The Weighted Generalized Moore-Penrose Inverse in Banach *-Algebras

1. Introduction

A Banach algebra is called a Banach *-algebra if there exists an involution $\ast :x\to x^{\ast }$ satisfying ${(x+y)}^{\ast }=x^{\ast }+y^{\ast },{\left(\lambda x\right)}^{\ast }=\overline{\lambda }{x}^{\ast },{\left(xy\right)}^{\ast }=y^{\ast }{x}^{\ast },{\left(x^{\ast }\right)}^{\ast }=x$. An element $a\in \mathcal{A}$ has Moore-Penrose inverse if there exist $x\in \mathcal{A}$ such that

$$axa=a,xax=x,{\left(ax\right)}^{\ast }=ax,{\left(xa\right)}^{\ast }=xa.$$

The preceding x is unique if it exists, and we denote it by $a^{†}$. The set of all Moore-Penrose invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}^{†}$. Moore-Penrose inverse is extensively studied in matrix and operator theory (see [9,11,12,13,15,16,23]).

Recently, Stojanović and Mosić introduced generalized Moore-Penrose inverse for Hilbert operators (see [18]). Afterwards, Mosić further studied generalized Moore-Penrose inverse with weights for an Hilbert operator (see [12]). Many properties of weighted generalized Moore-Penrose inverse for an Hilbert operator were presented by using the Hilbert operator technique. The generalized Moore-Penrose inverse for complex matrices has been investigated in [6,21,24].

The motivation of this paper is to introduce the weighted generalized Moore-Penrose inverse within the framework of a Banach *-algebra. Recall that an element $a\in \mathcal{A}$ has w-weighted core inverse if there exists an $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{\ast }=wawx,xw{\left(aw\right)}^2=aw.$$

If such x is unique if it exits, and we denote it by $a^{\#,w}$. Many elementary properties of the w-weighted core inverse are studied in [4]. In Section 2, we introduce and study a new generalized inverse based on the w-weighted core inverse. Some equivalent characterizations of such new generalized inverse are presented. In Section 3, we introduce the generalized weighted Moore-Penrose inverse by combing the w-weighted MP-inverse and quasinilpotents. This avoid to use the Hilbert operator technique and extend such kind of generalized inverse from Hilbert operator to the general Banach element. We characterize this new generalized inverse by employing the weighted MP-inverse and the generalized core-EP inverse. Finally, in Section 4, the connections with specific outer inverses are elucidated. Consequently, the attributes of the associated generalized inverses for complex matrices and bounded linear operators on Hilbert spaces are extended to the broader context of Banach algebras.

Throughout the paper, all Banach *-algebras are complex with an identity. Let ${\mathcal{A}}^{\#,w}$ denote the set of all w-weighted core invertible elements in $\mathcal{A}$. Let $a\in \mathcal{A}$. Set $im\left(a\right)=\left\{ax\right|x\in \mathcal{A}\}$ and $ker\left(a\right)=\{x\in \mathcal{A}|ax=0\}$. We use $p_{im\left(a\right)}$ to denote the projection p such that $im\left(p\right)=im\left(a\right)$.

2. Weighted MP-Core Inverse

The purpose of this section is to introduce a new generalized inverse with weights which is a natural generalization of Moore-Penrose inverse for a complex matrix. Our starting points is the following.

Definition 1.

An element $a\in \mathcal{A}$ has w-weighted MP-core inverse provided that $a\in {\mathcal{A}}^{\#,w}$ and $a^{\#,w}waw\in {\mathcal{A}}^{†}$. We denote the w-weighted MP-core inverse of a by

$$a^{\square ,w}={\left[a^{\#,w}waw\right]}^{†}{a}^{\#,w}.$$

Let ${\mathcal{A}}^{\square ,w}$ denote the set of all w-weighted MP invertible elements in $\mathcal{A}$. We now derive

Theorem 1.

Let $a,x\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{\square ,w}$.

(2)

There exists $x\in \mathcal{A}$ such that $xwawx=x,{\left(wawx\right)}^{\ast }=wawx,xwaw\in {\mathcal{A}}^{†},x\mathcal{A}=\left(aw\right)\mathcal{A},x^{\ast }\mathcal{A}=\left(waw\right)\mathcal{A}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ It is easy to verify that

$$xwawx=xwawa{\left(wx\right)}^2=xw{\left(aw\right)}^2\left(xwx\right)=awxwx=a{\left(wx\right)}^2=x.$$

Moreover, $x=a{\left(wx\right)}^2=\left(aw\right)xwx\in \left(aw\right)\mathcal{A}$. $aw=xw{\left(aw\right)}^2\in x\mathcal{A}$. Hence, $x\mathcal{A}=\left(aw\right)\mathcal{A}$. On the other hand, $x^{\ast }={\left[x\left(wawx\right)\right]}^{\ast }=wawx{x}^{\ast }=wawa{\left(wx\right)}^2\in \left(waw\right)\mathcal{A}$. Then $x^{\ast }\mathcal{A}\subseteq \left(waw\right)\mathcal{A}$. Moreover, we have

$$\begin{array}{ccc}\hfill waw& =\hfill & w\left(aw\right)=wxw{\left(aw\right)}^2=wa{\left(wx\right)}^{2}w{\left(aw\right)}^2\hfill \\ & =\hfill & \left(wawx\right)\left[wxw{\left(aw\right)}^2\right]={\left(wawx\right)}^{\ast }\left[wxw{\left(aw\right)}^2\right]\hfill \\ & \in \hfill & x^{\ast }\mathcal{A}.\hfill \end{array}$$

Therefore $x^{\ast }\mathcal{A}=\left(waw\right)\mathcal{A}$.

$\left(2\right)\Rightarrow \left(1\right)$ Since $1-xwaw{\in }^0\left(x\right)$, we see that $1-xwaw{\in }^0\left(aw\right)$. Thus, $aw=xw{\left(aw\right)}^2$. Moreover, we have ${\left(xwawx\right)}^{\ast }=x^{\ast }$, and so $1-wxwa{\in }^0\left(x^{\ast }\right)$; hence, $1-wawx{\in }^0\left(waw\right)$. This implies that $(w-wawx)aw=0$, and then $w-wawx{\in }^0\left(aw\right){=}^0\left(x\right)$. It follows that $wx=wa{\left(wx\right)}^2$. Therefore we deduce that

$$x=xwawx=\left(xwa\right)\left[wa{\left(wx\right)}^2\right]=x{\left(wa\right)}^2{\left(wx\right)}^2=\left[xw{\left(aw\right)}^2\right]xwx=a{\left(wx\right)}^2,$$

as required. □

Corollary 1.

Let $a,x\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{\square ,w}$.

(2)

$a{\left(wx\right)}^2=x=xwawx,{\left(wawx\right)}^{\ast }=wawx,\left(waw\right)x\left(waw\right)=waw,xw{\left(aw\right)}^2$$=aw,xwaw\in {\mathcal{A}}^{†}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.2, we have $x=xwawx$. Then we derive that

$$\begin{array}{ccc}\hfill \left(waw\right)x\left(waw\right)& =\hfill & wawxw\left(aw\right)=waw{\left(xw\right)}^2{\left(aw\right)}^2\hfill \\ & =\hfill & w\left[a{\left(wx\right)}^2\right]w{\left(aw\right)}^2=w\left(xw\right){\left(aw\right)}^2=waw,\hfill \end{array}$$

as required

$\left(2\right)\Rightarrow \left(1\right)$ This is trivial. □

An element $a\in \mathcal{A}$ has group inverse if there exist $x\in \mathcal{A}$ such that

$$axa=a,xax=x,ax=xa.$$

The preceding x is unique if it exists, and we denote it by $a^{\#}$. The set of all group invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}^{\#}$.

Theorem 2.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{\square ,w}$.

(2)

$aw\in {\mathcal{A}}^{\#}$ and there exists $x\in \mathcal{A}$ such that ${\left(wawx\right)}^{\ast }=wawx,$$xwaw=aw{\left(aw\right)}^{\#}\in {\mathcal{A}}^{†}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $x=a^{\square ,w}$. Then

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{\ast }=wawx,xw{\left(aw\right)}^2=aw,xwaw\in {\mathcal{A}}^{†}.$$

In view of [8] (Theorem 2.3), $wa\in {\mathcal{A}}^d$. By using Cline’s formula, $aw\in {\mathcal{A}}^d$. Since $\mathcal{A}aw=\mathcal{A}{\left(aw\right)}^2$, we can find some $z\in \mathcal{A}$ such that $aw=z{\left(aw\right)}^2$. For any $n\in \mathbb{N}$, we check that

$$\begin{array}{ccc}\hfill \left|\right|aw-{\left(aw\right)}^2{\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & \left|\right|z^{n-1}{\left(aw\right)}^n-z^{n-1}{\left(aw\right)}^{n+1}{\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & \le \hfill & {\left|\right|z\left|\right|}^{1-{\displaystyle \frac{1}{n}}}\left|\right|{[aw-{\left(aw\right)}^2{\left(aw\right)}^d]}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

This implies that $\underset{n\to \infty }{lim}\left|\right|aw-{\left(aw\right)}^2{\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ and then $aw={\left(aw\right)}^2{\left(aw\right)}^d\in {\left(aw\right)}^2\mathcal{A}\bigcap \mathcal{A}{\left(aw\right)}^2$. Thus $aw\in {\mathcal{A}}^{\#}$. Since $xw{\left(aw\right)}^2=aw$, we deduce that $xw{\left(aw\right)}^2{\left(aw\right)}^{\#}=aw{\left(aw\right)}^{\#}$. Therefore $xwaw=aw{\left(aw\right)}^{\#}\in {\mathcal{A}}^{†}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that ${\left(wawx\right)}^{\ast }=wawx,$$xwaw=aw{\left(aw\right)}^{\#}\in {\mathcal{A}}^{†}$. Hence, $xw{\left(aw\right)}^2=\left(aw\right){\left(aw\right)}^{\#}\left(aw\right)$, as required. □

Set $a^{\square }=a^{\square ,1}$ and ${\mathcal{A}}^{\square }={\mathcal{A}}^{\square ,1}$. We now derive

Corollary 2.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{\square }$.

(2)

$a\in {\mathcal{A}}^{\#}$ and there exists $x\in \mathcal{A}$ such that ${\left(ax\right)}^{\ast }=ax,xa=a{a}^{\#}\in {\mathcal{A}}^{†}$

Proof. 

This is obvious by Theorem 2.4. □

Theorem 3.

Let $a,w\in \mathcal{A}$. If there exists $x\in \mathcal{A}$ such that

$${\left(aw\right)}^{2}x=a,awx=xwa,{\left(wawx\right)}^{\ast }=wawx,{\left(awxw\right)}^{\ast }=awxw$$

then $a\in {\mathcal{A}}^{\square ,w}$.

Proof. 

Let $z=a{\left(wx\right)}^2$. Since $awx=xwa$, we check that

$$\begin{array}{ccc}\hfill awzwa& =\hfill & awa{\left(wx\right)}^{2}wa=\left[{\left(aw\right)}^{2}x\right]wxwa=awxwa={\left(aw\right)}^{2}x=a,\hfill \\ \hfill zwawz& =\hfill & a{\left(wx\right)}^{2}wawa{\left(wx\right)}^2=a{\left(wx\right)}^{2}wxw\left[{\left(aw\right)}^{2}x\right]\hfill \\ & =\hfill & a{\left(wx\right)}^{2}wxwa=\left[{\left(aw\right)}^{2}x\right]wxwx=a{\left(wx\right)}^2=z,\hfill \\ \hfill awz& =\hfill & awa{\left(wx\right)}^2=\left[{\left(aw\right)}^{2}x\right]wx=awx\hfill \\ & =\hfill & \left[{\left(aw\right)}^{2}x\right]wx=a{\left(wx\right)}^{2}wa=zwa.\hfill \end{array}$$

Hence $aw\in {\mathcal{A}}^{\#}$ and ${\left(aw\right)}^{\#}=zw$. Hence, $aw{\left(aw\right)}^{\#}=awzw=awa{\left(wx\right)}^{2}w=awawxwxw=\left[{\left(aw\right)}^{2}x\right]wxw=awxw=xwaw$. Since ${\left(xwaw\right)}^2=xw\left[{\left(aw\right)}^{2}x\right]w$$=xwaw$ and ${\left(awxw\right)}^{\ast }=awxw$, we see that ${\left(awxw\right)}^{†}=awxw$. Hence, $xwaw=aw{\left(aw\right)}^{\#}\in {\mathcal{A}}^{†}$. Therefore we complete the proof by Theorem 2–4. □

Recall that an element a in $\mathcal{A}$ is an EP element if there exists an $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{\ast }=ax=xa$ (see [19]).

Corollary 3.

Let $a\in \mathcal{A}$ be an EP element. Then $a\in {\mathcal{A}}^{\square }$. In this case, $a^{\square }=a^{\#}$.

Proof. 

This is an immediate consequence of Theorem 2.6. □

3. Weighted Generalized Moore-Penrose Inverse

The aim of this section is to introduce the notion of the generalized weighted core inverse in a Banach *-algebra. We begin with

Definition 2.

An element $a\in \mathcal{A}$ has generalized w-weighted Moore-Penrose decomposition if there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,ywx=0,x\in {\mathcal{A}}^{\square ,w},y\in {\mathcal{A}}^{qnil}.$$

Here,

$${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{{\displaystyle \frac{1}{n}}}=0\}.$$

It is well established that $x\in {\mathcal{A}}^{qnil}$ if and only if $1+\lambda x\in \mathcal{A}$ is invertible for any $\lambda \in C$. Let ${\mathcal{A}}^{⊠,w}$ denote the set of all generalized w-weighted Moore-Penrose invertible elements in $\mathcal{A}$.

Recall that an element $a\in \mathcal{A}$ has w-weighted generalized core-EP if there exist $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{\ast }=wawx,\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$$

(see [4]). The preceding x is unique if exists, and we denote it by $a^{ⓓ,w}$. Let ${\mathcal{A}}^{ⓓ,w}$ denote the set of all w-weighted generalized core-EP invertible elements in $\mathcal{A}$. We are ready to prove:

Theorem 4.

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{⊠,w}$.

(2)

$a\in {\mathcal{A}}^{ⓓ,w}$ and $a^{ⓓ,w}waw\in {\mathcal{A}}^{†}$.

In this case,

$$a^{⊠,w}={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}^{⊠,w}$, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,ywx=0,x\in {\mathcal{A}}^{\square ,w},y\in {\mathcal{A}}^{qnil}.$$

Then $x\in {\mathcal{A}}^{\#,w}$. In light of [4] (Theorem 3.1), $a\in {\mathcal{A}}^{ⓓ,w}$ and $a^{ⓓ,w}=x^{\#,w}$.

Additionally, $x^{\#,w}wxw\in {\mathcal{A}}^{†}$. It is easy to verify that

$$\begin{array}{ccc}\hfill a^{ⓓ,w}waw& =\hfill & x^{\#,w}w(x+y)w\hfill \\ & =\hfill & x^{\#,w}{\left(wxw{x}^{\#,w}\right)}^{\ast }w(x+y)w\hfill \\ & =\hfill & x^{\#,w}{\left(w{x}^{\#,w}\right)}^{\ast }{\left(wx\right)}^{\ast }w(x+y)w\hfill \\ & =\hfill & x^{\#,w}wxw\hfill \\ & \in \hfill & {\mathcal{A}}^{†}.\hfill \end{array}$$

Moreover, we check that

$$\begin{array}{ccc}\hfill a^{⊠,w}& =\hfill & {\left[x^{\#,w}wxw\right]}^{†}{x}^{\#,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w},\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}^{ⓓ,w}$, by virtue of [4] (Theorem 3.1), there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,{\left(wx\right)}^{\ast }\left(wy\right)=0,ywx=0,x\in {\mathcal{A}}^{\#,w},y\in {\mathcal{A}}^{qnil}.$$

In this case, $a^{ⓓ,w}=x^{\#,w}$. Moreover, we have

$$\begin{array}{ccc}\hfill x^{\#,w}wxw& =\hfill & x^{\#,w}w(x+y)w\hfill \\ & =\hfill & a^{ⓓ,w}waw\hfill \\ & \in \hfill & {\mathcal{A}}^{†}.\hfill \end{array}$$

Therefore $x\in {\mathcal{A}}^{\square ,w}$. Accordingly, $a\in {\mathcal{A}}^{⊠,w}$. □

Corollary 4.

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{⊠,w}$.

(2)

The system of conditions

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{\ast }=wawx,xwaw\in {\mathcal{A}}^{†},\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$$

is consistent and it has the unique solution.

In this case, $a^{⊠,w}={\left(xwaw\right)}^{†}x.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a^{ⓓ,w}$. In view of Theorem 3.2, we have $xwaw\in {\mathcal{A}}^{†}$. By virtue of [4], $a{\left(wx\right)}^2=x,{\left(wawx\right)}^{\ast }=wawx,{lim}_{n\to \infty }{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ as required.

$\left(2\right)\Rightarrow \left(1\right)$ In light of [4], $a^{ⓓ,w}=x$. Moreover, we have $xwaw=a^{ⓓ,w}waw\in {\mathcal{A}}^{†}$, as desired. □

Let $A^{\diamond }$ denote the generalized Moore-Penrose inverse of a bounded linear operator A over an Hilbert space H (see [18]). We deduce that

Corollary 5.

Let $A\in \mathcal{B}{\left(H\right)}^d$ and $X\in \mathcal{B}\left(H\right)$. Then the following are equivalent:

(1)

$A^{\diamond }=X$.

(2)

There exist $Y,Z\in \mathcal{B}\left(H\right)$ such that $A=Z+Y$, where $YZ=0,Z\in \mathcal{B}{\left(H\right)}^{\square }$ and $Y\in \mathcal{B}{\left(H\right)}^{qnil}$.

In this case, $X=Z^{\square }.$

Proof. 

This is obvious by Theorem 3.2 and [18]. □

Consider the system given by

$$\begin{array}{c}xwawx=x,xwa={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wa,\\ awx=aw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(2.1\right)\end{array}$$

Theorem 5.

If the system $\left(2.1\right)$ of equations has a solution then it is unique.

Proof. 

Assume that $x_1,x_2$ satisfy $\left(2.1\right)$. Then

$$\begin{array}{c}{x}_{i}waw{x}_i=x_i,x_{i}wa={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wa,\\ aw{x}_i=aw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\end{array}$$

for $i=1,2$. Therefore

$$\begin{array}{ccc}\hfill x_1& =\hfill & \left(x_{1}wa\right)w{x}_1={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wa\left(w{x}_1\right)\hfill \\ & =\hfill & \left(x_{2}wa\right)\left(w{x}_1\right)=x_{2}w\left(aw{x}_1\right)=x_{2}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & \left(x_{2}w\right)aw{x}_2=x_2,\hfill \end{array}$$

as required. □

Corollary 6.

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a^{⊠,w}=x$.

(2)

The system of equations

$$\begin{array}{c}xwawx=x,xwa={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wa,\\ awx=aw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\end{array}$$

is consistent and it has the unique solution x.

In this case, $a^{⊠,w}=x.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Taking $x={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$. Then

$$\begin{array}{ccc}\hfill xwawx& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & x,\hfill \\ \hfill xwa& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wa,\hfill \\ \hfill awx& =\hfill & aw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}.\hfill \end{array}$$

By virtue of Theorem 3.5, x is the unique solution of the preceding equations, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By the argument above, we have $x={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$. Therefore $a^{⊠,w}=x$, as asserted. □

We come now to consider the following equation in $\mathcal{A}$:

$$a^{ⓓ,w}wawx=a^{ⓓ,w}b\left(2.2\right)$$

where $a,w,b\in \mathcal{A}$.

Theorem 6.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then the equation $\left(2.2\right)$ is consistent and its general solution is

$$x=a^{⊠,w}b+[1-a^{⊠,w}waw]y,$$

where $y\in \mathcal{A}$ is arbitrary.

Proof. 

Let $x=a^{⊠,w}b+[1-a^{⊠,w}waw]y,$ where $y\in \mathcal{A}$. Then

$$\begin{array}{ccc}\hfill a^{ⓓ,w}waw{a}^{⊠,w}& =\hfill & a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{dag}{a}^{ⓓ,w}\hfill \\ & =\hfill & a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{dag}\left[a^{ⓓ,w}waw\right]a^{ⓓ,w}\hfill \\ & =\hfill & a^{ⓓ,w}\hfill \end{array}$$

Thus we verify that

$$\begin{array}{cc}& a^{ⓓ,w}wawx\hfill \\ \hfill =& a^{ⓓ,w}waw[a^{⊠,w}b+[1-a^{⊠,w}waw]y]\hfill \\ \hfill =& a^{ⓓ,w}b+[a^{ⓓ,w}waw-a^{ⓓ,w}waw{a}^{⊠,w}waw]y\hfill \\ \hfill =& a^{ⓓ,w}b.\hfill \end{array}$$

Let x be the solution of the Eq. (2.2). Then

$$a^{ⓓ,w}wawx=a^{ⓓ,w}b.$$

Hence,

$$\begin{array}{ccc}\hfill a^{⊠,w}wawx& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}wawx\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}b\hfill \\ & =\hfill & a^{⊠,w}b.\hfill \end{array}$$

Accordingly,

$$x=a^{⊠,w}b+[1-a^{⊠,w}waw]x,$$

as required. □

Corollary 7.

Let $a\in {\mathcal{A}}^{⊠,w}$. If x is the solution of the Eq. (2.2) and $im\left(x\right)\subseteq im{\left(a^{ⓓ,w}waw\right)}^{\ast }$, then

$$x=a^{⊠,w}b.$$

Proof. 

In view of Theorem 3.7, $x=a^{⊠,w}b$ is the solution of the Eq. (2.2). Let $x_1,x_2$ be the solution of the Eq. (2.2) and $im\left(x_1\right),im\left(x_2\right)\subseteq im{\left(a^{ⓓ,w}waw\right)}^{\ast }$. Then we have

$$z:=x_1-x_2\in ker\left[a^{ⓓ,w}waw\right]\bigcap im{\left(a^{ⓓ,w}waw\right)}^{\ast }.$$

Write $z=(a^{ⓓ,w}waw{)}^{\ast }t$ for some $t\in \mathcal{A}$. Then

$$a^{ⓓ,w}wawz=a^{ⓓ,w}waw(a^{ⓓ,w}waw{)}^{\ast }t=0,$$

and so

$${\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw(a^{ⓓ,w}waw{)}^{\ast }t=0.$$

Hence, we have

$${\left({\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw\right)}^{\ast }(a^{ⓓ,w}waw{)}^{\ast }t=0.$$

This implies that $(a^{ⓓ,w}waw{)}^{\ast }t=0.$ Therefore $x_1=x_2$. That is $a^{⊠,w}b$ is the unique solution of $\left(2.2\right)$ in $im{\left(a^{ⓓ,w}waw\right)}^{\ast }$, as desired. □

Consider the following matrix equation:

$$A^{Ⓓ,W}WAWX=A^{Ⓓ,W}B\left(2.3\right)$$

where $A\in C^{q\times n},W\in C^{n\times q},B\in C^{n\times p}$ and $m\in N$.

Corollary 8.$\left(1\right)$ The general solution of the Eq. (2.3) is

$$X=A^{⊠,w}B+[I-A^{⊠,w}WAW]Y,$$

where $Y\in C^{q\times p}$ is arbitrary.

Proof. 

This is obvious by Theorem 3.7. □

4. Connections Between Weighted Generalized Core-EP Inverses

An element $a\in \mathcal{A}$ has weighted generalized Drazin inverse x if there exists unique $x\in \mathcal{A}$ such that

$$awx=xwa,xwawx=xanda-awxwa\in {\mathcal{A}}^{qnil}.$$

We denote x by $a^{d,w}$ (see [14]). In this section, we establish the relations between weighted generalized core-EP inverse and weighted generalized Drazin inverse for an element in a Banach *-algebra.

Theorem 7.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{⊠,w}$.

(2)

$a\in {\mathcal{A}}^{d,w}$ and there exists $x\in \mathcal{A}$ such that

$$xwawx=x,xwaw\in {\mathcal{A}}^{†},im\left(xw\right)=im{\left(aw\right)}^{d}andim{\left(wx\right)}^{\ast }=im{\left(wa\right)}^d.$$

In this case, $a^{⊠,w}={\left[xwaw\right]}^{†}a{\left(wx\right)}^2.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a^{ⓓ,w}.$ Then $x=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2$ by [3]. Hence, $wx=wa{\left[{\left(wa\right)}^{ⓓ}\right]}^2={\left(wa\right)}^{ⓓ}$. In view of [1] (Theorem 1.2), we have

$$wxwawx=wx,im\left(wx\right)=im{\left(wa\right)}^{d}andim{\left(wx\right)}^{\ast }=im{\left(wa\right)}^d.$$

By virtue of [1] (Theorem 1.2), we have

$$xwawx=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2{\left(wa\right)}^2{\left[{\left(wa\right)}^{ⓓ}\right]}^2=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2\left(wa\right){\left(wa\right)}^{ⓓ}=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2=x.$$

By using [1] (Theorem 1.2), we can find a $q\in \mathcal{A}$ such that

$$xw=a{\left[{\left(wa\right)}^{ⓓ}\right]}^{2}w=\left[a{\left(wa\right)}^{ⓓ}\right]\left[{\left(wa\right)}^{ⓓ}w\right]=a{\left(wa\right)}^{d}q=aw{\left[{\left(aw\right)}^d\right]}^{2}aq,$$

and so $im\left(xw\right)\subseteq im{\left(aw\right)}^d.$ In view of [1], ${\left(wa\right)}^d\in {\mathcal{A}}^{(1,3)}$ and ${\left(wa\right)}^{ⓓ}$$={\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}$. Then we check that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^d& =\hfill & a{\left[{\left(wa\right)}^d\right]}^{2}w=a{\left(wa\right)}^d{\left(wa\right)}^d{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left(wa\right)}^{d}w\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left(wa\right)}^{d}w\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left[{\left(wa\right)}^d\right]}^2\left(wa\right)w\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left[wa\right)}^d{\left(wa\right)}^d{\left({\left(wa\right)}^d\right)}^{(1,3)}{\left(wa\right)}^{d}waw\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left[{\left(wa\right)}^d\right]}^2{\left[{\left(wa\right)}^d\right]}^{(1,3)}{\left(wa\right)}^{d}waw\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^{ⓓ}\right]}^2{\left(wa\right)}^{d}waw\hfill \\ & =\hfill & xwa{\left(wa\right)}^{d}w.\hfill \end{array}$$

Hence, $im{\left(aw\right)}^d\subseteq im\left(xw\right)$. Therefore $im\left(xw\right)=im{\left(aw\right)}^d$, as desired.

By hypothesis, $xwaw=a^{ⓓ,w}waw\in {\mathcal{A}}^{†}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that

$$xwawx=x,xwaw\in {\mathcal{A}}^{†},im\left(xw\right)=im{\left(aw\right)}^{d}andim{\left(wx\right)}^{\ast }=im{\left(wa\right)}^d.$$

Then $xw={\left(aw\right)}^{d}y$ for some $y\in \mathcal{A}$. Hence $wx=wxwawx=w\left({\left(aw\right)}^{d}y\right)awx=w\left(xw\right)awx=wa{\left[{\left(wa\right)}^d\right]}^{2}wyawx$. This implies that $wx\mathcal{A}\subseteq {\left(wa\right)}^d\mathcal{A}$. On the other hand, we can find some $z\in \mathcal{A}$ such that ${\left(wa\right)}^d=w{\left[{\left(aw\right)}^d\right]}^{2}a\in wxw\mathcal{A}{\left(aw\right)}^{d}a$, and then ${\left(wa\right)}^d\mathcal{A}\subseteq \left(wx\right)\mathcal{A}$. Therefore $im\left(wx\right)=im{\left(wa\right)}^d$. Obviously, $wx\left(wa\right)wx=wx$. According to [1], $wa\in {\mathcal{A}}^{ⓓ}$ and $wx={\left(wa\right)}^{ⓓ}$. In light of[3], $a\in {\mathcal{A}}^{ⓓ,w}.$ Moreover, $a^{ⓓ,w}=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2=a{\left(wx\right)}^2,$ as desired.

$$\begin{array}{ccc}\hfill a^{ⓓ,w}waw& =\hfill & \left(awxw\right)x\left(waw\right)=\left(awxw\right)\left(xwawx\right)\left(waw\right)\hfill \\ & =\hfill & \left(awxw\right)xw\left(awxwaw\right)=\left(awxw\right){\left(aw\right)}^2{\left[{\left(aw\right)}^d\right]}^{3}y\left(awxwaw\right)\hfill \\ & =\hfill & {\left(aw\right)}^{d}y\left(awxwaw\right)=\left(xwawx\right)waw=xwaw\hfill \\ & \in \hfill & {\mathcal{A}}^{†},\hfill \end{array}$$

and therefore $a^{ⓓ,w}waw\in {\mathcal{A}}^{†}.$ Therefore $a\in {\mathcal{A}}^{ⓓ,w}$. In this case,

$$\begin{array}{ccc}\hfill a^{⊠,w}& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w},\hfill \\ & =\hfill & {\left[xwaw\right]}^{†}a{\left(wx\right)}^2,\hfill \end{array}$$

as asserted. □

Let p and q are projections in $\mathcal{A}$. We note that $p=q$ if and only if $im\left(p\right)=im\left(q\right)$. We now derive

Corollary 9.

Let $a\in {\mathcal{A}}^{ⓓ,w}$ and $x\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{⊠,w}$.

(2)

The system of conditions

$$wawx=p_{im{\left(wa\right)}^d},xwaw\in {\mathcal{A}}^{†},im\left(x\right)\subseteq im{\left(aw\right)}^d$$

is consistent and it has the unique solution x.

Proof. 

Let $x=a^{ⓓ,w}$. Then $x=a{\left[{\left(wa\right)}^{ⓓ}\right]}^2.$ Then $wawx=wawa{\left[{\left(wa\right)}^{ⓓ}\right]}^2=wa{\left(wa\right)}^{ⓓ}$. Set $p=\left(wa\right)\left(wx\right)$. Then $p^2=p=p^{\ast }$, i.e., $p\in \mathcal{A}$ is a projection. Clearly, $im\left(p\right)=im{\left(wa\right)}^d$; whence, $wawx=p_{im{\left(wa\right)}^d}$. In light Theorem 4.1, we have $im\left(x\right)\subseteq im{\left(aw\right)}^d$. As in the proof in Theorem 4.1, we check that $xwaw\in {\mathcal{A}}^{†}$, as required.

$\left(1\right)\Rightarrow \left(2\right)$ Assume that there exists $y\in \mathcal{A}$ such that

$$wawy=p_{im{\left(wa\right)}^d},im\left(y\right)\subseteq im{\left(aw\right)}^d.$$

Then we check that

$$waw(x-y)=0,$$

and so $im(x-y)\subseteq ker\left(waw\right)\subseteq ker{\left(aw\right)}^d$. Since $im\left(x\right),im\left(y\right)\subseteq im{\left(aw\right)}^d$, we have $im(x-y)\subseteq im{\left(aw\right)}^d$. Therefore $im(x-y)\subseteq ker{\left(aw\right)}^d\bigcap im{\left(aw\right)}^d=0$, and so $x=y$. The uniqueness is proved.

$\left(2\right)\Rightarrow \left(1\right)$ By the preceding discussion, the system of conditions

$$wawx=p_{im{\left(wa\right)}^d},im\left(x\right)\subseteq im{\left(aw\right)}^d$$

has the solution $a^{ⓓ,w}$. Thus, $a^{ⓓ,w}=x$ by the uniqueness. Therefore the result follows by Theorem 4.1. □

We are ready to prove:

Theorem 8.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then the following are equivalent:

(1)

$a^{⊠,w}=x$.

(2)

${\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(wawx\right)=x,wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$.

(3)

$\left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}=x,xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

(4)

${\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(wawx\right)=x,a^{ⓓ,w}\left(wawx\right)=a^{ⓓ,w}$.

(5)

$\left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}=x,\left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}={\left[a^{ⓓ,w}waw\right]}^{†}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a^{⊠,w}=x$, we see that ${\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}=x$. Moreover, we have $xwawx=x$, and then ${\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(wawx\right)=x$.

In view of Corollary 4.2, $awx=aw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$. Hence, $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$.

$\left(2\right)\Rightarrow \left(4\right)$ By hypothesis, $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$. Then

$$\begin{array}{ccc}\hfill a^{ⓓ,w}\left(wawx\right)& =\hfill & a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]\hfill \\ & =\hfill & \left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)a^{ⓓ,w}\hfill \\ & =\hfill & \left[a^{ⓓ,w}waw\right]a^{ⓓ,w}\hfill \\ & =\hfill & a^{ⓓ,w}.\hfill \end{array}$$

as required.

$\left(4\right)\Rightarrow \left(1\right)$ By hypothesis, we have

$$\begin{array}{ccc}\hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(wawx\right)\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}\left(wawx\right)\right]\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & a^{⊠,w},\hfill \end{array}$$

as required. □

Corollary 10.

Let $a\in {\mathcal{A}}^{⊠}$. Then the following are equivalent:

(1)

$a^{⊠}=x$.

(2)

${\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}a=x,ax=a{\left[a^{ⓓ,w}a\right]}^{†}{a}^{ⓓ}$.

(3)

$xa{\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}=x,xa={\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}a$.

(4)

${\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}ax=x,a^{ⓓ}\left(ax\right)=a^{ⓓ}$.

(5)

$xa{\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}=x,\left(xa\right){\left[a^{ⓓ}a\right]}^{†}={\left[a^{ⓓ}a\right]}^{†}$.

Proof. 

This is obtained by choosing $w=1$ in Theorem 4.3. □

Theorem 9.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then the following are equivalent:

(1)

$a^{⊠,w}=x$.

(2)

$xwawx=x,wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w},xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

(3)

${\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}wawx={\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w},{\left[a^{d,w}waw\right]}^{†}{a}^{ⓓ,w}wawx=x.$

(4)

$xwaw{a}^{ⓓ,w}=x,xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a^{⊠,w}=x$, it follows by Theorem 4.1 that $x=xwawx$. By Theorem 4.3, $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w},xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we derive that

$$\begin{array}{ccc}\hfill x& =\hfill & xwawx=x\left(wawx\right)\hfill \\ & =\hfill & x\left[waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\right]\hfill \\ & =\hfill & \left[xwaw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}.\hfill \end{array}$$

Therefore $a^{⊠,w}=x$, as desired.

$\left(1\right)\Rightarrow \left(3\right)$ By virtue of Theorem 4.3, we have ${\left[a^{d,w}waw\right]}^{†}{a}^{ⓓ,w}wawx=x.$ Moreover, we check that

$$\begin{array}{cc}& {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}wawx\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{\left(a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w},\hfill \end{array}$$

as required.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, ${\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}wawx={\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}$. Then ${\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}wawx={\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}$. This implies that

$$a^{ⓓ,w}wawx=a^{ⓓ,w}.$$

By using Theorem 4.3, $a^{⊠,w}=x$.

$\left(1\right)\Rightarrow \left(4\right)$ By the preceding argument, we have $xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$. In view of Theorem 4.3, we verify that

$$\begin{array}{ccc}\hfill xwaw{a}^{ⓓ,w}& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}=x,\hfill \end{array}$$

as required.

$\left(4\right)\Rightarrow \left(1\right)$ By hypothesis, we check that

$$\begin{array}{cc}& \left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ \hfill =& {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ \hfill =& {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ \hfill =& {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]\hfill \\ \hfill =& \left({\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw\right)a^{ⓓ,w}\hfill \\ \hfill =& \left(xwaw\right)a^{ⓓ,w}=x.\hfill \end{array}$$

Therefore we complete the result by Theorem 4.3. □

Set $a^{⊠}=a^{⊠,1}$ and ${\mathcal{A}}^{⊠}={\mathcal{A}}^{⊠,1}$. We deduce that

Corollary 11.

Let $a\in {\mathcal{A}}^{⊠}$. Then the following are equivalent:

(1)

$a^{⊠}=x$.

(2)

$xax=x,ax=a{\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ},xa={\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}a$.

(3)

${\left(a^{ⓓ}a\right)}^{\ast }{a}^{ⓓ}ax={\left(a^{ⓓ}a\right)}^{\ast }{a}^{ⓓ},{\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}ax=x$.

(4)

$xa{a}^{ⓓ}=x,xa={\left[a^{ⓓ}a\right]}^{†}{a}^{ⓓ}a$.

Proof. 

This is obvious by choosing $w=1$ in Theorem 4.5. □

Theorem 10.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then the following are equivalent:

(1)

$a^{⊠,w}=x$.

(2)

$im\left(x\right)=im{\left(a^{ⓓ,w}waw\right)}^{\ast },wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$.

(3)

$im\left(x\right)=im{\left(a^{ⓓ,w}waw\right)}^{\ast },a^{ⓓ,w}\left(wawx\right)=a^{ⓓ,w}$.

(4)

$ker\left(x\right)=ker{\left(a^{ⓓ,w}\right)}^{\ast },xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

(5)

$ker\left(x\right)=ker{\left(a^{ⓓ,w}\right)}^{\ast },\left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}={\left[a^{ⓓ,w}waw\right]}^{†}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a^{⊠,w}=x$, it follows by Theorem ??? that

$$\begin{array}{ccc}\hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & \in \hfill & im{\left(a^{ⓓ,w}waw\right)}^{\ast };\hfill \\ \hfill a^{ⓓ,w}waw& =\hfill & \left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\hfill \\ & =\hfill & \left[a^{ⓓ,w}waw\right]{\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast };\hfill \\ \hfill {\left(a^{ⓓ,w}waw\right)}^{\ast }& =\hfill & {\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(waw\right){\left[a^{ⓓ,w}waw\right]}^{\ast }\hfill \\ & \in \hfill & im\left(x\right).\hfill \end{array}$$

Therefore $im\left(x\right)=im{\left(a^{ⓓ,w}waw\right)}^{\ast }$. By virtue of Theorem 4.3, $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$, as desired.

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$. As in the proof of Theorem 4.3, we prove that $a^{ⓓ,w}\left(wawx\right)=a^{ⓓ,w},$ as required.

$\left(3\right)\Rightarrow \left(1\right)$ Write $x={\left(a^{ⓓ,w}waw\right)}^{\ast }z$ for some $z\in \mathcal{A}$. Then it is easy to verify that

$$\begin{array}{ccc}\hfill {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left(wawx\right)& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{\ast }z\hfill \\ & =\hfill & {\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }\left[a^{ⓓ,w}\right]{waw]}^{\ast }z\hfill \\ & =\hfill & {\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }z\hfill \\ & =\hfill & \left[a^{ⓓ,w}\right]{waw]}^{\ast }z\hfill \\ & =\hfill & x.\hfill \end{array}$$

By virtue of Theorem 4.7, $a^{⊠,w}=x$, as asserted. □

Corollary 12.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then the following are equivalent:

(1)

$a^{⊠,w}=x$.

(2)

$im\left(x\right)\subseteq im{\left(a^{ⓓ,w}waw\right)}^{\ast },wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$.

(3)

$im\left(x\right)\subseteq im{\left(a^{ⓓ,w}waw\right)}^{\ast },a^{ⓓ,w}\left(wawx\right)=a^{ⓓ,w}$.

(4)

$ker{\left(a^{ⓓ,w}\right)}^{\ast }\subseteq ker\left(x\right),xwaw={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}waw$.

(5)

$ker{\left(a^{ⓓ,w}\right)}^{\ast }\subseteq ker\left(x\right),\left(xwaw\right){\left[a^{ⓓ,w}waw\right]}^{†}={\left[a^{ⓓ,w}waw\right]}^{†}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is trivial by Theorem 4.7.

$\left(2\right)\Rightarrow \left(3\right)$ Since $wawx=waw{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}$, analogously to Theorem 4.7, we verify that $a^{ⓓ,w}\left(wawx\right)=a^{ⓓ,w}$.

$\left(3\right)\Rightarrow \left(1\right)$ This is proved as in Theorem 4.7.

$\left(1\right)\Rightarrow \left(4\right)\Rightarrow \left(5\right)\Rightarrow \left(1\right)$ These are similar to the preceding discussion. □

5. Relations Involving Certain Outer Inverses

Our main concern in this section is to find relations between weighted generalized core-EP inverse with various outer inverses.

Theorem 11.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then

$$\begin{array}{ccc}\hfill a^{⊠,w}& =\hfill & {\left(waw{a}^{ⓓ,w}waw\right)}^{†}\hfill \\ & =\hfill & (p_{im{\left(wa\right)}^d}{waw)}^{†}\hfill \\ & =\hfill & p_{im{\left(a^{ⓓ,w}waw\right)}^{\ast }}{a}^{ⓓ,w}.\hfill \end{array}$$

Proof. 

Set $x=a^{⊠,w}$. Then $x={\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}$. By using [?, Theorem 3.1], we check that

$$\begin{array}{ccc}\hfill \left[waw{a}^{ⓓ,w}waw\right]x& =\hfill & waw{a}^{ⓓ,w}\left(wawx\right)=waw{a}^{ⓓ,w},\hfill \\ \hfill x\left[waw{a}^{ⓓ,w}waw\right]& =\hfill & {\left(a^{ⓓ,w}waw\right)}^{†}\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]waw\hfill \\ & =\hfill & {\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}waw,\hfill \\ \hfill \left(waw{a}^{ⓓ,w}waw\right)x\left(waw{a}^{ⓓ,w}waw\right)& =\hfill & \left(waw{a}^{ⓓ,w}\right)\left(waw{a}^{ⓓ,w}waw\right)\hfill \\ & =\hfill & waw\left(a^{ⓓ,w}waw{a}^{ⓓ,w}\right)waw\hfill \\ & =\hfill & waw{a}^{ⓓ,w}waw,\hfill \\ \hfill x\left[waw{a}^{ⓓ,w}waw\right]x& =\hfill & \left[{\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}waw\right]\left[{\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}\right]\hfill \\ & =\hfill & \left[{\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}waw{\left(a^{ⓓ,w}waw\right)}^{†}\right]a^{ⓓ,w}\hfill \\ & =\hfill & {\left(a^{ⓓ,w}waw\right)}^{†}{a}^{ⓓ,w}=x.\hfill \end{array}$$

Hence, we verify that

$$\begin{array}{ccc}\hfill {\left(\left[waw{a}^{ⓓ,w}waw\right]x\right)}^{\ast }& =\hfill & \left[waw{a}^{ⓓ,w}waw\right]x,\hfill \\ \hfill {\left(x\left[waw{a}^{ⓓ,w}waw\right]\right)}^{\ast }& =\hfill & x\left[waw{a}^{ⓓ,w}waw\right].\hfill \end{array}$$

Therefore ${\left(waw{a}^{ⓓ,w}waw\right)}^{†}=x$.

Case 1. Let $p=waw{a}^{ⓓ,w}$. In view of [3] (Theorem 2.1 and Theorem 3.1), $p^2=p^{\ast }=p\in \mathcal{A}$. Since $p=wawa{\left[{\left(wa\right)}^d\right]}^2=\left(wa\right){\left(wa\right)}^d$, we see that $im\left(p\right)=im{\left(wa\right)}^d$. Hence, $p=p_{im{\left(wa\right)}^d}$. Hence, $a^{⊠,w}=(p_{im{\left(wa\right)}^d}{waw)}^{†}$.

Case 2. Let $p=\left(a^{ⓓ,w}waw\right){\left(a^{ⓓ,w}waw\right)}^{†}$. Then we have

$$\begin{array}{ccc}\hfill a^{⊠,w}& =\hfill & {\left(waw{a}^{ⓓ,w}waw\right)}^{†}\hfill \\ & =\hfill & {\left(waw\left(a^{ⓓ,w}waw\right){\left(a^{ⓓ,w}waw\right)}^{†}\left(a^{ⓓ,w}waw\right)\right)}^{†}\hfill \\ & =\hfill & {\left(wawp\right)}^{†}.\hfill \end{array}$$

Clearly, $p^2=p^{\ast }=p\in \mathcal{A}$. Moreover, we have

$$\begin{array}{ccc}\hfill im\left(p\right)& =\hfill & im{\left(a^{ⓓ,w}waw\right)}^{†}\left(a^{ⓓ,w}waw\right)\hfill \\ & =\hfill & im{\left[{\left(a^{ⓓ,w}waw\right)}^{†}\left(a^{ⓓ,w}waw\right)\right]}^{\ast }\hfill \\ & =\hfill & im{\left(a^{ⓓ,w}waw\right)}^{\ast }.\hfill \end{array}$$

Therefore $a^{⊠,w}=p_{im{\left(a^{ⓓ,w}waw\right)}^{\ast }}{a}^{ⓓ,w}$. □

Corollary 13.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then $a^{⊠,w}\in {\mathcal{A}}^{†}$ and ${\left(a^{⊠,w}\right)}^{†}=waw{a}^{ⓓ,w}waw.$

Proof. 

By virtue of Theorem 5.1, $a^{⊠,w}={\left(waw{a}^{ⓓ,w}waw\right)}^{†}$. Accordingly,

$${\left(a^{⊠,w}\right)}^{†}={\left[{\left(waw{a}^{ⓓ,w}waw\right)}^{†}\right]}^{†}=waw{a}^{ⓓ,w}waw.$$

□

Let $a,d\in \mathcal{A}$. We say that $x\in \mathcal{A}$ is the Mary inverse of a relatively to d provided that

$$xad=d=dax,x\in d\mathcal{A}\bigcap \mathcal{A}d.$$

We denote x by $a^{\left|\right|d}$ (see [10]). We now derive

Theorem 12.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then

$$a^{⊠,w}={\left(waw\right)}^{\left|\right|{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}}.$$

Proof. 

Obviously, we have $a\in {\mathcal{A}}^{ⓓ,w}$. Let $x=a^{⊠,w}$. In view of [3] (Theorem 2.1 and Theorem 3.1), we derive

$$\begin{array}{ccc}\hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{\ast }{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}(\left({\left[a^{ⓓ,w}waw\right]}^{†}\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{a}^{ⓓ,w}waw{\left[a^{ⓓ,w}waw\right]}^{†}(\left({\left[a^{ⓓ,w}waw\right]}^{†}\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \end{array}$$

Thus,

$$x\in {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}\mathcal{A}\bigcap \mathcal{A}{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}.$$

Moreover, we check that

$$\begin{array}{cc}& x\left(waw\right){\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w},\hfill \\ & {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}wawx\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }{a}^{ⓓ,w}\hfill \\ \hfill =& {\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}.\hfill \end{array}$$

Therefore

$$x={\left(waw\right)}^{\left|\right|{\left(a^{ⓓ,w}waw\right)}^{\ast }{a}^{ⓓ,w}}.$$

□

Corollary 14.

Let $a\in {\mathcal{A}}^{\square }$. Then

$$a^{\square }=a^{\left|\right|{\left(a^{ⓓ}a\right)}^{\ast }{a}^{ⓓ}}.$$

Proof. 

This is obvious by choosing $w=1$ in Theorem 5.3. □

Let $a,b,c\in \mathcal{A}$. An element a has $(b,c)$-inverse provide that there exists $x\in \mathcal{A}$ such that

$$xab=b,cax=candx\in b\mathcal{A}x\bigcap x\mathcal{A}c.$$

If such x exists, it is unique and denote it by $a^{(b,c)}$. Evidently, $a^{†}$ is the $(a^{\ast },a^{\ast })$ inverse of a (see [7]). By virtue of Theorem 5.1, we claim that

$$a^{⊠,w}={\left(waw{a}^{ⓓ,w}waw\right)}^{\left({\left(waw\right)}^{\ast }waw{a}^{ⓓ,w},{\left(waw\right)}^{\ast }waw{a}^{ⓓ,w}\right)}.$$

Alternatively, we derive

Theorem 13.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then

$$a^{⊠,w}={\left(waw\right)}^{\left({\left(a^{ⓓ,w}waw\right)}^{\ast },waw{a}^{ⓓ,w}\right)}.$$

Proof. 

In view of Theorem 4.1, $a\in {\mathcal{A}}^{ⓓ,w}$. Let $x=a^{⊠,w}$. We verify that

$$\begin{array}{ccc}\hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & \in \hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }\mathcal{A}x,\hfill \\ \hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\left[waw{a}^{ⓓ,w}\right]\hfill \\ & \in \hfill & x\mathcal{A}\left[waw{a}^{ⓓ,w}\right],\hfill \\ \hfill x\left(waw\right){\left(a^{ⓓ,w}waw\right)}^{\ast }& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left(a^{ⓓ,w}waw\right)}^{\ast }\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left(a^{ⓓ,w}waw\right)}^{\ast }\hfill \\ & =\hfill & {\left(\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }\left[a^{ⓓ,w}waw\right]{)}^{\ast }\hfill \\ & =\hfill & {\left(a^{ⓓ,w}waw\right)}^{\ast },\hfill \\ \hfill waw{a}^{ⓓ,w}\left(waw\right)x& =\hfill & waw{a}^{ⓓ,w}\left(waw\right){\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & waw\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]a^{ⓓ,w}\hfill \\ & =\hfill & waw\left[a^{ⓓ,w}waw{a}^{ⓓ,w}\right]=waw{a}^{ⓓ,w}.\hfill \end{array}$$

Therefore

$$a^{⊠,w}={\left(waw\right)}^{\left({\left(a^{ⓓ,w}waw\right)}^{\ast },waw{a}^{ⓓ,w}\right)},$$

as asserted. □

As an immediate consequence, we now give a new presentation of the generalized Moore-Penrose inverse for a bounded linear operator over Hilbert spaces.

Corollary 15.

Let $A\in \mathcal{B}{\left(H\right)}^d$. Then $A^{\diamond }=A^{\left({\left(A^{ⓓ}A\right)}^{\ast },A{A}^{ⓓ}\right)}.$

Proof. 

This is obvious by choosing $w=1$ in Theorem 5.5. □

Let $a\in \mathcal{A}$. We say that a has $\left\{2\right\}$-inverse x provided that $x=xax$. We denote $a_{T,S}^{\left(2\right)}=\{x\in \mathcal{A}|xax=x,im\left(a\right)=T,ker\left(a\right)=S\}$. We now derive

Theorem 14.

Let $a\in {\mathcal{A}}^{⊠,w}$. Then

$$a^{⊠,w}={\left(waw\right)}_{im{\left(a^{ⓓ,w}waw\right)}^{\ast },ker{a}^{ⓓ,w}}^{\left(2\right)}.$$

Proof. 

Let $x=a^{⊠,w}$. In view of Theorem 4.1, we have $x=x\left(waw\right)x$.

Step 1. $im\left(x\right)=im{\left(a^{ⓓ,w}waw\right)}^{\ast }$. By virtue of [3] (Theorem 2.1 and Theorem 3.1), we have

$$\begin{array}{ccc}\hfill x& =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}\hfill \\ & =\hfill & {\left[a^{ⓓ,w}waw\right]}^{\ast }{\left({\left[a^{ⓓ,w}waw\right]}^{†}\right)}^{\ast }{\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}.\hfill \end{array}$$

Hence, $im\left(x\right)\subseteq im{\left(a^{ⓓ,w}waw\right)}^{\ast }$. One easily checks that

$$\begin{array}{ccc}\hfill a^{ⓓ,w}waw& =\hfill & \left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\hfill \\ & =\hfill & \left[a^{ⓓ,w}waw\right]{\left({\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]\right)}^{\ast }.\hfill \end{array}$$

Hence,

$${\left[a^{ⓓ,w}waw\right]}^{\ast }\subseteq {\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{\ast }.$$

This implies that ${\left[a^{ⓓ,w}waw\right]}^{\ast }\in im\left(x\right)$, as required.

Step 2. $ker\left(x\right)=ker{a}^{ⓓ,w}$. If $r\in ker\left(x\right)$, then $xr=0$ for a $r\in \mathcal{A}$. Hence,

$$\begin{array}{ccc}\hfill a^{ⓓ,w}r& =\hfill & \left[a^{ⓓ,w}waw\right]a^{ⓓ,w}r\hfill \\ & =\hfill & \left[a^{ⓓ,w}waw\right]{\left[a^{ⓓ,w}waw\right]}^{†}\left[a^{ⓓ,w}waw\right]a^{ⓓ,w}r\hfill \\ & =\hfill & \left[a^{ⓓ,w}waw\right]xr=0.\hfill \end{array}$$

This implies that $r\in ker{a}^{ⓓ,w}$. If $r\in ker{a}^{ⓓ,w}$, then $a^{ⓓ,w}r=0$. This implies that $xr={\left[a^{ⓓ,w}waw\right]}^{†}{a}^{ⓓ,w}r=0$; and so $r\in ker\left(x\right)$. Thus $ker\left(x\right)=ker{a}^{ⓓ,w}$.

Therefore we complete the proof. □

Corollary 16.

Let $a\in {\mathcal{A}}^{⊠}$. Then

$$a^{⊠}=a_{im{\left(a^{ⓓ}a\right)}^{\ast },ker{a}^{ⓓ}}^{\left(2\right)}.$$

We now present a new property of the weighted gMP inverse of a bounded operator on a Hilbert space.

Corollary 17.

Let $A,W\in \mathcal{B}\left(H\right)$. If $A\in \mathcal{B}{\left(H\right)}^{d,W}$, then

$$A^{\diamond ,W}={\left(WAW\right)}_{R{\left(A^{Ⓓ,W}WAW\right)}^{\ast },N\left(A^{Ⓓ,W}\right)}^{\left(2\right)}.$$

Proof. 

This is an immediate consequence of Theorem 5.7. □

Remark 1.

For a complex A, Stanimirovic et al. introduced and studied the $(M,N)$-weighted $(B,C)$-inverse in [17]. By using Theorem 5.7, we prove that $A^{\diamond ,W}=A_{\left({\left(A^{Ⓓ,W}WAW\right)}^{\ast },A^{Ⓓ,W}\right)}^{(2,W,W)}.$