generalize the solutions to the system of operator equations $BXA=B=AXB$ in a new fashion.

2. The existence of solutions of the system $BXA=B=AXB$

We start our work with the celebrated Douglas lemma.

Lemma 2.1** (Douglas Lemma).**

[9*]** Let $A,C\in\mathbb{B}(\mathscr{H})$. Then, the following statements are equivalent:

1. (a)

$\mathscr{R}(C)\subseteq\mathscr{R}(A)$. 2. (b)

There exists $X\in\mathbb{B}(\mathscr{H})$ such that $AX=C$. 3. (c)

There exists a positive number $\lambda$ such that $CC^{*}\leq\lambda^{2}AA^{*}$.

If one of these conditions holds, then there exists a unique solution $\widetilde{X}\in\mathbb{B}(\mathscr{H})$ of the equation $AX=C$ such that $\mathscr{R}(\widetilde{X})\subseteq\overline{\mathscr{R}(A^{*})}$ and $\mathscr{N}(\widetilde{X})=\mathscr{N}(C)$.

Lemma 2.2**.**

Let $A,B\in\mathbb{B}(\mathscr{H})$. If $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$, then $B=B_{1}\bigoplus 0,$ where $B_{1}\in\mathbb{B}(\overline{\mathscr{R}(A^{*})},\overline{\mathscr{R}(A)})$.

Proof.

Let $A,B$ be operators from the decomposition $\mathscr{H}=\overline{\mathscr{R}(A^{*})}\bigoplus\mathscr{N}(A)$ into the decomposition $\mathscr{H}=\overline{\mathscr{R}(A)}\bigoplus\mathscr{N}(A^{*})$. If $\mathscr{R}(B)\subseteq\mathscr{R}(A)$, then, by Lemma 2.1, there exists $C\in\mathbb{B}(\mathscr{H})$ such that $B=AC$ and $\mathscr{N}(C)=\mathscr{N}(B)$. Since $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$, so $\mathscr{R}(C^{*})\subseteq\overline{\mathscr{R}(C^{*})}=\overline{\mathscr{R}(B^{*})}\subseteq\overline{\mathscr{R}(A^{*})}=\mathscr{N}(P_{\mathscr{N}(A)})$. Hence, $P_{\mathscr{N}(A)}C^{*}=0$ and so $CP_{\mathscr{N}(A)}=0$. It follows from $\mathscr{N}(C)=\mathscr{N}(B)$ that $BP_{\mathscr{N}(A)}=0$.

If $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$, then a similar reasoning shows that $P_{\mathscr{N}(A^{*})}B=0$. Therefore, $P_{\overline{\mathscr{R}(A)}}BP_{\mathscr{N}(A)}=P_{\mathscr{N}(A^{*})}BP_{\overline{\mathscr{R}(A^{*})}}=P_{\mathscr{N}(A^{*})}BP_{\mathscr{N}(A)}=0$. Hence, $B=B_{1}\bigoplus 0$, where $B_{1}=P_{\overline{\mathscr{R}(A)}}BP_{\overline{\mathscr{R}(A^{*})}}$. ∎

Theorem 2.3**.**

Let $A\in\mathbb{B}(\mathscr{H})$ and $B\in\mathscr{S}(\mathscr{H})$. If $A$ has closed range, then the following statements are equivalent:

1. (1)

The system of operator equations $BXA=B=AXB$ is solvable; 2. (2)

$AA^{\dagger}BA^{\dagger}A=B$; 3. (3)

$\mathscr{R}(B)\subseteq\mathscr{R}(A)$* and $\mathscr{R}(B)\subseteq\mathscr{R}(A^{*})$.*

Proof.

$(1)\longrightarrow(2):$ Using (1) and $B=BXA,$ we get that $\mathscr{R}(B)\subseteq\mathscr{R}(A^{*})=\mathscr{R}(A^{\dagger}A)$. Hence, by Lemma 2.1, there exists $C^{*}\in\mathbb{B}(\mathscr{H})$ such that $B=A^{\dagger}AC^{*}$. Hence, $B=CA^{\dagger}A$. Applying (1) and $AXB=B,$ we derive that $\mathscr{R}(B)\subseteq\mathscr{R}(A)=\mathscr{R}(AA^{\dagger})$. Thus, by Lemma 2.1, there exists $\widetilde{C}\in\mathbb{B}(\mathscr{H})$ such that $B=AA^{\dagger}\widetilde{C}.$ It follows that

[TABLE]

(2) $\longrightarrow$ (3): Let $AA^{\dagger}BA^{\dagger}A=B$. Then, $\mathscr{R}(B)\subseteq\mathscr{R}(A)$. It follows from $B=B^{*}=(AA^{\dagger}BA^{\dagger}A)^{*}=A^{\dagger}ABAA^{\dagger}$ and (1) that $\mathscr{R}(B)\subseteq\mathscr{R}(A^{\dagger})=\mathscr{R}(A^{*})$.

(3) $\longrightarrow$ (1): Let $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(B)\subseteq\mathscr{R}(A^{*})$. Upon applying Lemma 2.2, $B=B_{1}\bigoplus 0$, where $B_{1}=P_{\overline{\mathscr{R}(A)}}BP_{\overline{\mathscr{R}(A^{*})}}$. Since $A$ has closed rang, so by using (1.3) and (1.4) we have

[TABLE]

Hence, $AA^{\dagger}B=B$ and $BA^{\dagger}A=B$. Thus $X=A^{\dagger}$ is a solution of the system $BXA=B=AXB$. ∎

Proposition 2.4**.**

*Let $A,B,X\in\mathbb{B}(\mathscr{H})$. Then,

[TABLE]

if and only if

[TABLE]

Proof.

$(\Longrightarrow):$ Suppose that $\mathscr{R}(A)\subseteq\mathscr{R}(B),\mathscr{N}(B)\subseteq\mathscr{N}(A)$ and $BXA=B=AXB$. It follows from $BXA=B$ and $\mathscr{N}(B)\subseteq\mathscr{N}(A)$ that $\mathscr{N}(A)\subseteq\mathscr{N}(B)\subseteq\mathscr{N}(A)$. Hence, $\mathscr{N}(A)=\mathscr{N}(B)$. It follows from $AXB=B$ and $\mathscr{R}(A)\subseteq\mathscr{R}(B)$ that $\mathscr{R}(A)\subseteq\mathscr{R}(B)\subseteq\mathscr{R}(A)$. Therefore, $\mathscr{R}(A)=\mathscr{R}(B)$. Moreover, $(I-AX)B=0$ and $\mathscr{R}(A)\subseteq\mathscr{R}(B)$ Hence, we derive that $(I-AX)A=0$. So, $AXA=A$.

$(\Longleftarrow)$: Suppose that $\mathscr{N}(B)=\mathscr{N}(A),\mathscr{R}(B)=\mathscr{R}(A)$ and $AXA=A$. Hence,

[TABLE]

∎

3. System of operator equations $BXA=B=AXB$ via $*$-order

We know that $(\mathbb{B}(\mathscr{H}),\stackrel{{\scriptstyle*}}{{\leq}})$ is a partially ordered set; see [2]. Let $G_{1},G_{2}\in\mathbb{B}(\mathscr{H})$ be invertible and $G_{1}\stackrel{{\scriptstyle*}}{{\leq}}A,G_{2}\stackrel{{\scriptstyle*}}{{\leq}}A$. Then, $G_{1}G_{1}^{*}=AG_{1}^{*}$ and $G_{2}G_{2}^{*}=AG_{2}^{*}$. Hence, we obtain $G_{1}=G_{2}=A$. This fact leads us to consider the characterizations of $A\stackrel{{\scriptstyle*}}{{\leq}}B$. Now we state the necessary and sufficient conditions in which the common $*-$ lower or $*-$ upper bounds of $A$ and $B$ exist.

We need the following essential lemmas.

Lemma 3.1**.**

[18, Lemma 2.1]** Let $A,B\in\mathbb{B}(\mathscr{H})$ and $\overline{\mathscr{M}}$ denote the closure of a space $\mathscr{M}$.

1. (a)

$AA^{*}=BA^{*}\Longleftrightarrow A=BP_{\overline{\mathscr{R}(A^{*})}}\Longleftrightarrow A=BQ$* for some $Q\in\mathscr{OP}(\mathscr{H})$;* 2. (b)

$A^{*}A=A^{*}B\Longleftrightarrow A=P_{\overline{\mathscr{R}(A)}}B\Longleftrightarrow A=PB$* for some $P\in\mathscr{OP}(\mathscr{H})$;* 3. (c)

$A\stackrel{{\scriptstyle*}}{{\leq}}B\Longleftrightarrow B=A+P_{\mathscr{N}(A^{*})}BP_{\mathscr{N}(A)}$; 4. (d)

$A\stackrel{{\scriptstyle*}}{{\leq}}B\Longleftrightarrow A=P_{\overline{\mathscr{R}(A)}}B=BP_{\overline{\mathscr{R}(A^{*})}}=P_{\overline{\mathscr{R}(A)}}BP_{\overline{\mathscr{R}(A^{*})}}$; 5. (e)

$A\stackrel{{\scriptstyle*}}{{\leq}}B\Longleftrightarrow A=A_{1}\bigoplus 0,B=A_{1}\bigoplus B_{1}$;

where $A_{1}\in\mathbb{B}(\overline{\mathscr{R}(A^{*})},\overline{\mathscr{R}(A)}),B_{1}\in\mathbb{B}(\mathscr{N}(A),\mathscr{N}(A^{*}))$ and $A\bigoplus B$ means the block matrix \left[\begin{array}[]{cc}A&0\\ 0&B\end{array}\right].

The following Lemma is a version of Lemma 2.1 when the operator $A$ has closed range.

Lemma 3.2**.**

[4*, Theorem 3.1]**. Let $A\in\mathbb{B}(\mathscr{H})$ have closed range. Then, the equation $AX=C$ has a solution $X\in\mathbb{B}(\mathscr{H})$ if and only if $AA^{\dagger}C=C$, and this if and only if $\mathscr{R}(C)\subseteq\mathscr{R}(A)$. In this case, the general solution is $X=A^{\dagger}C+(I-A^{\dagger}A)T,$ where $T\in\mathbb{B}(\mathscr{H})$ is arbitrary.

Proposition 3.3**.**

*Let $A,B\in\mathbb{B}(\mathscr{H})$. Then,

1. (a)

If $A$ has closed range and $B\stackrel{{\scriptstyle*}}{{\leq}}A$, then $X=A^{\dagger}$ is a solution of the system $BXA=B=AXB$. 2. (b)

If $B$ has closed range and $B\stackrel{{\scriptstyle*}}{{\leq}}A$, then $X=B^{\dagger}$ is a solution of the system $BXA=B=AXB$.

Proof.

(a) Let $A$ be a closed range operator and $B\stackrel{{\scriptstyle*}}{{\leq}}A$. It follows from Lemma 3.1(d) that $B=AP_{\overline{\mathscr{R}(B^{*})}}$ and $B=P_{\overline{\mathscr{R}(B)}}A$. Hence, $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$. It follows from $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and Lemma 3.2 that $AA^{\dagger}B=B$. It follows from $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$ and Lemma 3.2 that $BA^{\dagger}A=\left((A^{\dagger}A)^{*}B^{*}\right)^{*}=\left(A^{*}{A^{\dagger}}^{*}B^{*}\right)^{*}=B$. Hence, $X=A^{\dagger}$ is a solution of the system of operator equations $BXA=B=AXB$.

(b) Let $B$ be a closed range operator and $B\stackrel{{\scriptstyle*}}{{\leq}}A$. It follows from Lemma 3.1 that $B=AP_{\overline{\mathscr{R}(B^{*})}}$ and $B=P_{\mathscr{R}(B)}A$. Applying (1), we conclude that $AB^{\dagger}B=B$ and $BB^{\dagger}A=B$. Hence, $X=B^{\dagger}$ is a solution of the system $BXA=B=AXB$. ∎

Proposition 3.4**.**

*Let $A,B,X\in\mathbb{B}(\mathscr{H})$.

If $A\stackrel{{\scriptstyle*}}{{\leq}}B$ and $BXA=B=AXB$, then $\mathscr{N}(B)=\mathscr{N}(A),\mathscr{R}(B)=\mathscr{R}(A){\rm~{}and~{}}AXA=A$.*

Proof.

Let $A\stackrel{{\scriptstyle*}}{{\leq}}B$ and $BXA=B=AXB$. Applying Lemma 3.1(d) we have $A=P_{\overline{\mathscr{R}(A)}}B=BP_{\overline{\mathscr{R}(A^{*})}}$. Hence, $\mathscr{R}(A)\subseteq\mathscr{R}(B)$ and $\mathscr{N}(B)\subseteq\mathscr{N}(A)$. Using Proposition 2.4,

[TABLE]

∎

Remark 3.5*.*

Note that the converse of Proposition 3.4 is not true, in general. Set $A^{\dagger},A^{*},A$ instead of $A,B,X$. If $A\in\mathbb{B}(\mathscr{H})$ has closed range, then, by (1), we have $\mathscr{R}(A^{*})=\mathscr{R}(A^{\dagger}),\mathscr{N}(A^{*})=\mathscr{N}(A^{\dagger})$ and $A^{\dagger}AA^{\dagger}=A^{\dagger}$ but not $A^{\dagger}\stackrel{{\scriptstyle*}}{{\leq}}A^{*}$. Indeed, if $A^{\dagger}\stackrel{{\scriptstyle*}}{{\leq}}A^{*}$, then by utilizing Lemma 3.1(d), we have $A^{\dagger}=P_{\mathscr{R}(A^{\dagger})}A^{*}$. It follows from $\mathscr{R}(A^{\dagger})=\mathscr{R}(A^{*})$ that $A^{\dagger}=P_{\mathscr{R}(A^{*})}A^{*}=A^{*}$.

Theorem 3.6**.**

*Let $A,B\in\mathbb{B}(\mathscr{H})$ and $B\stackrel{{\scriptstyle*}}{{\leq}}A$. Then, the following statements are equivalent:

1. (a)

There exists a solution $X\in\mathbb{B}(\mathscr{H})$ of the system $BXA=B=AXB$; 2. (b)

$B\stackrel{{\scriptstyle*}}{{\leq}}AXA$

Proof.

$(a)\Longrightarrow(b)$: Let $X\in\mathbb{B}(\mathscr{H})$ is a solution of the system $BXA=B=AXB$. Hence, $B-BXA=0$ and $B-AXB=0$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and Lemma 3.1(d) that $B=P_{\overline{\mathscr{R}(B)}}A$ and $B=AP_{\overline{\mathscr{R}(B^{*})}}$. Hence,

[TABLE]

and

[TABLE]

Therefore, $B\stackrel{{\scriptstyle*}}{{\leq}}AXA$.

$(b)\Longrightarrow(a)$: Suppose that $B\stackrel{{\scriptstyle*}}{{\leq}}AXA$. Applying Lemma 3.1(d), we infer that $P_{\overline{\mathscr{R}(B)}}(B-AXA)=0$ and $(B-AXA)P_{\overline{\mathscr{R}(B^{*})}}=0$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and Lemma 3.1(d) that $B=P_{\overline{\mathscr{R}(B)}}A$ and $B=AP_{\overline{\mathscr{R}(B^{*})}}$, whence

[TABLE]

and

[TABLE]

Therefore, $X$ is a solution of the system $BXA=B=AXB$. ∎

Let $A,B\in\mathbb{B}(\mathscr{H})$ have closed ranges. It follows from Proposition 3.3 that $A^{\dagger}$ and $B^{\dagger}$ are solutions of the system $BXA=B=AXB$. Therefore, we are interested in the study of the following system of operator equations:

[TABLE]

[TABLE]

Let $A,B\in\mathbb{B}(\mathscr{H})$. An operator $C\in\mathbb{B}(\mathscr{H})$ is said to be an inverse of $A$ along $B$ if it fulfills one of the equations (3.1) or (3.2). If $A\in\mathbb{B}(\mathscr{H})$ is invertible, then $X=A^{-1}$ is a solution of the system $XA=I=AX$. Hence, $A^{-1}$ is an inverse of $A$ along $I,$ where $I$ is the identity of $\mathbb{B}(\mathscr{H})$.

Let $A\in\mathbb{B}(\mathscr{H})$ have closed range. Using (1), we have $AA^{\dagger}A=A=AA^{\dagger}A$. Hence, $A^{\dagger}$ satisfies Eq. (3.1). Therefore, $A^{\dagger}$ is the inverse of $A$ along $A$.

It follows from (1) that $A^{*}AA^{\dagger}=A^{*}=A^{\dagger}AA^{*}$. Hence, $A^{\dagger}$ satisfies Eq. (3.2). Therefore, $A$ is the inverse of $A$ along $A^{*}$.

Lemma 3.7**.**

[11, Theorem 2.1]** Let $C\in\mathbb{B}(\mathscr{H})$ and $A,B\in\mathbb{B}(\mathscr{H})$ have closed ranges. Then, the equation $AXB=C$ has a solution $X\in\mathbb{B}(\mathscr{H})$ if and only if $\mathscr{R}(C)\subseteq\mathscr{R}(A),\mathscr{R}(C^{*})\subseteq\mathscr{R}(B^{*})$, and this if and only if $AA^{\dagger}CB^{\dagger}B=C$. In this case, $X=A^{\dagger}CB^{\dagger}+U-A^{\dagger}AUBB^{\dagger}$, where $U\in\mathbb{B}(\mathscr{H})$ is arbitrary.

In the next result we provide a general solution of the system $BXA=B=AXB$.

Theorem 3.8**.**

Let $A,B\in\mathbb{B}(\mathscr{H})$ have closed ranges and $B\stackrel{{\scriptstyle*}}{{\leq}}A$. Then, the general solution of the system of operator equations $BXA=B=AXB$ is

[TABLE]

where $S,T\in\mathbb{B}(\mathscr{H}).$

Proof.

Let $A,B$ have closed ranges. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and Lemma 3.1(d) that $B=AP_{\mathscr{R}(B^{*})}$. Hence, $\mathscr{R}(B)\subseteq\mathscr{R}(A)$. Using Lemma 3.2, we have $AA^{\dagger}B=B$. It follows from $AA^{\dagger}BB^{\dagger}B=B$ and Lemma 3.7 that the equation $AXB=B$ is solvable. In this case, the general solution is

[TABLE]

where $W\in\mathbb{B}(\mathscr{H})$ is arbitrary. If $X$ satisfies the equation $BXA=B$, then

[TABLE]

It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and Lemma 3.1(d) that $B=P_{\mathscr{R}(B)}A$. Applying (1), $BB^{\dagger}A=B$. Hence,

[TABLE]

Therefore, $B(A^{\dagger}B+WA-A^{\dagger}AWB)=B$. So, $A^{\dagger}B+WA-A^{\dagger}AWB$ is a solution of the equation $BX=B$. Utilizing Lemma 3.2 again, we have

[TABLE]

where $S\in\mathbb{B}(\mathscr{H})$ is arbitrary. Multiply the left hand side of Eq. (3.4) by $A$, to get

[TABLE]

It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and Lemma 3.1(d) that $B=AP_{\mathscr{R}(B^{*})}$. Applying (1), $AB^{\dagger}B=B$. We derive that

[TABLE]

Now, we get $AW(A-B)=B(I-AA^{\dagger})+(A-B)S$. So, $W$ is a solution of the equation $AX(A-B)=B(I-AA^{\dagger})+(A-B)S$. Using Lemma 3.2, we get that

[TABLE]

where $T\in\mathbb{B}(\mathscr{H})$ is arbitrary. By putting $W$ in Eq. (3.3), we reach

[TABLE]

∎

Theorem 3.9**.**

Let $A,B\in\mathbb{B}(\mathscr{H})$ where $A$ has closed range. If the system $BXA=B=AXB$ is solvable, then the system $XB=A^{\dagger}B,BX=BA^{\dagger}$ is solvable. Conversely, If $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and the system $XB=A^{\dagger}B,BX=BA^{\dagger}$ is solvable, then the system $BXA=B=AXB$ is solvable.

Proof.

($\Longrightarrow$): Let $\widetilde{X}$ be a solution of the system $BXA=B=AXB$. It follows from $B=A\widetilde{X}B$ that $\mathscr{R}(B)\subseteq\mathscr{R}(A)$. Using Lemma 3.2, $AA^{\dagger}B=B$. It follows from (1) that

[TABLE]

So, $P_{\overline{\mathscr{R}(A^{*})}}\widetilde{X}AA^{\dagger}$ is a solution of the equation $XB=A^{\dagger}B$. Since $B^{*}=(B\widetilde{X}A)^{*}=A^{*}\widetilde{X}^{*}B^{*}$, we have $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$. Applying Lemma 2.1, there exists $Y\in\mathbb{B}(\mathscr{H})$ such that $B=YA$. Hence,

[TABLE]

Therefore, $P_{\overline{\mathscr{R}(A^{*})}}\widetilde{X}AA^{\dagger}$ is a solution of the equation $B=BA^{\dagger}$. Thus $P_{\overline{\mathscr{R}(A^{*})}}\widetilde{X}AA^{\dagger}$ is a solution of the system $XB=A^{\dagger}B,BX=BA^{\dagger}$.

($\Longleftarrow$): Suppose that $\widetilde{X}$ is a solution of the system $XB=A^{\dagger}B,BX=BA^{\dagger}$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ that $B=AP_{\overline{\mathscr{R}(B^{*})}}$ and $B=P_{\overline{\mathscr{R}(B)}}A$. Hence, $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$. It follows from $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ to Lemma 3.2 that $AA^{\dagger}B=B$. Hence, $A\widetilde{X}B=A(A^{\dagger}B)=AA^{\dagger}B=B$. It follows from $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$ and Lemma 2.1 that there exists $Z^{*}\in\mathbb{B}(\mathscr{H})$ such that $B=ZA$. Hence,

[TABLE]

Therefore, $\widetilde{X}$ is a solution of the system $BXA=B=AXB$. ∎

Lemma 3.10**.**

[4, Theorem 4.2]** Let $A,B,C,D\in\mathbb{B}(\mathscr{H})$ and $A,B,M=B^{*}(I-A^{\dagger}A)$ have closed ranges. Then, the system $AX=C,\quad XB=D$ have a hermitian solution $X\in\mathbb{B}(\mathscr{H})$ if and only if

[TABLE]

and $AC^{*}$ and $B^{*}D$ are hermitian. In this case, the general hermitian solution is

[TABLE]

where $W\in\mathbb{B}(\mathscr{H})$ is hermitian and $s(T)=D^{*}-B^{*}A^{\dagger}C$ is the so-called Schur complement of the block matrix T=\left[\begin{array}[]{cc}A&C\\ B^{*}&D^{*}\end{array}\right].

Theorem 3.11**.**

Suppose that $A,B\in\mathbb{B}(\mathscr{H})$ have closed ranges. If $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and $B^{*}A^{\dagger}B,B{A^{\dagger}}^{*}B^{*}$ are hermitian, then the system $BXA=B=AXB$ has a hermitian solution.

Proof.

Replace $A,B,C,D$ in Lemma 3.10 by $B,B,BA^{\dagger},A^{\dagger}B$ to get

[TABLE]

and

[TABLE]

Using Lemma 3.10, the system $XB=A^{\dagger}B,BX=BA^{\dagger}$ has a hermitian solution, say, $\widetilde{X}$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A$ that $B=AP_{\overline{\mathscr{R}(B^{*})}}$ and $B=P_{\overline{\mathscr{R}(B)}}A$. Hence, $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$. It follows from $\mathscr{R}(B)\subseteq\mathscr{R}(A)$ and Lemma 3.2 that $AA^{\dagger}B=B$. Hence, $A\widetilde{X}B=A(A^{\dagger}B)=AA^{\dagger}B=B$. It follows from $\mathscr{R}(B^{*})\subseteq\mathscr{R}(A^{*})$ and Lemma 2.1 that there exists $Z\in\mathbb{B}(\mathscr{H})$ such that $B=ZA$. Hence,

[TABLE]

Therefore, $\widetilde{X}$ is a hermitian solution of the system $BXA=B=AXB$. ∎

4. $*$-order via other operator equations

Generally speaking, the inequality $PB\stackrel{{\scriptstyle*}}{{\leq}}B$ dose not hold for any $P\in\mathscr{P}(\mathscr{H})$ even if $\mathscr{R}(P)\subseteq\overline{\mathscr{R}(B)}$. In [2, Lemma 2.6], some conditions are mentioned which give a one-sided description of the relation $A\stackrel{{\scriptstyle*}}{{\leq}}B$ regarding (1.2).

The next result is known.

Proposition 4.1**.**

[2*, Proposition 2.6]** Let $B\in\mathbb{B}(\mathscr{H})$.

1. (a)

If $P\in\mathscr{OP}(\mathscr{H})$ and $\mathscr{R}(P)\subseteq\overline{\mathscr{R}(B)},$ then $PB\stackrel{{\scriptstyle*}}{{\leq}}B$ if and only if $PBB^{*}=BB^{*}P$. 2. (b)

If $Q\in\mathscr{OP}(\mathscr{H})$ and $\mathscr{R}(Q)\subseteq\overline{\mathscr{R}(B^{*})},$ then $BQ\stackrel{{\scriptstyle*}}{{\leq}}B$ if and only if $QB^{*}B=B^{*}BQ$.

In the following, we state a generalization of Proposition 4.1.

Proposition 4.2**.**

Let $B\in\mathbb{B}(\mathscr{H})$. If there exist $P,Q\in\mathscr{OP}(\mathscr{H})$ such that $\mathscr{R}(P)\subseteq\overline{\mathscr{R}(B)}$and $\mathscr{R}(Q)\subseteq\overline{\mathscr{R}(B^{*})},$ then $PBQ\stackrel{{\scriptstyle*}}{{\leq}}B$ if and only if $PBQB^{*}=BQB^{*}P$ and $QB^{*}PB=B^{*}PBQ$.

Proof.

$(\Longrightarrow)$: Let $PBQ\stackrel{{\scriptstyle*}}{{\leq}}B$. Applying (1.2), we get that

[TABLE]

and

[TABLE]

$(\Longleftarrow)$: Let $PBQB^{*}=BQB^{*}P$ and $QB^{*}PB=B^{*}PBQ$. Applying (1.2), we obtain that

[TABLE]

and

[TABLE]

∎

The next known theorem gives a characterization of the order $\stackrel{{\scriptstyle*}}{{\leq}}$.

Theorem 4.3**.**

[6, Theorem 2.3]** Let $A\in\mathbb{B}(\mathscr{H})$ and $C\in\mathscr{Q}(\mathscr{H})$. Then, $C\stackrel{{\scriptstyle*}}{{\leq}}A$ if and only if there exists $X\in\mathbb{B}(\mathscr{H})$ such that $A=C+(I-C^{*})X(I-C^{*})$.

In the following, we establish an analogue of Theorem 4.3 for generalized projections on a Hilbert space. Recall that an operator $A\in\mathbb{B}(\mathscr{H})$ is a generalized projection if $A^{2}=A^{*}$.

Lemma 4.4**.**

[14, Theorem A.2]** Let $A\in\mathbb{B}(\mathscr{H})$ be a generalized projection.Then, $A$ is a closed range operator and $A^{3}$ is an orthogonal projection on $\mathscr{R}(A)$. Moreover, $\mathscr{H}$ has decomposition

[TABLE]

and $A$ has the following matrix representation

[TABLE]

where the restriction $A_{1}=A|_{\mathscr{R}(A)}$ is unitary on $\mathscr{R}(A)$.

Theorem 4.5**.**

Let $A\in\mathbb{B}(\mathscr{H})$ and $B\in\mathscr{GP}(\mathscr{H})$. Then, $B\stackrel{{\scriptstyle*}}{{\leq}}A$ if and only if there exists $X\in\mathbb{B}(\mathscr{H})$ such that $A=B+(I-BB^{*})X(I-B^{*}B)$.

Proof.

$(\Longrightarrow)$: Let $B\in\mathscr{GP}(\mathscr{H})$ and $B\stackrel{{\scriptstyle*}}{{\leq}}A$. Employing Lemma 4.4, we infer that $B$ has closed range and $B^{3}=P_{\mathscr{R}(B)}$. It follows from (1) that

[TABLE]

Hence, $P_{\mathscr{R}(B)}=P_{\mathscr{R}(B^{*})}=BB^{*}=B^{*}B$. Therefore, $P_{\mathscr{N}(B)}=P_{\mathscr{N}(B^{*})}=I-BB^{*}=I-B^{*}B$. Applying Lemma 3.1(c), we get $A=B+P_{\mathscr{N}(B^{*})}AP_{\mathscr{N}(B)}$. Hence, $A=B+(I-BB^{*})A(I-B^{*}B)$.

$(\Longleftarrow)$: Let $X\in\mathbb{B}(\mathscr{H})$ be a solution of the equation $A=B+(I-BB^{*})X(I-B^{*}B)$. Since $B$ is a generalized projection, so $B^{*}BB^{*}=B^{*}$. Hence,

[TABLE]

and

[TABLE]

Therefore, $B\stackrel{{\scriptstyle*}}{{\leq}}A$ by (1.2). ∎

In the next result, we show that if $A$ is a generalized projection and $B\stackrel{{\scriptstyle*}}{{\leq}}A\stackrel{{\scriptstyle*}}{{\wedge}}A^{*}$, then $AA^{*}$ can be written as the sum of two idempotents.

Theorem 4.6**.**

Let $A\in\mathscr{GP}(\mathscr{H})$ and $B\in\mathbb{B}(\mathscr{H})$. If $B\stackrel{{\scriptstyle*}}{{\leq}}A\stackrel{{\scriptstyle*}}{{\wedge}}A^{*}$, then $B$ is an idempotent and there exist an idempotent $X$ such that $AA^{*}=B+X$ and $B^{*}X=XB^{*}=0$.

Proof.

Let $B\stackrel{{\scriptstyle*}}{{\leq}}A\stackrel{{\scriptstyle*}}{{\wedge}}A^{*}$. It follows from the assumption $A^{2}=A^{*}$ and Lemma 3.1(d) that

[TABLE]

Using Lemma 3.1, we get that

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Let $X=AA^{*}-B$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}A\stackrel{{\scriptstyle*}}{{\wedge}}A^{*}$ that

[TABLE]

Hence, $X$ is an idempotent. Applying (1.2), we have

[TABLE]

and

[TABLE]

∎

Lemma 4.7**.**

Let $A\in\mathscr{Q}(\mathscr{H})$ and $B\in\mathbb{B}(\mathscr{H})$. Then, $B\stackrel{{\scriptstyle*}}{{\leq}}A$ if and only if $B$ is an idempotent and there exists an idempotent $X$ such that $A=B+X$ and $B^{*}X=XB^{*}=0$.

Proof.

$(\Longrightarrow)$: Let $B\stackrel{{\scriptstyle*}}{{\leq}}A$. It follows from the assumption $A^{2}=A$ and Lemma 3.1(d) that

[TABLE]

Utilizing Lemma 3.1(d), we obtain that

[TABLE]

and

[TABLE]

Hence, $X=A-B$ is an idempotent and $B^{*}X=B^{*}(A-B)=0$ and $XB^{*}=(A-B)B^{*}=0$.

$(\Longleftarrow)$: Let $A=B+X$ and $B^{*}X=XB^{*}=0$ for some idempotent $X$. Then, $B^{*}(A-B)=B^{*}X=0$ and $(A-B)B^{*}=XB^{*}=0$. Therefore, $B\stackrel{{\scriptstyle*}}{{\leq}}A$ by (1.2). ∎

Corollary 4.8**.**

Let $A\in\mathscr{GP}(\mathscr{H})$ and $B\in\mathbb{B}(\mathscr{H})$. Then, $B\stackrel{{\scriptstyle*}}{{\leq}}AA^{*}$ if and only if $B$ is an idempotent and there exists an idempotent $X$ such that $AA^{*}=B+X$ and $B^{*}X=XB^{*}=0$.

Proof.

Let $A\in\mathscr{GP}(\mathscr{H})$. Then, $(AA^{*})^{2}=AA^{*}AA^{*}=AA^{*}$. Hence, $AA^{*}$ is an idempotent. Now apply Lemma 4.7. ∎

We end our work with the following result.

Proposition 4.9**.**

Let $A\in\mathbb{B}(\mathscr{H})$ and $C\in\mathscr{GP}(\mathscr{H})$. Then, $B\in\mathbb{B}(\mathscr{H})$ is common $*-$ lower bound of $A$ and $CC^{*}$ if and only if $B$ is an idempotent and there exist $X,Y\in\mathbb{B}(\mathscr{H})$ such that

[TABLE]

where $B^{*}Y=YB^{*}=0$.

Proof.

$(\Longrightarrow)$: If $B$ be a common $*-$ lower bound of $A$ and $CC^{*}$, then $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and $B\stackrel{{\scriptstyle*}}{{\leq}}CC^{*}$. It follows from the assumption $B\stackrel{{\scriptstyle*}}{{\leq}}CC^{*}$ and Lemma 4.7 that $B$ is an idempotent and there exists an idempotent $Y\in\mathbb{B}(\mathscr{H})$ such that $CC^{*}=B+R$, where $B^{*}R=RB^{*}=0$. Since $B$ is an idempotent and $B\stackrel{{\scriptstyle*}}{{\leq}}A$, by Theorem 4.3, there exists $S\in\mathbb{B}(\mathscr{H})$ such that $A=B+(I-B^{*})S(I-B^{*})$.

$(\Longleftarrow)$: If there exists an idempotent $Y$ such that $CC^{*}=B+Y$ with $B^{*}Y=0$ and $YB^{*}=0$, then $B\stackrel{{\scriptstyle*}}{{\leq}}CC^{*}$. The assumption $A=B+(I-B^{*})S(I-B^{*})$ and the fact that $B$ is an idempotent yield $B^{*}(A-B)=0$ and $(A-B)B^{*}=0$. Hence, $B\stackrel{{\scriptstyle*}}{{\leq}}A$ and $B$ is a common $*-$ lower bound of $A$ and $CC^{*}$. ∎