Generalized weighted group inverse in Banach algebras with proper involution

1. Introduction

An involution of a Banach algebra $\mathcal{A}$ is an anti-automorphism whose square is the identity map 1. A Banach algebra $\mathcal{A}$ with involution * is called a Banach *-algebra. Let $a,w\in \mathcal{A}$. Recall that $a\in \mathcal{A}$ has generalized w-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 [10]). Here, ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid 1+\lambda x\in {\mathcal{A}}^{-1}\}.$ As is well known, $x\in {\mathcal{A}}^{qnil}\iff \underset{n\to \infty }{lim}{\Vert x^n\Vert }^{\frac{1}{n}}=0.$ If we replace ${\mathcal{A}}^{qnil}$ by the set ${\mathcal{A}}^{nil}$ of all nilpotents in $\mathcal{A}$, $a^{d,w}$ is called the w-Drazin inverse of a, and denote it by $a^{D,w}$. If the weight $w=1$, we call $a^{d,w}\left(a^{D,w}\right)$ the g-Drazin (Drazin) inverse of a, and denote it by $a^d\left(a^D\right)$. Evidently, $a^{d,w}={\left[{\left(aw\right)}^d\right]}^{2}a=a{\left[{\left(wa\right)}^d\right]}^2={\left(aw\right)}^{d}a{\left(wa\right)}^d$.

Definition 1.1.(see [4]) An element $a\in \mathcal{A}$ has w-group inverse if there exist $x\in \mathcal{A}$ such that

$$awxwa=a,xwawx=x,awx=xwa.$$

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

An element a in a Banach algebra $\mathcal{A}$ has group inverse provided that it has w-group inverse for the weight $w=1$, i.e., there exists $x\in \mathcal{A}$ such that

$$x{a}^2=a,a{x}^2=x,ax=xa.$$

Such x is unique if exists, denoted by $a^{\#}$, and called the group inverse of a. As is well known, a square complex matrix A has group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$.

A square complex matrix X is the core-EP inverse of A if

$$XAX=X,Im\left(X\right)=Im{X}^{*}=Im\left(A^m\right),$$

where $m=ind\left(A\right)$ (see [8]). Recently, Wang and Chen (see [15]) introduced and studied a weak group inverse for square complex matrices. A square complex matrix A has weak group inverse X if it satisfies the equations

$$A{X}^2=X,AX=A^{D}A.$$

The involution * is proper, that is, $x^{*}x=0⟹x=0$, e.g., in a Rickart *-ring, the involution is always proper. Let $C^{n\times n}$ be the Banach algebra of all $n\times n$ complex matrices, with conjugate transpose * as the involution. Then the involution * is proper. Zou et al. extended the notion of weak group inverse to a proper *-Banach algebra element. We refer the reader for weak group inverse in [3,13,18,19]. In [2], the generalized group inverse in a Banach algebra with proper involution was introduced. An element a in a Banach *-algebra has generalized group inverse if there exists $x\in \mathcal{A}$ such that

$$x=a{x}^2,{\left(a^{*}{a}^{2}x\right)}^{*}=a^{*}{a}^{2}x,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.$$

Such x is unique if it exists and is called the generalized group inverse of a. We denote it by $a^g$. Many properties of generalized group inverse were presented in [2]. Recently, Mosić and Zhang introduced and studied weighted weak group inverse for a Hilbert space operator A in $\mathcal{B}{\left(X\right)}^d$ (see [12]).

The motivation of this paper is to introduce and study a new kind of generalized weighted inverse as a natural generalization of generalized inverses mentioned above.

In Section 2, we introduce generalized weighted group inverse by using a new kind of weighted group decomposition. Let ${\mathcal{A}}_w^{qnil}=\{x\in \mathcal{A}\mid xw\in {\mathcal{A}}^{qnil}\}.$

Definition 1.2.

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

$$a=x+y,x^{*}y=ywx=0,x\in {\mathcal{A}}_w^{\#},y\in {\mathcal{A}}_w^{qnil}.$$

We prove that $a\in \mathcal{A}$ has generalized w-group decomposition if and only if $aw\in {\mathcal{A}}^d$ and there exists unique $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

The preceding x is called generalized w-group inverse of a.

In Section 3, we establish elementary properties of generalized w-group inverse. Many properties of the weak group inverse generalized group inverse are thereby generalized to wider cases.

In Section 3, we investigate equivalences between generalized w-group inverse and weighted g-Drazin inverse for a Banach algebra element. We prove that $a\in {\mathcal{A}}_w^g$ if and only if $a\in {\mathcal{A}}^{d,w}$ and ${\left(a^{d,w}\right)}^{*}{a}^{d,w}x={\left(a^{d,w}\right)}^{*}a$ for some $x\in \mathcal{A}$.

Throughout the paper, all Banach algebras are complex with a proper involution *. We use ${\mathcal{A}}^{-1},{\mathcal{A}}^{\#},{\mathcal{A}}^d$ and ${\mathcal{A}}^D$ to denote the sets of all invertible, group invertible, g-Drazin invertible and Drazin invertible elements in $\mathcal{A}$, respectively. We use $\ell \left(a\right)$ to stand for the left annihilator of a in $\mathcal{A}$.

2. Generalized w-Group Inverse

The purpose of this section is to introduce a new generalized inverse which is a natural generalization of (weak) group inverse in a *-Banach algebra.

Lemma 2.1.

Let $a\in {\mathcal{A}}_w^{\#}$. Then $aw,wa\in {\mathcal{A}}^{\#}$. In this case,

$$a_w^{\#}={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#},{\left(aw\right)}^2{a}_w^{\#}=aanda{\left[w{a}_w^{\#}\right]}^2=a_w^{\#}.$$

Proof. 

Let $x=a_w^{\#}$. Then

$$awxwa=a,xwawx=x,awx=xwa.$$

Hence, ${\left(aw\right)}^{2}x=aw\left(awx\right)=\left(aw\right)x\left(wa\right)=a$, and so ${\left(aw\right)}^{2}xw=aw$. Likewise, $xw{\left(aw\right)}^2=aw$. This implies that $aw\in {\mathcal{A}}^{\#}$. Moreover, we have

$$a{\left[w{a}_w^{\#}\right]}^2=a{\left(wx\right)}^2=\left(awx\right)wx=\left(xwa\right)wx=x,$$

as desired. □

Theorem 2.2.

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

(1)

$a\in \mathcal{A}$ has generalized w-group decomposition.

(2)

$aw\in {\mathcal{A}}^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, a has the generalized w-group decomposition $a=a_1+a_2$. Let $x={\left(a_1\right)}_w^{\#}$. By virtue of Lemma 2.1, we have

$$\begin{array}{ccc}\hfill awx& =\hfill & (a_{1}w+a_{2}w){\left(a_1\right)}_w^{\#}=a_{1}w{\left(a_1\right)}_w^{\#},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & a_1{\left[w{\left(a_1\right)}_w^{\#}\right]}^2={\left(a_1\right)}_w^{\#}=x,\hfill \end{array}$$

Since $a_{2}w{a}_1=0$, we have

$$\begin{array}{cc}& aw-xw{\left(aw\right)}^2\hfill \\ \hfill =& (a_{1}w+a_{2}w)-[{\left(a_1\right)}_w^{\#}w{a}_{1}w+{\left(a_1\right)}_w^{\#}w{a}_{2}w](a_{1}w+a_{2}w)\hfill \\ \hfill =& [1-{\left(a_1\right)}_w^{\#}w{a}_{1}w-{\left(a_1\right)}_w^{\#}w{a}_{2}w]a_{2}w,\hfill \end{array}$$

and then

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill =& \left|\right|(aw-xw{\left(aw\right)}^2){\left(aw\right)}^{n-1}{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill =& \left|\right|[1-{\left(a_1\right)}_w^{\#}w{a}_{1}w-{\left(a_1\right)}_w^{\#}w{a}_{2}w]a_{2}w{\left(aw\right)}^{n-1}{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill =& \left|\right|[1-{\left(a_1\right)}_w^{\#}w{a}_{1}w-{\left(a_1\right)}_w^{\#}w{a}_{2}w]{\left(a_{2}w\right)}^n{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill \le & \left|\right|1-{\left(a_1\right)}_w^{\#}w{a}_{1}w-{\left(a_1\right)}_w^{\#}w{a}_2{w\left|\right|}^{\frac{1}{n}}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{\frac{1}{n}}.\hfill \end{array}$$

Since $a_2\in {\mathcal{A}}_w^{qnil}$, we have $\underset{n\to \infty }{lim}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{\frac{1}{n}}=0$. Accordingly, we have

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.$$

Since $a_{2}w{a}_1=0,\left(a_{1}w\right){\left(a_{1}w\right)}^{\pi }=0$ and ${\left(a_{2}w\right)}^d=0$, Then $aw\in {\mathcal{A}}^d$ and

$${\left(aw\right)}^d={\left(a_{1}w\right)}^{\#}+\sum _{n=1}^{\infty }{\left({\left(a_{1}w\right)}^{\#}\right)}^{n+1}{\left(a_{2}w\right)}^n.$$

Then

$$\begin{array}{cc}& {\left({\left(aw\right)}^d\right)}^{*}{a}_2\hfill \\ \hfill =& {\left({\left(a_{1}w\right)}^{\#}\right)}^{*}{a}_2+\sum _{n=1}^{\infty }{\left[{\left({\left(a_{1}w\right)}^{\#}\right)}^{n+1}{\left(a_{2}w\right)}^n\right]}^{*}{a}_2\hfill \\ \hfill =& {\left[{\left({\left(a_{1}w\right)}^{\#}\right)}^2\right]}^{*}\left({\left(a_{1}w\right)}^{*}{a}_2\right)+\sum _{n=1}^{\infty }{\left[{\left({\left(a_{1}w\right)}^{\#}\right)}^{n+2}{\left(a_{2}w\right)}^n\right]}^{*}\left({\left(a_{1}w\right)}^{*}{a}_2\right)\hfill \\ \hfill =& 0;\hfill \end{array}$$

hence, ${\left({\left(aw\right)}^d\right)}^{*}{a}_1={\left({\left(aw\right)}^d\right)}^{*}a$. Accordingly,

$$\begin{array}{ccc}\hfill {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x& =\hfill & {\left({\left(aw\right)}^d\right)}^{*}(a_{1}w+a_{2}w)(a_{1}w+a_{2}w){\left(a_1\right)}_w^{\#}\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{*}{\left(a_{1}w\right)}^2{\left(a_1\right)}_w^{\#}={\left({\left(aw\right)}^d\right)}^{*}{a}_1\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{*}a.\hfill \end{array}$$

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

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Then

$$\begin{array}{ccc}\hfill x& =\hfill & \left(aw\right)x\left(wx\right)=\left(aw\right)\left[\left(aw\right)x\left(wx\right)\right]\left(wx\right)={\left(aw\right)}^{2}x{\left(wx\right)}^2\hfill \\ & =\hfill & \cdots ={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}.\hfill \end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. Then we check that

$$\begin{array}{ccc}\hfill \left|\right|a_{2}w{a}_1\left|\right|& =\hfill & \left|\right|[a-{\left(aw\right)}^{2}x]w{\left(aw\right)}^{2}x\left|\right|\hfill \\ & =\hfill & \left|\right|[{\left(aw\right)}^2-{\left(aw\right)}^{2}xw\left(aw\right)]\left(aw\right)x\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^2[1-xw\left(aw\right)]\left(aw\right)x\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^2[1-xw\left(aw\right)]{\left(aw\right)}^{n}x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^2{\left|\right|\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|\left|\right|x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

Hence

$$\left|\right|a_{2}w{a}_1{\left|\right|}^{\frac{1}{n}}\le {\left|\right|\left(aw\right)}^2{\left|\right|}^{\frac{1}{n}}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}{\left|\right|x\left(wx\right)}^{n-1}{\left|\right|}^{\frac{1}{n}}.$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0,$ we deduce that

$$\underset{n\to \infty }{lim}\left|\right|a_{2}w{a}_1{\left|\right|}^{\frac{1}{n}}=0.$$

Thus, $a_{2}w{a}_1=0$.

Moreover, we have

$$\begin{array}{cc}& {\left(aw\right)}^{2}x\hfill \\ \hfill =& {\left(aw\right)}^2\left[{\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}\right]={\left(aw\right)}^{n+1}x{\left(wx\right)}^{n-1}\hfill \\ \hfill =& [{\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2}]x{\left(wx\right)}^{n-1}+{\left(aw\right)}^d{\left(aw\right)}^{n+2}{]x\left(wx\right)}^{n-1}\hfill \\ \hfill =& [{\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2}]x{\left(wx\right)}^{n-1}+{\left(aw\right)}^d{\left(aw\right)}^3\left[{\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}\right]\hfill \\ \hfill =& [{\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2}]x{\left(wx\right)}^{n-1}+{\left(aw\right)}^d{\left(aw\right)}^{3}x.\hfill \end{array}$$

Then we verify that

$$\begin{array}{cc}& \left|\right|a_1^{*}{a}_2\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ \hfill =& \left|\right|{\left([{\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2}]x{\left(wx\right)}^{n-1}+{\left(aw\right)}^d{\left(aw\right)}^{3}x\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}\left|\right|\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|a-{\left(aw\right)}^{2}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^d{\left(aw\right)}^{3}x\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}\left|\right|\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|a-{\left(aw\right)}^{2}x\left|\right|\hfill \\ \hfill +& {\left|\right|\left(aw\right)}^{3}x\left|\right|\left|\right|{\left({\left(aw\right)}^d\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ \hfill =& \left|\right|{({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}\left|\right|\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|a-{\left(aw\right)}^{2}x\left|\right|.\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill \left|\right|a_1^{*}{a}_2{\left|\right|}^{\frac{1}{n+1}}& \le \hfill & \left|\right|{\left({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2}\right)}^{*}{\left|\right|}^{\frac{1}{n+1}}\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}{\left|\right|}^{\frac{1}{n+1}}\hfill \\ & & \left|\right|a-{\left(aw\right)}^{2}x{\left|\right|}^{\frac{1}{n+1}}.\hfill \end{array}$$

Therefore $\underset{n\to \infty }{lim}\left|\right|a_1^{*}{a}_2{\left|\right|}^{\frac{1}{n+1}}=0$, and so $a_1^{*}{a}_2=0$. Clearly, we have

$$\begin{array}{ccc}\hfill \left|\right|awx-xw{\left(aw\right)}^{2}x\left|\right|& =\hfill & \left|\right|(aw-xw{\left(aw\right)}^2)x\left|\right|\hfill \\ & =\hfill & \left|\right|[{\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}]x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|\left|\right|x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

Hence, $\underset{n\to \infty }{lim}\left|\right|awx-xw{\left(aw\right)}^{2}x{\left|\right|}^{\frac{1}{n}}=0$, and then $awx=xw{\left(aw\right)}^{2}x$. Moreover, we check that

$$\begin{array}{ccc}\hfill a_{1}wx& =\hfill & {\left(aw\right)}^{2}xwx=awawxwx=awa{\left(wx\right)}^2\hfill \\ & =\hfill & awx=xw{\left(aw\right)}^{2}x=xw{a}_1,\hfill \\ \hfill a_{1}wxw{a}_1& =\hfill & {\left(aw\right)}^{2}xwxw{\left(aw\right)}^{2}x=awa{\left(wx\right)}^{2}wawawx\hfill \\ & =\hfill & awxw{\left(aw\right)}^{2}x={\left(aw\right)}^{2}x=a_1,\hfill \\ \hfill xw{a}_{1}wx& =\hfill & xw{\left(aw\right)}^{2}xwx=xwawa{\left(wx\right)}^2=xwawx=x.\hfill \end{array}$$

Thus, $a_1\in {\mathcal{A}}_w^{\#}$. By hypothesis, we have $aw\in {\mathcal{A}}^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}xw=aw{\left(xw\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}xw={\left({\left(aw\right)}^d\right)}^{*}aw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

As in the proof of ([2] Theorem 2.2), we have $aw-{\left(aw\right)}^{2}xw\in {\mathcal{A}}^{qnil}$, and then $a_{2}w=(a-{\left(aw\right)}^{2}x)w\in {\mathcal{A}}^{qnil}$. This implies that $a_2\in {\mathcal{A}}_w^{qnil}$. Accordingly, we have the generalized w-group inverse $a=a_1+a_2$, as required. □

Corollary 2.3.

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ has generalized w-group decomposition.

(2)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, a has generalized w-group decomposition $a=a_1+a_2$. Let $x={\left(a_1\right)}_w^{\#}$. By virtue of Theorem 2.2, we have $aw\in {\mathcal{A}}^d$ and

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Moreover, we check that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^2\left(xw\right)& =\hfill & {(a_{1}w+a_{2}w)}^2{\left(a_1\right)}_w^{\#}w={\left(a_{1}w\right)}^2{\left(a_1\right)}_w^{\#}w=a_{1}w,\hfill \\ \hfill {\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right)& =\hfill & {(a_{1}w+a_{2}w)}^{*}{a}_{1}w={\left(a_{1}w\right)}^{*}{a}_{1}w.\hfill \end{array}$$

Therefore

$${\left[{\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right)\right]}^{*}={\left[{\left(a_{1}w\right)}^{*}{a}_{1}w\right]}^{*}={\left(a_{1}w\right)}^{*}{a}_{1}w={\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right),$$

as required.

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

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. As is the proof of $"\left(3\right)\Rightarrow \left(2\right)"$, we verify that $a_{2}w{a}_1=0$ and ${\left(a_1\right)}_w^{\#}=x$. As in the proof of Theorem 2.2, $x={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}$. Obviously, we verify that

$$\begin{array}{ccc}\hfill \left|\right|x-\left(xw\right)\left(awx\right)\left|\right|& =\hfill & \left|\right|[{\left(aw\right)}^{n-1}-xwaw{\left(aw\right)}^{n-1}]x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^{n-1}-xw{\left(aw\right)}^n{\left|\right|\left|\right|x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

By hypothesis, we have

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{n-1}-xw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n-1}}=0.$$

Hence,

$$\underset{n\to \infty }{lim}\left|\right|x-{\left(xw\right)\left(awx\right)\left|\right|}^{\frac{1}{n-1}}=0.$$

This implies that $x=\left(xw\right)\left(awx\right)$. Moreover, we have

$$\begin{array}{ccc}\hfill w^{*}{a}_1^{*}{a}_{2}w& =\hfill & w^{*}{\left[{\left(aw\right)}^{2}x\right]}^{*}[a-{\left(aw\right)}^{2}x]w\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{2}xw\right]}^{*}aw[1-\left(aw\right)xw]\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}[1-\left(aw\right)xw]\hfill \\ & =\hfill & \left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right][1-\left(aw\right)xw]\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\left(aw\right)xw\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}{\left(aw\right)}^2\left(xwawx\right)w\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Since $w\in {\mathcal{A}}^{-1}$, we have $w^{*}\in {\mathcal{A}}^{-1}$. Hence, $a_1^{*}{a}_2=0$.

By hypothesis, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}xw=\left(aw\right){\left(xw\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Then $xw={\left(aw\right)}^g$. As in the proof of ([2] Theorem 2.2), we have $aw-{\left(aw\right)}^2\left(xw\right)\in {\mathcal{A}}^{qnil}$. This implies that $a_2=a-{\left(aw\right)}^{2}x\in {\mathcal{A}}_w^{qnil}$. Therefore a has generalized w-group decomposition $a=a_1+a_2$, as asserted. □

Theorem 2.4.

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

(1)

$a\in \mathcal{A}$ has generalized w-group decomposition.

(2)

$aw\in {\mathcal{A}}^d$ and there exists unique $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.2, $aw\in {\mathcal{A}}^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Assume that there exists $y\in \mathcal{A}$ such that

$$\begin{array}{c}y=a{\left(wy\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}y={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-yw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Claim 1. $xw=yw$.

By hypothesis, we have

$$\begin{array}{c}xw=\left(aw\right){\left(xw\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^2\left(xw\right)={\left({\left(aw\right)}^d\right)}^{*}\left(aw\right),\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Then $xw$ is generalized group inverse of $aw$. Likewise, $yw$ is generalized group inverse of $aw$. By virtue of ([2] Corollary 2.3), we have $xw=yw$, as desired.

Claim 2. ${\left(aw\right)}^{2}x={\left(aw\right)}^{2}y$.

As in the proof of Theorem 2.2, $x={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}$, and so $awx={\left(aw\right)}^{n+1}x{\left(wx\right)}^{n-1}$.

$$\begin{array}{cc}& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}x-{\left({\left(aw\right)}^{2}x\right)}^{*}a\left|\right|\hfill \\ \hfill \le & \left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|{\left({\left(aw\right)}^{n+1}\right)}^{*}{\left(aw\right)}^{2}x-{\left({\left(aw\right)}^{n+1}\right)}^{*}a\left|\right|\hfill \\ \hfill \le & \left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|{({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}{\left(aw\right)}^{2}x\hfill \\ \hfill -& {({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}a\left|\right|\hfill \\ \hfill +& \left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{n+2}\left|\right|\left|\right|{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x-{\left({\left(aw\right)}^d\right)}^{*}a\left|\right|\hfill \\ \hfill =& \left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|{({\left(aw\right)}^{n+1}-{\left(aw\right)}^d{\left(aw\right)}^{n+2})}^{*}{\left|\right|\left|\right|\left(aw\right)}^{2}x-a\left|\right|.\hfill \end{array}$$

Therefore

$$\underset{n\to \infty }{lim}\left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}x-{\left({\left(aw\right)}^{2}x\right)}^{*}a{\left|\right|}^{\frac{1}{n}}=0.$$

Hence, ${\left[{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}x\right]}^{*}={\left({\left(aw\right)}^{2}x\right)}^{*}a$. Likewise, ${\left[{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}y\right]}^{*}={\left({\left(aw\right)}^{2}x\right)}^{*}a,{\left[{\left({\left(aw\right)}^{2}y\right)}^{*}{\left(aw\right)}^{2}x\right]}^{*}={\left({\left(aw\right)}^{2}x\right)}^{*}a$ and ${\left[{\left({\left(aw\right)}^{2}y\right)}^{*}{\left(aw\right)}^{2}y\right]}^{*}={\left({\left(aw\right)}^{2}x\right)}^{*}a$. Let $z={\left(aw\right)}^{2}x-{\left(aw\right)}^{2}y$. Then we check that

$$\begin{array}{ccc}\hfill z^{*}z& =\hfill & {({\left(aw\right)}^{2}x-{\left(aw\right)}^{2}y)}^{*}({\left(aw\right)}^{2}x-{\left(aw\right)}^{2}y)\hfill \\ & =\hfill & {\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}x-{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{2}y\hfill \\ & -\hfill & {\left({\left(aw\right)}^{2}y\right)}^{*}{\left(aw\right)}^{2}x+{\left({\left(aw\right)}^{2}y\right)}^{*}{\left(aw\right)}^{2}y\hfill \\ & =\hfill & {\left({\left(aw\right)}^{2}x\right)}^{*}a-{\left({\left(aw\right)}^{2}x\right)}^{*}a-{\left({\left(aw\right)}^{2}y\right)}^{*}{a}^2+{\left({\left(aw\right)}^{2}y\right)}^{*}{a}^2\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Since $\mathcal{A}$ is a proper *-Banach algebra, we have $z=0$; hence, ${\left(aw\right)}^{2}x={\left(aw\right)}^{2}y$.

Claim 3. $x=y$.

As in the proof of Corollary 2.3, we have $x=\left(xw\right)\left(awx\right)$. Also we check that

$$\begin{array}{cc}& \left|\right|\left(xw\right)\left(awx\right)-{\left(xw\right)}^2{\left(aw\right)}^{2}x\left|\right|\hfill \\ \hfill =& \left|\right|\left(xw\right)[aw-\left(xw\right){\left(aw\right)}^2]x\left|\right|\hfill \\ \hfill =& \left|\right|\left(xw\right)[aw-\left(xw\right){\left(aw\right)}^2]{\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ \hfill \le & {\left|\right|xw\left|\right|\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|\left|\right|x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

This implies that

$$\underset{n\to \infty }{lim}\left|\right|\left(xw\right)\left(awx\right)-{\left(xw\right)}^2{\left(aw\right)}^{2}x{\left|\right|}^{\frac{1}{n}}=0.$$

Hence, $\left(xw\right)\left(awx\right)={\left(xw\right)}^2{\left(aw\right)}^{2}x.$ Thus, $x={\left(xw\right)}^2{\left(aw\right)}^{2}x$. Likewise, $y={\left(yw\right)}^2{\left(aw\right)}^{2}y$. By the preceding discussion, we have

$$xw=ywand{\left(aw\right)}^{2}x={\left(aw\right)}^{2}y.$$

Therefore $x=y$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ This is obvious by Theorem 2.1. □

We denote x in Corollary 2.3 by $a_w^g$, and call it the generalized w-group inverse of a. ${\mathcal{A}}_w^g$ denotes the sets of all generalized w-group invertible elements in $\mathcal{A}$.

Corollary 2.5.

Let $a\in {\mathcal{A}}_w^g$. Then the following hold.

(1)

$a_w^g=a_w^{g}waw{a}_w^g$.

(2)

$a_w^{g}w={\left(aw\right)}^g$.

Proof. 

$\left(1\right)$ These are obvious by the proof of Theorem 2.4. □

Let $C^{n\times n}$ be the Banach algebra of all $n\times n$ complex matrices, with conjugate transpose * as the involution. As is well known, ${\left(C^{n\times n}\right)}^{qnil}={\left(C^{n\times n}\right)}^{nil}$. For a complex $A\in C^{n\times n}$, the weak W-group inverse and generalized W-group inverse coincide with each other, i.e., $A_W^g=A_W^W$. We come now to provide a new characterization for the W-weak group inverse of a complex matrix.

Corollary 2.6.

Let $A,W\in C^{n\times n}$. Then the following are equivalent:

(1)

A has weak W-group inverse.

(2)

There exist $k\in N$ and $X\in C^{n\times n}$ such that $X=A{\left(WX\right)}^2,$${\left({\left(AW\right)}^D\right)}^{*}$${\left(AW\right)}^{2}X={\left({\left(AW\right)}^D\right)}^{*}A,{\left(AW\right)}^k=XW{\left(AW\right)}^{k+1}.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Since A has weak W-group inverse, there exist $k\in N$ and $X\in C^{n\times n}$ such that $X=A{\left(WX\right)}^2,{\left({\left(AW\right)}^D\right)}^{*}{\left(AW\right)}^{2}X={\left({\left(AW\right)}^D\right)}^{*}A,{\left(AW\right)}^k=XW{\left(AW\right)}^{k+1}.$ Hence,

$$\begin{array}{ccc}\hfill {\left({\left(AW\right)}^D\right)}^{*}{\left(AW\right)}^{2}X& =\hfill & {\left[{\left(AW\right)}^k{\left({\left(AW\right)}^D\right)}^{k+1}\right]}^{*}{\left(AW\right)}^{2}X\hfill \\ & =\hfill & {\left[{\left({\left(AW\right)}^D\right)}^{k+1}\right]}^{*}{\left({\left(AW\right)}^k\right)}^{*}{\left(AW\right)}^{2}X\hfill \\ & =\hfill & {\left[{\left({\left(AW\right)}^D\right)}^{k+1}\right]}^{*}{\left({\left(AW\right)}^k\right)}^{*}A\hfill \\ & =\hfill & {\left[{\left(AW\right)}^k{\left({\left(AW\right)}^D\right)}^{k+1}\right]}^{*}A\hfill \\ & =\hfill & {\left({\left(AW\right)}^D\right)}^{*}A.\hfill \end{array}$$

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exist $k\in N$ and $X\in C^{n\times n}$ such that

$$\begin{array}{c}{\left(AW\right)}^k=XW{\left(AW\right)}^{k+1},X=A{\left(WX\right)}^2,\\ {\left({\left(AW\right)}^D\right)}^{*}{\left(AW\right)}^{2}X={\left({\left(AW\right)}^D\right)}^{*}A.\end{array}$$

Hence,

$$\underset{n\to \infty }{lim}{\left|\right|\left(AW\right)}^n-XW{\left(AW\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.$$

In view of Theorem 2.4, $A\in C^{n\times n}$ has generalized W-group inverse. Therefore A has weak W-group inverse. □

3. Elementary Properties

In this section we establish some elementary properties of generalized group inverses. We now derive

Theorem 3.1.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)xw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

In this case, $a_w^g=wxw$.

Proof. 

⟹ In view of Theorem 2.4, $a\in {\mathcal{A}}^{d,w}$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{n-1}-xw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Obviously, we have

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

Then $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)xw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n}}=0$, as desired.

⟸ By hypothesis, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}\hfill x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \hfill \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. Then we check that

$$\begin{array}{ccc}\hfill \left|\right|a_{2}w{a}_1\left|\right|& =\hfill & \left|\right|[a-{\left(aw\right)}^{2}x]w{\left(aw\right)}^{2}x\left|\right|\hfill \\ & =\hfill & \left|\right|[aw-{\left(aw\right)}^{2}xw]{\left(aw\right)}^{2}x\left|\right|\hfill \\ & =\hfill & \left|\right|\left(aw\right)[aw-(aw\left)xw\right(aw\left)\right]\left(aw\right)x\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)[aw-\left(aw\right)xw\left(aw\right)]\left(aw\right)}^{k-1}{x}^{k-1}\left|\right|\hfill \\ & =\hfill & \left|\right|\left(aw\right)[{\left(aw\right)}^k-\left(aw\right)xw{\left(aw\right)}^k]x^{k-1}\left|\right|.\hfill \end{array}$$

Hence

$$\left|\right|a_{2}w{a}_1{\left|\right|}^{\frac{1}{k}}\le {\left|\right|aw\left|\right|}^{\frac{1}{k}}{\left|\right|\left(aw\right)}^k-\left(aw\right)xw{\left(aw\right)}^k{\left|\right|}^{\frac{1}{k}}{\left|\right|x\left|\right|}^{1-\frac{1}{k}}.$$

Since $\underset{k\to \infty }{lim}{\left|\right|\left(aw\right)}^k-\left(aw\right)xw{\left(aw\right)}^k{\left|\right|}^{\frac{1}{k}}=0,$ we deduce that

$$\underset{k\to \infty }{lim}\left|\right|a_{2}w{a}_1{\left|\right|}^{\frac{1}{k}}=0.$$

Thus, $a_{2}w{a}_1=0$.

As in the proof of Theorem 2.2, we check that $a_1^{*}{a}_2=0$.

Let $y=wxw$. Then we verify that

$$\begin{array}{ccc}\hfill a_{1}wy& =\hfill & {\left(aw\right)}^{2}xwxwx=awa{\left(wx\right)}^{2}wx=a{\left(wx\right)}^2\hfill \\ & =\hfill & x=xwxw{\left(aw\right)}^{2}x=xwxw{\left(aw\right)}^{2}x=yw{a}_1,\hfill \\ \hfill a_{1}wyw{a}_1& =\hfill & {\left(aw\right)}^{2}xwxw{\left(aw\right)}^{2}x=awa{\left(wx\right)}^{2}wawawx\hfill \\ & =\hfill & \left(awxw\right){\left(aw\right)}^{2}x={\left(aw\right)}^{2}x=a_1,\hfill \\ \hfill yw{a}_{1}wy& =\hfill & a_1{\left(wx\right)}^2={\left(aw\right)}^{2}x{\left(wx\right)}^2=awa{\left(wx\right)}^{2}wx\hfill \\ & =\hfill & a{\left(wx\right)}^2=y,\hfill \end{array}$$

Hence ${\left(a_1\right)}_w^{\#}=y$. Similarly to ([2] Theorem 2.2), $a_2=a-{\left(aw\right)}^{2}x\in {\mathcal{A}}_w^{qnil}$. Therefore $x=a_w^g$, as required. □

Corollary 3.2.

Let $A,W\in C^{n\times n}$. Then the following are equivalent:

(1)

A has weak W-group inverse.

(2)

There exist $k\in N$ and $X\in C^{n\times n}$ such that $X=A{\left(WX\right)}^2,$${\left(AW\right)}^k=AWXW{A}^k,{\left({\left(AW\right)}^D\right)}^{*}{\left(AW\right)}^{2}X={\left({AW)}^D\right)}^{*}A.$

Proof. 

This is clear by Theorem 3.3. □

We are ready to prove:

Theorem 3.3.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x\mathcal{A}=a^{d,w}\mathcal{A},{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

In this case, $a_w^g=x$.

Proof. 

⟹ In view of Theorem 2.4, $aw\in {\mathcal{A}}^d$ and there exists unique $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Hence, $a\in {\mathcal{A}}^{d,w}$ by 3.1. Obviously, $xw={\left(aw\right)}^g$. In light of ([2] Theorem 3.4),

$$\begin{array}{ccc}\hfill x& =\hfill & a{\left(wx\right)}^2=aw{\left(aw\right)}^{g}x={\left(aw\right)}^2{\left(aw\right)}^d{\left(aw\right)}^{g}x\hfill \\ & =\hfill & {\left[{\left(aw\right)}^d\right]}^2{\left(aw\right)}^3{\left(aw\right)}^{g}x=a^{d,w}\left[w{\left(aw\right)}^2{\left(aw\right)}^{g}x\right].\hfill \end{array}$$

One easily checks that

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

We infer that

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^d-xw\left(aw\right){\left(aw\right)}^d{\left|\right|}^{\frac{1}{n}}=0.$$

Hence, ${\left(aw\right)}^d=xw\left(aw\right){\left(aw\right)}^d$. Therefore $x\mathcal{A}=a^{d,w}\mathcal{A}$, as required.

⟸ We directly check that

$$\begin{array}{cc}& {\left|\right|awxw\left(aw\right)}^d-{\left(aw\right)}^d{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill =& {\left|\right|awxw\left(aw\right)}^{n+1}{\left({\left(aw\right)}^d\right)}^{n+2}-aw{\left(aw\right)}^n{\left({\left(aw\right)}^d\right)}^{n+2}{\left|\right|}^{\frac{1}{n}}\hfill \\ \hfill \le & {\left|\right|aw|}^{\frac{1}{n}}{\left|\right||\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{|}^{\frac{1}{n}}\left|\right||{\left({\left(aw\right)}^d\right)}^{n+2}{\left|\right|}^{\frac{1}{n}}.\hfill \end{array}$$

Accordingly,

$$\underset{n\to \infty }{lim}{\left|\right|awxw\left(aw\right)}^d-{\left(aw\right)}^d{\left|\right|}^{\frac{1}{n}}=0.$$

Therefore we have $awxw{\left(aw\right)}^d={\left(aw\right)}^d$, and so $1-awxw\in \ell {\left(aw\right)}^d$. Since $x\mathcal{A}={\left(aw\right)}^d\mathcal{A}$, we see that $1-awxw\in \ell \left(x\right)$. Therefore $x=awxwx=a{\left(wx\right)}^2$. In light of Theorem 2.2, $a\in {\mathcal{A}}_w^g$. In this case, $a_w^g=x$, as asserted. □

Corollary 3.4.

Let $A,W\in C^{n\times n}$. Then the following are equivalent:

(1)

A has weak W-group inverse.

(2)

There exist $k\in N$ and $X\in C^{n\times n}$ such that

$$\begin{array}{c}{\left(AW\right)}^k=XW{A}^{k+1},Im\left(X\right)=Im\left(A^{D,W}\right),\\ {\left({\left(AW\right)}^D\right)}^{*}{\left(AW\right)}^{2}X={\left({AW)}^D\right)}^{*}A.\end{array}$$

Proof. 

This is obvious by Theorem 3.3. □

We next investigate the polar-like property of generalized weighted group inverse.

Lemma 3.5.

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

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists an idempotent $p\in \mathcal{A}$ such that

$$wa+p\in {\mathcal{A}}^{-1},{\left({\left(aw\right)}^d\right)}^{*}ap=0andpwa=pwap\in {\mathcal{A}}^{qnil}.$$

Proof. 

In view of Theorem 3.1, $a\in {\mathcal{A}}^{d,w}$. Since $a\in R_w^g$, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}y=ywz=0,z\in {\mathcal{A}}_w^{\#},y\in {\mathcal{A}}_w^{qnil}.$$

Set $x=z_w^{\#}$. The we check that

$$\begin{array}{ccc}\hfill awx& =\hfill & (z+y)w{z}_w^{\#}=zw{z}_w^{\#},\hfill \\ \hfill {\left(wawx\right)}^2& =\hfill & wzw{z}_w^{\#}wzw{z}_w^{\#}=wzw{z}_w^{\#}=wawx,\hfill \end{array}$$

Let $p=1-wzw{z}_w^{\#}$. Then $p=1-wawx=p^2\in \mathcal{A}$. Furthermore, $apw=a(1-wzw{z}_w^{\#})w=(z+y)(1-wzw{z}_w^{\#})w=yw+z(1-wzw{z}_w^{\#})w\in {\mathcal{A}}^{qnil}$, and so $pwa\in {\mathcal{A}}^{qnil}$ by Cline’s formula (see ([7] Theorem 2.1)). Since $pwa(1-p)=(1-wzw{z}_w^{\#})w(z+y)wzw{z}_w^{\#}=(1-wzw{z}_w^{\#})wzwzw{z}_w^{\#}=wzwzw{z}_w^{\#}-wzw{z}_w^{\#}wzwzw{z}_w^{\#}=wzwzw{z}_w^{\#}-wzwzw{z}_w^{\#}=0$, we get $pwa=pwap$.

By Theorem 2.4, ${\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^2{a}_w^g={\left({\left(aw\right)}^d\right)}^{*}a$, and so ${\left({\left(aw\right)}^d\right)}^{*}awaw{z}_w^{\#}={\left({\left(aw\right)}^d\right)}^{*}a$. This implies that ${\left({\left(aw\right)}^d\right)}^{*}a[1-wzw{z}_w^{\#}]={\left({\left(aw\right)}^d\right)}^{*}a[1-w(z+y)w{z}_w^{\#}]={\left({\left(aw\right)}^d\right)}^{*}a[1-waw{z}_w^{\#}]=0$, i.e., ${\left({\left(aw\right)}^d\right)}^{*}ap=0$. Obviously, $wz+1-wzw{z}_w^{\#}={(w{z}_w^{\#}+1-wzw{z}_w^{\#})}^{-1}\in {\mathcal{A}}^{-1}$. Since $y(w{z}_w^{\#}+1-wzw{z}_w^{\#})=y$, we have $y{(w{z}_w^{\#}+1-wzw{z}_w^{\#})}^{-1}=y\in {\mathcal{A}}_w^{qnil}$. It follows by ([7] Theorem 2.1) that ${(w{z}_w^{\#}+1-wzw{z}_w^{\#})}^{-1}wy\in {\mathcal{A}}^{qnil}$. Hence $1+{(w{z}_w^{\#}+1-wzw{z}_w^{\#})}^{-1}wy\in {\mathcal{A}}^{-1}$. This implies that

$$\begin{array}{ccc}\hfill wa+p& =\hfill & wz+wy+1-wzw{z}_w^{\#}\hfill \\ & =\hfill & (wz+1-wzw{z}_w^{\#})[1+{(w{z}_w^{\#}+1-wzw{z}_w^{\#})}^{-1}wy]\hfill \\ & \in \hfill & {\mathcal{A}}^{-1}.\hfill \end{array}$$

This completes the proof. □

Lemma 3.6.

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_w^{\#}$ if and only if $aw,wa\in {\mathcal{A}}^{\#}$. In this case,

$$a_w^{\#}={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}.$$

Proof. 

⟹ This is clear by Lemma 2.1.

⟸ Let $x={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}$. Then we directly check that $x=a_w^{\#}$, as desired. □

Theorem 3.7.

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists an idempotent $p\in \mathcal{A}$ such that

$$a+w^{-1}p\in {\mathcal{A}}^{-1},{\left({\left(aw\right)}^d\right)}^{*}ap=0andpwa=pwap\in {\mathcal{A}}^{qnil}.$$

Proof. 

⟹ This is obvious by Lemma 3.5.

⟸ By hypothesis, there exists an idempotent $p\in \mathcal{A}$ such that

$$wa+p\in {\mathcal{A}}^{-1},{\left({\left(aw\right)}^d\right)}^{*}ap=0andpwa=pwap\in {\mathcal{A}}^{qnil}.$$

Set $x=a(1-p)$ and $y=ap$. Then $a=x+y$. Since $pwa=pwap\in {\mathcal{A}}^{qnil}$, we have $yw=apw\in {\mathcal{A}}^{qnil}$ by ([7] Theorem 2.1). Hence, $y\in {\mathcal{A}}_w^{qnil}$. We also see that $pwa(1-p)=0$, and then $ywx=apwa(1-p)=0$. Clearly, $wx=wa(1-p)=u(1-p)$. Let $z=(1-p)u^{-1}$. Then we check that $zwx=1-p$; hence, $z{\left(wx\right)}^2=(1-p)wx=(1-p)wa(1-p)=wa(1-p)=wx$. Since $a\in {\mathcal{A}}^d,ap\in {\mathcal{A}}^d$ and $pwa(1-p)=0$, it follows by ([17] Lemma 2.2) that $wx=wa(1-p)\in {\mathcal{A}}^d$. Therefore $wx=z{\left(wx\right)}^2=z^n{\left(wx\right)}^{n+1}=z^n({\left(wx\right)}^n-{\left(wx\right)}^d{\left(wx\right)}^{n+1})wx+z^n{\left(wx\right)}^d{\left(wx\right)}^{n+2}$ and ${\left(wx\right)}^2{\left(wx\right)}^d=\left(z{\left(wx\right)}^2\right)wx{\left(wx\right)}^d=z^n{\left(wx\right)}^{n+2}{\left(wx\right)}^d$. Then

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

Accordingly,

$$\underset{n\to \infty }{lim}\left|\right|wx-{\left(wx\right)}^2{\left(wx\right)}^d{\left|\right|}^{\frac{1}{n}}=0.$$

Therefore we have $wx={\left(wx\right)}^2{\left(wx\right)}^d$. By using Cline’s formula (see ([7] Theorem 2.1)), we have

$$\begin{array}{ccc}\hfill xw& =\hfill & w^{-1}\left(wx\right)w=w^{-1}\left[{\left(wx\right)}^2{\left(wx\right)}^d\right]w\hfill \\ & =\hfill & w^{-1}wxwxw{\left[{\left(xw\right)}^d\right]}^{2}xw\hfill \\ & =\hfill & xwxw{\left[{\left(xw\right)}^d\right]}^{2}xw\hfill \\ & =\hfill & {\left(xw\right)}^2{\left(xw\right)}^d.\hfill \end{array}$$

Hence, $x\in {\mathcal{A}}_w^{\#}$. Therefore

$$\begin{array}{c}\hfill w^{*}{x}^{*}yw={\left[xw\right]}^{*}apw={\left({\left(xw\right)}^2\right)}^{*}{\left({\left(xw\right)}^d\right)}^{*}apw=0.\end{array}$$

Since $w\in {\mathcal{A}}^{-1}$, we have $w^{*}\in {\mathcal{A}}^{-1}$. Thus, $x^{*}y=0$. Then $a=x+y$ is a generalized w-group decomposition of a. In light of Theorem 2.2, $a\in {\mathcal{A}}_w^g$. □

Corollary 3.8.

Let $A\in C^{n\times n}$ and $W\in C^{n\times n}$ be invertible, with conjugate transpose * as the involution. Then the following are equivalent:

(1)

A has weak W-group inverse.

(2)

There exists an idempotent $E\in C^{n\times n}$ such that $A+W^{-1}E$ is invertible, ${\left({\left(AW\right)}^D\right)}^{*}AE=0andEWA=EWAE$ is nilpotent.

Proof. 

As is well known, every complex has Drazin inverse. This completes the proof by Theorem 3.7. □

Example 3.9.

Let $C^{2\times 2}$ be the Banach algebra of all $2\times 2$ complex matrices, with transpose * as the involution. Let $A=\left(\begin{array}{cc}1& 0\\ -i& 0\end{array}\right),$$W=\left(\begin{array}{cc}1& 0\\ 0& -i\end{array}\right)\in C^{2\times 2}$. Then the equations

$$\begin{array}{c}A{\left(WX\right)}^2=X,{\left({\left(AW\right)}^{*}{\left(AW\right)}^{2}X\right)}^{*}={\left(AW\right)}^{*}{\left(AW\right)}^{2}X,\\ \underset{n\to \infty }{lim}{\left|\right|\left(AW\right)}^n-XW{\left(AW\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0\end{array}$$

has two solutions

$$X_1=\left(\begin{array}{cc}1& 0\\ -i& 0\end{array}\right),X_2=\left(\begin{array}{cc}0& -1\\ 0& i\end{array}\right).$$

Then the solution of the preceding equations is not unique. Choose an idempotent $E=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$. Then $A+W^{-1}E$ is invertible, ${\left({\left(AW\right)}^D\right)}^{*}AE=0,EWA=EWAE$ is nilpotent. In this case, the involution * is not proper.

4. Relations with Weighted g-Drazin Inverses

In this section we discuss the relations between generalized weighted group and weighted g-Drazin inverses. We now derive

Theorem 4.1.

Let $a\in \mathcal{A},w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists an idempotent $q\in \mathcal{A}$ such that

$$a^{d,w}\mathcal{A}=q\mathcal{A}and{\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$$

In this case, $a_w^g=a^{d,w}wq{w}^{-1}.$

Proof. 

⟹ By virtue of Theorem 3.1, $a\in {\mathcal{A}}^{d,w}$. Let $q=aw{a}_w^{g}w$. Since $a_w^g=a_w^{g}waw{a}_w^g.$ Then $q^2=q$, i.e., $q\in \mathcal{A}$ is an idempotent. We check that

$${\left(aw\right)}^{*}\left(aw\right)q={\left(aw\right)}^{*}{\left(aw\right)}^2{a}_w^{g}w.$$

In view of Corollary 2.3, ${\left(aw\right)}^{*}\left(aw\right)q={\left[{\left(aw\right)}^{*}\left(aw\right)q\right]}^{*}=q^{*}{\left(aw\right)}^{*}\left(aw\right)$. As in the proof of Theorem 3.3, we have

$$a_w^g=a^{d,w}\left[w{\left(aw\right)}^2{\left(aw\right)}^g{a}_w^g\right].$$

Then

$$q=aw{a}_w^{g}w=aw{a}^{d,w}\left[w{\left(aw\right)}^2{\left(aw\right)}^g{a}_w^g\right]w.$$

Hence, $q\mathcal{A}\subseteq a^{d,w}\mathcal{A}$. In light of Theorem 3.3, $a_w^g\mathcal{A}=a^{d,w}\mathcal{A}$. Write $a^{d,w}=a_w^{g}z$ for some $z\in \mathcal{A}$. Then $a^{d,w}=a{\left(w{a}_w^g\right)}^{2}z=aw{a}_w^{g}waw{\left(a_w^g\right)}^{2}z=qaw{\left(a_w^g\right)}^{2}z$. Hence $a^{d,w}\mathcal{A}\subseteq q\mathcal{A}.$ Therefore $q\mathcal{A}=a^{d,w}A$, as required.

⟸ By hypothesis, there exists an idempotent $q\in \mathcal{A}$ such that $a^{d,w}\mathcal{A}=q\mathcal{A}$ and ${\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$ Hence, $q\mathcal{A}=aw{a}^{d,w}w\mathcal{A}$. Set $x=a^{d,w}wq{w}^{-1}$. Then $awxw=aw{a}^{d,w}wq=q$. We verify that

$$\begin{array}{ccc}\hfill {\left({\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right)}^{*}& =\hfill & {\left({\left(aw\right)}^{*}\left(aw\right)awxw\right)}^{*}={\left({\left(aw\right)}^{*}\left(aw\right)q\right)}^{*}\hfill \\ & =\hfill & q^{*}{\left(aw\right)}^{*}\left(aw\right)={\left(aw\right)}^{*}\left(aw\right)q\hfill \\ & =\hfill & \left({\left(aw\right)}^{*}\left(aw\right)\right)\left(awxw\right)={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw.\hfill \end{array}$$

Moreover, we have

$$a{\left(wx\right)}^2=\left(awxw\right)x=qx=q{a}^{d,w}wq{w}^{-1}=a^{d,w}wq{w}^{-1}=x.$$

We verify that

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}\left|\right|\hfill \\ \hfill =& \left|\right|[{\left(aw\right)}^n-a^{d,w}wq{a}^{d,w}w{\left(aw\right)}^{n+2}]\hfill \\ \hfill -& \left[xw({\left(aw\right)}^{n+1}-xw{a}^{d,w}w{\left(aw\right)}^{n+2})\right]\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(a^{d,w}w\right)}^2{\left(aw\right)}^{n+2}\left|\right|+\left|\right|xw\left|\right|\left|\right|{\left(aw\right)}^{n+1}-a^{d,w}w{\left(aw\right)}^{n+2}\left|\right|\hfill \\ \hfill =& {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|+{\left|\right|xw\left|\right|\left|\right|\left(aw\right)}^{n+1}-a^{d,w}w{\left(aw\right)}^{n+2}\left|\right|\hfill \\ \hfill =& \left(1+\left|\right|xw\left|\right|\left|\right|aw\left|\right|\right){\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|.\hfill \end{array}$$

Since $aw-a^{d,w}w{\left(aw\right)}^2\in {\mathcal{A}}^{qnil}$, we have

$$\underset{n\to \infty }{lim}\left|\right|{(aw-a^{d,w}w{\left(aw\right)}^2)}^n{\left|\right|}^{\frac{1}{n}}=0.$$

Therefore

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.$$

In view of Corollary 2.3, $a\in {\mathcal{A}}_w^g$. In this case, $a_w^g=a^{d,w}wq{w}^{-1}.$ □

Corollary 4.2.

Let $a\in \mathcal{A},w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists a unique idempotent $q\in \mathcal{A}$ such that

$$a^{d,w}\mathcal{A}=q\mathcal{A}and{\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$$

Proof. 

⟹ By hypothesis, there exists an idempotent $q\in \mathcal{A}$ such that

$$a^{d,w}\mathcal{A}=q\mathcal{A}and{\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$$

In this case, $q=aw{a}_w^{g}w$. In view of Theorem 2.4, $q\in \mathcal{A}$ is unique.

⟸ This is obvious by Theorem 4.1. □

Corollary 4.3.

Let $a\in \mathcal{A},w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists an idempotent $q\in \mathcal{A}$ such that

$$\ell \left(a^{d,w}\right)=\ell \left(q\right)and{\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$$

In this case, $a_w^g=a^{d,w}wq{w}^{-1}.$

Proof. 

⟹ In view of Theorem 4.1, $a\in {\mathcal{A}}^{d,w}$ and there exists an idempotent $q\in \mathcal{A}$ such that

$$a^{d,w}\mathcal{A}=q\mathcal{A}and{\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$$

We infer that $\ell \left(a^{d,w}\right)=\ell \left(q\right)$, as desired.

⟸ By hypothesis, there exists an idempotent $q\in \mathcal{A}$ such that $\ell \left(a^{d,w}\right)=\ell \left(q\right)$ and ${\left(aw\right)}^{*}\left(aw\right)q=q^{*}{\left(aw\right)}^{*}aw.$ Clearly, $1-aw{a}^{d,w}w\in \ell \left(a^{d,w}\right)\subseteq \ell \left(q\right)$; hence, $q=aw{a}^{d,w}wq$. This implies that $q\mathcal{A}\subseteq a^{d,w}\mathcal{A}.$ Also we have $1-q\in \ell \left(q\right)\subseteq \ell \left(a^{d,w}\right)$, and then $a^{d,w}=q{a}^{d,w}$. We infer that $a^{d,w}\mathcal{A}\subseteq q\mathcal{A}$. Therefore $a^{d,w}\mathcal{A}=q\mathcal{A}$. In light of Theorem 4.1, $a\in {\mathcal{A}}_w^g$. In this case, $a_w^g=a^{d,w}wq{w}^{-1}.$ □

We are ready to prove:

Theorem 4.4.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_w^g$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists some $x\in \mathcal{A}$ such that ${\left(a^{d,w}\right)}^{*}{a}^{d,w}x={\left(a^{d,w}\right)}^{*}a.$

In this case, $a_w^g={\left({\left(aw\right)}^d\right)}^{4}ax.$

Proof. 

⟹ By virtue of Theorem 2.4, $a\in {\mathcal{A}}^{d,w}$ and there exists $z\in \mathcal{A}$ such that

$$\begin{array}{c}z=a{\left(wz\right)}^2,{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}z={\left({\left(aw\right)}^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-zw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Here, $z=a_w^g$. Let $y={\left(aw\right)}^{3}z$. Obviously, $z=\left(aw\right)z\left(wz\right)={\left(aw\right)}^{2}z{\left(wz\right)}^2={\left(aw\right)}^{k-2}z{\left(wz\right)}^{k-2}$. Then we verify that

$$\begin{array}{cc}& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}a-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}y\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}a-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^d{\left(aw\right)}^{3}z\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}a-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^d{\left(aw\right)}^3\left[{\left(aw\right)}^{k-2}z{\left(wz\right)}^{k-2}\right]\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}z-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^d{\left(aw\right)}^{k+1}{\left(zw\right)}^{k-2}z\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^2\left[{\left(aw\right)}^{k-2}z{\left(wz\right)}^{k-2}\right]\hfill \\ \hfill -& {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^d{\left(aw\right)}^{k+1}{\left(zw\right)}^{k-2}z\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^k{\left(zw\right)}^{k-2}z-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^d{\left(aw\right)}^{k+1}{\left(zw\right)}^{k-2}z\left|\right|\hfill \\ \hfill \le & \left|\right|{\left({\left(aw\right)}^d\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^k-{\left(aw\right)}^d{\left(aw\right)}^{k+1}{\left|\right|\left|\right|\left(zw\right)}^{k-2}z\left|\right|.\hfill \end{array}$$

Since $\underset{k\to \infty }{lim}{\left|\right|\left(aw\right)}^k-{\left(aw\right)}^d{\left(aw\right)}^{k+1}{\left|\right|}^{\frac{1}{k}}=0$, we derive that

$$\underset{k\to \infty }{lim}\left|\right|{\left({\left(aw\right)}^d\right)}^{*}a-{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}y{\left|\right|}^{\frac{1}{k}}=0.$$

Hence ${\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}y={\left({\left(aw\right)}^d\right)}^{*}a.$

Set $x=waw{\left(aw\right)}^{d}y$. Since $a^{d,w}={\left[{\left(aw\right)}^d\right]}^{2}a$, we see that

$$\begin{array}{ccc}\hfill {\left(a^{d,w}\right)}^{*}{a}^{d,w}x& =\hfill & {\left({\left[{\left(aw\right)}^d\right]}^{2}a\right)}^{*}{\left[{\left(aw\right)}^d\right]}^{2}ax\hfill \\ & =\hfill & a^{*}{\left({\left(aw\right)}^d\right)}^{*}{\left({\left(aw\right)}^d\right)}^{*}{\left[{\left(aw\right)}^d\right]}^{2}ax\hfill \\ & =\hfill & a^{*}{\left({\left(aw\right)}^d\right)}^{*}{\left({\left(aw\right)}^d\right)}^{*}{\left[{\left(aw\right)}^d\right]}^{2}a\left(waw{\left(aw\right)}^{d}y\right)\hfill \\ & =\hfill & a^{*}{\left({\left(aw\right)}^d\right)}^{*}\left[{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}y\right]\hfill \\ & =\hfill & a^{*}{\left({\left(aw\right)}^d\right)}^{*}{\left({\left(aw\right)}^d\right)}^{*}a\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^d\right]}^{2}a\right)}^{*}a\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}a,\hfill \end{array}$$

as required.

⟸ By hypothesis, ${\left(a^{d,w}\right)}^{*}{a}^{d,w}x={\left(a^{d,w}\right)}^{*}a.$ for some $x\in \mathcal{A}$. Set $z={\left(aw\right)}^{d}ax$. Then ${\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}z={\left({\left(aw\right)}^d\right)}^{*}a$. We verify that

$$\begin{array}{ccc}\hfill {\left[aw{\left(aw\right)}^d\right]}^{*}aw{\left(aw\right)}^d& =\hfill & {\left(aw\right)}^{*}\left[{\left({\left(aw\right)}^d\right)}^{*}a\right]w{\left(aw\right)}^d\hfill \\ & =\hfill & {\left(aw\right)}^{*}\left[{\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}z\right]w{\left(aw\right)}^d\hfill \\ & =\hfill & {\left[aw{\left(aw\right)}^d\right]}^{*}{\left(aw\right)}^{d}zw{\left(aw\right)}^d.\hfill \end{array}$$

Since the involution * is proper, we get $aw{\left(aw\right)}^d={\left(aw\right)}^{d}zw{\left(aw\right)}^d$. Let $y={\left({\left(aw\right)}^d\right)}^{3}z$. Then we verify that

$$\begin{array}{ccc}\hfill a{\left(wy\right)}^2& =\hfill & aw{\left({\left(aw\right)}^d\right)}^{3}zw{\left({\left(aw\right)}^d\right)}^{3}z\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{2}zw{\left({\left(aw\right)}^d\right)}^{3}z\hfill \\ & =\hfill & {\left(aw\right)}^d\left[{\left(aw\right)}^{d}zw{\left(aw\right)}^d\right]{\left({\left(aw\right)}^d\right)}^{2}z\hfill \\ & =\hfill & {\left(aw\right)}^{d}aw{\left(aw\right)}^d{\left({\left(aw\right)}^d\right)}^{2}z\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{3}z=y;\hfill \\ \hfill {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{2}y& =\hfill & {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^2{\left({\left(aw\right)}^d\right)}^{3}z\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{d}z\hfill \\ & =\hfill & {\left({\left(aw\right)}^d\right)}^{*}a.\hfill \end{array}$$

Moreover, we see that

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-yw{\left(aw\right)}^{n+1}\left|\right|\hfill \\ \hfill =& \left|\right|[{\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}]+[{\left(aw\right)}^d{\left(aw\right)}^{n+1}-yw{\left(aw\right)}^{n+1}]\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|+{\left|\right|\left(aw\right)}^d{\left(aw\right)}^{n+1}-{\left({\left(aw\right)}^d\right)}^{3}zw{\left(aw\right)}^{n+1}\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|+\left|\right|{\left({\left(aw\right)}^d\right)}^{3}zw[1-{\left(aw\right)}^d\left(aw\right)]{\left(aw\right)}^{n+1}\left|\right|\hfill \\ \hfill +& {\left|\right|\left(aw\right)}^d{\left(aw\right)}^{n+1}-{\left({\left(aw\right)}^d\right)}^{3}zw{\left(aw\right)}^d{\left(aw\right)}^{n+2}\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|+{\left|\right|\left(aw\right)}^d{)}^{3}zw\left|\right||\hfill \\ & {|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|\left|\right|\left(aw\right)}^{n+1}\left|\right|\hfill \\ \hfill +& {\left|\right|\left(aw\right)}^d{\left(aw\right)}^{n+1}-{\left(aw\right)}^d{)}^2\left[{\left(aw\right)}^{d}zw{\left(aw\right)}^d\right]{\left(aw\right)}^{n+2}\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|+{\left|\right|\left(aw\right)}^d{)}^3{zw\left|\right|\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|\hfill \\ & {\left|\right|\left(aw\right)}^{n+1}\left|\right|+{\left|\right|\left(aw\right)}^d{\left(aw\right)}^{n+1}-{\left({\left(aw\right)}^d\right)}^{2}aw{\left(aw\right)}^d{\left(aw\right)}^{n+2}\left|\right|\hfill \\ \hfill =& {[1+||\left(aw\right)}^d{)}^3{zw\left|\right|\left|\right|\left(aw\right)}^{n+1}{\left|\right|\left]\right||\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}\left|\right|.\hfill \end{array}$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0$, we deduce that

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-yw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.$$

In light of Theorem 2.4, $a\in {\mathcal{A}}_w^g$. In this case, $a_w^g=y={\left({\left(aw\right)}^d\right)}^{3}z={\left({\left(aw\right)}^d\right)}^3{\left(aw\right)}^{d}ax={\left({\left(aw\right)}^d\right)}^{4}ax.$ This completes the proof. □

Recall that $a\in R$ has weak w-group inverse provide that there exist $x\in R$ and $n\in N$ such that

$$x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^n\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}^n\right)}^{*}a,{\left(aw\right)}^n=xw{\left(aw\right)}^{n+1}.$$

If x exists, it is unique, and we denote it by $a_w^W$.

Lemma 4.5.

Let $a\in \mathcal{A},w\in {\mathcal{A}}^{-1}$. Then $a\in R_w^W$ if and only if $a\in R^{D,w}\bigcap R_w^g$. In this case, $a_w^W=a_w^g$.

Proof. 

⟹ Obviously, $a\in R_w^g$. We easily check that $aw\in R_w^W$. Then $a\in R^{D,w}$ by ([19] Proposition 3.4).

⟸ Let $x=a_w^g$. Then there exist $x\in R$ and $m\in N$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{\frac{1}{n}}=0.\end{array}$$

Since $a\in R^{D,w}$, we can find some $y\in R$ such that

$$yw{\left(aw\right)}^{k+1}={\left(aw\right)}^k,ywawy=y,awy=ywa$$

for some $k\in N$. Set $z={\left(aw\right)}^{k}y{\left(wx\right)}^k$. Then we verify that

$$\begin{array}{ccc}\hfill awz& =\hfill & ({\left(aw\right)}^{k+1}yw{\left(xw\right)}^{k-1}x={\left(aw\right)}^k{\left(xw\right)}^{k-1}xw\hfill \\ & =\hfill & {\left(aw\right)}^k{\left(xw\right)}^{k-1}x={\left(aw\right)}^{2}xwx.\hfill \end{array}$$

Claim 1. $z=a{\left(wz\right)}^2$.

$$\begin{array}{ccc}\hfill a{\left(wz\right)}^2& =\hfill & aw{\left(aw\right)}^{k}y{\left(wx\right)}^{k}w{\left(aw\right)}^{k}y{\left(wx\right)}^k\hfill \\ & =\hfill & {\left(aw\right)}^{k+1}yw{\left(xw\right)}^k{\left(aw\right)}^{k}y{\left(wx\right)}^k\hfill \\ & =\hfill & {\left(aw\right)}^k{\left(xw\right)}^k{\left(aw\right)}^{k}y{\left(wx\right)}^k=\left(aw\right)xw{\left(aw\right)}^{k}y{\left(wx\right)}^k\hfill \\ & =\hfill & \left(aw\right)xw{\left(aw\right)}^k\left(yw\right){\left(xw\right)}^{k-1}x={\left(aw\right)}^k\left(yw\right){\left(xw\right)}^{k-1}x\hfill \\ & =\hfill & {\left(aw\right)}^{k}y{\left(wx\right)}^k=z.\hfill \end{array}$$

Claim 2. ${\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}zw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}zw.$

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{*}{\left(aw\right)}^{2}zw& =\hfill & {\left(aw\right)}^{*}\left(aw\right)\left(awz\right)w={\left(aw\right)}^{*}\left(aw\right){\left(aw\right)}^{2}xwxw\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}a{\left(wx\right)}^{2}w={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw;\hfill \end{array}$$

hence, ${\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}zw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}zw.$

Claim 3. ${\left(aw\right)}^k=\left(zw\right){\left(aw\right)}^{k+1}.$

Since $\left(aw\right)\left(yw\right)=\left(yw\right)\left(aw\right)$, we see that

$$\begin{array}{ccc}\hfill \left(zw\right){\left(aw\right)}^{k+n+1}& =\hfill & \left[{\left(aw\right)}^{k}y{\left(wx\right)}^{k}w\right]{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & {\left(aw\right)}^k\left(yw\right){\left(xw\right)}^k{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left(yw\right)\left[{\left(aw\right)}^k{\left(xw\right)}^k\right]{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left(yw\right)\left[\left(aw\right)\left(xw\right)\right]{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left(aw\right)\left(yw\right)\left(xw\right){\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left(aw\right)\left(yw\right)\left[{\left(aw\right)}^k{\left(xw\right)}^{k+1}\right]{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left[\left(yw\right){\left(aw\right)}^{k+1}\right]{\left(xw\right)}^{k+1}{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & {\left(aw\right)}^k{\left(xw\right)}^{k+1}{\left(aw\right)}^{k+n+1}\hfill \\ & =\hfill & \left(xw\right){\left(aw\right)}^{k+n+1}.\hfill \end{array}$$

Hence, we deduce that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}& =\hfill & {\left(aw\right)}^{k+n}{\left(yw\right)}^n-\left(zw\right){\left(aw\right)}^{k+n+1}{\left(yw\right)}^n\hfill \\ & =\hfill & [{\left(aw\right)}^{k+n}-xw{\left(aw\right)}^{k+n+1}]{\left(yw\right)}^n.\hfill \end{array}$$

Then

$${\left|\right|\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}\left|\right|\le {\left|\right|\left(aw\right)}^{k+n}-xw{\left(aw\right)}^{k+n+1}{\left|\right|\left|\right|yw\left|\right|}^n.$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{k+n}-xw{\left(aw\right)}^{k+n+1}{\left|\right|}^{\frac{1}{n}}=0,$ we see that

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}{\left|\right|}^{\frac{1}{n}}=0.$$

Hence, ${\left(aw\right)}^k=\left(zw\right){\left(aw\right)}^{k+1}$. Therefore $a\in R_w^W$. □

Theorem 4.6.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_w^W$ if and only if

(1)

$a\in {\mathcal{A}}^{D,w}$;

(2)

There exists some $x\in \mathcal{A}$ such that ${\left(a^{D,w}\right)}^{*}{a}^{D,w}x={\left(a^{D,w}\right)}^{*}a.$

In this case, $a_w^W={\left({\left(aw\right)}^D\right)}^{4}ax.$

Proof. 

This is an immediate consequence of Theorem 4.4 and Lemma 4.5. □

Corollary 4.7.

Let $A,W\in C^{n\times n}$, with conjugate transpose * as the involution. Then the following are equivalent:

(1)

A has weak W-group inverse.

(2)

There exists some $X\in C^{n\times n}$ such that ${\left(A^{D,W}\right)}^{*}{A}^{D,W}X={\left(A^{D,W}\right)}^{*}A.$

In this case, $A_W^W={\left({\left(AW\right)}^D\right)}^{4}AX.$

Proof. 

This is an immediate consequence of Theorem 4.6. □