Weighted w-EP Element and Its Associated Decomposition

1. Introduction

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

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

Such an x is unique, if it exists, and is denoted by $a^{\#}$, termed the group inverse of a (see [21]).

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

$$x{a}^2=a,a{x}^2=x,{\left(ax\right)}^{*}=ax.$$

If such x exists, it is unique, and denote it by $a^{\#}$ (see [7,24,26]). Dually, an element $a\in \mathcal{A}$ has dual core inverse if and only if there exist $x\in \mathcal{A}$ such that

$$a^{2}x=a,x^{2}a=x,{\left(xa\right)}^{*}=xa.$$

If such x exists, it is unique, and denote it by $a_{\#}$.

Rakí introduced the EP element in rings and algebras, integrating concepts from both core and dual-core elements (see [21]). An element $a\in \mathcal{A}$ is an EP element if $a\in {\mathcal{A}}^{\#}\bigcap {\mathcal{A}}_{\#}$ and $a^{\#}=a_{\#}$. Evidently, $a\in \mathcal{A}$ is an EP element if and only if there exists $x\in \mathcal{A}$ such that

$$a^{2}x=a,x^{2}a=x,{\left(ax\right)}^{*}=xa$$

(see [25]). The EP element has been applied to many fields such as control theory and image processing (see [1,10]). It has been extensively studied by numerous authors from various perspectives, e.g., [2,9,12,13,16,17,18,22,23,27].

Recently, many authors studied various weighted generalized inverses, e.g., [11,14,15,20]. An element $a\in \mathcal{A}$ has w-group inverse if there exists $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awx=xwa.$$

We denote the preceding x by $a_w^{\#}$. The set of all generalized w-group invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_w^{\#}$. In [6], Chen and Sheibani introduced a new weighted generalized inverse as a natural generalization of core inverse and w-group inverse.

Definition 1. 

An element $a\in \mathcal{A}$ has weighted w-core inverse if there exists $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,{\left(awxw\right)}^{*}=awxw.$$

If such an x exists, we denote it by $a^{w,\#}$. Let ${\mathcal{A}}^{w,\#}$ denote the set of all weighted w-core invertible elements in $\mathcal{A}$. We use ${\mathcal{A}}_w^{(1,3)}$ to stand for sets of all weighted w-$(1,3)$-invertible elements in $\mathcal{A}$. Here we list some characterizations of weighted w-core inverse.

Theorem 1.  (see [6]) Let $\mathcal{A}$ be a Banach *-algebra, and let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

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

(2)

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

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awxwa=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.$$

(3)

$a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$.

(4)

$w\in {\mathcal{A}}^{\left|\right|a}$ and $a\in {\mathcal{A}}_w^{(1,3)}$.

(5)

$a\in w^{*}\mathcal{A}{\left(awa\right)}^{*}a\bigcap \mathcal{A}\left(awa\right)$.

(6)

There exists a projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wandaw+p\in {\mathcal{A}}^{-1}.$$

Dually, we have

Definition 2. 

An element $a\in \mathcal{A}$ has dual weighted w-core inverse if there exist $x\in \mathcal{A}$ such that

$${\left(aw\right)}^{2}x=a,{\left(xw\right)}^{2}a=x,{\left(wxwa\right)}^{*}=wxwa.$$

If such an x exists, we denote it by $a_{w,\#}$. Let ${\mathcal{A}}_{w,\#}$ denote the set of all dual weighted w-core invertible elements in $\mathcal{A}$. Evidently, a has dual weighted w-core inverse if and only if $a^{*}$ has weighted $w^{*}$-core inverse. However, the (dual) weighted w-core inverse may not be unique. This motivates us to consider a subclass that contains the set of all elements whose weighted w-core inverse is uniquely determined. Consequently, we extend the concept of an EP element to a bordered context and study the weighted w-EP property.

Definition 3. 

An element $a\in \mathcal{A}$ is a weighted w-EP element if $a\in {\mathcal{A}}^{w,\#}\bigcap {\mathcal{A}}_{w,\#}$ and $a^{w,\#}=a_{w,\#}$.

In Section 2, we establish fundamental properties of the weighted w-EP element. We prove that $a\in {\mathcal{A}}^{w,e}$ if and only if there exists some $x\in \mathcal{A}$ such that ${\left(aw\right)}^{2}x=a,awx=xwa,{\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.$

In Section 3, we investigate algebraic properties of the weighted w-EP element. Additive and multiplicative results for this new generalized inverse are proved, and the weighted inverse of an anti-triangular matrix is thereby obtained.

Let ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{{\displaystyle \frac{1}{n}}}=0\}$ and ${\mathcal{A}}_w^{qnil}=\{x\in \mathcal{A}\mid wx\in {\mathcal{A}}^{qnil}\}.$ In Section 4, we consider a decomposition associated with the weighted w-EP element.

Definition 4. 

An element $a\in \mathcal{A}$ is a generalized weighted w-EP element if there exist $z,y\in \mathcal{A}$ such that

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

We prove that $a\in \mathcal{A}$ is a generalized weighted w-EP element if and only if there exists $x\in \mathcal{A}$ such that

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

If such x exists, it is unique, and we denote it by $a^{w,e}$. We use ${\mathcal{A}}^{w,e}$ to stand for the set of all generalized weighted w-EP elements in $\mathcal{A}$. Finally, we characterize the generalized weighted w-EP element using its associated weight generalized Drazin inverse.

2. Weighted w-EP Inverse

In this section, we delineate the foundational properties of the weighted w-EP inverse. We begin our examination with

Lemma 1. 

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

(1)

$a\in {\mathcal{A}}_{w,e}$.

(2)

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

$$\begin{array}{c}x{\left(wa\right)}^2={\left(aw\right)}^{2}x=a,a{\left(wx\right)}^2={\left(xw\right)}^{2}a=x,\\ {\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.\end{array}$$

Proof. 

Straightforward. □

If such an x exists, we denote it by $a^{w,e}$. Let ${\mathcal{A}}^{w,e}$ denote the set of all weighted w-EP invertible elements in $\mathcal{A}$. The weighted w-EP inverse $a^{w,e}$ of a is unique as the following shows.

Theorem 2. 

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

(1)

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

(2)

$aw\in {\mathcal{A}}^{\#},wa\in {\mathcal{A}}_{\#}$ and

$$\begin{array}{c}{\left(aw\right)}^{\pi }a=0,{\left(aw\right)}^{\#}a=a{\left(wa\right)}_{\#},\\ {\left(aw\right)}^{\#}\left(aw\right)=\left(aw\right){\left(aw\right)}^{\#},wa{\left(wa\right)}_{\#}={\left(wa\right)}_{\#}wa.\end{array}$$

In this case,

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

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Lemma 2.1, there exists some $x\in \mathcal{A}$ such that

$$\begin{array}{c}x{\left(wa\right)}^2={\left(aw\right)}^{2}x=a,a{\left(wx\right)}^2={\left(xw\right)}^{2}a=x,\\ {\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.\end{array}$$

Then

$$\begin{array}{c}aw{\left(xw\right)}^2=xw,xw{\left(aw\right)}^2=aw,{\left(awxw\right)}^{*}=awxw,\\ {\left(wx\right)}^{2}wa=wx,{\left(wa\right)}^{2}wx=wa,{\left(wxwa\right)}^{*}=wxwa.\end{array}$$

Therefore

$$aw\in {\mathcal{A}}^{\#},wa\in {\mathcal{A}}_{\#}$$

and ${\left(aw\right)}^{\#}=xw,{\left(wa\right)}_{\#}=wx.$ We verify that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{\#}a& =\hfill & xwa=\left[a{\left(wx\right)}^2\right]wa=aw\left[{\left(xw\right)}^{2}a\right]=awx=a{\left(wa\right)}_{\#},\hfill \\ \hfill {\left(aw\right)}^{\#}\left(aw\right)& =\hfill & aw\left[{\left(xw\right)}^{2}a\right]w=\left(awx\right)w=\left(aw\right){\left(aw\right)}^{\#},\hfill \\ \hfill {\left(wa\right)}_{\#}wa& =\hfill & wxwa=w\left(xw\right)a=w{\left(aw\right)}^{\#}a\hfill \\ & =\hfill & waw{\left[{\left(aw\right)}^{\#}\right]}^{2}a=waw{\left(xw\right)}^{2}a=wawx=wa{\left(wa\right)}_{\#}.\hfill \end{array}$$

Then ${\left(aw\right)}^{\#}={\left(aw\right)}^{\#}$ and ${\left(wa\right)}_{\#}={\left(wa\right)}^{\#}$. Since $aw{\left(aw\right)}^{\#}a=aw\left(xw\right)a=a\left(wx\right)\left(wa\right)=a\left(w{a}_{\#}\right)\left(wa\right)=a\left(wa\right)\left(w{a}_{\#}\right)={\left(aw\right)}^{2}x=a$, we deduce that $[1-aw{\left(aw\right)}^{\#}]a=0$. Thus $[1-aw{\left(aw\right)}^{\#}]a=0$. That is, ${\left(aw\right)}^{\pi }a=0$. Moreover, we derive that

$$\begin{array}{ccc}\hfill x& =\hfill & a{\left(wx\right)}^2=a{\left[{\left(wa\right)}_{\#}\right]}^2\hfill \\ & =\hfill & {\left(xw\right)}^{2}a={\left[{\left(aw\right)}^{\#}\right]}^{2}a\hfill \\ & =\hfill & \left(xw\right)a\left(wx\right)={\left(aw\right)}^{\#}a{\left(wa\right)}_{\#}.\hfill \end{array}$$

$\left(2\right)\Rightarrow \left(1\right)$ Set $x={\left(aw\right)}^{\#}a{\left(wa\right)}_{\#}$. Since ${\left(aw\right)}^{\#}a=a{\left(wa\right)}_{\#}$, we check that $x={\left[{\left(aw\right)}^{\#}\right]}^{2}a=a{\left[{\left(wa\right)}_{\#}\right]}^2.$ Moreover, we have

$$\begin{array}{ccc}\hfill awxw& =\hfill & aw{\left[{\left(aw\right)}^{\#}\right]}^{2}aw\hfill \\ & =\hfill & {\left(aw\right)}^{\#}aw=aw{\left(aw\right)}^{\#},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(awxw\right)x={\left(aw\right)}^{\#}aw{\left[{\left(aw\right)}^{\#}\right]}^{2}a=x,\hfill \\ \hfill x{\left(wa\right)}^2& =\hfill & a{\left[{\left(wa\right)}_{\#}\right]}^2{\left(wa\right)}^2=a{\left(wa\right)}_{\#}wa={\left(aw\right)}^{\#}awa=a,\hfill \\ \hfill {\left(awxw\right)}^{*}& =\hfill & awxw.\hfill \end{array}$$

Moreover, we verify that

$$\begin{array}{ccc}\hfill wxwa& =\hfill & wa{\left[{\left(wa\right)}_{\#}\right]}^{2}wa=wa{\left(wa\right)}_{\#}={\left(wa\right)}_{\#}wa,\hfill \\ \hfill {\left(xw\right)}^{2}a& =\hfill & xwxwa={\left[{\left(aw\right)}^{\#}\right]}^{2}aw{\left[{\left(aw\right)}^{\#}\right]}^{2}awa\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{\#}\right]}^{3}awa=aw{\left[{\left(aw\right)}^{\#}\right]}^{3}a={\left[{\left(aw\right)}^{\#}\right]}^{2}a=x,\hfill \\ \hfill {\left(aw\right)}^{2}x& =\hfill & a\left(wawx\right)=awa{\left(wa\right)}_{\#}=aw{\left(aw\right)}^{\#}a=a,\hfill \\ \hfill {\left(wxwa\right)}^{*}& =\hfill & wxwa.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}^{w,e}$ by Lemma 2.1. □

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 $a^{(b,c)}$ (see [8]). We now derive

Corollary 1. 

Let $a\in {\mathcal{A}}^{w,e}$. Then

$$a^{w,e}={\left(waw\right)}^{\left({\left(aw\right)}^{\#},{\left(wa\right)}_{\#}\right)}.$$

Proof. 

In view of Theorem 2.2, $aw\in {\mathcal{A}}^{\#},wa\in {\mathcal{A}}_{\#}$. Let $x=a^{w,e}$. Then we verify that

$$\begin{array}{ccc}\hfill x\left(waw\right){\left(aw\right)}^{\#}& =\hfill & ={\left[{\left(aw\right)}^{\#}\right]}^{2}a\left(waw\right){\left(aw\right)}^{\#}={\left(aw\right)}^{\#},\hfill \\ \hfill {\left(wa\right)}_{\#}\left(waw\right)x& =\hfill & {\left(wa\right)}_{\#}\left(waw\right)a{\left[{\left(wa\right)}_{\#}\right]}^2={\left(wa\right)}_{\#},\hfill \\ \hfill x& =\hfill & xwawx={\left(aw\right)}^{\#}a{\left(wa\right)}_{\#}\in {\left(aw\right)}^{\#}\mathcal{A}x\bigcap x\mathcal{A}{\left(wa\right)}_{\#}.\hfill \end{array}$$

Therefore, $waw$ has $\left({\left(aw\right)}^{\#},{\left(wa\right)}_{\#}\right)$-inverse x, as asserted. □

Theorem 3. 

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

(1)

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

(2)

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

$$a{\left(wx\right)}^2=x,x{\left(wa\right)}^2=a,{\left(wxwa\right)}^{*}=wxwa,{\left(xwaw\right)}^{*}=xwaw.$$

In this case, $a^{w,e}={\left(xw\right)}^{2}a.$

(3)

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

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

In this case, $a^{w,e}=a{\left(wx\right)}^2.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ We verify that

$$xwaw={\left(aw\right)}^{\#}aw=aw{\left(aw\right)}^{\#}=awxw.$$

Then ${\left(xwaw\right)}^{*}=xwaw$, as required.

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

$$a{\left(wx\right)}^2=x,x{\left(wa\right)}^2=a,{\left(xwaw\right)}^{*}=xwaw.$$

Set $z={\left(xw\right)}^{2}a$. Then we verify that

$$\begin{array}{ccc}\hfill awz& =\hfill & aw{\left(xw\right)}^{2}a=\left[a{\left(wx\right)}^2\right]wa=xwa,\hfill \\ \hfill zwa& =\hfill & {\left(xw\right)}^{2}awa=xw\left[x{\left(wa\right)}^2\right]=xwa.\hfill \end{array}$$

Thus, $awz=zwa$. Accordingly, we have

$$\begin{array}{ccc}\hfill awzw& =\hfill & aw{\left(xw\right)}^{2}aw=\left[a{\left(wx\right)}^2\right]waw=xwaw,\hfill \\ \hfill {\left(awzw\right)}^{*}& =\hfill & awzw,\hfill \\ \hfill z{\left(wa\right)}^2& =\hfill & {\left(xw\right)}^{2}a{\left(wa\right)}^2=xw\left[x{\left(wa\right)}^2\right]wa=x{\left(wa\right)}^2=a,\hfill \\ \hfill a{\left(wz\right)}^2& =\hfill & \left(awzw\right)z=\left(xwaw\right)z=xw\left(awz\right)=xwaw{\left(xw\right)}^{2}a\hfill \\ & =\hfill & xw\left[a{\left(wx\right)}^2\right]wa=xwxwa={\left(xw\right)}^{2}a=z.\hfill \end{array}$$

In light of Theorem 1.2, we have $awzwa=a$ and $zwawz=z$. Then we verify that

$$\begin{array}{ccc}\hfill wzwa& =\hfill & w{\left(xw\right)}^{2}awa=wxw\left[x{\left(wa\right)}^2\right]=wxwa,\hfill \\ \hfill {\left(aw\right)}^{2}z& =\hfill & aw\left(awz\right)=aw\left(zwa\right)=awzwa=a,\hfill \\ \hfill {\left(zw\right)}^{2}a& =\hfill & zw\left(zwa\right)=zw\left(awz\right)=zwawz=z,\hfill \\ \hfill {\left(wzwa\right)}^{*}& =\hfill & {\left(wxwa\right)}^{*}=wxwa=wzwa.\hfill \end{array}$$

This implies that $a\in {\mathcal{A}}^{w,e}$. Accordingly, $a^{w,e}=z={\left(xw\right)}^{2}a$, as required.

$\left(1\right)⟺\left(3\right)$ Obviously, $a\in {\mathcal{A}}^{w,e}$ if and only if $a^{*}\in {\mathcal{A}}^{w^{*},e}$. By applying the preceding equivalence to $a^{*}$ and $w^{*}$, we obtain the result. □

We use $a^0$ and $0^a$ to denote the right and left annihilator of a in $\mathcal{A}$, respectively. We now derive

Corollary 2. 

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

(1)

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

(2)

There exists some $x\in \mathcal{A}$ such that $a=awxwa,x=xwawx,a\mathcal{A}=x\mathcal{A}={\left(xw\right)}^{*}\mathcal{A},a^{*}\mathcal{A}=\left(wx\right)\mathcal{A}$.

(3)

There exists some $x\in \mathcal{A}$ such that $a=awxwa,x=xwawx{,}^{0}a{=}^{0}x{=}^0{\left(xw\right)}^{*}{,}^0\left(a^{*}\right){=}^0\left(wx\right)$.

In this case, $a^{w,e}=x.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a^{w,e}$. Then there exists some $x\in \mathcal{A}$ such that

$$\begin{array}{c}x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awxwa=a,xwawx=x,{\left(awxw\right)}^{*}=awxw,\\ {\left(aw\right)}^{2}x=a,{\left(xw\right)}^{2}a=x,{\left(wxwa\right)}^{*}=wxwa,awxwa=a,xwawx=x.\end{array}$$

Then $a=x{\left(wa\right)}^2\in x\mathcal{A}$, and so $a\mathcal{A}\subseteq x\mathcal{A}$. As $x=a{\left(wx\right)}^2\in a\mathcal{A}$, we have $x\mathcal{A}\subseteq a\mathcal{A}$. Hence, $a\mathcal{A}=x\mathcal{A}$.

Since $a=awxwa={\left(awxw\right)}^{*}a={\left(xw\right)}^{*}{\left(aw\right)}^{*}a\in {\left(xw\right)}^{*}\mathcal{A}$; hence, $a\mathcal{A}\subseteq {\left(xw\right)}^{*}\mathcal{A}$. As $xw=xwawxw$, we have ${\left(xw\right)}^{*}={\left(awxw\right)}^{*}{\left(xw\right)}^{*}=awxw{\left(xw\right)}^{*}\in a\mathcal{A}$; hence, ${\left(xw\right)}^{*}\mathcal{A}\subseteq a\mathcal{A}$. Thus, $a\mathcal{A}={\left(xw\right)}^{*}\mathcal{A}$.

Clearly, $a=awxwa$, and so $a^{*}={\left(wxwa\right)}^{*}{a}^{*}=xwaw{\left(aw\right)}^{*}=wxwa{a}^{*}\in wx\mathcal{A}$. This implies that $a^{*}\mathcal{A}\subseteq wx\mathcal{A}$. On the other hand, $wx=wxwawx={\left(wxwa\right)}^{*}wx=a^{*}{\left(wxw\right)}^{*}wx\in a^{*}\mathcal{A}$; hence, $wx\mathcal{A}\subseteq a^{*}\mathcal{A}$. Therefore $a^{*}\mathcal{A}=\left(wx\right)\mathcal{A}$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ This is obvious.

$\left(3\right)\Rightarrow \left(1\right)$ Since $xwawx=x$, we see that $1-xwaw{\in }^{0}x$, and so $1-xwaw{\in }^{0}a$. Hence, $a=x{\left(wa\right)}^2$. On the other hand, $awxwa=a$; whence, $1-awxw{\in }^{0}a{\subseteq }^{0}x$. Thus, $x=\left(awxw\right)x=a{\left(wx\right)}^2$. Since $a=awxwa$, we have $a^{*}={\left(wxwa\right)}^{*}{a}^{*}$, and so $1-{\left(wxwa\right)}^{*}{\in }^0\left(a^{*}\right){\subseteq }^0\left(wx\right)$, and so $wx={\left(wxwa\right)}^{*}wx$. Then $wxwa={\left(wxwa\right)}^{*}wxwa$. This implies that ${\left(wxwa\right)}^{*}={\left(wxwa\right)}^{*}wxwa=wxwa$. Since $0^a{=}^0{\left(xw\right)}^{*}$, we see that ${\left(a^{*}\right)}^0={\left(xw\right)}^0$. Since $a=awxwa$, we have $a^{*}={\left(awxwa\right)}^{*}=a^{*}{\left(awxw\right)}^{*}$; hence, ${(1-awxw)}^{*}\in {\left(a^{*}\right)}^0\subseteq {\left(xw\right)}^0$. This shows that $xw=xw{\left(awxw\right)}^{*}$. This implies that $awxw=awxw{\left(awxw\right)}^{*}$. Therefore ${\left(awxw\right)}^{*}=awxw{\left(awxw\right)}^{*}=awxw.$ This completes the proof by Theorem 2.4. □

Theorem 4. 

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

(1)

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

(2)

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

$${\left(aw\right)}^{2}x=a,awx=xwa,{\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.$$

In this case, $a^{w,e}=a{\left(wx\right)}^2$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $x=a^{w,e}$. Then ${\left(aw\right)}^{2}x=a,{\left(awxw\right)}^{*}=awxw$ and ${\left(wxwa\right)}^{*}=wxwa$. In view of Theorem 2.2, we derive

$$awx=aw{a}^{w,e}=aw{\left({\left(aw\right)}^{\#}\right)}^{2}a={\left[{\left(aw\right)}^{\#}\right]}^2\left(aw\right)a=a^{w,e}wa=xwa,$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ Set $z=a{\left(wx\right)}^2$. Since $awx=xwa$, we see that

$$\begin{array}{ccc}\hfill {\left(zw\right)}^{2}a& =\hfill & {\left[a{\left(wx\right)}^{2}w\right]}^{2}a=awxwxw\left[{\left(aw\right)}^{2}x\right]wx\hfill \\ & =\hfill & \left[{\left(aw\right)}^{2}x\right]wxwx=a{\left(wx\right)}^2=z,\hfill \\ \hfill {\left(aw\right)}^{2}z& =\hfill & awawa{\left(wx\right)}^2=aw\left[{\left(aw\right)}^{2}x\right]wx={\left(aw\right)}^{2}x=a,\hfill \\ \hfill awzw& =\hfill & awa{\left(wx\right)}^{2}w=\left[{\left(aw\right)}^{2}x\right]wxw=awxw,\hfill \\ \hfill {\left(awzw\right)}^{*}& =\hfill & awzw,\hfill \\ \hfill wzwa& =\hfill & wa{\left(wx\right)}^{2}wa=wawxw\left(xwa\right)=wawxw\left(awx\right)=waw\left(xwa\right)wx\hfill \\ & =\hfill & waw\left(awx\right)wx=w\left[{\left(aw\right)}^{2}x\right]wx=w\left(awx\right)=w\left(xwa\right)=wxwa,\hfill \\ \hfill {\left(wzwa\right)}^{*}& =\hfill & wzwa.\hfill \end{array}$$

In light of Theorem 2.4, $a\in {\mathcal{A}}^{w,e}$. In this case, $a^{w,e}=a{\left(wx\right)}^2.$ □

Corollary 3. 

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

(1)

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

(2)

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

$$a={\left(aw\right)}^{2}x=x{\left(wa\right)}^2,{\left(awxw\right)}^{*}=xwaw,{\left(wxwa\right)}^{*}=wawx.$$

In this case, $a^{w,e}=a{\left(wx\right)}^2$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.6, there exists some $x\in \mathcal{A}$ such that

$${\left(aw\right)}^{2}x=a,awx=xwa,{\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.$$

Hence,

$$\begin{array}{ccc}\hfill x{\left(wa\right)}^2& =\hfill & \left(xwa\right)wa=\left(awx\right)wa\hfill \\ & =\hfill & aw\left(xwa\right)=aw\left(awx\right)={\left(aw\right)}^{2}x=a.\hfill \end{array}$$

and

$${\left(awxw\right)}^{*}=\left(awx\right)w=\left(xwa\right)w=xwaw,{\left(wxwa\right)}^{*}=w\left(xwa\right)=w\left(awx\right)=wawx,$$

as desired.

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

$$a={\left(aw\right)}^{2}x=x{\left(wa\right)}^2,{\left(awxw\right)}^{*}=xwaw,{\left(wxwa\right)}^{*}=wawx.$$

Thus, we have $a=x{\left(wa\right)}^2=\left(xwaw\right)a={\left(awxw\right)}^{*}a$, and then $awxw={\left(awxw\right)}^{*}awxw$. This implies that $xwaw={\left(awxw\right)}^{*}={\left(awxw\right)}^{*}awxw=awxw$. Thus, we have

$$\begin{array}{ccc}\hfill awx& =\hfill & \left[{\left(aw\right)}^{2}x\right]wx=aw\left[awxw\right]x\hfill \\ & =\hfill & aw\left[xwaw\right]x=\left[awxw\right]awx=xw{\left(aw\right)}^{2}x=xwa.\hfill \end{array}$$

Moreover, ${\left(awxw\right)}^{*}=xwaw=awxw$ and ${\left(wxwa\right)}^{*}={\left(wawx\right)}^{*}=wxwa$. Therefore we complete the proof by Lemma 2.6. □

As is well known, a is EP if and only if $a\in {\mathcal{A}}^{\#}$ and ${\left(a{a}^{\#}\right)}^{*}=a{a}^{\#}$ (see [25]). We now extend this result to the broader context of the weighted w-EP inverse.

Corollary 4. 

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

(1)

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

(2)

$a\in {\mathcal{A}}^{\#,w},{\left(aw{a}^{\#,w}w\right)}^{*}=aw{a}^{\#,w}w$ and ${\left(waw{a}^{\#,w}\right)}^{*}=waw{a}^{\#,w}$.

In this case, $a^{w,e}=a^{\#,w}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a^{w,e}$. Then we have

$${\left(aw\right)}^{2}x=a,{\left(xw\right)}^{2}a=x,awx=xwa,{\left(awxw\right)}^{*}=awxw,{\left(wxwa\right)}^{*}=wxwa.$$

Hence, $awxwa=aw\left(awx\right)={\left(aw\right)}^{2}x=a,xwawx=xw\left(xwa\right)={\left(xw\right)}^{2}a=x,awx=xwa.$ Therefore $a\in {\mathcal{A}}^{\#,w}$ and $a^{\#,w}=x=a^{w,e}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, ${\left(aw{a}^{\#,w}w\right)}^{*}=aw{a}^{\#,w}w=a^{\#,w}waw$ and ${\left(waw{a}^{\#,w}\right)}^{*}$$=waw{a}^{\#,w}=w{a}^{\#,w}wa$. Obviously, we have $a={\left(aw\right)}^2{a}^{\#,w}=a^{\#,w}{\left(wa\right)}^2$. In light of Corollary 2.7, $a\in {\mathcal{A}}^{w,e}$ and $a^{w,e}=a^{\#,w}$, as asserted. □

3. Algebraic Properties

In this section we establish the multiplicative and additive properties of generalized weighted core inverse in a Banach algebra with involution.

Lemma 2. 

Let $a\in {\mathcal{A}}^{w,e},x\in \mathcal{A}$. If $xa=ax$ and $xw=wx$, then $x{a}^{w,e}=a^{w,e}x.$

Proof. 

Since $xa=ax$ and $xw=wx$,, we have $x{a}^{\#,w}=a^{\#,w}x$. In view of Corollary 2.8, $a^{w,e}=a^{\#,w}$. This completes the proof. □

We are now ready to prove:

Theorem 5. 

Let $a,b\in {\mathcal{A}}^{w,e}$. If $ab=ba,aw=wa$ and $bw=wb$, then $awb\in {\mathcal{A}}^{w,e}$. In this case,

$${\left(awb\right)}^{w,e}=a^{w,e}w{b}^{w,e}.$$

Proof. 

Since $ab=ba,aw=wa$, by virtue of Lemma 3.1, $a{b}^{w,e}=b^{w,e}a$. Likewise, $w{b}^{w,e}=b^{w,e}w$. By using Lemma 3.1 again, we derive $a^{w,e}{b}^{w,e}=b^{w,e}{a}^{w,e},w{a}^{w,e}=a^{w,e}w$ and $w{b}^{w,e}=b^{w,e}w.$

Set $x=a^{w,e}w{b}^{w,e}.$ Then we check that

$$\begin{array}{ccc}\hfill awb{\left(wx\right)}^2& =\hfill & awb{\left[w{a}^{w,e}w{b}^{w,e}\right]}^2\hfill \\ & =\hfill & \left[a{\left(w{a}^{w,e}\right)}^2\right]w\left[b{\left(w{b}^{w,e}\right)}^2\right]\hfill \\ & =\hfill & a^{w,e}w{b}^{w,e}=x,\hfill \\ \hfill x{\left(wawb\right)}^2& =\hfill & a^{w,e}w{b}^{w,e}{\left(wawb\right)}^2\hfill \\ & =\hfill & \left[a^{w,e}{\left(wa\right)}^2\right]w\left[b^{w,e}{\left(wb\right)}^2\right]\hfill \\ & =\hfill & awb,\hfill \\ \hfill wxw\left(awb\right)& =\hfill & w{a}^{w,e}w{b}^{w,e}wawb\hfill \\ & =\hfill & \left[w{a}^{w,e}wa\right]\left[w{b}^{w,e}wb\right],\hfill \\ \hfill {\left(wxw\left(awb\right)\right)}^{*}& =\hfill & \left[w{b}^{w,e}wb\right]\left[w{a}^{w,e}wa\right]=\left[w{a}^{w,e}wa\right]\left[w{b}^{w,e}wb\right]\hfill \\ & =\hfill & w\left[a^{w,e}w{b}^{w,e}\right]wawb=wxw\left(awb\right),\hfill \\ \hfill xw\left(awb\right)w& =\hfill & a^{w,e}w{b}^{w,e}w\left(awb\right)w=\left[a^{w,e}waw\right]\left[b^{w,e}wbw\right],\hfill \\ \hfill {\left(xw\left(awb\right)w\right)}^{*}& =\hfill & \left[b^{w,e}wbw\right]\left[a^{w,e}waw\right]=\left[a^{w,e}waw\right]\left[b^{w,e}wbw\right]=xw\left(awb\right)w.\hfill \end{array}$$

By virtue of Theorem 2.4, $awb\in {\mathcal{A}}^{w,e}$ and

$$\begin{array}{ccc}\hfill {\left(awb\right)}^{w,e}& =\hfill & {\left(awb\right)}^{\#,w}\hfill \\ & =\hfill & a^{\#,w}w{b}^{\#,w}\hfill \\ & =\hfill & a^{w,e}w{b}^{w,e},\hfill \end{array}$$

as asserted. □

It is convenient at this stage to include the following additive result for the generalized weighted core inverse.

Theorem 6. 

Let $a,b\in {\mathcal{A}}^{w,e}$. If $awb=bwa=0$, then $a+b\in {\mathcal{A}}^{w,e}$. In this case,

$${(a+b)}^{w,e}=a^{w,e}+b^{w,e}.$$

Proof. 

Set $x=a^{w,e}+b^{w,e}$. Then we check that

$$\begin{array}{ccc}\hfill (a+b){\left(wx\right)}^2& =\hfill & [aw{a}^{w,e}+bw{b}^{w,e}]wx\hfill \\ & =\hfill & [aw{a}^{w,e}+bw{b}^{w,e}]w[a^{w,e}+b^{w,e}]\hfill \\ & =\hfill & aw{a}^{w,e}w{a}^{w,e}+bw{b}^{w,e}w{b}^{w,e}\hfill \\ & =\hfill & a{\left(w{a}^{w,e}\right)}^2+b{\left(w{b}^{w,e}\right)}^2\hfill \\ & =\hfill & a^{w,e}+b^{w,e}=x,\hfill \\ \hfill x{\left[w(a+b)\right]}^2& =\hfill & [a^{w,e}w+b^{w,e}w](awa+bwb)\hfill \\ & =\hfill & [a^{w,e}w+b^{w,e}w](awa+bwb)\hfill \\ & =\hfill & a^{w,e}wawa+b^{w,e}wbwb\hfill \\ & +\hfill & a^{w,e}wbwb+b^{w,e}wawa\hfill \\ & =\hfill & a^{w,e}{\left(wa\right)}^2+b^{w,e}{\left(wb\right)}^2\hfill \\ & =\hfill & a+b,\hfill \\ \hfill wxw(a+b)& =\hfill & w(a^{w,e}+b^{w,e})(wa+wb)\hfill \\ & =\hfill & w{a}^{w,e}wa+w{b}^{w,e}wb,\hfill \\ \hfill {\left(wxw(a+b)\right)}^{*}& =\hfill & wxw(a+b),\hfill \\ \hfill xw(a+b)w& =\hfill & (a^{w,e}+b^{w,e})w(aw+bw)=a^{w,e}waw+b^{w,e}wbw,\hfill \\ \hfill {\left(xw(a+b)w\right)}^{*}& =\hfill & xw(a+b)w.\hfill \end{array}$$

Therefore $a+b\in {\mathcal{A}}^{w,e}$. In this case,

$$\begin{array}{ccc}\hfill {(a+b)}^{w,e}& =\hfill & {(a+b)}^{\#,w}={\left[{\left((a+b)w\right)}^{\#}\right]}^2(a+b)\hfill \\ & =\hfill & [{\left({\left(aw\right)}^{\#}\right)}^2+{\left({\left(bw\right)}^{\#}\right)}^2](a+b)\hfill \\ & =\hfill & a^{\#,w}+b^{\#,w}=a^{w,e}+b^{w,e},\hfill \end{array}$$

This completes the proof. □

Theorem 7. 

Let $a\in {\mathcal{A}}^{w,e}$. Then $a^{w,e}\in {\mathcal{A}}^{w,e}$. In this case, ${\left(a^{w,e}\right)}^{w,e}=a$.

Proof. 

We easily check that

$$a^{w,e}{\left(wa\right)}^2=a,a{\left(w{a}^{w,e}\right)}^2=a^{w,e},{\left(aw{a}^{w,e}w\right)}^{*}=aw{a}^{w,e}w,{\left(waw{a}^{w,e}\right)}^{*}=waw{a}^{w,e}.$$

In view of Theorem 3.4, $a^{w,e}\in {\mathcal{A}}^{w,e}$. In this case, ${\left(a^{w,e}\right)}^{w,e}=a,$ thus yielding the result. □

Corollary 5. 

Let $a\in {\mathcal{A}}^{w,e}$. Then ${\left[{\left(a^{w,e}\right)}^{w,e}\right]}^{w,e}=a^{w,e}.$

Proof. 

Straightforward by Theorem 3.4. □

Lemma 3. 

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

(1)

$a^{\pi }b=0$.

(2)

$(1-a^{\#}a)b=0$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a^{\pi }b=0$, we have that $b=a^{\#}ab$. Then $a^{\#}ab=\left[a^{\#}a{a}^{(1,3)}\right]ab=a^{\#}ab$. Hence $(1-a^{\#}a)b=0$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Since $(1-a^{\#}a)b=0$, we have $b=a^{\#}ab$. Thus, $b=a^{\#}a{a}^{(1,3)}ab=a^{\#}ab$. This implies that $a^{\pi }b=0$, as desired. □

Theorem 8. 

Let $\alpha =\left(\begin{array}{cc}a& c\\ 0& b\end{array}\right)$ with $a,b\in {\mathcal{A}}^{w,e}$. If ${\left(aw\right)}^{\pi }c=0$ and $c{\left(wb\right)}^{\pi }=0$, then $\alpha \in M_2{\left(\mathcal{A}\right)}^{e,w{I}_2}$ and

$${\alpha }^{e,w{I}_2}=\left(\begin{array}{cc}{a}^{w,e}& -a^{w,e}wcw{b}^{w,e}\\ 0& b^{w,e}\end{array}\right).$$

Proof. 

In view of Theorem 2.2, $aw,bw\in {\mathcal{A}}^{\#}$. Since ${\left(aw\right)}^{\pi }cw=0$, by virtue of Lemma 3.6, we have $[1-\left(aw\right){\left(aw\right)}^{\#}]cw=0$. By using [24], we have

$${\left(\alpha w{I}_2\right)}^{\#}=\left(\begin{array}{cc}{\left(aw\right)}^{\#}& s\\ 0& {\left(bw\right)}^{\#}\end{array}\right),$$

where

$$s=-{\left(aw\right)}^{\#}cw{\left(bw\right)}^{\#}.$$

On the other hand, ${\left(wa\right)}^{\pi }wc=[1-wa{\left(wa\right)}^{\#}]wc=[1-waw{\left({\left(aw\right)}^{\#}\right)}^{2}a]wc=w{\left(aw\right)}^{\pi }c=0$. Dually, we have

$${\left(w{I}_2\alpha \right)}_{\#}=\left(\begin{array}{cc}{\left(wa\right)}_{\#}& t\\ 0& {\left(wb\right)}_{\#}\end{array}\right),$$

where $t=-{\left(wa\right)}_{\#}wc{\left(wb\right)}_{\#}.$

Set $\beta =\left(\begin{array}{cc}{\left(aw\right)}^{\#}& s\\ 0& {\left(bw\right)}^{\#}\end{array}\right)\left(\begin{array}{cc}a& c\\ 0& b\end{array}\right)\left(\begin{array}{cc}{\left(wa\right)}_{\#}& t\\ 0& {\left(wb\right)}_{\#}\end{array}\right).$ Then

$$\beta =\left(\begin{array}{cc}{a}^{w,e}& z\\ 0& b^{w,e}\end{array}\right),$$

where

$$\begin{array}{ccc}\hfill z& =\hfill & {\left(aw\right)}^{\#}at+sb{\left(wb\right)}_{\#}+{\left(aw\right)}^{\#}c{\left(wb\right)}_{\#}\hfill \\ & =\hfill & -a^{w,e}wc{\left(wb\right)}_{\#}-{\left(aw\right)}^{\#}cw{b}^{w,e}+{\left(aw\right)}^{\#}c{\left(wb\right)}_{\#}\hfill \\ & =\hfill & -a^{w,e}wc{\left(wb\right)}_{\#}-{\left(aw\right)}^{\#}c{\left(wb\right)}_{\#}+{\left(aw\right)}^{\#}c{\left(wb\right)}_{\#}\hfill \\ & =\hfill & -a^{w,e}wc{\left(wb\right)}_{\#}\hfill \\ & =\hfill & -a^{w,e}wcw{b}^{w,e}.\hfill \end{array}$$

Since $z\in a\mathcal{A}$ and $aw{a}^{w,e}wa=a$, we have $aw{a}^{w,e}wz=z$. Moreover, we have

$$\begin{array}{cc}& awzw{b}^{w,e}+c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& -aw{a}^{w,e}wc{\left(wb\right)}_{\#}wb{\left[{\left(wb\right)}_{\#}\right]}^2+c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& -awa{\left({\left(wa\right)}_{\#}\right)}^{2}wc{\left({\left(wb\right)}_{\#}\right)}^2+c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& [1-a{\left(wa\right)}_{\#}w]c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& [1-aw{\left({\left(aw\right)}^{\#}\right)}^{2}aw]c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& [1-{\left(aw\right)}^{\#}aw]c{\left({\left(wb\right)}_{\#}\right)}^2=0.\hfill \end{array}$$

Then we easily verify that

$$\begin{array}{cc}& a[w{a}^{w,e}wz+wzw{b}^{w,e}]+c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& aw{a}^{w,e}wz+awzw{b}^{w,e}+c{\left({\left(wb\right)}_{\#}\right)}^2\hfill \\ \hfill =& z,\hfill \end{array}$$

Thus we have

$$\begin{array}{ccc}\hfill \alpha {\left(w{I}_2\beta \right)}^2& =\hfill & \left(\begin{array}{cc}a& c\\ 0& b\end{array}\right){\left(\begin{array}{cc}w{a}^{w,e}& wz\\ 0& w{b}^{w,e}\end{array}\right)}^2\hfill \\ & =\hfill & \left(\begin{array}{cc}a{\left[w{a}^{w,e}\right]}^2& a[w{a}^{w,e}wz+wzw{b}^{w,e}]+c{\left({\left(wb\right)}_{\#}\right)}^2\\ 0& b{\left[w{b}^{w,e}\right]}^2\end{array}\right)\hfill \\ & =\hfill & \beta .\hfill \end{array}$$

It is easy to verify that

$$\begin{array}{ccc}\hfill a^{w,e}{\left(wa\right)}^2& =\hfill & a^{w,e}wawa\hfill \\ & =\hfill & \left[a^{w,e}waw{a}^{w,e}\right]a^{w,e}wa\hfill \\ & =\hfill & {\left[a^{w,e}\right]}^{2}wa\hfill \\ & =\hfill & a.\hfill \end{array}$$

Likewise, $b^{w,e}{\left(wb\right)}_2=b$. Furthermore, we check that

$$\begin{array}{cc}& a^{w,e}[\left(wa\right)\left(wc\right)+\left(wc\right)\left(wb\right)]-a^{w,e}wcw{b}^{w,e}{\left(wb\right)}_2\hfill \\ \hfill =& {\left[{\left(aw\right)}^{\#}\right]}^2{\left(aw\right)}^{2}c+a^{w,e}\left(wc\right)\left(wb\right)-a^{w,e}wcwb{\left[{\left(wb\right)}_{\#}\right]}^2{\left(wb\right)}_2\hfill \\ \hfill =& {\left(aw\right)}^{\#}\left(aw\right)c+a^{w,e}\left(wc\right)\left(wb\right)-a^{w,e}wc\left[wb{\left(wb\right)}_{\#}wb\right]\hfill \\ \hfill =& c.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}& \beta {\left(w{I}_2\alpha \right)}^2\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^{w,e}& -a^{w,e}wcw{b}^{w,e}\\ 0& b^{w,e}\end{array}\right){\left(\begin{array}{cc}wa& wc\\ 0& wb\end{array}\right)}^2\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^{w,e}{\left(wa\right)}^2& a^{w,e}[\left(wa\right)\left(wc\right)+\left(wc\right)\left(wb\right)]-a^{w,e}wcw{b}^{w,e}{\left(wb\right)}_2\\ 0& b^{w,e}{\left(wb\right)}_2\end{array}\right)\hfill \\ \hfill =& \alpha .\hfill \end{array}$$

Clearly, we have

$$\begin{array}{cc}& aw{a}^{w,e}wcw{b}^{w,e}w-cw{b}^{w,e}w\hfill \\ \hfill =& aw{a}^{\#,w}wcw{b}^{w,e}w-cw{b}^{w,e}w\hfill \\ \hfill =& aw{\left[{\left(aw\right)}^{\#}\right]}^{2}awcw{b}^{w,e}w-cw{b}^{w,e}w\hfill \\ \hfill =& [{\left(aw\right)}^{\#}aw-1]cw{b}^{w,e}w\hfill \\ \hfill =& -\left[{\left(aw\right)}^{\pi }c\right]w{b}^{w,e}w=0.\hfill \end{array}$$

Thus, we derive that

$$\begin{array}{ccc}\hfill \alpha \left(w{I}_2\right)\beta \left(w{I}_2\right)& =\hfill & \left(\begin{array}{cc}aw& cw\\ 0& bw\end{array}\right)\left(\begin{array}{cc}{a}^{w,e}w& -a^{w,e}wcw{b}^{w,e}w\\ 0& b^{w,e}w\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}aw{a}^{w,e}w& 0\\ 0& bw{b}^{w,e}w\end{array}\right).\hfill \end{array}$$

Therefore

$${\left(\alpha \left(w{I}_2\right)\beta \left(w{I}_2\right)\right)}^{*}=\alpha \left(w{I}_2\right)\beta \left(w{I}_2\right).$$

Analogously,

$${\left(\left(w{I}_2\right)\beta \left(w{I}_2\right)\alpha \right)}^{*}=\left(w{I}_2\right)\beta \left(w{I}_2\right)\alpha .$$

In this case,

$$\begin{array}{cc}& {\alpha }^{w,e}\hfill \\ \hfill =& {\left[\beta \left(w{I}_2\right)\right]}^2\alpha \hfill \\ \hfill =& {\left(\begin{array}{cc}{a}^{w,e}w& -a^{w,e}wcw{b}^{w,e}w\\ 0& b^{w,e}w\end{array}\right)}^2\left(\begin{array}{cc}a& c\\ 0& b\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^{w,e}w& -a^{w,e}wcw{b}^{w,e}w\\ 0& b^{w,e}w\end{array}\right)\left(\begin{array}{cc}{a}^{w,e}wa& a^{w,e}wc-a^{w,e}wcw{b}^{w,e}wb\\ 0& b^{w,e}wb\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^{w,e}& {\left(a^{w,e}w\right)}^{2}c[1-{\left(wb\right)}_{\#}wb]-a^{w,e}wcw{b}^{w,e}\\ 0& b^{w,e}\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^{w,e}& -a^{w,e}wcw{b}^{w,e}\\ 0& b^{w,e}\end{array}\right),\hfill \end{array}$$

as asserted. □

Corollary 6. 

Let $\alpha =\left(\begin{array}{cc}a& c\\ 0& b\end{array}\right)\in M_2\left(\mathcal{A}\right)$, and let $a,b\in \mathcal{A}$ be EP. If $a^{\pi }c=0$ and $c{b}^{\pi }=0$, then $\alpha \in M_2\left(\mathcal{A}\right)$ is EP.

Proof. 

Straightforward by choosing $w=1$ in Theorem 3.7. □

4. The Associated Decomposition

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

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

The preceding x is denoted by $a^{w,d}$ (see [6]). Dually, $a\in {\mathcal{A}}_{w,d}$ if and only if $a^{*}\in {\mathcal{A}}^{w^{*},d}$. We use $a_{w,d}$ to stand for ${\left[{\left(a^{*}\right)}^{w^{*},d}\right]}^{*}$. We now derive

Theorem 9. 

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

(1)

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

(2)

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

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

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that

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

Set $x=z^{w,e}$. Then $x=z^{w,e}=z^{w,\#}=z^{\#,w}$. As in the proof of [6], we check that

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

Moreover, we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & (z+y)w{z}^{w,\#}=zw{z}^{w,\#}=zw{z}^{\#,w},\hfill \\ \hfill xwa& =\hfill & z^{w,\#}w(z+y)=z^{w,\#}w(z+y)=z^{w,\#}wzw{z}^{w,\#}w(z+y)\hfill \\ & =\hfill & z^{w,\#}w{\left(zw{z}^{w,\#}w\right)}^{*}(z+y)=z^{w,\#}w{\left(w{z}^{w,\#}w\right)}^{*}{z}^{*}(z+y)\hfill \\ & =\hfill & z^{w,\#}w{\left(w{z}^{w,\#}w\right)}^{*}{z}^{*}z=z^{w,\#}w{\left(zw{z}^{w,\#}w\right)}^{*}z=z^{\#}wz.\hfill \end{array}$$

Then $awx=xwa$. Hence, we have $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ Since $wxwa=w\left(xwa\right)=w{z}^{w,\#}w$, we see that ${\left(wxwa\right)}^{*}=wxwa$, as required.

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

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

Then $xw{\left(aw\right)}^{2}x=aw\left(xwa\right)wx=aw\left(awx\right)\left(wx\right)=aw\left[a{\left(wx\right)}^2\right]=awx$. Moreover, we have $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. This implies that $a\in {\mathcal{A}}^{w,d}$. Set $z=awxwa$ and $y=a-awxwa$. As in the proof of [6], we have

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

Explicitly, $z^{w,\#}=x$. This implies that ${\left(aw\right)}^{2}x=a,{\left(awxw\right)}^{*}=awxw.$ Moreover, we verify that $wxwz=wxw\left(awxwa\right)=wxwawxwa=wxwa$; hence, ${\left(wxwz\right)}^{*}=wxwz$. Since $awx=xwa$, it is easy to check that

$$\begin{array}{ccc}\hfill zwx& =\hfill & \left(awxwa\right)wx=\left(awxw\right)\left(awx\right)=\left(awxw\right)\left(xwa\right)=\left[a{\left(wx\right)}^2\right]wa\hfill \\ & =\hfill & xwa=awx=awa{\left(wx\right)}^2=aw\left(awx\right)wx=aw\left(xwa\right)wx\hfill \\ & =\hfill & \left(awx\right)w\left(awx\right)=xw\left(awxwa\right)=xwz.\hfill \end{array}$$

Therefore $z\in {\mathcal{A}}^{w,e}$ by Theorem 2.6. This completes the proof. □

We denote x in Theorem 4.1 by $a^{w,e}$, and call it a generalized weighted w-EP inverse of a.

Corollary 7. 

Let $a\in {\mathcal{A}}^{w,e}$. Then

$$a^{w,e}=a^{w,e}waw{a}^{w,e}.$$

Proof. 

By virtue of Theorem 4.1, we have $a^{w,e}wa=aw{a}^{w,e}wa$. Hence, $a^{w,e}waw{a}^{w,e}=a{\left(w{a}^{w,e}\right)}^2=a^{w,e}$, as asserted. □

Corollary 8. 

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

(1)

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

(2)

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

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

In this case, $a^{w,e}=x.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 4.1.

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

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

Then we verify that

$$awx=aw\left[{\left(xw\right)}^{2}a\right]=\left[a{\left(wx\right)}^2\right]wa=xwa.$$

By virtue of Theorem 4.1, $a\in {\mathcal{A}}^{w,e}$ and $a^{w,e}=x$ by Theorem 4.1. □

Theorem 10. 

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

(1)

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

(2)

$a\in {\mathcal{A}}^{w,d}\bigcap {\mathcal{A}}_{w,d}$ and $a^{w,d}=a_{w,d}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $x=a^{w,e}$. In view of Theorem 4.1, there exists $x\in \mathcal{A}$ such that

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

According, we verify that the equalities

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

Therefore $a\in {\mathcal{A}}^{w,d}\bigcap {\mathcal{A}}_{w,d}$ and $a^{w,d}=a_{w,d}$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that the preceding equalities hold. Then we verify that

$$awx=aw\left[{\left(xw\right)}^{2}a\right]=\left[a{\left(wx\right)}^2\right]wa=xwa.$$

Therefore $a^{w,e}=x$, as desired. □

Corollary 9. 

Let $a\in {\mathcal{A}}^{w,e}$. Then $aw\in {\mathcal{A}}^d,wa\in {\mathcal{A}}_d$ and

$$a^{w,e}={\left[{\left(aw\right)}^d\right]}^{2}a=a{\left[{\left(wa\right)}_d\right]}^2={\left(aw\right)}^{d}a{\left(wa\right)}_d.$$

Proof. 

By virtue of Theorem 4.4, $a\in {\mathcal{A}}^{w,d}\bigcap {\mathcal{A}}_{w,d}$. It follows by [6] that $aw\in {\mathcal{A}}^d$. Dually, we have $wa\in {\mathcal{A}}^d$. Set $x=a^{w,e}$. Then we derive that

$$\begin{array}{ccc}\hfill x& =\hfill & a{\left(wx\right)}^2=a{\left[{\left(wa\right)}_d\right]}^2\hfill \\ & =\hfill & {\left(xw\right)}^{2}a={\left[{\left(aw\right)}^d\right]}^{2}a\hfill \\ & =\hfill & \left(xw\right)a\left(wx\right)={\left(aw\right)}^{d}a{\left(wa\right)}_d.\hfill \end{array}$$

□

We are ready to prove:

Theorem 11. 

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

(1)

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

(2)

$aw\in {\mathcal{A}}^d,wa\in {\mathcal{A}}_d$ and

$${\left(aw\right)}^{d}a=a{\left(wa\right)}_d,{\left(aw\right)}^d\left(aw\right)=\left(aw\right){\left(aw\right)}^d,wa{\left(wa\right)}_d={\left(wa\right)}_{d}wa.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a^{w,e}$. In view of Corollary 4.5, $aw\in {\mathcal{A}}^d,wa\in {\mathcal{A}}_d$. Moreover, we have $a{\left(wx\right)}^2=x={\left(xw\right)}^{2}a.$ Hence

$$aw{\left(xw\right)}^2=xw,{\left(wx\right)}^{2}wa=wx.$$

Thus

$${\left(aw\right)}^d=xw,{\left(wa\right)}_d=wx.$$

We verify that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{d}a& =\hfill & xwa=\left[a{\left(wx\right)}^2\right]wa=aw\left[{\left(xw\right)}^{2}a\right]=awx=a{\left(wa\right)}_d,\hfill \\ \hfill {\left(aw\right)}^d\left(aw\right)& =\hfill & aw\left[{\left(xw\right)}^{2}a\right]w=\left(awx\right)w=\left(aw\right){\left(aw\right)}^d,\hfill \\ \hfill {\left(wa\right)}_{d}wa& =\hfill & w\left(xw\right)a=w\left(xwa\right)=w\left(awx\right)=wa\left(wx\right)wa{\left(wa\right)}_d,\hfill \end{array}$$

as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x={\left(aw\right)}^{d}a{\left(wa\right)}_d$. Since ${\left(aw\right)}^{d}a=a{\left(wa\right)}_d$, we check that $x={\left[{\left(aw\right)}^d\right]}^{2}a=a{\left[{\left(wa\right)}_d\right]}^2.$ Moreover, we have

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

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

Furthermore, we verify that

$$\begin{array}{ccc}\hfill xwaw& =\hfill & wa{\left[{\left(wa\right)}_d\right]}^{2}wa=wa{\left(wa\right)}_d={\left(wa\right)}_{d}wa,\hfill \\ \hfill {\left(xw\right)}^{2}a& =\hfill & xwxwa={\left[{\left(aw\right)}^d\right]}^{2}aw{\left[{\left(aw\right)}^d\right]}^{2}awa\hfill \\ & =\hfill & {\left[{\left(aw\right)}^d\right]}^{3}awa=aw{\left[{\left(aw\right)}^d\right]}^{3}a={\left[{\left(aw\right)}^d\right]}^{2}a=x,\hfill \\ \hfill {\left(xwaw\right)}^{*}& =\hfill & xwaw.\hfill \end{array}$$

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

Thus, $a\in {\mathcal{A}}^{w,d}\bigcap {\mathcal{A}}_{w,d}$ and $a^{w,d}=x=a_{w,d}$. Therefore we complete the proof by Theorem 4.4. □

Corollary 10. 

Let $a\in {\mathcal{A}}^{w,e}$. Then

$$a^{w,d}={\left(waw\right)}^{\left({\left(aw\right)}^d,{\left(wa\right)}_d\right)}.$$

Proof. 

In view of Theorem 4.6, $aw\in {\mathcal{A}}^d,wa\in {\mathcal{A}}_d$. Let $x=a^{w,d}$. Then we verify that

$$\begin{array}{ccc}\hfill x\left(waw\right){\left(aw\right)}^d& =\hfill & ={\left[{\left(aw\right)}^d\right]}^{2}a\left(waw\right){\left(aw\right)}^d={\left(aw\right)}^d,\hfill \\ \hfill {\left(wa\right)}_d\left(waw\right)x& =\hfill & {\left(wa\right)}_d\left(waw\right)a{\left[{\left(wa\right)}_d\right]}^2={\left(wa\right)}_d,\hfill \\ \hfill x& =\hfill & xwawx={\left(aw\right)}^{d}a{\left(wa\right)}_d\in {\left(aw\right)}^d\mathcal{A}x\bigcap x\mathcal{A}{\left(wa\right)}_d.\hfill \end{array}$$

Therefore, $waw$ has $\left({\left(aw\right)}^d,{\left(wa\right)}_d\right)$-inverse x, as asserted. □

An element $a\in \mathcal{A}$ has generalized w-Drazin inverse x if x satisfies the equations

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

We denote such x by $a^{d,w}$ (see [14]).

Theorem 12. 

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

(1)

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

(2)

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

(3)

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

In this case, $a^{w,e}=a^{d,w}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 4.6, $aw\in {\mathcal{A}}^d$. Hence $a\in {\mathcal{A}}^{d,w}$ and $a^{d,w}={\left(aw\right)}^{d}a{\left(wa\right)}^d={\left[{\left(aw\right)}^d\right]}^{2}a=a{\left[{\left(wa\right)}^d\right]}^2$.

Let $x=a^{w,e}$. Then we have

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

We verify that

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

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

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

Hence, $awxw\left(aw\right){\left(aw\right)}^d=aw{\left(aw\right)}^d$. Let $z={\left(aw\right)}^{2}x$. Then

$$\begin{array}{ccc}\hfill a^{d,w}wz& =\hfill & a^{d,w}w{\left(aw\right)}^{2}x={\left[{\left(aw\right)}^d\right]}^{2}aw{\left(aw\right)}^{2}x={\left(aw\right)}^d{\left(aw\right)}^{2}x=awx,\hfill \\ \hfill a^{d,w}{\left(wz\right)}^2& =\hfill & \left(a^{d,w}wz\right)wz=\left(awx\right)w{\left(aw\right)}^{2}x=xw{\left(aw\right)}^{3}x={\left(aw\right)}^{2}x=z,\hfill \\ \hfill z{\left(w{a}^{d,w}\right)}^2& =\hfill & {\left(aw\right)}^{2}xw{a}^{d,w}w{a}^{d,w}=\left[{\left(aw\right)}^{2}xw{\left(aw\right)}^2\right]{\left[{\left(aw\right)}^d\right]}^2{a}^{d,w}w{a}^{d,w}\hfill \\ & =\hfill & \left[{\left(aw\right)}^2\left(aw\right)\right]{\left[{\left(aw\right)}^d\right]}^2{a}^{d,w}w{a}^{d,w}\hfill \\ & =\hfill & \left[{\left(aw\right)}^2{\left(aw\right)}^d\right]a^{d,w}w{a}^{d,w}=a^{d,w},\hfill \\ \hfill {\left(a^{d,w}wzw\right)}^{*}& =\hfill & {\left(awxw\right)}^{*}=awxw=a^{d,w}wzw,\hfill \\ \hfill wzw{a}^{d,w}& =\hfill & w{\left(aw\right)}^{2}xw{a}^{d,w}=wxw{\left(aw\right)}^2{\left[{\left(aw\right)}^d\right]}^{2}a=wxwa,\hfill \\ \hfill {\left(wzw{a}^{d,w}\right)}^{*}& =\hfill & wzw{a}^{d,w}.\hfill \end{array}$$

Accordingly,

$$a^{d,w}{\left(wz\right)}^2=z,z{\left(w{a}^{d,w}\right)}^2=a^{d,w},{\left(a^{d,w}wzw\right)}^{*}=a^{d,w}wzw,{\left(wzw{a}^{d,w}\right)}^{*}=wzw{a}^{d,w}.$$

Then $a^{d,w}\in {\mathcal{A}}^{w,e}$ and ${\left(a^{d,w}\right)}^{w,e}=z={\left(aw\right)}^2{a}^{w,e},$ as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Let $z={\left(a^{d,w}\right)}^{w,e}$. Then

$${\left(a^{d,w}w\right)}^{2}z=a^{d,w},a^{d,w}wz=zw{a}^{d,w},{\left(a^{d,w}wzw\right)}^{*}=a^{d,w}wzw,{\left(w{a}^{d,w}wz\right)}^{*}=w{a}^{d,w}wz.$$

Moreover, we verify that

$$\begin{array}{ccc}\hfill aw{a}^{d,w}& =\hfill & aw\left[{\left(a^{d,w}w\right)}^{2}z\right]=aw[\left(a^{d,w}w\right)\left(a^{d,w}wz\right)\hfill \\ & =\hfill & aw\left[\left(a^{d,w}w\right)\left(zw{a}^{d,w}\right)\right]=aw{\left[{\left(aw\right)}^d\right]}^{2}awzw{a}^{d,w}\hfill \\ & =\hfill & zw{a}^{d,w}=zw{a}^{d,w}\left(w{a}^{d,w}\right)wa)=a^{d,w}wz\left(w{a}^{d,w}\right)wa)\hfill \\ & =\hfill & a^{d,w}w\left(zw{a}^{d,w}\right)wa=a^{d,w}w\left(a^{d,w}wz\right)wa\hfill \\ & =\hfill & {\left(a^{d,w}w\right)}^{2}zwa=a^{d,w}wa.\hfill \end{array}$$

By hypothesis, we have

$$a^{d,w}wzw=a{\left(w{a}^{d,w}\right)}^{2}wzw=aw\left[{\left(a^{d,w}w\right)}^{2}z\right]w=aw{a}^{d,w}w.$$

Likewise, we have $waw{a}^{d,w}=w{a}^{d,w}wz$; hence, $w{a}^{d,w}wa=w{a}^{d,w}wz$. Accordingly, we have

$$\begin{array}{ccc}\hfill a{\left(w{a}^{d,w}\right)}^2& =\hfill & a^{d,w},\hfill \\ \hfill {\left(aw{a}^{d,w}w\right)}^{*}& =\hfill & aw{a}^{d,w}w,\hfill \\ \hfill {\left(w{a}^{d,w}wa\right)}^{*}& =\hfill & w{a}^{d,w}wa.\hfill \end{array}$$

Obviously, we have

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

Therefore $a^{w,e}=a^{d,w}$, as required.

$\left(1\right)\Rightarrow \left(3\right)$ Set $x=a^{w,e}$. By virtue of Theorem 4.1, we have

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

Therefore $a\in {\mathcal{A}}^{d,w}$ and $a^{d,w}=x=a^{w,e}$, as required.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, ${\left(aw{a}^{d,w}w\right)}^{*}=aw{a}^{d,w}w$ and ${\left(w{a}^{d,w}wa\right)}^{*}=w{a}^{d,w}wa$. Obviously, we have

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

In light of Theorem 4.1, $a\in {\mathcal{A}}^{w,e}$ and $a^{w,e}=a^{d,w}$, as asserted. □