Weighted EP Elements in Rings

1. Introduction

Let R be a ring with involution *. An element a in R is an EP element if there exists some $x\in R$ such that $a{x}^2=x,x{a}^2=a,{\left(ax\right)}^{*}=ax,ax=xa$ (see [15]). Let $C^{n\times n}$ be the ring of all $n\times n$ complex matrices with conjugate transpose *. Evidently, $A\in C^{n\times n}$ is an EP matrix if and only if $rank\left(A\right)=rank\left(A^{*}\right)$ if and only if $A{A}^{*}=A^{*}A$. EP elements are a specialized class of elements within ring and algebra, characterized by a specific relationship between the element and its Moore-Penrose inverse or generalized inverses in structured settings such as rings, Banach algebras and $C^{*}$-algebras. EP elements in a ring and algebra are expensively studied by many authors. Their study bridges abstract algebra and operator theory, with applications in matrix analysis, functional analysis, and beyond (see [1,3,5,6,7,8,9,10,13,14,16,18,20,21]).

Zhu et al. introduced and studied w-core inverse for a ring element (see [22]). Let $a,w\in \mathcal{A}$. An element $a\in \mathcal{A}$ has w-core inverse if there exists $x\in \mathcal{A}$ such that $aw{x}^2=x,{\left(awx\right)}^{*}=awx,xawa=a.$ If such x exists, it is unique, and denote it by $a_w^{\overline{)\#}}$. Let $R_w^{\overline{)\#}}$ denote the set of all w-core invertible elements in R. We refer the reader to [11,19] for more results on the w-core inverse. Following Lu, an element $a\in R$ is a w-EP element if there exists $x\in R$ such that $aw{x}^2=x,xawaaa$ and ${\left(xaw\right)}^{*}=xaw$. Let $R_w^{EP}$ denote the set of all w-EP elements in R. Evidently, $a\in R^{EP}$ if and only if $a\in R_w^{\overline{)\#}}$ and $aw{a}_w^{\overline{)\#}}=a_w^{\#}aw$ (see [12][Theorem 2.10 and Theorem 2.12]).

Let $a,v,w\in R$. We consider the solution of a system of equations involving $a,v,w$, which aims to extend EP and w-EP elements to a broader class of elements with two weights. We adopt

Definition 1.1.

An element $a\in R$ is a $(v,w)$-weighted EP element if there exists $x\in R$ such that

$$xvawa=a,awxvx=x,{\left(awxv\right)}^{*}=awxv,awxv=xvaw.$$

In Section 2, we present the fundamental properties of $(v,w)$-weighted EP elements. Let $R_{v,w}^{EP}$ be the set of all $(v,w)$-weighted EP elements in R. We prove that $a\in R_{v,w}^{EP}$ if and only if there exists $x\in R$ such that $xvawa=a,awxvx=x,{\left(xvaw\right)}^{*}=xvaw$, if and only if there exists a projection $p\in R$ (i..e., $p^2=p=p^{*}$) such that

$$aw+p\in R^{-1},1-p\in Rv,awp=pa=0.$$

Here, $R^{-1}$ denotes the set of all invertible elements in R.

Definition 1.2.

An element $a\in R$ has $(v,w)$-weighted group inverse if there exists $x\in R$ such that

$$xvawa=a,awxvx=x,awxv=xvaw.$$

The preceding x is unique if it exists, and we denote it by $a_{v,w}^{\#}$.

The set of all $(v,w)$-weighted group invertible elements in R is denoted by $R_{v,w}^{\#}$.

Definition 1.3.

An element $a\in R$ is $(v,w)$-weighted MP invertible if there exists $x\in R$ such that

$$awxva=a,xvawx=x,{\left(awxv\right)}^{*}=awxv,{\left(xvaw\right)}^{*}=xvaw.$$

We use $R_{v,w}^{†}$ to stand for the set of all $(v,w)$-weighted MP invertible elements in R. In Section 3, we provide characterizations of $(v,w)$-weighted EP elements in terms of related images. It is shown that $a\in R_{v,w}^{EP}$ if and only if $a\in R_{v,w}^{\#}$ and $aR={\left(aw\right)}^{*}R$, and respectively, if and only if $a\in R_{v,w}^{†}$ and $aR={\left(aw\right)}^{*}R$.

Finally, Section 4 further explore $(v,w)$-weighted EP elements via its weighted MP invertibility, proving that $a\in R_{v,w}^{EP}$ if and only if $a\in R_{v,w}^{†}$ and $a_{v,w}^{†}v=u\left(aw\right)$ for some $u\in R^{-1}$. These elements are also characterized using related weighted core inverse.

2. Weighted EP Elements

The aim of this section is to present fundamental properties of weighted EP elements in a ring. We begin with

Theorem 2.1.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $x\in R$ such that

$$xvawx=x,awxva=a,{\left(awxv\right)}^{*}=xvaw.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exists $x\in R$ such that

$$xvawx=x,awxva=a,awxv=xvaw,{\left(awxv\right)}^{*}=awxv.$$

Then ${\left(awxv\right)}^{*}=xvaw$. Moreover, we have

$$\begin{array}{ccc}\hfill xvawa& =\hfill & awxva=a,\hfill \\ \hfill awxvx& =\hfill & xvawx=x,\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in R$ such that

$$xvawa=a,awxvx=x,{\left(awxv\right)}^{*}=xvaw.$$

Then $ax{\left(xv\right)}^2=xv$. This implies that $awxv=xvaw$ by [16][Lemma 2.1]. Hence $xvawx=\left(awxv\right)x=awxvx=x$ and $awxva=xvawa=a$, as required. □

Corollary 2.2.

Let $a,v,w\in R$. Then $a\in R_{v,w}^{EP}$ if and only if there exists $a_{v,w}^{\#}$ such that ${\left(aw{a}_{v,w}^{\#}v\right)}^{*}=aw{a}_{v,w}^{\#}v$.

Proof. 

⟹ Set $x=a\in R_{v,w}^{EP}$. Then $x\in R_{v,w}^{\#}$. Moreover, we see that ${\left(awxv\right)}^{*}=awxv$, as required.

⟸ By hypothesis, we have $x\in R_{v,w}^{\#}$ such that ${\left(awxv\right)}^{*}=awxv$. Furthermore, we have $xvawa=a,awxvx=x,awxv=xvaw.$ Thus we check that

$$xvawx=awxvx=x,awxva=xvawa=a,{\left(awxv\right)}^{*}=awxv=xvaw.$$

By virtue of Theorem 2.1, $a\in R_{v,w}^{EP}$. □

We are ready to prove:

Theorem 2.3.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $x\in R$ such that

$$awxva=a,awxvx=x,{\left(awxv\right)}^{*}=xvaw.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exists $x\in R$ such that

$$xvawx=x,awxva=a,awxv=xvaw,{\left(awxv\right)}^{*}=awxv.$$

Hence, ${\left(awxv\right)}^{*}=xvaw$. Moreover, we check that $awxvx=\left(awxv\right)x=\left(xvaw\right)x=x$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in R$ such that

$$awxva=a,awxvx=x,{\left(awxv\right)}^{*}=xvaw.$$

Suppose that there exists $y\in R$ such that

$$awyva=a,awyvy=y,{\left(awyv\right)}^{*}=yvaw.$$

In view of [16][Lemma 2.1], $awxv=xvaw,awyv=yvaw$. Hence,

$$x=awxvx=\left[awyva\right]wxvx=yv\left[awa\right]wxvx=yv\left[awxva\right]wx=yvawx.$$

Likewise, we have $y=xvawy$. Then

$$\begin{array}{ccc}\hfill y& =\hfill & yvawx={\left(yvaw\right)}^{*}x\hfill \\ & =\hfill & {\left(yvawxvaw\right)}^{*}x={\left(xvaw\right)}^{*}x\hfill \\ & =\hfill & awxvx=x.\hfill \end{array}$$

Therefore $x=y$.

We check that

$$\begin{array}{ccc}\hfill aw\left[xvawx\right]va& =\hfill & \left[awxva\right]wxva=awxva=a,\hfill \\ \hfill aw\left[xvawx\right]v\left[xvawx\right]& =\hfill & \left(awxva\right)wxvxvawx\hfill \\ & =\hfill & \left(awxvx\right)vawx=xvawx,\hfill \\ \hfill aw\left(xvawx\right)v& =\hfill & \left[awxva\right]wxv=awxv,\hfill \\ \hfill \left(xvawx\right)vaw& =\hfill & xv\left[awxva\right]w=xvaw,\hfill \\ \hfill {\left(aw\left(xvawx\right)v\right)}^{*}& =\hfill & \left(xvawx\right)vaw.\hfill \end{array}$$

By the uniqueness of x, we obtain

$$xvawx=x,awxva=a,awxv=xvaw,{\left(awxv\right)}^{*}=awxv$$

Therefore $a\in R_{v,w}^{EP}$. □

Corollary 2.4.(see [18][Theorem 2.7 and Theorem 2.11]) Let $a\in R$. Then the following are equivalent:

(1)

$a\in R^{EP}$.

(2)

There exists $x\in R$ such that $a^{2}x=a,ax=xa,{\left(ax\right)}^{*}=ax.$

(3)

There exists $x\in R$ such that $a^{2}x=a,x^{2}a=x,{\left(ax\right)}^{*}=ax.$

Proof. 

This is obvious by choosing $v=w=1$ in Theorem 2.3. □

Theorem 2.5.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $x\in R$ such that $xvawa=a,{\left(awxv\right)}^{*}=xvaw.$

(3)

There exists $a_{v,w}^{\#}$ such that $aw{a}_{v,w}^{\#}v=a_{v,w}^{\#}vaw.$

Proof.$\left(1\right)\Rightarrow \left(3\right)$ Let $x=a_w^{EP}$. Obviously, we have $x=a_{v,w}^{\#}$ and $awxv=xvaw,$ as desired.

$\left(3\right)\Rightarrow \left(2\right)$ Let $x=a_{v,w}^{\#}$. Then $xvawa=a,{\left(awxv\right)}^{*}=awxv$. By hypothesis, $awxv=xvaw$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in R$ such that

$$xvawa=a,{\left(awxv\right)}^{*}=xvaw.$$

Then $xv{\left(aw\right)}^2=aw$. In view of [16][Lemma 2.1], we have $xvaw=awxw.$ Hence ${\left(awxv\right)}^{*}=xvaw=awxv$, and so $a=xvawa=\left(xvaw\right)a={\left(awxv\right)}^{*}a=awxva.$

Let $z=xvawx$. Then

$$\begin{array}{ccc}\hfill zvawz& =\hfill & xv\left(awxva\right)wxvawx=xv\left(awxva\right)wx=xvawx=z,\hfill \\ \hfill zvawa& =\hfill & xvaw\left[xvawa\right]=xvaw\left(xvawa\right)=xvawa=a,\hfill \\ \hfill zvaw& =\hfill & xv\left(awxva\right)w=xvaw={\left(awxv\right)}^{*},\hfill \\ \hfill {\left(zvaw\right)}^{*}& =\hfill & awxv={\left(awxv\right)}^{*}=zvaw,\hfill \\ \hfill awzvz& =\hfill & \left(awxva\right)wxvxvawx={\left(xv\right)}^3{\left(aw\right)}^{3}x\hfill \\ & =\hfill & \left(xv\right)\left(aw\right)x=z.\hfill \end{array}$$

Therefore $a\in R_{v,w}^{EP}$ by Theorem 2.1. □

Corollary 2.6.

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

(1)

$a\in R_{w,w}^{EP}$.

(2)

There exists $x\in R$ such that ${\left(aw\right)}^{2}x=a,{\left(awxw\right)}^{*}=xwaw.$

Proof. 

This is proved by Theorem 2.5 and [16][Theorem 2.3]. □

Theorem 2.7.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $x\in R$ such that $awxvx=x,xvawa=a,{\left(xvaw\right)}^{*}=xvaw$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exists $x\in R$ such that $awxvx=x,xvawa=a,awxv=xvaw,{\left(awxv\right)}^{*}=awxv$. Thus ${\left(xvaw\right)}^{*}=xvaw$, as required.

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

$$\begin{array}{ccc}\hfill {\left(xvaw\right)}^{*}& =\hfill & {\left(\left(awxvx\right)vaw\right)}^{*}={\left(\left(awxv\right)\left(xvaw\right)\right)}^{*}\hfill \\ & =\hfill & {\left(xvaw\right)}^{*}{\left(awxv\right)}^{*}=\left(xvaw\right)\left(awxv\right)\hfill \\ & =\hfill & \left(xvawa\right)wxv=awxv.\hfill \end{array}$$

Hence ${\left(awxv\right)}^{*}=xvaw$. Moreover, we have

$${\left(awxv\right)}^{*}=xvaw=\left(awxvx\right)vaw=\left(awxv\right)\left(xvaw\right)=\left(awxv\right){\left(awxv\right)}^{*}.$$

Hence, ${\left(awxv\right)}^{*}=awxv$. Therefore we complete the proof. □

Recall that an element $a\in R$ has $(v,w)$-weighted core inverse if there exists $x\in R$ such that $xvawa=a,awxvx=x,{\left(awxv\right)}^{*}=awxv.$ Let $R_{v,w}^{\#}$ be the set of all $(v,w)$-weighted core invertible elements in R.

Corollary 2.8.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $a_{v,w}^{\#}$ such that ${\left(a_{v,w}^{\#}\right)}_v^{\#}=aw.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{v,w}^{\#}$. By hypothesis, we verify that

$$\begin{array}{ccc}\hfill xv{\left(aw\right)}^2& =\hfill & \left(awxva\right)w=aw,\hfill \\ \hfill \left(aw\right)xvx& =\hfill & xvawx=x,\hfill \\ \hfill {\left(xvaw\right)}^{*}& =\hfill & {\left(awxv\right)}^{*}=awxv=xv\left(aw\right).\hfill \end{array}$$

Therefore $x_v^{\overline{)\#}}=aw$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x=a_{v,w}^{\overline{)\#}}$. Then $awxvx=x,xvawa=a$. Since $x_v^{\overline{)\#}}=aw$, we have ${\left(xvaw\right)}^{*}=xvaw$. Accordingly, we obtain the result. □

As is well known, every EP element possess the polar-like property, i.e., there exists a projection $p\in R$ such that $a+p\in R^{-1},ap=pa=0$ (see [18][Theorem 3.2]). We next extend this result and establish the polar-like property of weighted EP elements.

Theorem 2.9.

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists a projection $p\in R$ such that

$$aw+p\in R^{-1},1-p\in Rv,awp=pa=0.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{v,w}^{\overline{)\#}}$ and $p=1-awxv$. Then $p^2=p=p^{*}$. Moreover, we check that

$$\begin{array}{ccc}\hfill aw+p& =\hfill & aw+1-awxv={(xv+1-awxv)}^{-1}\in R^{-1},\hfill \\ \hfill 1-p& =\hfill & awxv\in Rv,\hfill \\ \hfill awp& =\hfill & aw(1-awxv)=aw-awxvaw=0,pa=(1-awxv)a=0,\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists a projection $p\in R$ such that

$$aw+p\in R^{-1},1-p\in Rv,awp=pa=0.$$

Set $x=(1-p){(aw+p)}^{-1}$. Then $x={(aw+p)}^{-1}(1-p)\in \mathcal{A}v$. Write $x=zv$ with $(1-p)z=z$. Then

$$\begin{array}{ccc}\hfill awzvz& =\hfill & awzvz=awxz=aw(1-p){(aw+p)}^{-1}z\hfill \\ & =\hfill & (1-p)(aw+p){(aw+p)}^{-1}z=(1-p)z=z,\hfill \\ \hfill zvawa& =\hfill & xawa=(1-p){(aw+p)}^{-1}awa\hfill \\ & =\hfill & (1-p){(aw+p)}^{-1}(aw+p)a\hfill \\ & =\hfill & (1-p)a=a,\hfill \\ \hfill zvaw& =\hfill & xaw=(1-p){(aw+p)}^{-1}aw\hfill \\ & =\hfill & {(aw+p)}^{-1}aw(1-p)\hfill \\ & =\hfill & {(aw+p)}^{-1}(aw+p)(1-p)=1-p,\hfill \\ \hfill awzv& =\hfill & awx=aw(1-p){(aw+p)}^{-1}\hfill \\ & =\hfill & (aw+p){(aw+p)}^{-1}(1-p)=1-p.\hfill \end{array}$$

Therefore $a\in R_{v,w}^{EP}$. □

Corollary 2.10.

Let $a\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

$a\in R_{v,w}^{\overline{)\#}}$ and $awp=0$, where $p=1-aw{a}_{v,w}^{\overline{)\#}}v.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $aw{a}_{v,w}^{\overline{)\#}}v=a_{v,w}^{\overline{)\#}}vaw$, we deduce that

$$awp=aw[1-aw{a}_{v,w}^{\overline{)\#}}v]=aw-\left[aw{a}_{v,w}^{\overline{)\#}}va\right]w=0,$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ Clearly, $p^2=p=p^{*}$. Moreover, we check that

$$aw+p\in R^{-1},1-p\in Rv,awp=pa=0.$$

Therefore we obtain the result by Theorem 2.9. □

3. Image-Based Characterizations

In this section, we establish certain equivalent conditions on weighted group elements to be the weighted EP element. We now derive

Theorem 3.1.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

$a\in R_{v,w}^{\#}$ and $aR={\left(aw\right)}^{*}R$.

(3)

$a\in R_{v,w}^{\#}$ and $aR\subseteq {\left(aw\right)}^{*}R$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Clearly, $a\in R_{v,w}^{\#}$. Since $a=awxva=\left(awxv\right)a={\left(awxv\right)}^{*}a={\left(xvaw\right)}^{*}a={\left(aw\right)}^{*}{\left(xv\right)}^{*}a$, we have $aR\subseteq {\left(aw\right)}^{*}R$. As $aw=\left(awxva\right)w=aw\left(xvaw\right)$, we see that ${\left(aw\right)}^{*}={\left(xvaw\right)}^{*}{\left(aw\right)}^{*}=\left(xvaw\right)\left(aw\right)*=\left(aw\right)\left(xv\right){\left(aw\right)}^{*}\in aR$; whence, ${\left(aw\right)}^{*}R\subseteq aR$. Therefore $aR={\left(aw\right)}^{*}R$.

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

$\left(3\right)\Rightarrow \left(1\right)$ Set $x=a_{v,w}^{\#}$. Then $awxva=a$ and $xvaw=awxv$. Write $a={\left(aw\right)}^{*}z$ for a $z\in R$. Then ${\left(xvaw\right)}^{*}={\left(awxv\right)}^{*}={\left(xv\right)}^{*}{\left(aw\right)}^{*}$. We easily see that $awxvaw=\left(awxva\right)w=aw$; hence, ${\left(xvaw\right)}^{*}{\left(aw\right)}^{*}={\left(aw\right)}^{*}$. Thus ${\left(xvaw\right)}^{*}{\left(aw\right)}^{*}z={\left(aw\right)}^{*}z$, and so ${\left(xvaw\right)}^{*}a=a$. We deduce that ${\left(xvaw\right)}^{*}awxv=awxv$; whence ${\left(xvaw\right)}^{*}\left(xvaw\right)=xvaw$. Then ${\left(xvaw\right)}^{*}={\left(xvaw\right)}^{*}\left(xvaw\right)=xvaw$. Clearly, $awxvx=xvawx=x$ and $xvawa=awxva=a$. Therefore $a\in R_{v,w}^{EP}$. □

Corollary 3.2.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

$a\in R_{v,w}^{\#}$ and $aR={\left(vaw\right)}^{*}R$.

(3)

$a\in R_{v,w}^{\#}$ and $aR\subseteq {\left(vaw\right)}^{*}R$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Obviously, $a\in R_{v,w}^{\#}$. Since $a=xvawa={\left(xvaw\right)}^{*}a={\left(vaw\right)}^{*}{x}^{*}a$, we have $aR\subseteq {\left(vaw\right)}^{*}R$. As $vaw=v\left(awxva\right)w=vaw\left(xvaw\right)$, it follows that ${\left(vaw\right)}^{*}={\left(xvaw\right)}^{*}{\left(vaw\right)}^{*}=\left(xvaw\right)\left(vaw\right)*=\left(aw\right)\left(xv\right){\left(vaw\right)}^{*}\in aR$, and then ${\left(vaw\right)}^{*}R\subseteq aR$. We infer that $aR={\left(vaw\right)}^{*}R$, as required.

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

$\left(3\right)\Rightarrow \left(1\right)$ Since $aR\subseteq {\left(vaw\right)}^{*}R={\left(aw\right)}^{*}{v}^{*}R\subseteq {\left(aw\right)}^{*}R$. This completes the proof by Theorem 3.1. □

Corollary 3.3.

Let $a,w\in R$. Then the following are equivalent:

(1)

$a\in R_{w,w}^{EP}$.

(2)

$a\in R_{w,w}^{\#}$ and $aR={\left(aw\right)}^{*}R$.

(3)

$a\in R_{w,w}^{\#}$ and $aR\subseteq {\left(aw\right)}^{*}R$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Clearly, $a\in R_{w,w}^{\#}$. By virtue of Theorem 3.1, $aR={\left(aw\right)}^{*}R$.

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

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in R$ such that $xwawa=a,awxwx=x,{\left(awxw\right)}^{*}=awxw.$ Set $z={\left(xw\right)}^{2}a$. Then we verify that

$$\begin{array}{ccc}\hfill awzw& =\hfill & aw{\left(xw\right)}^{2}aw=xwaw,\hfill \\ \hfill zwaw& =\hfill & {\left(xw\right)}^2{\left(aw\right)}^2=xwaw\hfill \\ \hfill awzw& =\hfill & zwaw,\hfill \\ \hfill zwawa& =\hfill & \left(xwaw\right)a=a,\hfill \\ \hfill awzwz& =\hfill & \left(xwaw\right){\left(xw\right)}^{2}a={\left(xw\right)}^{2}a=z.\hfill \end{array}$$

This implies that $a_{w,w}^{\#}=x$. Therefore we complete the proof by Theorem 3.1. □

Theorem 3.4.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

$a\in R_{v,w}^{†}$ and $aR={\left(aw\right)}^{*}R$.

(3)

$a\in R_{v,w}^{†}$ and $aR={\left(vaw\right)}^{*}R$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Obviously, $a\in R_{v,w}^{†}$. By virtue of Corollary 3.3, $aR={\left(aw\right)}^{*}R$.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x=a_{v,w}^{†}$. Then we have $awxva=a,xvawx=x,{\left(awxv\right)}^{*}=awxv$ and ${\left(xvaw\right)}^{*}=xvaw$. Since $aR={\left(aw\right)}^{*}R$, we see that $R{a}^{*}=R\left(aw\right)$. As $awxvaw=aw$, we have ${\left(xvaw\right)}^{*}{\left(aw\right)}^{*}={\left(aw\right)}^{*}$, and so $xvawa=a$.

As $awxva=a$, we get $awxv{\left(aw\right)}^{*}={\left(aw\right)}^{*}$. Then $\left(aw\right)\left(awxv\right)=aw$. This implies that $\left(xvaw\right)\left(awxv\right)=xvaw$. Therefore $\left(xvaw\right)\left(awxv\right)x=\left(xvaw\right)x=x$. Accordingly, we deduce that

$$awxvx=\left[xvawa\right]wxvx=\left(xvaw\right)\left(awxv\right)x=x,$$

as required.

$\left(3\right)\Rightarrow \left(2\right)$ Clearly, $aR={\left(vaw\right)}^{*}R={\left(aw\right)}^{*}{v}^{*}R\subseteq {\left(aw\right)}^{*}R$. Set $x=a_{v,w}^{†}$. Then $aw=awxvaw=\left(awx\right)\left(vaw\right)$. Hence, ${\left(aw\right)}^{*}R={\left(vaw\right)}^{*}{\left(awx\right)}^{*}R\subseteq {\left(vaw\right)}^{*}R=aR$. Thus, $aR={\left(aw\right)}^{*}R$, as asserted. □

We use $[x,y]$ to stand the commutator of x and y, i.e., $[x,y]=xy-yx$. We derive

Corollary 3.5.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $a_{v,w}^{†}$ such that $[aw,a_{v,w}^{†}v]=0$.

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

$\left(2\right)\Rightarrow \left(1\right)$ Since $[aw,a_{v,w}^{†}v]=0$, we see that $aw{a}_{v,w}^{†}v=a_{v,w}^{†}vaw$. Clearly, $a=\left[aw{a}_{v,w}^{†}v\right]a=\left[a_{v,w}^{†}vaw\right]a={\left[a_{v,w}^{†}vaw\right]}^{*}a={\left(aw\right)}^{*}{\left[a_{v,w}^{†}v\right]}^{*}a$. On the other hand, we have

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{*}& =\hfill & {\left(aw{a}_{v,w}^{†}vaw\right)}^{*}\hfill \\ & =\hfill & {\left(a_{v,w}^{†}vaw\right)}^{*}{\left(aw\right)}^{*}\hfill \\ & =\hfill & \left(a_{v,w}^{†}vaw\right){\left(aw\right)}^{*}\hfill \\ & =\hfill & aw{a}_{v,w}^{†}v{\left(aw\right)}^{*}.\hfill \end{array}$$

Therefore $aR={\left(aw\right)}^{*}R$, thus yielding the result by Theorem 3.4. □

Lemma 3.6.

Let $a\in R_{v,w}^{EP}$. Then $xvawx=x$.

Proof. 

By hypothesis, we can find $x\in R$ such that $awxvx=x,xvawa=a,{\left(awxw\right)}^{*}=awxw,{\left(xwaw\right)}^{*}=xwaw.$ Then $xvawx=xvaw\left[awxvx\right]=\left[xvawa\right]wxvx=awxvx=x.$ □

We use $\ell \left(x\right)$ and $r\left(x\right)$ to denote the left and right annihilator of x in R, respectively. We deduce that

Theorem 3.7.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $x\in R$ such that $awxva=a,xR=aR,R{x}^{*}=Raw.$

(3)

There exists $x\in R$ such that $xvawx=x,xR=aR,R{x}^{*}=Raw.$

(4)

There exists $x\in R$ such that $awxva=a,\ell \left(x\right)=\ell \left(a\right),r\left(x^{*}\right)=r\left(aw\right).$

(5)

There exists $x\in R$ such that $xvawx=x,\ell \left(x\right)=\ell \left(aw\right),r\left(x^{*}\right)=r\left(aw\right).$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $x=a_{v,w}^{EP}$. In view of Theorem 2.1 and Theorem 2.7, $awxva=a,awxvx=x,xvawa=a,{\left(xvaw\right)}^{*}=xvaw.$ Hence, $x=a\left(wxvx\right)\in aR$ and $a=xvawa\in xR$. This implies that $xR=aR$. By virtue of Lemma 3.6, $x=xvawx$, and so $x^{*}=x^{*}{\left(xvaw\right)}^{*}=x^{*}xvaw\in Raw$. Since $a=awxva$, we have $aw=aw\left(xvaw\right)=aw{\left(xvaw\right)}^{*}=aw{\left(vaw\right)}^{*}{x}^{*}\in R{x}^{*}$. Therefore $R{x}^{*}=Raw$.

$\left(2\right)\Rightarrow \left(4\right)$ If $rx=0$ for $r\in R$, then $ra=0$. If $ra=0$ for $r\in R$, then $rx=0$. Hence $\ell \left(x\right)=\ell \left(a\right)$. Likewise, we have $r\left(x^{*}\right)=r\left(aw\right),$ as desired.

$\left(4\right)\Rightarrow \left(1\right)$ Since $awxva=a$, we have $1-awxv\in \ell \left(a\right)$, and then $1-awxv\in \ell \left(x\right)$. Hence $x=awxvx$, and so $x^{*}=x^{*}{\left(awxv\right)}^{*}$. This implies that $1-{\left(awxv\right)}^{*}\in r\left(x^{*}\right)$; and then $1-{\left(awxw\right)}^{*}\in r\left(aw\right)$. We infer that $aw={\left(awxv\right)}^{*}aw$; whence, $awxv={\left(awxv\right)}^{*}awxv$. Therefore ${\left(awxw\right)}^{*}={\left(awxv\right)}^{*}awxv=awxv$. As $1-xvaw\in r\left(aw\right)$, we have $1-xvaw\in r\left(x^{*}\right)$. it follows that $x^{*}=x^{*}xwaw$. Then ${\left(vaw\right)}^{*}{x}^{*}={\left(vaw\right)}^{*}{x}^{*}xvaw$, i.e., ${\left(xvaw\right)}^{*}={\left(xvaw\right)}^{*}xvaw$. Thus ${\left(xvaw\right)}^{*}=xvaw$.

Since $a=awxva$, we have $aw=awxvaw$; hence, $1-xvaw\in r\left(aw\right)$. This implies that $1-xvaw\in r\left(x^{*}\right)$; whence $x^{*}=x^{*}\left(xvaw\right)$. Then $x={\left(xwaw\right)}^{*}x=xvawx$. We infer that $1-xvaw\in \ell \left(x\right)=\ell \left(a\right)$. It follows that $a=xvawa$. Therefore $a\in R_{v,w}^{EP}$ by Theorem 2.7.

$\left(1\right)\Rightarrow \left(3\right)$ By virtue of Lemma 3.6, $x=xwawx$. As in the argument above, we prove that $xR=awR,R{x}^{*}=Raw,$ as required.

$\left(3\right)\Rightarrow \left(5\right)$ This is obvious.

$\left(5\right)\Rightarrow \left(1\right)$ This is similar to the discussion in $\left(3\right)\Rightarrow \left(1\right)$. □

Corollary 3.8.

Let $a,w\in R$. Then the following are equivalent:

(1)

$a\in R_w^{EP}$.

(2)

There exists $x\in R$ such that $awxa=a,xR=aR,R{x}^{*}=Raw.$

(3)

There exists $x\in R$ such that $xawx=x,xR=aR,R{x}^{*}=Raw.$

(4)

There exists $x\in R$ such that $awxa=a,\ell \left(x\right)=\ell \left(a\right),r\left(x^{*}\right)=r\left(aw\right).$

(5)

There exists $x\in R$ such that $xawx=x,\ell \left(x\right)=\ell \left(aw\right),r\left(x^{*}\right)=r\left(aw\right).$

Proof. 

This is obvious by choosing $v=1$ in Theorem 3.7. □

4. Association with Related Generalized Inverses

The goal of this section is to characterize weighted EP elements by using the related weight Moore-Penrose invertibility. We come now to the demonstration for which this section has been developed.

Theorem 4.1.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exists $a_{v,w}^{†}$ such that $a_{v,w}^{†}v=u\left(aw\right)$ for some $u\in R^{-1}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Set $x=a_{v,w}^{\#}$. As in the proof of Theorem 3.4, we see that $x=a_{v,w}^{†}$. Let $u={\left(xv\right)}^2+1-awxv$. Then $u[{\left(aw\right)}^2+1-awxv]=[{\left(xv\right)}^2+1-awxv][{\left(aw\right)}^2+1-awxv]={\left(xv\right)}^2{\left(aw\right)}^2+1-\left(aw\right)\left(xv\right)=1.$ Likewise, $[{\left(aw\right)}^2+1-awxv]u=1$. Hence, $u={[{\left(aw\right)}^2+1-awxv]}^{-1}\in R^{-1}$. We verify that

$$u\left(aw\right)=[{\left(xv\right)}^2+1-awxv]aw={\left(xv\right)}^{2}aw=\left(xvawx\right)v=xv,$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists some $u\in R^{-1}$ such that $a_{v,w}^{†}v=u\left(aw\right)$. Then

$$\begin{array}{ccc}\hfill aw& =\hfill & u^{-1}{a}_{v,w}^{†}v=u^{-1}{a}_{v,w}^{†}v\hfill \\ & =\hfill & u^{-1}\left[a_{v,w}^{†}vaw{a}_{v,w}^{†}v\right]\hfill \\ & =\hfill & \left[u^{-1}{a}_{v,w}^{†}v\right]\left[aw{a}_{v,w}^{†}v\right].\hfill \end{array}$$

Hence ${\left(aw\right)}^{*}={\left[aw{a}_{v,w}^{†}v\right]}^{*}{\left[u^{-1}{a}_{v,w}^{†}v\right]}^{*}=a\left[w{a}_{v,w}^{†}v\right]{\left[u^{-1}{a}_{v,w}^{†}v\right]}^{*}.$

On the other hand, we have

$$\begin{array}{ccc}\hfill a& =\hfill & aw{a}_{v,w}^{†}va={\left(aw{a}_{v,w}^{†}v\right)}^{*}a={\left(a_{v,w}^{†}v\right)}^{*}{\left(aw\right)}^{*}a\hfill \\ & =\hfill & {\left(u\left(aw\right)\right)}^{*}{\left(aw\right)}^{*}a={\left(aw\right)}^{*}{u}^{*}{\left(aw\right)}^{*}a.\hfill \end{array}$$

Therefore $aR={\left(aw\right)}^{*}R$. By virtue of Theorem 3.4, $a\in R_{v,w}^{EP}$, as asserted. □

Corollary 4.2.( [18][Theorem 3.12]) Let $a\in R$. Then the following are equivalent:

(1)

$a\in R^{EP}$.

(2)

$a\in R^{†}$ and $a^{†}=ua$ for some $u\in R^{-1}$.

Proof. 

This is obtained by choosing $v=w=1$ in Theorem 4.1. □

Lemma 4.3.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exist $a_{v,w}^{†}$ and $a_{v,w}^{\#}$ such that $a_{v,w}^{†}=a_{v,w}^{\#}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 3.1 and Theorem 3.4, $a\in R_{v,w}^{†}\bigcap R_{v,w}^{\#}$. Moreover, we have $a_{v,w}^{†}=a_{v,w}^{EP}=a_{v,w}^{\#}$.

$\left(2\right)\Rightarrow \left(1\right)$ Since $a_{v,w}^{†}=a_{v,w}^{\#}$, we see that ${\left(aw{a}_{v,w}^{\#}v\right)}^{*}=aw{a}_{v,w}^{\#}v$. This complete the proof by Corollary 2.2. □

Theorem 4.4.

Let $a,v,w\in R$. Then the following are equivalent:

(1)

$a\in R_{v,w}^{EP}$.

(2)

There exist $a_{v,w}^{†}$ and $a_{v,w}^{\#}$ such that ${\left(a_{v,w}^{†}\right)}_{[w,v]}^{\overline{)\#}}=a$.

(3)

There exist $a_{v,w}^{†}$ and $a_{v,w}^{\#}$ such that ${\left(a_{v,w}^{†}\right)}_{[w,v]}^{\overline{)\#}}={\left(a_{v,w}^{\overline{)\#}}\right)}_{[w,v]}^{†}$.

Proof.$\left(1\right)\Rightarrow \left(3\right)$ In view of Lemma 4.3, $a\in R_{v,w}^{†}\bigcap R_{v,w}^{\#}$ and $a_{v,w}^{EP}=a_{v,w}^{†}=a_{v,w}^{\#}$. Thus, ${\left(a_{v,w}^{†}\right)}_{[w,v]}^{\overline{)\#}}={\left(a_{v,w}^{\#}\right)}_{[w,v]}^{\overline{)\#}}=a={\left(a_{v,w}^{\overline{)\#}}\right)}_{[w,v]}^{†},$ as required.

$\left(3\right)\Rightarrow \left(2\right)$ It is easy to verify that ${\left(a_{v,w}^{\overline{)\#}}\right)}_{[w,v]}^{†}=a$, and then ${\left(a_{v,w}^{†}\right)}_{[w,v]}^{\overline{)\#}}=a$.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x=a_{v,w}^{†}$. Then we have

$$awxva=a,xvawx=x,{\left(awxv\right)}^{*}=awxv,{\left(xvaw\right)}^{*}=xvaw.$$

By hypothesis, we have $x_{[w,v]}^{\overline{)\#}}={\left(a_{v,w}^{\overline{)\#}}\right)}_{[w,v]}^{†}=a.$ Hence

$$awxvx=x,xvawa=a.$$

According to Theorem 2.7, we prove that $a\in R_{v,w}^{EP}$, as asserted. □

Lemma 4.5.

Let $a,v,w\in R$ and $n\in N$. $a\in R_{v,w}^{EP}$ if and only if

(1)

$a\in R_{v,w}^{\#}$;

(2)

there exists $a_{v,w}^{†}$ such that $[{\left(aw\right)}^n,a_{v,w}^{†}v]=0$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Obviously, $a\in R_{v,w}^{\#}$. Choose $x=a_{v,w}^{EP}$. Then $x=a{\in }_{v,w}^{†}$. Moreover, we see that $\left(aw\right)xv=xv\left(aw\right)$. Then ${\left(aw\right)}^{n}xv=xv{\left(aw\right)}^n$. That is, $[{\left(aw\right)}^n,xv]=0.$

$\left(2\right)\Rightarrow \left(1\right)$ If $n=1$, then $a\in R_{v,w}^{EP}$ by Corollary 3.5.

If $n>1$, then there exists $a_{v,w}^{†}$ such that ${\left(aw\right)}^n{a}_{v,w}^{†}v=a_{v,w}^{†}v{\left(aw\right)}^n$. Hence, ${\left[a_{v,w}^{\#}v\right]}^{n-1}{\left(aw\right)}^n{a}_{v,w}^{†}v={\left[a_{v,w}^{\#}v\right]}^{n-1}{a}_{v,w}^{†}v{\left(aw\right)}^n$, and then

$$\left(aw\right)a_{v,w}^{†}v={\left[a_{v,w}^{\#}v\right]}^{n-1}{a}_{v,w}^{†}v{\left(aw\right)}^n.$$

This implies that

$$\begin{array}{ccc}\hfill \left(aw\right)a_{v,w}^{†}v{\left[a_{v,w}^{\#}v\right]}^{n-1}{\left(aw\right)}^{n-1}& =\hfill & {\left[a_{v,w}^{\#}v\right]}^{n-1}\left[a_{v,w}^{†}v{\left(aw\right)}^n\right]\hfill \\ & =\hfill & {\left[a_{v,w}^{\#}v\right]}^{n-1}\left[{\left(aw\right)}^n{a}_{v,w}^{†}v\right]\hfill \\ & =\hfill & aw{a}_{v,w}^{†}v.\hfill \end{array}$$

Since $\left(aw\right)a_{v,w}^{†}v{\left[a_{v,w}^{\#}v\right]}^{n-1}{\left(aw\right)}^{n-1}=\left[aw{a}_{v,w}^{†}va\right]w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v$, we deduce that $aw{a}_{v,w}^{†}v=aw{a}_{v,w}^{\#}v$. Thus ${\left[aw{a}_{v,w}^{\#}v\right]}^{*}=aw{a}_{v,w}^{\#}v$. By virtue of Corollary 2.2, $a\in R_{v,w}^{EP}$, as asserted. □

Theorem 4.6.

Let $a,v,w\in R$ and $n\in N$. Then $a\in R_{v,w}^{EP}$ if and only if

(1)

$a\in R_{v,w}^{\#}$;

(2)

there exists $a_{v,w}^{†}$ such that $[{\left(aw\right)}^n{\left(aw\right)}_{v,w}^{†}v,a_{v,w}^{†}v{\left(aw\right)}^n]=0$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Clearly, $a\in R_{v,w}^{\#}$. In view of Lemma 4.3, we have

$$[\left(aw\right)a_{v,w}^{†}v,a_{v,w}^{†}v\left(aw\right)]=0.$$

Then $[{\left(aw\right)}^n{a}_{v,w}^{†}v,a_{v,w}^{†}v{\left(aw\right)}^n]=0$.

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

$${\left(aw\right)}^n{a}_{v,w}^{†}v{a}_{v,w}^{†}v{\left(aw\right)}^n=a_{v,w}^{†}v{\left(aw\right)}^n{\left(aw\right)}^n{a}_{v,w}^{†}v,$$

i..e.,

$${\left(aw\right)}^n{a}_{v,w}^{†}v{a}_{v,w}^{†}v{\left(aw\right)}^n=a_{v,w}^{†}v{\left(aw\right)}^{2n}{a}_{v,w}^{†}v,$$

Thus, we derive

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{2n}{a}_{v,w}^{†}v& =\hfill & {\left(aw\right)}^{n+1}{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^n,\hfill \\ \hfill a_{v,w}^{†}v{\left(aw\right)}^{2n}& =\hfill & {\left(aw\right)}^n{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^{n+1}.\hfill \end{array}$$

Accordingly, we have

$$\begin{array}{ccc}\hfill \left(a_{v,w}^{\#}v\right){\left(aw\right)}^{2n}{a}_{v,w}^{†}v& =\hfill & \left(a_{v,w}^{\#}v\right){\left(aw\right)}^{n+1}{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^n,\hfill \\ \hfill a_{v,w}^{†}v{\left(aw\right)}^{2n}\left(a_{v,w}^{\#}v\right)& =\hfill & {\left(aw\right)}^n{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^{n+1}\left(a_{v,w}^{\#}v\right).\hfill \end{array}$$

Since $\left(a_{v,w}^{\#}v\right){\left(aw\right)}^{2n}={\left(aw\right)}^{n+1}\left(a_{v,w}^{\#}v\right)={\left(aw\right)}^n$, we deduce that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{2n-1}{a}_{v,w}^{†}v& =\hfill & {\left(aw\right)}^n{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^n,\hfill \\ \hfill a_{v,w}^{†}v{\left(aw\right)}^{2n-1}& =\hfill & {\left(aw\right)}^n{\left(a_{v,w}^{†}v\right)}^2{\left(aw\right)}^n.\hfill \end{array}$$

Therefore ${\left(aw\right)}^{2n-1}{a}_{v,w}^{†}v=a_{v,w}^{†}v{\left(aw\right)}^{2n-1}$. That is, $[{\left(aw\right)}^{2n-1},a_{v,w}^{†}v]=0$. In light of Lemma 4.5, $a\in R_{v,w}^{EP}$, as asserted. □

Corollary 4.7.

Let $a,v,w\in R$. Then $a\in R_{v,w}^{EP}$ if and only if

(1)

$a\in R_{v,w}^{\#}$;

(2)

there exists $a_{v,w}^{†}$ such that $[{\left(aw\right)}^2{a}_{v,w}^{†}v,a_{v,w}^{†}v{\left(aw\right)}^2]=0$.

Proof. 

This is obvious by choosing $n=2$ in Theorem 4.6. □

We come now to characterize the weighted EP element by using the core inverse with two weights.

Theorem 4.8.

Let $a,v,w\in R$. Then $a\in R_{v,w}^{EP}$ if and only if $a\in R_{v,w}^{\#}$ and there exists $x\in R$ such that

(1)

$a=awxva$;

(2)

$a_{v,wxv}^{\#}=aw{a}_{v,w}^{\#}$.

Proof. 

⟹ By virtue of Theorem 3.1, $a\in R_{v,w}^{\#}$. Choose $x=a_{v,w}^{\#}$. Then $a=awxva$ and we further verify that

$$\begin{array}{ccc}\hfill a\left(wxv\right)\left(aw{a}_{v,w}^{\#}\right)v& =\hfill & \left(awxva\right)w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v,\hfill \\ \hfill {\left(a\left(wxv\right)\left(aw{a}_{v,w}^{\#}\right)v\right)}^{*}& =\hfill & a\left(wxv\right)\left(aw{a}_{v,w}^{\#}\right)v,\hfill \\ \hfill a\left(wxv\right)\left(aw{a}_{v,w}^{\#}\right)v\left(aw{a}_{v,w}^{\#}\right)& =\hfill & (awxvaw{a}_{v,w}^{\#}v\left(aw{a}_{v,w}^{\#}\right)\hfill \\ & =\hfill & \left(aw{a}_{v,w}^{\#}va\right)\left(w{a}_{v,w}^{\#}\right)=\left[aw{a}_{v,w}^{\#}va\right]w{a}_{v,w}^{\#}\hfill \\ & =\hfill & aw{a}_{v,w}^{\#},\hfill \\ \hfill \left(aw{a}_{v,w}^{\#}\right)va\left(wxv\right)a& =\hfill & \left(aw{a}_{v,w}^{\#}v\right)\left(awxva\right)=awxva=a.\hfill \end{array}$$

Therefore $a_{v,wxv}^{\overline{)\#}}=aw{a}_{v,w}^{\#}$, as required.

⟸ By hypothesis, we have $a=awxva$ and $a_{v,wxv}^{\overline{)\#}}=aw{a}_{v,w}^{\#}$. It follows from $a\left(wxv\right)\left(aw{a}_{v,w}^{\#}\right)v=\left(awxva\right)w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v$ that ${\left(aw{a}_{v,w}^{\#}v\right)}^{*}=aw{a}_{v,w}^{\#}v$. Therefore we complete the proof by Corollary 2.2. □

Corollary 4.9.

Let $a,w\in R$. Then $a\in R_w^{EP}$ if and only if $a\in R_w^{\#}$ and there exists $x\in R$ such that $a=awxa$ and $a_{wx}^{\#}=aw{a}_w^{\#}$.

Proof. 

This is immediate by choosing $v=1$ in Theorem 4.8. □

Theorem 4.10.

Let $a,v,w\in R$. Then $a\in R_{v,w}^{EP}$ if and only if $a\in R_{v,w}^{\#}$ and there exists $x\in R$ such that

(1)

$a=awxva$;

(2)

$a_{v,wxvaw}^{\overline{)\#}}=a_{v,w}^{\#}$.

Proof. 

⟹ Set $x=a_{v,w}^{\#}$. Then $a=awxva$. We verify that

$$\begin{array}{ccc}\hfill a\left(wxvaw\right)a_{v,w}^{\#}v& =\hfill & \left(awxva\right)w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v,\hfill \\ \hfill {\left(a\left(wxvaw\right)a_{v,w}^{\#}v\right)}^{*}& =\hfill & a\left(wxvaw\right)a_{v,w}^{\#}v,\hfill \\ \hfill a\left(wxvaw\right)a_{v,w}^{\#}v{a}_{v,w}^{\#}& =\hfill & \left[aw{a}_{v,w}^{\#}v\right]a_{v,w}^{\#}=a_{v,w}^{\#},\hfill \\ \hfill a_{v,w}^{\#}va\left(wxvaw\right)a& =\hfill & a_{v,w}^{\#}v\left(awxva\right)wa=a_{v,w}^{\#}vawa=a.\hfill \end{array}$$

Accordingly, $a_{v,wxvaw}^{\overline{)\#}}=a_{v,w}^{\#}$, as desired.

⟸ By hypothesis, we have $a=awxva$ and $a_{v,wxvaw}^{\overline{)\#}}=a_{v,w}^{\#}$. It is easy to verify that $a\left(wxvaw\right)a_{v,w}^{\#}v=\left(awxva\right)w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v$, and then ${\left(aw{a}_{v,w}^{\#}v\right)}^{*}=aw{a}_{v,w}^{\#}v$. In light of Corollary 2.2, $a\in R_{v,w}^{EP}$, as asserted. □

Corollary 4.11.

Let $a,w\in R$. Then $a\in R_w^{EP}$ if and only if $a\in R_w^{\#}$ and there exists $x\in R$ such that $a=awxa$ and $a_{xaw}^{\overline{)\#}}=a_w^{\#}$.

Proof. 

This is obvious by choosing $v=1$ in Theorem 4.10. □

Theorem 4.12.

Let $a,v,w\in R$. Then $a\in R_{v,w}^{EP}$ if and only if $a\in R_{v,w}^{\#}\bigcap R_{v,w}^{†}$ and there exists $x\in R$ such that

(1)

$a=awxva$;

(2)

${\left(aw{a}_{v,w}^{†}\right)}_{v,vawxv}^{\overline{)\#}}=aw{a}_{v,w}^{\overline{)\#}}$.

Proof. 

⟹ By virtue of Theorem 3.1 and Theorem 3.4, $a\in R_{v,w}^{\#}\bigcap R_{v,w}^{†}$. Set $x=a_{v,w}^{\#}$. One directly checks that

$$\begin{array}{ccc}\hfill \left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{\#}\right)v& =\hfill & aw{a}_{v,w}^{†}v\left(awxva\right)w{a}_{v,w}^{\#}v\hfill \\ & =\hfill & \left(aw{a}_{v,w}^{†}va\right)w{a}_{v,w}^{\#}v=aw{a}_{v,w}^{\#}v,\hfill \\ \hfill {\left(\left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{\#}\right)v\right)}^{*}& =\hfill & \left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{\#}\right)v,\hfill \\ \hfill \left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{\#}\right)v\left(aw{a}_{v,w}^{\#}\right)& =\hfill & aw{a}_{v,w}^{\#}v\left(aw{a}_{v,w}^{\#}\right)\hfill \\ & =\hfill & \left(aw{a}_{v,w}^{\#}va\right)w{a}_{v,w}^{\#}=aw{a}_{v,w}^{\#},\hfill \\ \hfill \left(aw{a}_{v,w}^{\#}\right)v\left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{†}\right)& =\hfill & aw{a}_{v,w}^{\#}vaw{a}_{v,w}^{†}v\left(awxva\right)w{a}_{v,w}^{†}\hfill \\ & =\hfill & \left(aw{a}_{v,w}^{\#}va\right)w\left(a_{v,w}^{†}vaw{a}_{v,w}^{†}\right)\hfill \\ & =\hfill & (aw\left(a_{v,w}^{†}va\right)w{a}_{v,w}^{†}=aw{a}_{v,w}^{†}.\hfill \end{array}$$

Therefore ${\left(aw{a}_{v,w}^{†}\right)}_{v,vawxv}^{\overline{)\#}}=aw{a}_{v,w}^{\#}$.

⟸ By the assumption, there exists $x\in R$ such that

$$a=awxva{\left(aw{a}_{v,w}^{†}\right)}_{v,vawxv}^{\overline{)\#}}=aw{a}_{v,w}^{\#}.$$

Since $\left(aw{a}_{v,w}^{†}\right)\left(vawxv\right)\left(aw{a}_{v,w}^{\#}\right)v=aw{a}_{v,w}^{\#}v$, we deduce that ${\left(aw{a}_{v,w}^{\#}v\right)}^{*}=aw{a}_{v,w}^{\#}v$. This completes the proof by Corollary 2.2. □

As an immediate consequence, we derive

Corollary 4.13.

Let $a,w\in R$. Then $a\in R_w^{EP}$ if and only if $a\in R_w^{\#}\bigcap R_w^{†}$ and there exists $x\in R$ such that $a=awxa$ and ${\left(aw{a}_w^{†}\right)}_{awx}^{\overline{)\#}}=aw{a}_w^{\#}$.

Note that each element in the set $\{a_{v,w}^{\#},a_{v,w}^{†},a_{v,w}^{\#}\}$ can be viewed as the element x in Theorem 4.8, Theorem 4.10 and Theorem 4.12.