Core-EP Inverses with Respect to Two Ring Elements

1. Introduction

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

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

Such an x is unique if it exists, is denoted by $a^{\#}$, and is called the group inverse of a. As is well known, a square complex matrix A has a group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$ (see [2,7,21]).

An element a in a *-ring R has core inverse if and only if there exist $x\in R$ such that

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

If such x exists, it is unique, and denote it by $a^{\#⃝}$. It is well known that

$$x=a^{\#⃝}\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if}a=axa,xR=aR,\mathrm{and}Rx=R{a}^{\ast }.$$

A square complex matrix A has core inverse if and only if $AX=P_A,R\left(X\right)\subseteq R\left(A\right)$, where $P_A$ is a projection on $R\left(A\right)$ (see [1,17,32]).

Gao and Chen extended the concept of the core inverse and introduced the notion of core-EP inverse (i.e., pseudo core inverse) (see [12]). An element $a\in R$ has core-EP inverse if there exist $x\in R$ and $m\in \mathbb{N}$ such that

$$a{x}^2=x,{\left(ax\right)}^{\ast }=ax,x{a}^{m+1}=a^m.$$

If such x exists, it is unique, and denote it by $a^{Ⓓ}$. Note that

$$x=a^{Ⓓ}\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if}x=xax,xR=a^{k}R,\mathrm{and}Rx=R{\left(a^k\right)}^{\ast }$$

for some $m\in \mathbb{N}$ (see [22,27,33]).

An element $a\in R$ has $(b,c)$-inverse if there exists an element $x\in R$ such that

$$x\in bRx\bigcap xRc,xab=b,cax=c.$$

If such x exists, it is unique and is denoted by $a^{(b,c)}.$ Evidently,

$$x=a^{(b,c)}\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if}x=xax,xR=bR,\mathrm{and}Rx=Rc.$$

(see [8,11,14,16,23,26]).

Given a ring R and any $a,y\in R$ such that the Moore-Penrose inverse $a^{†}$ of a exists, Rakić proved that a has y as $(a{a}^{†},a{a}^{†})$-inverse if and only if a has y as $(a{a}^{\ast },a{a}^{\ast })$-inverse, if and only if a has y as $(a,a^{\ast })$-inverse if and only if a has core inverse y (see [25]. This motivates us to explore when the core inverse of a ring element a coincides with its $(b,c)$-inverse. For this purpose, we adopt the following definition.

Definition 1. 

An element $a\in R$ has core-$(b,c)$-inverse if $a\in R^{\#⃝}$ and $a^{\#⃝}=a^{(b,c)}$. We use $a_{b,c}^{\#⃝}$ to stands for $a^{\#⃝}$. The set of all $(b,c)$-core invertible elements in R is denoted by $R_{b,c}^{\#⃝}$.

In Section 2, we establish elementary properties of the core-$(b,c)$-inverse in a ring. In particular, we present characterizations of the core-$(b,b)$-inverse.

Next, we explore when the core-EP inverse of a ring element coincides with its $(b,c)$-inverse. To this end, we introduce the following definition.

Definition 2. 

An element $a\in R$ has core-EP-$(b,c)$-inverse if $a\in R^{Ⓓ}$ and $a^{Ⓓ}=a^{(b,c)}$. We use $a_{b,c}^{Ⓓ}$ to stands for $a^{Ⓓ}$. The set of all core-EP-$(b,c)$-invertible elements in R is denoted by $R_{b,c}^{Ⓓ}$.

In Section 3, this generalized inverse is examined through a novel limit-based approach. We characterize this new generalized inverse via a combination of core-$(b,c)$-inverses and quasinilpotent elements. We present polar-like properties for the core-EP-$(b,c)$-inverse. Notably, we obtain a new characterization of the core-EP-$(b,b)$ inverse.

Finally, in Section 4, we are concerned with the presentations of core-EP-$(b,c)$-inverse. In particular, we investigate the core-EP-$(b,b)$-inverse and then derive the reverse law for the core-EP inverse.

Throughout the paper, all rings are associative with an identity. Let ${\mathbb{C}}^{n\times n}$ be the ring of all $n\times n$ complex matrices with conjugate transpose *. $\ell \left(x\right)$ and $r\left(x\right)$ stand for the left and right annihilators of $x\in R$, respectively. The inner inverse of a, denoted by $a^{-}$, satisfies $a=a{a}^{-}a$.

2. Core-$(b,c)$-Inverse

The purpose of this section is to investigate elementary properties of the core-$(b,c)$-inverse in a *-Banach algebra. Our starting points is the following.

Theorem 1. 

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

(1)

$a\in R_{b,c}^{\#⃝}$.

(2)

$a\in R^{\#⃝}$ and $aR=bR,R{a}^{\ast }=Rc.$

(3)

$a\in R^{(b,c)}$ and $aR=bR,R{a}^{\ast }=Rc.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, $a\in R^{\#⃝}$ and $x:=a^{\#⃝}=a^{(b,c)}$. Then $a=x{a}^2$ and $x=a{x}^2$; hence, $xR=aR$. Since $x=xax$ and ${\left(ax\right)}^{\ast }=ax$, we see that $x^{\ast }=\left(ax\right)x^{\ast }$. Since $a=axa={\left(ax\right)}^{\ast }a=x^{\ast }\left(a^{\ast }a\right)$. This implies that $x=x{x}^{\ast }{a}^{\ast }$ and $a^{\ast }=\left(a^{\ast }a\right)x$. Hence, $R{a}^{\ast }=Rx$.

Since $x=a^{(b,c)}$, we see that $xax=x,xR=bR,Rx=Rc.$ Therefore

$$aR=xR=bR,R{a}^{\ast }=Rx=Rc.$$

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, $a\in R^{\#⃝}$. Set $x=a^{\#⃝}$. As in the preceding discussion, $xax=x,xR=aR$ and $R{a}^{\ast }=Rx$. Thus we see that $xax=x,xR=bR,Rx=Rc.$ Therefore $x=a^{(b,c)}$, as required.

$\left(3\right)\Rightarrow \left(1\right)$ Let $x=a^{(b,c)}$. Then $xax=x,xR=bR,Rx=Rc.$ By hypothesis, we have $xax=x,xR=aR,Rx=R{a}^{\ast }.$ Since $x(1-ax)=0$, we have $a^{\ast }(1-ax)=0$, and then $a^{\ast }=a^{\ast }ax$. This implies that ${\left(ax\right)}^{\ast }={\left(ax\right)}^{\ast }ax$. Hence, $ax={\left(ax\right)}^{\ast }ax={\left(ax\right)}^{\ast }$. As $x(1-ax)=0$, we see that $a^{\ast }(1-ax)=0$, and so $(1-ax)a=0$. Hence $a=axa$. Therefore $a\in R^{\#⃝}$ and $a^{\#⃝}=x=a^{(b,c)}$, as asserted. □

We denote $a^{\#⃝}$ in Theorem 2.1 by $a_{b,c}^{\#⃝}$, and call it the core-$(b,c)$-inverse of a. As an immediate consequence, we prove that the core inverse $a^{\#⃝}$ of a can be characterized as its $(b,c)$-inverse, namely with $(b,c)=(a,a^{\ast })$ (see [25]).

Corollary 1. 

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

(1)

$a\in R_{b,c}^{\#⃝}$.

(2)

$a\in R^{(b,c)}\bigcap bR\bigcap c^{\ast }R$ has Drazin inverse.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, $a\in R$ has core inverse and $a^{\#⃝}=a^{(b,c)}.$ Then $a\in R$ has g-Drazin inverse. By virtue of Theorem 2.1, $a\in R^{(b,c)}\bigcap bR\bigcap c^{\ast }R$

$\left(2\right)\Rightarrow \left(1\right)$ Set $x=a^{(b,c)}$. Then $xab=b$ and $cax=c$. Hence, $1-xa\in \ell \left(b\right)$; and so $1-xa\in \ell \left(a\right)$. We have $a=x{a}^2$. Furthermore, $1-ax\in r\left(c\right)$, and then $a^{\ast }(1-ax)=0$. This implies that $a^{\ast }=a^{\ast }\left(ax\right)$. Hence, $a={\left(ax\right)}^{\ast }a$; hence, $ax={\left(ax\right)}^{\ast }ax$. Therefore ${\left(ax\right)}^{\ast }=ax$; whence, $a^{\ast }=a^{\ast }{\left(ax\right)}^{\ast }$. We infer that $a=axa$. On the other hand, we see that $a^2{a}^D=\left(x^k{a}^{k+2}\right)a^D=x^k\left(a^{k+2}{a}^D\right)=x^k{a}^{k+1}=a$. Therefore $a\in R$ has group inverse. Hence

$$b=xab=x\left(a^2{a}^D\right)b=\left(x{a}^2\right)a^{d}b=a{a}^{d}b,$$

and so $x\in bR\subseteq aR$. Write $x=az$ for a $z\in R$. Then $a{x}^2=ax\left(az\right)=\left(axa\right)z=az=x$, and so a has core inverse x. Therefore $a_{b,c}^{\#⃝}=x$, as asserted. □

Corollary 2. 

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

(1)

$a\in R_{b,c}^{\#⃝}$.

(2)

$a\in R^{(b,c)}$ and $\ell \left(a\right)=\ell \left(b\right),r\left(a^{\ast }\right)=r\left(c\right).$

Proof. 

⟹ By virtue of Theorem 2.1, we have $a\in R^{(b,c)}$ and $aR=bR,R{a}^{\ast }=Rc.$ Therefore we verify that $\ell \left(a\right)=\ell \left(b\right),r\left(a^{\ast }\right)=r\left(c\right).$

⟸ Set $x=a^{(b,c)}$. Then $xab=b$ and $cax=c$. Hence, $1-xa\in \ell \left(b\right)$; and so $1-xa\in \ell \left(a\right)$. This implies that $a=x{a}^2$. Moreover, $1-ax\in r\left(c\right)$, and then $a^{\ast }(1-ax)=0$. This implies that $a^{\ast }=a^{\ast }\left(ax\right)$. We infer that $a={\left(ax\right)}^{\ast }a$; hence, $ax={\left(ax\right)}^{\ast }ax$. Therefore ${\left(ax\right)}^{\ast }=ax$. As $cax=c$, we have $1-ax\in r\left(c\right)=r\left(a^{\ast }\right)$. Hence, $a^{\ast }=a^{\ast }\left(ax\right)$. Thus $a={\left(ax\right)}^{\ast }a=axa$. Since $x\in bR$ and $a=x{a}^2\in bR$; hence, $aR\subseteq bR$. As $axa=a$, we have $1-ax\in \ell \left(a\right)\subseteq \ell \left(b\right)$; hence, $b=axb\in aR$. Hence $bR\subseteq aR$, and so $aR=bR$.

Since $axa=a$, we have $a^{\ast }=a^{\ast }{(1-ax)}^{\ast }$, and then ${(1-ax)}^{\ast }\in r\left(a^{\ast }\right)\subseteq r\left(c\right)$. Thus, $c{(1-ax)}^{\ast }=0$. We infer that $c=c{\left(ax\right)}^{\ast }=c{x}^{\ast }{a}^{\ast }$. Hence, $Rc\subseteq R{a}^{\ast }$. On the other hand, $a^{\ast }={\left(axa\right)}^{\ast }=a^{\ast }{\left(ax\right)}^{\ast }=a^{\ast }ax$ and $x\in Rc$, and so $R{a}^{\ast }\subseteq Rc$. Thus $R{a}^{\ast }=Rc$.

Accordingly, $a\in R_{b,c}^{\#⃝}$ by Theorem 2.1. □

Let ${\mathbb{C}}^{n\times n}$ be the ring of all $n\times n$ complex matrices, with conjugate transpose as the involution.

Corollary 3. 

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

(1)

A has core-$(B,C)$-inverse.

(2)

$rank\left(A\right)=rank\left(B\right)=rank\left(C\right)=rank\left(C^{\ast }\right)=rank\left(CAB\right)$.

In this case, $A_{B,C}^{\#⃝}=B{\left(CAB\right)}^{†}C.$

Proof. 

In view of [24], A has $(B,C)$-inverse if and only if $rank\left(B\right)=rank\left(C\right)=rank\left(CAB\right)$. In this case, $A^{(B,C)}=B{\left(CAB\right)}^{-}C.$ On the other hand, $R\left(A\right)=R\left(B\right)$ and $R\left(A\right)=R\left(C^{\ast }\right)$ if and only if $rank\left(A\right)=rank\left(B\right)=rank\left(C^{\ast }\right)$. Therefore we complete the proof by Theorem 2.1. □

Corollary 4. 

Let $A\in C^{n\times n}$ have a group inverse with full-rank factorization $BC$. Then

$$A_{B,B^{\ast }}^{\#⃝}=B{\left(CB\right)}^{-1}{\left(B^{\ast }B\right)}^{-1}{B}^{\ast }.$$

Proof. 

One directly verify that $A^{\#⃝}=B{\left(CB\right)}^{-1}{\left(B^{\ast }B\right)}^{-1}{B}^{\ast }.$ Clearly, $R\left(A\right)\subseteq R\left(B\right)$. Since $AB{\left(CB\right)}^{-1}=B\left(CB\right){\left(CB\right)}^{-1}=B$, we have $R\left(B\right)\subseteq R\left(A\right)$. Hence, $R\left(A\right)=R\left(B\right)$. This implies that $R\left(A\right)=R\left({\left(B^{\ast }\right)}^{\ast }\right)$. According to Theorem 2.1, A has $(B,B^{\ast })$-core inverse, as required. □

Recall that y is the inverse of a along an element b if $y\in bR\bigcap Rb,yab=b=bay$ (see [19]). Evidently, $y=a^{(b,b)}$. We next consider when the core inverse of a ring element is its inverse along an element b. This is just the case of the core-$(b,b)$-inverse.

Theorem 2. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$a,b\in R^{\#⃝}$ and $a{a}^{\#⃝}=b^{\#⃝}b.$

(3)

$ba\in R^{\#⃝}$ and $aR=bR=b^{\ast }R=baR.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 2.1, $a\in R^{\#⃝}$ and $aR=bR.$ Then $R{a}^{\ast }=Rb$. Write $a=bs$ and $b=t{a}^{\ast }$ for some $s,t\in R$. Then $b=t{a}^{\ast }=t{\left(a{a}^{\#⃝}a\right)}^{\ast }=\left(t{a}^{\ast }\right)a{a}^{\#⃝}=ba{a}^{\#⃝}=b\left(bs\right)a^{\#⃝}=b^{2}s{a}^{\#⃝}.$ Write $b=av$ and $a^{\ast }=wb$ for some $v,w\in R$. Then we have

$$b=av=\left(a^{\#⃝}{a}^2\right)v=a^{\#⃝}ab.$$

Then $a^{\#}a{a}^{(1,3)}ab=b$. This implies that $a{a}^{\#}b=b$. Then we have $b=a{a}^{\#}b=\left(a{a}^{\#⃝}a{a}^{\#}\right)b=a{a}^{\#⃝}\left(a{a}^{\#}b\right)=a{a}^{\#⃝}b={\left(a{a}^{\#⃝}\right)}^{\ast }b={\left(a^{\#⃝}\right)}^{\ast }{a}^{\ast }b={\left(a^{\#⃝}\right)}^{\ast }w{b}^2.$ Therefore $b\in R^{\#}$ and $b^{\#}=\left[s{a}^{\#⃝}\right]b\left[{\left(a^{\#⃝}\right)}^{\ast }w\right]$.

Let $e=a{a}^{\#⃝}$. Then $eR=bR$. Write $e=by$ and $b=ez$ for $y,z\in R$. Then $b=eb=byb$ and ${\left(by\right)}^{\ast }=by$. Hence, $b\in R^{(1,3)}$. By virtue of [32], $b\in R^{\#⃝}$.

By using Theorem 2.1, $b\in R_{(a,a^{\ast })}^{\#⃝}$. Then $a^{\#⃝}\in bRb$ and $b^{\#⃝}\in aR{a}^{\ast }$. Write $a^{\#⃝}=bvb$ and $b^{\#⃝}=aw{a}^{\ast }$ for some $w,v\in R$. Furthermore, we check that $a{a}^{\#⃝}=abvb=abvb{b}^{\#⃝}b=a{a}^{\#⃝}{b}^{\#⃝}b=a{a}^{\#⃝}aw{a}^{\ast }b=aw{a}^{\ast }b=b^{\#⃝}b.$

$\left(2\right)\Rightarrow \left(1\right)$ Since $a{a}^{\#⃝}=b^{\#⃝}b,$ we have $b=b^{\#⃝}{b}^2=\left(b^{\#⃝}b\right)b=\left(a{a}^{\#⃝}\right)b\in aR$. Further, we see that $a=a{a}^{\#⃝}a=b^{\#⃝}ba=b{\left(b^{\#⃝}\right)}^{2}ba\in bR$. Hence, $aR=bR$. Moreover, we see that $b=b\left(b^{\#⃝}b\right)=ba{a}^{\#⃝}=b{\left(a{a}^{\#⃝}\right)}^{\ast }=b{\left(a^{\#⃝}\right)}^{\ast }{a}^{\ast }\in R{a}^{\ast }$. On the other hand, $a^{\ast }={\left(a{a}^{\#⃝}a\right)}^{\ast }=a^{\ast }\left(a{a}^{\#⃝}\right)=a^{\ast }\left(b^{\#⃝}b\right)\in Rb$. Thus, $R{a}^{\ast }=Rb$. By virtue of Theorem 2.1, $a\in R_{b,b}^{\#⃝}$, as desired.

$\left(1\right)\Rightarrow \left(3\right)$ In view of Theorem 2.1, $a\in R^{(b,b)}$ and $aR=bR,R{a}^{\ast }=Rb.$ By virtue of [14], Theorem 2.13, we have $ba\in R^{\#}$ and $baR=bR$. Hence, $aR=bR=b^{\ast }R=baR$.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, $ba\in R^{\#}$ and $baR=bR$. In light of [14], $a\in R^{(b,b)}$. Moreover, we see that $aR=bR$ and $R{a}^{\ast }=Rb$. According to Theorem 2.1, $a\in R_{b,c}^{\#⃝}$, as asserted. □

Corollary 5. 

Let $a,b\in R$. If $a\in R_{b,b}^{\#⃝}$, then b is EP.

Proof. 

In view of Theorem 2.6, $b\in R^{\#⃝}$ and $a{a}^{\#⃝}=b^{\#⃝}b.$ Then ${\left(b^{\#⃝}b\right)}^{\ast }={\left(a{a}^{\#⃝}\right)}^{\ast }=a{a}^{\#⃝}=b^{\#⃝}b.$ In light of [30], b is EP. □

We are ready to prove:

Theorem 3. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$a,b,ba\in R^{\#⃝}$, $a^{\#⃝}={\left(ba\right)}^{\#⃝}b,b^{\#⃝}=a{\left(ba\right)}^{\#⃝}.$

(3)

$a,b,ba\in R^{\#⃝}$, $aR=bR,{\left(ba\right)}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}.$

(4)

$a,b,ba\in R^{\#⃝}$, $a^{\pi }b=b^{\pi }a=0,{\left(ba\right)}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}.$

Proof. $\left(1\right)\Rightarrow \left(3\right)$ In view of Theorem 2.1, $a,b\in R^{\#⃝}$ and $R{a}^{\ast }=Rb$. Set $x=a^{\#⃝}{b}^{\#⃝}$. Then we verify that $bax=ba{a}^{\#⃝}{b}^{\#⃝}=b{b}^{\#⃝},{\left(bax\right)}^{\ast }=bax.$ In view of Theorem 2.1, we have $aR=bR$, and then $b^{\#⃝}ba=a$. Hence $b^{\#⃝}b{a}^{\#⃝}{b}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}$. This implies that $b{b}^{\#⃝}{a}^{\#⃝}{b}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}.$ Hence, we get

$$\begin{array}{ccc}\hfill ba{x}^2& =\hfill & b{b}^{\#⃝}{a}^{\#⃝}{b}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}=x;\hfill \\ \hfill x{\left(ba\right)}^2& =\hfill & a^{\#⃝}{b}^{\#⃝}baba=a^{\#⃝}aba=\left(a^{\#⃝}ab\right)a=ba.\hfill \end{array}$$

Therefore $ba\in R^{\#⃝}$ and ${\left(ba\right)}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}$.

$\left(3\right)\Rightarrow \left(2\right)$ Since $aR=bR$, we have $a{a}^{\#⃝}R=b{b}^{\#⃝}R$. Since $(1-a{a}^{\#⃝})a{a}^{\#⃝}=0$, we see that $(1-a{a}^{\#⃝})b{b}^{\#⃝}=0$. Thus, $b{b}^{\#⃝}=a{a}^{\#⃝}b{b}^{\#⃝}.$ Therefore $b^{\#⃝}=\left(b{b}^{\#⃝}\right)b^{\#⃝}=\left[a{a}^{\#⃝}b{b}^{\#⃝}\right]b^{\#⃝}=a\left[a^{\#⃝}{b}^{\#⃝}\right]=a{\left(ba\right)}^{\#⃝}.$

On the other hand, $(1-b^{\#⃝}b)b=0$, we have $(1-b^{\#⃝}b)a=0$. Hence $a=b^{\#⃝}ba.$ Since $a{a}^{\#⃝}R=b{b}^{\#⃝}R$, we have $a{a}^{\#⃝}b{b}^{\#⃝}=b{b}^{\#⃝}.$ Hence, ${\left[a{a}^{\#⃝}b{b}^{\#⃝}\right]}^{\ast }={\left[b{b}^{\#⃝}\right]}^{\ast }.$ Hence, $b{b}^{\#⃝}a{a}^{\#⃝}=b{b}^{\#⃝}.$ This implies that $b^{\#⃝}ba{a}^{\#⃝}=b^{\#⃝}b.$ Accordingly, we derive that $a^{\#⃝}=a^{\#⃝}a{a}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}ba{a}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}b={\left(ba\right)}^{\#⃝}b,$ as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, $a^{\#⃝}={\left(ba\right)}^{\#⃝}b,b^{\#⃝}=a{\left(ba\right)}^{\#⃝}.$ Then $a{a}^{\#⃝}=a{\left(ba\right)}^{\#⃝}b=b^{\#⃝}b.$ According to Theorem 2.1, $a\in R_{b,c}^{\#⃝}$.

$\left(3\right)\iff \left(4\right)$ Since $a,b\in R^{\#}$, we see that $a^{\pi }a=b^{\pi }b=0$. Thus, $aR=bR$ if and only if $a^{\pi }b=b^{\pi }a=0$, as asserted. □

Corollary 6. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$a\in R^{(b,b)},b\in aR,{\left(ba\right)}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 2.1 and Theorem 2.6.

$\left(2\right)\Rightarrow \left(1\right)$ As $a\in R^{(b,b)}$, we have $a\in Rb$, and then $aR=bR$. This completes the proof by Theorem 2.8. □

Corollary 7. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$a,b,ba\in R^{\#⃝}$, $aR=bR=baR,a^{\ast }{b}^{\ast }b(1-a{a}^{\#⃝})b=0$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.8, $a,b,ba\in R^{\#⃝},aR=bR$ and ${\left(ba\right)}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}$. By hypothesis, we have $b=ba{a}^{\#⃝}\in baR$, and then $bR=baR$. It is easy to verify that ${\left(ba\right)}^{\ast }ba{\left(ba\right)}^{\#⃝}={\left(ba\right)}^{\ast }.$ Then $a^{\ast }{b}^{\ast }ba{a}^{\#⃝}{b}^{\#⃝}=a^{\ast }{b}^{\ast }.$ This implies that $a^{\ast }{b}^{\ast }ba{a}^{\#⃝}b=a^{\ast }{b}^{\ast }ba{a}^{\#⃝}{b}^{\#⃝}{b}^2=a^{\ast }{b}^{\ast }{b}^2.$ Therefore $a^{\ast }{b}^{\ast }b(1-a{a}^{\#⃝})b=0$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we have $a^{\ast }{b}^{\ast }{b}^2=a^{\ast }{b}^{\ast }ba{a}^{\#⃝}b$. Hence $a^{\ast }{b}^{\ast }=a^{\ast }{b}^{\ast }b{b}^{\#⃝}=a^{\ast }{b}^{\ast }{b}^2{\left(b^{\#⃝}\right)}^2=a^{\ast }{b}^{\ast }ba{a}^{\#⃝}b{\left(b^{\#⃝}\right)}^2=a^{\ast }{b}^{\ast }ba{a}^{\#⃝}{b}^{\#⃝}.$ Hence, $ba=ba{a}^{\#⃝}{b}^{\#⃝}.$ Set $x=a^{\#⃝}{b}^{\#⃝}$. Then we verify that

$$\begin{array}{ccc}\hfill bax& =\hfill & ba{a}^{\#⃝}{b}^{\#⃝}\hfill \\ & =\hfill & {\left[{\left(ba\right)}^{\#⃝}\right]}^{\ast }\left[{\left(ba\right)}^{\ast }ba{a}^{\#⃝}{b}^{\#⃝}\right]\hfill \\ & =\hfill & {\left[{\left(ba\right)}^{\#⃝}\right]}^{\ast }{\left(ba\right)}^{\ast }=\left(ba\right){\left(ba\right)}^{\#⃝},\hfill \\ \hfill {\left(bax\right)}^{\ast }& =\hfill & {\left[\left(ba\right){\left(ba\right)}^{\#⃝}\right]}^{\ast }=\left(ba\right){\left(ba\right)}^{\#⃝}=bax,\hfill \\ \hfill \left(ba\right)x\left(ba\right)& =\hfill & \left(bax\right)\left(ba\right)=\left(ba\right){\left(ba\right)}^{\#⃝}\left(ba\right)=ba.\hfill \end{array}$$

Write $a=\left(ba\right)y$ for a $y\in R$. Then we check that

$$\begin{array}{cc}& \left(ba\right){\left(ba\right)}^{\#}{a}^{\#⃝}{b}^{\#⃝}=\left(ba\right){\left(ba\right)}^{\#}a{\left(a^{\#⃝}\right)}^2{b}^{\#⃝}\hfill \\ \hfill =& \left[\left(ba\right){\left(ba\right)}^{\#}\left(ba\right)\right]y{\left(a^{\#⃝}\right)}^2{b}^{\#⃝}=\left(bay\right){\left(a^{\#⃝}\right)}^2{b}^{\#⃝}\hfill \\ \hfill =& a{\left(a^{\#⃝}\right)}^2{b}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}.\hfill \end{array}$$

We infer that $[1-\left(ba\right){\left(ba\right)}^{\#}]a^{\#⃝}{b}^{\#⃝}=0$. Then $a^{\#⃝}{b}^{\#⃝}=\left(ba\right){\left(ba\right)}^{\#}{a}^{\#⃝}{b}^{\#⃝}$. This implies that

$$\begin{array}{cc}& [1-\left(ba\right){\left(ba\right)}^{\#⃝}]a^{\#⃝}{b}^{\#⃝}\hfill \\ \hfill =& [1-\left(ba\right){\left(ba\right)}^{\#⃝}]\left(ba\right){\left(ba\right)}^{\#⃝}{a}^{\#⃝}{b}^{\#⃝}\hfill \\ \hfill =& \left(ba\right){\left(ba\right)}^{\#}{a}^{\#⃝}{b}^{\#⃝}-\left[\left(ba\right){\left(ba\right)}^{\#⃝}\left(ba\right)\right]{\left(ba\right)}^{\#}{a}^{\#⃝}{b}^{\#⃝}=0.\hfill \end{array}$$

Thus, $\left(ba\right)x^2=\left(bax\right)x=\left(ba\right){\left(ba\right)}^{\#⃝}{a}^{\#⃝}{b}^{\#⃝}=a^{\#⃝}{b}^{\#⃝}=x$. Since $ba$ has group inverse, we have $baR={\left(ba\right)}^{2}R$. Hence ${\left(ba\right)}^{\#⃝}=x=a^{\#⃝}{b}^{\#⃝}$. This completes the proof by Theorem 2.8. □

An element a in R is EP (i.e., an EP element) if there exists some $x\in R$ such that $a^{2}x=a,ax=xa,{\left(ax\right)}^{\ast }=ax$. Evidently, $a\in R$ is EP if and only if $a\in R^{\#}$ and ${\left(a{a}^{\#}\right)}^{\ast }=a{a}^{\#}$ (see [20,31]).

Theorem 4. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$ab\in R$ is EP, $r\left(a\right)\bigcap bR=0,{\left(ba\right)}^{\pi }a=0,{\left(ab\right)}^{\pi }b=0.$

(3)

$ab\in R$ is EP, $\ell \left(a\right)\bigcap Rb=0,{\left(ba\right)}^{\pi }a=0,{\left(ab\right)}^{\pi }b=0.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By virtue of [14], $ab,ba\in R^{\#}$, $r\left(a\right)\bigcap bR=0$. Set $x=b{\left(ab\right)}^{\#}$. Then $x=a^{(b,c)}=a^{\#⃝}$ by [14], Theorem 2.13. As $x{a}^2=a$, we have $b{\left(ab\right)}^{\#}{a}^2=a$, and so $ba{\left({\left(ba\right)}^{\#⃝}\right)}^{2}b{a}^2=a$. This implies that ${\left(ba\right)}^{\#}b{a}^2=a$. Hence ${\left(ba\right)}^{\pi }a=0$. Clearly, $a{x}^2=x$, and then $ab{\left(ab\right)}^{\#⃝}x=x$. Since $xab=b$, we have $ab{\left(ab\right)}^{\#}b=b$. This implies that ${\left(ba\right)}^{\pi }b=0$.

Since $ab{\left(ab\right)}^{\#}=ax$, we have ${\left(ab{\left(ab\right)}^{\#}\right)}^{\ast }={\left(ax\right)}^{\ast }=ax=ab{\left(ab\right)}^{\#}$. According to [30], Theorem 3.6, $ab\in R$ is EP.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, $ab\in R$ is EP, $r\left(a\right)\bigcap bR=0$. Hence $ab\in R^{\#}$. In view of [14], Theorem 2.13, $a\in R^{(b,b)}$ and $a^{(b,b)}=b{\left(ab\right)}^{\#}$. Set $x=a^{(b,b)}$. Then $ax=ab{\left(ab\right)}^{\#}$, and so ${\left(ax\right)}^{\ast }=ax$. Furthermore, we verify that

$$\begin{array}{ccc}\hfill a{x}^2& =\hfill & \left[ab{\left(ab\right)}^{\#}\right]b{\left(ab\right)}^{\#}=[1-{\left(ab\right)}^{\pi }]b{\left(ab\right)}^{\#}=b{\left(ab\right)}^{\#}=x,\hfill \\ \hfill x{a}^2& =\hfill & b{\left(ab\right)}^{\#}{a}^2=ba\left({\left(ba\right)}^D\right){)}^{2}b{a}^2=ba{\left(ba\right)}^{D}a=[1-{\left(ba\right)}^{\pi }]a=a.\hfill \end{array}$$

Therefore $a\in R^{\#⃝}$ and $a^{\#⃝}=x=a^{(b,b)}$, as required.

$\left(1\right)\Rightarrow \left(3\right)$ By the argument above, ${\left(ba\right)}^{\pi }a=0$. In view of Theorem 2.6, $aR=bR=b^{\ast }R=baR,$ and then ${\left(ba\right)}^{\pi }b=0$. Let $x\in \ell \left(a\right)\bigcap Rb=0$. Then $xa=0$ and $x=rb$ for some $r\in R$. Hence $x=r\left[\left(ba\right){\left(ba\right)}^{\#}\right]b=\left(xa\right){\left(ba\right)}^{\#}b=0$. Thus $\ell \left(a\right)\bigcap Rb=0$, as desired.

$\left(3\right)\Rightarrow \left(1\right)$ Since $ab$ is EP, it has Drazin inverse. By using Cline’s formula, $ba$ has Drazin inverse. It is easy to verify that ${\left(ba\right)}^D{\left(ba\right)}^2=b{\left({\left(ab\right)}^{\#}\right)}^{2}a{\left(ba\right)}^2=b{\left({\left(ab\right)}^{\#}\right)}^{2}ababa=b{\left(ab\right)}^{\#}aba=b[1-{\left(ab\right)}^{\pi }]a=ba.$ Thus $ba\in R^{\#}$. Moreover, ${\left(ba\right)}^{\pi }b\in \ell \left(a\right)\bigcap Rb=0$; hence, ${\left(ba\right)}^{\pi }b=0$. We infer that $bR=baR$. By virtue of [14], Theorem 2.13, $a\in R^{(b,b)}$ and $a^{(b,b)}={\left(ba\right)}^{\#⃝}b$. Set $x=a^{(b,b)}$. Then $ax=a{\left(ba\right)}^{\#}b=ab{\left({\left(ab\right)}^{\#}\right)}^{2}ab=ab{\left(ab\right)}^{\#}$. Hence, ${\left(ax\right)}^{\ast }=ax$. Since ${\left(ab\right)}^{\pi }b=0$ and ${\left(ba\right)}^{\pi }a=0,$ we check that

$$\begin{array}{ccc}\hfill a{x}^2& =\hfill & a{\left(ba\right)}^{\#}b{\left(ba\right)}^{\#}b=ab{\left[{\left(ab\right)}^{\#}\right]}^{2}ab{\left(ba\right)}^{\#}b\hfill \\ & =\hfill & ab{\left(ab\right)}^{\#}{\left(ba\right)}^{\#}b=[1-{\left(ab\right)}^{\pi }]{\left(ba\right)}^{\#}b={\left(ba\right)}^{\#}b=x,\hfill \\ \hfill x{a}^2& =\hfill & {\left(ba\right)}^{\#}b{a}^2=[1-{\left(ba\right)}^{\pi }]a=a.\hfill \end{array}$$

Therefore $a^{(b,b)}=a^{\#⃝}$, as required. □

Corollary 8. 

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

(1)

$a\in R_{b,b}^{\#⃝}$.

(2)

$ab\in R$ is EP, $Rb=Rab,{\left(ba\right)}^{\pi }a=0,{\left(ab\right)}^{\pi }b=0.$

(3)

$ab\in R$ is EP, $bR=baR,{\left(ba\right)}^{\pi }a=0,{\left(ab\right)}^{\pi }b=0.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.11, $ab\in R$ is EP, ${\left(ba\right)}^{\pi }a=0$ and ${\left(ab\right)}^{\pi }b=0.$ Clearly, $Rab\subseteq Rb$. Since $[{\left(ba\right)}^{\#}bab-b]a=0$, we see that ${\left(ba\right)}^{\#}bab-b\in \ell \left(a\right)\bigcap Rb$. In light of Theorem 2.11, $\ell \left(a\right)\bigcap Rb=0$, and then ${\left(ba\right)}^{\#}bab=b$. This implies that $Rb\subseteq Rab$. Therefore $Rb=Rab$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Let $x\in r\left(a\right)\bigcap bR$. Then $ax=0$ and $x=br$ for some $r\in R$. Hence $abr=ax=0$. Since $Rb=Rab$, we can find $s\in R$ such that $b=sab$. Then $br=sabr=s\left(abr\right)=0$. Thus $r\left(a\right)\bigcap bR=0$. Accordingly, $a\in R_{b,b}^{\#⃝}$ by Theorem 2.11.

$\left(1\right)\iff \left(3\right)$ Similarly to the argument above, we obtain the result by Theorem 2.11. □

Corollary 9. 

Let $a\in R_{b,b}^{\#⃝}$. Then ${\left(ab\right)}^n\in R$ is EP for any $n\in N$.

Proof. 

By the argument in Theorem 2.11, we have $a{a}_{b,b}^{\#⃝}=ab{\left(ab\right)}^{\#}={\left(ab\right)}^n{\left[{\left(ab\right)}^n\right]}^{\#}$ for any $n\in N$. Then ${\left[{\left(ab\right)}^n{\left({\left(ab\right)}^n\right)}^{\#}\right]}^{\ast }={\left(a{a}_{b,b}^{\#⃝}\right)}^{\ast }=a{a}_{b,b}^{\#⃝}={\left(ab\right)}^n{\left({\left(ab\right)}^n\right)}^{\#}$. In view of [30], Theorem 3.6, ${\left(ab\right)}^n\in R$ is EP. □

3. Core-EP-$(b,c)$-Inverses

This section presents the necessary and sufficient conditions for the existence of the core-EP-$(b,c)$-inverse in a *-ring. We begin with

Theorem 5. 

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

(1)

$a\in R$ has core-EP-$(b,c)$-inverse.

(2)

There exist $x,y\in R$ such that

$$a=x+y,x^{\ast }y=yx=0,x\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

(3)

There exists $x\in R_{b,c}^{\#⃝}$ such that

$$x=a{x}^2,{\left(ax\right)}^{\ast }=ax,a^n=x{a}^{n+1}$$

for some $n\in N$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, $a\in R^{Ⓓ}$ and $a^{Ⓓ}=a^{(b,c)}$. Then we have

$$\begin{array}{ccc}\hfill a^{Ⓓ}& =\hfill & a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]a^{Ⓓ},\hfill \\ \hfill a^{Ⓓ}& \in \hfill & bR\bigcap Rc,\hfill \\ \hfill b& =\hfill & a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]b,\hfill \\ \hfill c& =\hfill & c\left[a{a}^{Ⓓ}a\right]a^{Ⓓ}.\hfill \end{array}$$

Hence, $a^{Ⓓ}={\left(a{a}^{Ⓓ}a\right)}^{(b,c)}$. Thus, $a{a}^{Ⓓ}a\in R^{(b,c)}$. In view of [4], Corollary 3.4, there exist $x,y\in R$ such that

$$a=x+y,x^{\ast }y=yx=0,x\in R^{\#⃝},y\in R^{nil}.$$

Moreover, we have $x^{\#⃝}=a^{Ⓓ}={\left(a{a}^{Ⓓ}a\right)}^{(b,c)}=x^{(b,c)}.$ Therefore $x\in R_{b,c}^{\#⃝}$, as desired.

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

$$a=x+y,x^{\ast }y=yx=0,x\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

Then $x\in R^{\#⃝}$. Then $a\in R^{Ⓓ}$ and $a^{Ⓓ}=x^{\#⃝}$ (see [4], Corollary 3.4). Hence $a^{Ⓓ}=x^{(b,c)}={\left(a{a}^{Ⓓ}a\right)}^{(b,c)}.$ We derive that

$$\begin{array}{ccc}\hfill a^{Ⓓ}& =\hfill & a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]a^{Ⓓ},\hfill \\ \hfill a^{Ⓓ}& \in \hfill & bR\bigcap Rc,\hfill \\ \hfill b& =\hfill & a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]b,\hfill \\ \hfill c& =\hfill & c\left[a{a}^{Ⓓ}a\right]a^{Ⓓ}.\hfill \end{array}$$

By virtue of [12], Theorem 2.10, we have

$$\begin{array}{ccc}\hfill a^{Ⓓ}& =\hfill & a^{Ⓓ}a{a}^{Ⓓ},\hfill \\ \hfill a^{Ⓓ}& \in \hfill & bR\bigcap Rc,\hfill \\ \hfill b& =\hfill & a^{Ⓓ}ab,\hfill \\ \hfill c& =\hfill & ca{a}^{Ⓓ}.\hfill \end{array}$$

Therefore $a^{Ⓓ}=a^{(b,c)}$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, there exist $z,y\in R$ such that

$$a=z+y,z^{\ast }y=yz=0,z\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

Set $x=z_{b,c}^{\#⃝}$. Then

$$x\in R^{\#⃝},xR=zR=bR,R{x}^{\ast }=R{z}^{\ast }=Rc.$$

By using Theorem 2.1, $x\in R_{b,c}^{\#⃝}$. One easily checks that

$$\begin{array}{ccc}\hfill ax& =\hfill & (z+y)z_{b,c}^{\#⃝}=z{z}_{b,c}^{\#⃝}+yz{\left(z_{b,c}^{\#⃝}\right)}^2=z{z}_{b,c}^{\#⃝},\hfill \\ \hfill a{x}^2& =\hfill & \left(ax\right)x=z{\left(z_{b,c}^{\#⃝}\right)}^2=z_{b,c}^{\#⃝}=x,\hfill \\ \hfill z_{b,c}^{\#⃝}y& =\hfill & xy=xzxy=x\left(zx\right)y=x{\left(zx\right)}^{\ast }y=x{x}^{\ast }\left(z^{\ast }y\right)=0.\hfill \end{array}$$

By using $z_{b,c}^{\#⃝}y=0$ and $x{x}^{\#⃝}x=x$, we derive that $axa=\left(ax\right)a=z{z}_{b,c}^{\#⃝}(z+y)=z{z}_{b,c}^{\#⃝}z=z.$ Then

$$\begin{array}{c}{\left(ax\right)}^{\ast }={\left(z{z}_{b,c}^{\#⃝}\right)}^{\ast }=z{z}_{b,c}^{\#⃝}=ax,\\ a(1-xa)=a-axa=a-z=y\in R^{nil}.\end{array}$$

Write $y^{n-1}=0$ for some $n\ge 2$. As $yz=0$, we see that $(a-x{a}^2)z=[z+y-z_{b,c}^{\#⃝}{(z+y)}^2]z=(z-z_{b,c}^{\#⃝}{z}^2)z=0.$ Thus we have $a^n-x{a}^{n+1}=(a-x{a}^2)a^{n-1}=(a-x{a}^2){(z+y)}^{n-1}=(a-x{a}^2)y^{n-1}=0.$ Therefore $a^n=x{a}^{n+1},$ as required.

$\left(3\right)\Rightarrow \left(2\right)$ By hypotheses, we have $z\in R_{b,c}^{\#⃝}$ such that

$$z=a{z}^2,{\left(az\right)}^{\ast }=az,a^n=z{a}^{n+1}.$$

For any $n\in N$, we have

$$\begin{array}{ccc}\hfill az& =\hfill & a\left(a{z}^2\right)=a^2{z}^2=a^2\left(a{z}^2\right)z=a^3{z}^3\hfill \\ & =\hfill & \dots =a^n{z}^n=\dots =a^{n+1}{z}^{n+1}.\hfill \end{array}$$

Hence

$$z-zaz=\left(az\right)z-zaz=\left(a^n{z}^n\right)z-z\left(a^{n+1}{z}^{n+1}\right)=(a^n-z{a}^{n+1})z^{n+1}=0.$$

Then $z=zaz$. Set $x=aza$ and $y=a-aza.$ Then $a=x+y$. We check that

$$\begin{array}{ccc}\hfill (a-z{a}^2)z& =\hfill & (a-z{a}^2)a{z}^2\hfill \\ & =\hfill & (a-z{a}^2)a^2{z}^3\hfill \\ & ⋮\hfill & \\ & =\hfill & (a-z{a}^2)a^{n-1}{z}^n\hfill \\ & =\hfill & (a^n-z{a}^{n+1})z^n=0.\hfill \end{array}$$

Therefore $(a-z{a}^2)z=0.$

We claim that x has core inverse. Evidently, we verify that

$$\begin{array}{ccc}\hfill z{x}^2& =\hfill & za\left(z{a}^{2}z\right)a=z{a}^{2}za=aza=x,\hfill \\ \hfill x{z}^2& =\hfill & aza{z}^2=a{z}^2=z,\hfill \\ \hfill {\left(xz\right)}^{\ast }& =\hfill & {\left(azaz\right)}^{\ast }={\left(az\right)}^{\ast }=az=\left(aza\right)z=xz.\hfill \end{array}$$

Thus, $x^{\#⃝}=z$. We verify that

$$\begin{array}{ccc}\hfill {(a-z{a}^2)}^{n+1}& =\hfill & {(a-z{a}^2)}^n(a-z{a}^2)\hfill \\ & =\hfill & {(a-z{a}^2)}^{n-1}(a-z{a}^2)a\hfill \\ & =\hfill & {(a-z{a}^2)}^{n-1}{a}^2\hfill \\ & ⋮\hfill & \\ & =\hfill & (a-z{a}^2)a^n=(a^n-z{a}^{n+1})a=0.\hfill \end{array}$$

This implies that $a-z{a}^2\in R^{nil}$. By using Cline’s formula (see [18], Theorem 2.1), $y=a-aza\in R^{nil}$.

Moreover, we see that

$$\begin{array}{ccc}\hfill x^{\ast }y& =\hfill & {\left(aza\right)}^{\ast }(1-az)a=a^{\ast }{\left(az\right)}^{\ast }(1-az)a=a^{\ast }{\left(az\right)}^{\ast }(1-az)a=0,\hfill \\ \hfill yx& =\hfill & (a-aza)aza=a(a-z{a}^2)za=0.\hfill \end{array}$$

Obviously, we have $z^{\#⃝}={\left[x^{\#⃝}\right]}^{\#⃝}=x^2{x}^{\#⃝}.$ By hypothesis, we have $x^2{x}^{\#⃝}=z^{(b,c)}.$

Claim 1. $z\in bRz\bigcap zRc.$

Obviously, $x^2{x}^{\#⃝}={\left(aza\right)}^{2}z=az{a}^{2}z=a^{2}z.$ Thus, $z=\left(a^{2}z\right)z^3=\left[x^2{x}^{\#⃝}\right]z^3\in bRz,z=zaz=z^2{a}^{2}z=z^2\left[x^2{x}^{\#⃝}\right]\in zRc.$

Claim 2. $zxb=b$.

We claim that $b=x^2{x}^{\#⃝}zb={\left(aza\right)}^2{z}^{2}b=azaz=azb.$

By hypothesis, we have $b=azb=a{x}^{\#⃝}b=(x+y)x^{\#⃝}b=x{x}^{\#⃝}b.$ Hence, $x^{\#⃝}xb=x^{\#⃝}x\left[x{x}^{\#⃝}b\right]=\left[x^{\#⃝}{x}^2{x}^{\#⃝}\right]b=x{x}^{\#⃝}b=b.$ Thus, we see that

$$\begin{array}{ccc}\hfill zab& =\hfill & x^{\#⃝}(x+y)b\hfill \\ & =\hfill & x^{\#⃝}x{x}^{\#⃝}(x+y)b\hfill \\ & =\hfill & x^{\#⃝}{\left(x{x}^{\#⃝}\right)}^{\ast }(x+y)b\hfill \\ & =\hfill & x^{\#⃝}xb=b.\hfill \end{array}$$

Therefore $zxb=z\left(aza\right)b=\left(zaz\right)ab=zab=b.$

Claim 3. $cxz=c$.

We verify that $c=cz{x}^2{x}^{\#⃝}=cz\left(aza\right)\left(aza\right)z=cz{a}^{2}z=caz.$ Hence, $cxz=c\left(aza\right)z=ca\left(zaz\right)=caz=c.$ Therefore $z=x^{(b,c)}$, and so $x\in R_{b,c}^{\#⃝}$ and $x_{b,c}^{\#⃝}=z$.

Accordingly, we have a generalized core-EP-$(b,c)$-decomposition $a=x+y$, thus yielding the result. □

Corollary 10. 

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

(1)

$a\in R$ has core-EP-$(b,c)$-inverse.

(2)

There exists a unique pair $x,y\in R$ such that

$$a=x+y,x^{\ast }y=yx=0,x\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 3.1, there exists $x\in R_{b,c}^{\#⃝}$ such that $x=a{x}^2,{\left(ax\right)}^{\ast }=ax,a^n=x{a}^{n+1}.$ Then $a\in R^{Ⓓ}$ and $x=a^{Ⓓ}$. Accordingly, x is unique by [12], Theorem 2.2.

$\left(2\right)\Rightarrow \left(1\right)$ This is clear by Theorem 3.1. □

Corollary 11. 

Let $a\in R_{b,c}^{Ⓓ}$. Then $a_{b,c}^{Ⓓ}\in R_{b,c}^{\#⃝}$. In this case, ${\left(a_{b,c}^{Ⓓ}\right)}_{b,c}^{\#⃝}=a^2{a}_{b,c}^{Ⓓ}.$

Proof. 

In view of Theorem 3.1, there exist $x,y\in R$ such that

$$a=x+y,x^{\ast }y=yx=0,x\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

Then $a_{b,c}^{Ⓓ}=x_{b,c}^{\#⃝}$. By virtue of Theorem 2.1, we have $xR=bR,R{x}^{\ast }=Rc.$ In view of [12], we have ${\left(x_{b,c}^{\#⃝}\right)}^{\#⃝}=x^2{x}_{b,c}^{\#⃝}.$ Moreover, we verify that $x_{b,c}^{\#⃝}R=xR=bR,R{\left(x_{b,c}^{\#⃝}\right)}^{\ast }=R{x}^{\ast }=Rc.$

By using Theorem 2.1 again, $a_{b,c}^{Ⓓ}\in R_{b,c}^{\#⃝}.$ Moreover, we have ${\left(a_{b,c}^{Ⓓ}\right)}_{b,c}^{\#⃝}=a^2{a}_{b,c}^{Ⓓ}.$ □

We next investigate the polar-like property for the core-EP-$(b,c)$-invertibility.

Theorem 6. 

Let $a\in R$. Then $a\in R_{b,c}^{Ⓓ}$ if and only if $a\in R^D$ and there exists a projection $p\in R$ such that

$$u:=a+p\in R^{-1},1-p\in R_{b,c}^{\#⃝},pa\in R^{nil},pu(1-p)=p{u}^{-1}(1-p)=0.$$

Proof. 

⟹ In view of [12], Theorem 2.3, $a\in R^D$. Since $a\in R_{b,c}^{Ⓓ}$, by using Theorem 3.1, there exist $x,y\in R$ such that

$$a=x+y,x^{\ast }y=yx=0,x\in R_{b,c}^{\#⃝},y\in R^{nil}.$$

By virtue of [32], Lemma 2.3, we have $x=x{x}_{b,c}^{\#⃝}x,x_{b,c}^{\#⃝}x{x}_{b,c}^{\#⃝}=x_{b,c}^{\#⃝},x_{b,c}^{\#⃝}{x}^2=x,x{\left(x_{b,c}^{\#⃝}\right)}^2=x_{b,c}^{\#⃝},{\left(x{x}_{b,c}^{\#⃝}\right)}^{\ast }=x{x}_{b,c}^{\#⃝}.$ Let $p=1-x{x}_{b,c}^{\#⃝}$. Then $p^2=p=p^{\ast }$ and $px=0$. We directly check that

$$\begin{array}{c}(x+1-x{x}_{b,c}^{\#⃝})(x_{b,c}^{\#⃝}+1-x_{b,c}^{\#⃝}x)=1\\ =(x_{b,c}^{\#⃝}+1-x_{b,c}^{\#⃝}x)(x+1-x{x}_{b,c}^{\#⃝}).\end{array}$$

Hence, ${(x+p)}^{-1}=x_{b,c}^{\#⃝}+1-x_{b,c}^{\#⃝}x.$ Since $y(x+p)=y(x+1-x{x}_{b,c}^{\#⃝})=y$, we see that $y{(x+p)}^{-1}=y\in R^{nil}$. By virtue of [18], Theorem 2.1, ${(x+p)}^{-1}y\in R^{nil}$. Hence, $1+{(x+p)}^{-1}y\in R^{-1}$. Therefore, we check that

$$\begin{array}{ccc}\hfill pa& =\hfill & p(x+y)=py=(1-x{x}_{b,c}^{\#⃝})y={(1-x{x}_{b,c}^{\#⃝})}^{\ast }y\hfill \\ & =\hfill & (1-{\left(x_{b,c}^{\#⃝}\right)}^{\ast }{x}^{\ast })y=y\in R^{nil},\hfill \\ \hfill pa(1-p)& =\hfill & yx{x}_{b,c}^{\#⃝}=0,pa=pap,\hfill \\ \hfill a+p& =\hfill & x+y+p=(x+p)[1+{(x+p)}^{-1}y]\in R^{-1}.\hfill \end{array}$$

Then $pu(1-p)=pa(1-p)=0$. Since ${(x+p)}^{-1}=x+1-x{x}_{b,c}^{\#⃝}$, we see that

$$\begin{array}{ccc}\hfill p{u}^{-1}(1-p)& =\hfill & (1-x{x}_{b,c}^{\#⃝})(x+1-x{x}_{b,c}^{\#⃝})x{x}_{b,c}^{\#⃝}\hfill \\ & =\hfill & (1-x{x}_{b,c}^{\#⃝})x^2{x}_{b,c}^{\#⃝}\hfill \\ & =\hfill & x^2{x}_{b,c}^{\#⃝}-x\left(x_{b,c}^{\#⃝}{x}^2\right)x_{b,c}^{\#⃝}=0.\hfill \end{array}$$

Obviously, ${(1-p)}^{\ast }={(1-p)}^2=1-p\in R^{\#⃝}.$ Moreover, we check that

Claim 1. $(1-p)R=x{x}^{\#⃝}R=xR=bR$.

Claim 2. $R{(1-p)}^{\ast }=R{\left(x{x}^{\#⃝}\right)}^{\ast }=R{x}^{\ast }=Rc$.

Therefore $1-p\in R_{b,c}^{\#⃝}$, as required.

⟸ By hypothesis, there exists a projection $p\in R$ such that

$$u:=a+p\in R^{-1},1-p\in R_{b,c}^{\#⃝},pa\in R^{nil},pu(1-p)=p{u}^{-1}(1-p)=0.$$

Then $pa(1-p)=p(a+p)(1-p)=pu(1-p)=0$. Set $x=(1-p)a$ and $y=pa$. Then

$$\begin{array}{ccc}\hfill x^{\ast }y& =\hfill & {\left[(1-p)a\right]}^{\ast }\left(pa\right)=\left[a^{\ast }{(1-p)}^{\ast }\right]p^{\ast }a=a^{\ast }{(1-p)}^{\ast }{p}^{\ast }a=0,\hfill \\ \hfill yx& =\hfill & pa(1-p)a=\left(pap\right)(1-p)a=0,\hfill \\ \hfill y& =\hfill & pa\in R^{nil}.\hfill \end{array}$$

Observing that

$$\begin{array}{ccc}\hfill u^{-1}{x}^2& =\hfill & u^{-1}(1-p)a(1-p)a=u^{-1}a(1-p)a-u^{-1}\left[pa(1-p)\right]a\hfill \\ & =\hfill & u^{-1}a(1-p)a={(a+p)}^{-1}(a+p)(1-p)a=(1-p)a=x,\hfill \end{array}$$

and then $Rx=R{x}^2$. Set $y=u^{-1}(1-p)$. Then $xy=x{u}^{-1}(1-p)=(1-p)a{(a+p)}^{-1}(1-p)=(1-p)(a+p){(a+p)}^{-1}(1-p)=1-p$. Then $xyx=(1-p)x=x$ and ${\left(xy\right)}^{\ast }={(1-p)}^{\ast }=1-p^{\ast }=1-p=xy$. Moreover, we see that $x{y}^2=\left(xy\right)y=(1-p)u^{-1}(1-p)=u^{-1}(1-p)=y$. According to [32], Theorem 3.3, $x\in R^{\#⃝}$.

Clearly, $xR=x{u}^{-1}R=(1-p)R=bR$ and $R{x}^{\ast }=R{u}^{\ast }{(1-p)}^{\ast }=R{(1-p)}^{\ast }=Rc$. In light of Theorem 2.1, $x\in R_{b,c}^{\#⃝}.$ Therefore, $a\in R_{b,c}^{Ⓓ}$ by Theorem 3.1. □

Theorem 7. 

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

(1)

$a\in R_{b,c}^{Ⓓ}$.

(2)

$a\in R^D$ and there exists $x\in R_{b,c}^{\#⃝}$ and $m\in N$ such that

$$xax=x,xR=x^{\ast }R=a^{m}R.$$

(3)

$a\in R^D$ and there exists $x\in R_{b,c}^{\#⃝}$ and $m\in N$ such that

$$xax=x,\ell \left(x\right)=\ell \left(x^{\ast }\right)=\ell \left(a^m\right).$$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ Let $x=a_{b,c}^{Ⓓ}$. Set $m=ind\left(a\right)$. By virtue of [12], Theorem 2.10, we have $xax=x,xR=x^{\ast }R=a^{m}R.$ In view of Corollary 3.3, $x\in R_{b,c}^{\#⃝}$, as desired.

$\left(2\right)\Rightarrow \left(3\right)$ If $rx=0$ for $r\in R$, then $r{a}^m=0$. If $r{a}^m=0$ for $r\in R$, then $rx=0$. Hence $\ell \left(x\right)=\ell \left(a^m\right)$. Likewise, we have $\ell \left(x^{\ast }\right)=\ell \left(a^m\right),$ as desired.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in R_{b,c}^{\#⃝}$ such that $xax=x,\ell \left(x\right)=\ell \left(x^{\ast }\right)=\ell \left(a^m\right).$ By virtue of [12], Theorem 2.10, $a\in R^{Ⓓ}$ and $a^{Ⓓ}=x$. Then we have

$${\left(a^{Ⓓ}\right)}^{\#⃝}={\left(a^{Ⓓ}\right)}^{(b,c)}.$$

This implies that $a^2{a}^{Ⓓ}={\left(a^{Ⓓ}\right)}^{(b,c)}.$ Hence

$$a^2{a}^{Ⓓ}\in bR\bigcap Rc,\left[a^2{a}^{Ⓓ}\right]a^{Ⓓ}b=b,c{a}^{Ⓓ}\left[a^2{a}^{Ⓓ}\right]=c.$$

This implies that

$$a^{Ⓓ}=\left[a^2{a}^{Ⓓ}\right]{\left[a^{Ⓓ}\right]}^3={\left[a^{Ⓓ}\right]}^2\left[a^2{a}^{Ⓓ}\right]\in bR\bigcap Rc.$$

Moreover, we have

$$a{a}^{Ⓓ}b=b,ca{a}^{Ⓓ}]=c.$$

We deduce that

$$a^{Ⓓ}ab=a^{Ⓓ}a\left[a{a}^{Ⓓ}b\right]=a{a}^{Ⓓ}b=b.$$

Accordingly, $a^{Ⓓ}=a^{(b,c)}$, as required. □

An element a in a *-ring R is *-DMP (i.e., *-DMP element) if there exist $m\in N$ and $x\in R$ such that $a{x}^2=x,{\left(ax\right)}^{\ast }=xa=ax,a^m=x{a}^{m+1}$ (see [13]). We next characterize *-DMP elements in a ring by using the core-EP-$(b,c)$-invertibility.

Lemma 1. 

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

(1)

$a\in R$ is *-DMP.

(2)

$a\in R_{a^n,a^n}^{Ⓓ}$ for some $n\in N$.

(3)

$a\in R_{{\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast }}^{Ⓓ}$ for some $n\in N$.

(4)

$a\in R_{{\left(a^n\right)}^{\ast },a^n}^{Ⓓ}$ for some $n\in N$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ Set $n=ind\left(A\right)$. In view of [12], Theorem 2.5, $a^{Ⓓ}={\left(a^n\right)}^{\#⃝}={\left(a^n\right)}^{\#}$. By virtue of [19], Theorem 11, ${\left(a^n\right)}^{\#}={\left(a^n\right)}^{(a^n,a^n)}$. Hence, $a^{Ⓓ}={\left(a^n\right)}^{(a^n,a^n)}$. This implies that

$$a^{Ⓓ}=a^{Ⓓ}{a}^n{a}^{Ⓓ},a^{Ⓓ}R=a^{n}R,R{a}^{Ⓓ}=R{a}^n.$$

Since $a^{Ⓓ}=a^{Ⓓ}a{a}^{Ⓓ}$, we deduce that $a^{Ⓓ}=a^{(a^n,a^n)}$. Thus $a\in R_{a^n,a^n}^{Ⓓ}$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we have $a^{Ⓓ}=a^{(a^n,a^n)}$. Then $a^{Ⓓ}R=a^{n}R,R{a}^{Ⓓ}=R{a}^n.$ Write $a^n=x{a}^{Ⓓ}$ for some $x\in R$. Then $a^{n+1}{a}^{Ⓓ}=a^n\left(a{a}^{Ⓓ}\right)=x\left(a^{Ⓓ}a{a}^{Ⓓ}\right)=x{a}^{Ⓓ}=a^n$. By using [12], Theorem 2.10, $a\in R$ is *-DMP.

$\left(1\right)\Rightarrow \left(3\right)$ Set $n=ind\left(a\right)$. Then $a^{Ⓓ}={\left(a^n\right)}^{\#}={\left(a^n\right)}^{\#⃝}.$ Thus ${\left(a^n\right)}^{\#⃝}={\left(a^n\right)}^{({\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast })}$. Hence we verify that $a^{Ⓓ}=a^{({\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast })}$. Therefore $a\in R_{{\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast }}^{Ⓓ}$.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, $a^{Ⓓ}=a^{({\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast })}$. Then ${\left(a^n\right)}^{\ast }=a^n{\left(a^{Ⓓ}\right)}^{n+1}a{\left(a^n\right)}^{\ast }$; hence, ${\left(a^n\right)}^{\ast }R\subseteq a^{n}R$. On the other hand, ${\left(a^n\right)}^{\ast }={\left(a^n\right)}^{\ast }a{a}^{Ⓓ}=\left(a{a}^{Ⓓ}{a}^n\right)$, and so $a^n=a{a}^{Ⓓ}{a}^n=a^{Ⓓ}{a}^2{a}^{Ⓓ}{a}^n\in {\left(a^n\right)}^{\ast }R$; whence, $a^{n}R\subseteq {\left(a^n\right)}^{\ast }R$. We infers that $a^{n}R={\left(a^n\right)}^{\ast }R$. By virtue of [31], Theorem 3.9, $a\in R$ is *-DMP.

$\left(1\right)\Rightarrow \left(4\right)$ Since $a\in R$ is *-DMP, we can find some $n\in N$ such that $a^n$ is EP. In view of [28], Corollary 5.1.12, we have ${\left(a^n\right)}^{\#⃝}={\left(a^n\right)}^{\#}={\left(a^n\right)}^{{\left(a^n\right)}^{\ast },a^n)}.$ Then we get ${\left(a^n\right)}^{\#}R={\left(a^n\right)}^{\ast }R$ and $R{\left(a^n\right)}^{\#}=R{a}^n$. Then $a^{n-1}{\left(a^n\right)}^{\#}R={\left(a^n\right)}^{\ast }R$ and $R{a}^{n-1}{\left(a^n\right)}^{\#}=R{a}^n$. In view of [12], Theorem 2.5, $a^{Ⓓ}=a^{n-1}{\left(a^n\right)}^{\#}=a^{n-1}{\left(a^n\right)}^{\#⃝}.$ By virtue of [12], Theorem 2.10, $a^{Ⓓ}=a^{Ⓓ}a{a}^{Ⓓ}$. Therefore $a^{Ⓓ}=a^{{\left(a^n\right)}^{\ast },a^n)}$, as desired.

$\left(4\right)\Rightarrow \left(1\right)$ By assumption, we have $a^{Ⓓ}=a^{({\left(a^n\right)}^{\ast },a^n)}$. Since $a^{Ⓓ}a{\left(a^n\right)}^{\ast }={\left(a^n\right)}^{\ast }$, we have ${\left(a^n\right)}^{\ast }R\subseteq a^{n}R$. Therefore we complete the proof by [31], Theorem 3.9. □

Theorem 8. 

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

(1)

$a\in R$ is *-DMP.

(2)

$a\in R_{a^D,a^D}^{Ⓓ}$.

(3)

$a\in R_{{\left(a^D\right)}^{\ast },{\left(a^D\right)}^{\ast }}^{Ⓓ}$.

(4)

$a\in R_{{\left(a^D\right)}^{\ast },a^D}^{Ⓓ}$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By virtue of [13], Lemma 2.3, $a^{Ⓓ}=a^D.$ Obviously, $a^D=a^{(a^D,a^D)}$. Hence, $a^{Ⓓ}=a^{(a^D,a^D)}$, and so $a\in R_{a^D,a^D}^{Ⓓ}$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, $a^{Ⓓ}=a^{(a^D,a^D)}=a^D$. This implies that $a\in R$ is *-DMP by [13], Lemma 2.3.

$\left(1\right)\Rightarrow \left(3\right)$ In view of [13], Lemma 2.3, we have $a^{Ⓓ}=a^D.$ We will suffice to verify that $a^{Ⓓ}=a^{{\left(a^D\right)}^{\ast },{\left(a^D\right)}^{\ast })}.$

Clearly, $a^{Ⓓ}=a^{Ⓓ}a{a}^{Ⓓ}$.

It is easy to verify that $a^{Ⓓ}a{\left(a^D\right)}^{\ast }={\left(a{a}^{Ⓓ}\right)}^{\ast }{\left(a^D\right)}^{\ast }={\left(a^{D}a{a}^{Ⓓ}\right)}^{\ast }={\left(a^D\right)}^{\ast }$. Moreover, ${\left(a^D\right)}^{\ast }a{a}^{Ⓓ}={\left(a^D\right)}^{\ast }$ and $a^{Ⓓ}\in {\left(a^D\right)}^{\ast }R{a}^{Ⓓ}\bigcap a^{Ⓓ}R{\left(a^D\right)}^{\ast }.$ Therefore $a\in R_{{\left(a^D\right)}^{\ast },{\left(a^D\right)}^{\ast }}^{Ⓓ}$.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, we have $a^{Ⓓ}=a^{({\left(a^D\right)}^{\ast },{\left(a^D\right)}^{\ast })}.$ We will suffice to verify that $a^{Ⓓ}=a^{({\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast })}.$

Clearly, $a^{Ⓓ}=a^{Ⓓ}a{a}^{Ⓓ}$.

Step 1. We verify that

$$\begin{array}{ccc}\hfill a^{Ⓓ}a{\left(a^n\right)}^{\ast }& =\hfill & a^{Ⓓ}a{\left(a^{n+1}{a}^D\right)}^{\ast }\hfill \\ & =\hfill & \left[a^{Ⓓ}a{\left(a^D\right)}^{\ast }\right]{\left(a^{n+1}\right)}^{\ast }={\left(a^D\right)}^{\ast }{\left(a^{n+1}\right)}^{\ast }={\left(a^n\right)}^{\ast }.\hfill \end{array}$$

Step 2. One directly checks that

$${\left(a^n\right)}^{\ast }a{a}^{Ⓓ}={\left(a{a}^{Ⓓ}{a}^n\right)}^{\ast }={\left(a{a}^{Ⓓ}{a}^{n+1}{a}^D\right)}^{\ast }={\left(a^{n+1}{a}^D\right)}^{\ast }={\left(a^n\right)}^{\ast }.$$

Step 3. Obviously, we have $a^{Ⓓ}=a^{Ⓓ}a{a}^{Ⓓ}\in R{a}^{Ⓓ}.$ On the other hand, we have $a^{Ⓓ}\in {\left(a^D\right)}^{\ast }R={\left({\left(a^D\right)}^{n+1}{a}^n\right)}^{\ast }R\subseteq {\left(a^n\right)}^{\ast }R$. Hence, $a^{Ⓓ}\in {\left(a^n\right)}^{\ast }R{a}^{Ⓓ}.$ Likewise, we have $a^{Ⓓ}\in a^{Ⓓ}R{\left(a^n\right)}^{\ast }.$ Therefore $a\in R_{{\left(a^n\right)}^{\ast },{\left(a^n\right)}^{\ast }}^{Ⓓ}$. By virtue of Lemma 3.6, $a\in R$ is *-DMP.

$\left(1\right)\iff \left(4\right)$ This can be proven in a similar way. □

For a complex $A\in {\mathbb{C}}^{n\times n}$, it follows by [12], Theorem 3.3 that A has core-EP-$(B,B)$-inverse, where $B=A^m{\left(A^m\right)}^{\ast }$ and $m=ind\left(A\right)$. We come now to characterize the core-EP-$(b,b)$-inverse in a ring.

Lemma 2. 

Let $a,b\in R$. Then $a\in R_{b,b}^{Ⓓ}$ if and only if

(1)

$a\in R^{Ⓓ},b\in R^{\#⃝}$;

(2)

$ba+1-b{b}^{\#⃝}\in R^{-1},ba{a}^{Ⓓ}=b$ and $b{b}^{\#⃝}{a}^{Ⓓ}=a^{Ⓓ}$.

Proof. 

⟹ Obviously, $a\in R^{Ⓓ}$ and $a^{Ⓓ}=a^{(b,b)}$. Then $ba{a}^{Ⓓ}=b$. In view of Corollary 3.3, $a^{Ⓓ}\in R_{b,b}^{\#⃝}$. According to Theorem 2.8, $b\in R^{\#⃝}$.

Set $u=ba+1-b{b}^{\#⃝}$. Since $a^{Ⓓ}ab=b=ba{a}^{Ⓓ}$ and $a^{Ⓓ}\in bR\bigcap Rb$, we check that $u^{-1}=a^{Ⓓ}{b}^{\#⃝}+1-b{b}^{\#⃝}.$ Write $a^{Ⓓ}=bx$ for some $x\in R$. Then $(1-b{b}^{\#⃝})a^{Ⓓ}=(1-b{b}^{\#⃝})bx=0,$ as required.

⟸ Set $u=ba+1-b{b}^{\#⃝}$. Then we verify that

$$\begin{array}{ccc}\hfill u^{-1}b& =\hfill & u^{-1}ba{u}^{-1}b,\hfill \\ \hfill u^{-1}bab& =\hfill & b,\hfill \\ \hfill ba{u}^{-1}b& =\hfill & b.\hfill \end{array}$$

This implies that $u^{-1}b=a^{(b,b)}$. It is easy to check that $u{a}^{Ⓓ}=[ba+1-b{b}^{\#⃝}]a^{Ⓓ}=ba{a}^{Ⓓ}=b$, and then $a^{Ⓓ}=u^{-1}b=a^{(b,b)}$, as desired. □

Theorem 9. 

Let $a,b\in R$. Then $a\in R_{b,b}^{Ⓓ}$ if and only if

(1)

$a\in R^{Ⓓ},b\in R^{\#⃝}$;

(2)

$ba+b^{\pi }\in R^{-1},b^{\pi }{a}^D=a^{\pi }{b}^{\ast }=0$.

Proof. 

⟹ By Theorem 2.8, $\left(1\right)$ holds. Moreover, $ba+1-b{b}^{\#⃝}\in R^{-1},ba{a}^{Ⓓ}=b$ and $b{b}^{\#⃝}{a}^{Ⓓ}=a^{Ⓓ}$. Since $1+b(a-b^{\#⃝})\in R^{-1}$, by using Jacobson’s Lemma (see [6]), $1+(a-b^{\#⃝})b\in R^{-1}$. Hence $1+(a-b^{\#⃝})b\in R^{-1}$. By using Jacobson’s Lemma again, $1+b(a-b^{\#⃝})\in R^{-1}$. Thus $ba+b^{\pi }\in R^{-1}$. Clearly, $(1-b{b}^{\#⃝})a^{Ⓓ}=0$. Since $a^{Ⓓ}{a}^m{\left(a^m\right)}^D=a^D$ for some $m\in N$, we have $(1-b{b}^{\#⃝})a^D=0$. Thus, $b{b}^{\#⃝}{a}^D=b{b}^{\#⃝}\left[b{b}^{\#⃝}{a}^D\right]=\left[\left(b{b}^{\#⃝}b\right)b^{\#⃝}\right]a^D=b{b}^{\#⃝}{a}^D=a^D$. Hence $b^{\pi }{a}^D=(1-b{b}^{\#⃝})a^D=0$. Since $ba{a}^{Ⓓ}=b$, we have $a{a}^{Ⓓ}{b}^{\ast }=b^{\ast }$. Set $n=ind\left(a\right)$. Then $a^D{a}^{n+1}=a^n$. Then

$$a{a}^D{a}^{Ⓓ}=a^D{a}^{n+1}{\left(a^{Ⓓ}\right)}^{n+1}=a^n{\left(a^{Ⓓ}\right)}^{n+1}=a^{Ⓓ}.$$

Hence $a{a}^D{b}^{\ast }=\left[a{a}^{D}a{a}^{Ⓓ}\right]b^{\ast }=a{a}^{Ⓓ}{b}^{\ast }=b^{\ast }$. Therefore $a^{\pi }{b}^{\ast }=(1-a{a}^D)b^{\ast }=0$.

⟸ By hypothesis, $1+b(a-b^{\#⃝})=ba+b^{\pi }\in R^{-1}$. Then $1+(a-b^{\#⃝})b\in R^{-1}$. Hence $1+(a-b^{\#⃝})b\in R^{-1}$. This implies that $ba+1-b{b}^{\#⃝}=1+b(a-b^{\#⃝})\in R^{-1}$. Since $b^{\pi }{a}^D=a^{\pi }{b}^{\ast }=0$, by the argument above, we have $(1-b{b}^{\#⃝})a^D=0$ and $(1-a{a}^{Ⓓ})b^{\ast }=0$. Then there exists $m\in N$ such that $(1-b{b}^{\#⃝})a^{Ⓓ}=(1-b{b}^{\#⃝})a^D{a}^m{\left(a^m\right)}^{(1,3)}=0$. This implies that $b{b}^{\#⃝}{a}^{Ⓓ}=a^{Ⓓ}$. Moreover, we have $b(1-a{a}^{Ⓓ})=0$, and so $ba{a}^{Ⓓ}=b$. This completes the proof by Lemma 3.8. □

Corollary 12. 

Let $a\in R_{b,b}^{Ⓓ}$. Then there exists a projection $p\in R$ such that $ba+p\in R^{-1}$ and $pb=p{a}^D=0$.

Proof. 

By virtue of Theorem 3.9, $ba+1-b{b}^{\#⃝}\in R^{-1}$. Set $p=1-b{b}^{\#⃝}$. Then $pb=0$ and $p^2=p^{\ast }=p\in R$. By virtue of Theorem 3.9, $b^{\pi }{a}^D=0$. Therefore we have $p{a}^D=(1-b{b}^{\#⃝})a^D=a^D-b{b}^{\#⃝}\left(b{b}^{\#⃝}{a}^D\right)=a^D-\left(b{b}^{\#⃝}b\right)b^{\#⃝}{a}^D=a^D-b{b}^{\#⃝}{a}^D=0$, as desired. □

4. Representations of Core-EP-$(b,c)$-Inverse

We next establish representations of core-EP-$(b,c)$-inverses by using other generalized inverses. We come now to the demonstration for which this section has been developed.

Theorem 10. 

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

(1)

$a\in R_{b,c}^{Ⓓ}.$

(2)

There exists $m\in N$ such that $a^k\in R_{b,c}^{\#⃝}$ for any $k\ge m$.

(3)

$a^m\in R_{b,c}^{\#⃝}$ for some $m\in N$.

In this case, $a_{b,c}^{Ⓓ}=a^{m-1}{\left(a^m\right)}_{b,c}^{\#⃝}$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ As in the proof of Theorem 3.1, $a\in R^{Ⓓ}$ and $a^{Ⓓ}={\left[a{a}^{Ⓓ}a\right]}_{b,c}^{\#⃝}$. Set $m=ind\left(a\right)$ and $k\ge m$. Then $a^k\in R^{\#⃝}$ and $a^{Ⓓ}=a^{k-1}{\left(a^k\right)}^{\#⃝}$. We verify that

$$\begin{array}{ccc}\hfill {\left(a^k\right)}^{\#⃝}{a}^k{\left(a^k\right)}^{\#⃝}& =\hfill & {\left(a^k\right)}^{\#⃝},\hfill \\ \hfill {\left(a^k\right)}^{\#⃝}& =\hfill & \left[a^{k-1}{\left(a^k\right)}^{\#⃝}\right]{\left(a^k\right)}^{2}a{\left[{\left(a^k\right)}^{\#⃝}\right]}^3\in bR,\hfill \\ \hfill {\left(a^k\right)}^{\#⃝}& =\hfill & {\left(a^k\right)}^{\#⃝}a\left[a^{k-1}{\left(a^k\right)}^{\#⃝}\right]\in Rc,\hfill \\ \hfill {\left(a^k\right)}^{\#⃝}{a}^{k}b& =\hfill & {\left(a^k\right)}^{\#⃝}{a}^k{a}^{Ⓓ}ab\hfill \\ & =\hfill & a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]b=b,\hfill \\ \hfill c{a}^k{\left(a^k\right)}^{\#⃝}& =\hfill & ca{a}^{Ⓓ}{a}^k{\left(a^k\right)}^{\#⃝}\hfill \\ & =\hfill & c\left[a{a}^{Ⓓ}a\right]a^{Ⓓ}=c.\hfill \end{array}$$

Thus ${\left(a^k\right)}^{\#⃝}={\left(a^k\right)}^{(b,b)}$. Therefore $a^k\in R_{b,c}^{\#⃝}$, as required.

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

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, $a^m\in R_{b,c}^{\#⃝}$ for some $m\in \mathbb{N}$. Then $a^m\in R^{\#⃝}$ and ${\left(a^m\right)}^{\#⃝}={\left(a^m\right)}^{(b,b)}$. Hence, we have

$${\left(a^m\right)}^{\#⃝}\in bR{\left(a^m\right)}^{\#⃝}\bigcap {\left(a^m\right)}^{\#⃝}Rc,{\left(a^m\right)}^{\#⃝}{a}^{m}b=b,c{a}^m{\left(a^m\right)}^{\#⃝}=c.$$

In light of [12], Theorem 2.5, $a\in R^{Ⓓ}$ and $a^{Ⓓ}=a^{m-1}{\left(a^m\right)}^{\#⃝}.$

Claim 1. $a^{m-1}{\left(a^m\right)}^{\#⃝}\in bR{a}^{m-1}{\left(a^m\right)}^{\#⃝}\bigcap a^{m-1}{\left(a^m\right)}^{\#⃝}Rc.$

We check that

$$\begin{array}{cc}& a^{m-1}{\left(a^m\right)}^{\#⃝}\hfill \\ \hfill =& a^{m-1}\left[{\left(a^m\right)}^{\#⃝}{a}^m{\left(a^m\right)}^{\#⃝}\right]\hfill \\ \hfill =& a^{m-1}{a}^m\left[{\left({\left(a^m\right)}^{\#⃝}\right)}^{2}a\right]\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]\hfill \\ \hfill =& a^m\left[a^{m-1}{\left({\left(a^m\right)}^{\#⃝}\right)}^{2}a\right]\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]\hfill \\ \hfill =& {\left(a^m\right)}^{\#⃝}{\left(a^m\right)}^2\left[a^{m-1}{\left({\left(a^m\right)}^{\#⃝}\right)}^{2}a\right]\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]\hfill \\ \hfill \in & bR\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right].\hfill \end{array}$$

Likewise, we claim that $a^{m-1}{\left(a^m\right)}^{\#⃝}\in \left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]Rc.$

Claim 2. $\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]ab=b$.

$$\begin{array}{cc}& \left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]ab\hfill \\ \hfill =& \left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]a\left[{\left(a^m\right)}^{\#⃝}{a}^{m}b\right]\hfill \\ \hfill =& a^{m-1}\left[{\left(a^m\right)}^{\#⃝}{\left(a^m\right)}^2\right]a{\left[{\left(a^m\right)}^{\#⃝}\right]}^3{a}^{m}b\hfill \\ \hfill =& a^{m-1}{a}^{m+1}{\left[{\left(a^m\right)}^{\#⃝}\right]}^3{a}^{m}b\hfill \\ \hfill =& {\left(a^m\right)}^2{\left[{\left(a^m\right)}^{\#⃝}\right]}^3{a}^{m}b\hfill \\ \hfill =& {\left(a^m\right)}^{\#⃝}{a}^{m}b=b.\hfill \end{array}$$

Claim 3. $ca\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]=c$. We verify that $ca\left[a^{m-1}{\left(a^m\right)}^{\#⃝}\right]=c{a}^m{\left(a^m\right)}^{\#⃝}=c$.

Therefore we have $a^{Ⓓ}\left[a{a}^{Ⓓ}a\right]b=a^{Ⓓ}ab=b$ and $c\left[a{a}^{Ⓓ}a\right]a^{Ⓓ}=ca{a}^{Ⓓ}=c$. Therefore $a^{Ⓓ}={\left(a{a}^{Ⓓ}a\right)}^{(b,b)}$. As in the proof, we verify that $a^{Ⓓ}=a^{(b,b)}$, thus yielding the result. □

Corollary 13. 

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

(1)

$a\in R_{b,b}^{Ⓓ}$.

(2)

$a^{m}b\in R$ is EP, $R{a}^{m}b=Rb,{\left(b{a}^m\right)}^{\pi }{a}^m=0$ and ${\left(a^{m}b\right)}^{\pi }b=0$ for some $m\in N$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 4.1, $a^m\in R_{b,b}^{\#⃝}$. In light of Corollary 2.12, $a^{m}b\in RisEP,R{a}^{m}b=Rb,{\left(b{a}^m\right)}^{\pi }{a}^m=0$ and ${\left(a^{m}b\right)}^{\pi }b=0$ for some $m\in N$.

$\left(2\right)\Rightarrow \left(1\right)$ In view of Corollary 2.12, $a^m\in R_{b,b}^{\#⃝}$. Therefore we complete the proof by Theorem 4.1. □

Theorem 11. 

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

(1)

$a\in R_{b,c}^{Ⓓ}$.

(2)

$a\in R^D$ and $a^D\in R_{b,c}^{\#⃝}$.

In this case, $a_{b,c}^{Ⓓ}={\left(a^D\right)}^2{\left(a^D\right)}_{b,c}^{\#⃝}.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 4.1, $a^m\in R^{\#}$, and then $a\in R^D$. Set $m=ind\left(a\right)$. By using Theorem 4.1 again, $a^m\in R_{b,c}^{\#⃝}$ and $a_{b,b}^{Ⓓ}=a^{m-1}{\left(a^m\right)}_{b,c}^{\#⃝}$. Since $a^{m+1}=a^D{a}^m$, we verify that

$$\begin{array}{ccc}\hfill a^D\left(a^{m+1}{\left(a^m\right)}_{b,c}^{\#⃝}\right)& =\hfill & a^m{\left(a^m\right)}_{b,c}^{\#⃝},\hfill \\ \hfill a^D{\left(a^{m+1}{\left(a^m\right)}_{b,c}^{\#⃝}\right)}^2& =\hfill & a^{m+1}{\left(a^m\right)}_{b,c}^{\#⃝},\hfill \\ \hfill a^{m+1}{\left(a^m\right)}_{b,c}^{\#⃝}{\left(a^D\right)}^2& =\hfill & a^D.\hfill \end{array}$$

Thus ${\left(a^D\right)}^{\#⃝}=a^{m+1}{\left(a^m\right)}_{b,c}^{\#⃝}.$ By hypothesis, we have

$${\left(a^m\right)}^{\#⃝}{a}^{m}b=b=b{a}^m{\left(a^m\right)}^{\#⃝},$$

and therefore ${\left(a^D\right)}^{\#⃝}{a}^{D}b=b=b{a}^D{\left(a^D\right)}^{\#⃝}.$ Accordingly, $a_{b,b}^{Ⓓ}={\left(a^D\right)}^2\left(a^{m+1}{\left(a^m\right)}^{{\#⃝}_{(b,b)}}\right)={\left(a^D\right)}^2{\left(a^D\right)}^{{\#⃝}_{(b,b)}}.$

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

$$\begin{array}{ccc}\hfill a^m{\left(a^D\right)}^{m+1}{\left(a^D\right)}^{\#⃝}& =\hfill & a^D{\left(a^D\right)}^{\#⃝},\hfill \\ \hfill a^m{\left({\left(a^D\right)}^{m+1}{\left(a^D\right)}^{\#⃝}\right)}^2& =\hfill & {\left(a^D\right)}^{m+1}{\left(a^D\right)}^{\#⃝},\hfill \\ \hfill \left({\left(a^D\right)}^{m+1}{\left(a^D\right)}^{\#⃝}\right){\left(a^m\right)}^2& =\hfill & a^m.\hfill \end{array}$$

Therefore ${\left(a^m\right)}^{\#⃝}={\left(a^D\right)}^{m+1}{\left(a^D\right)}^{\#⃝}$, and thus yielding the result. □

By using Theorem 4.3, we prove that $a\in R$ has core-EP inverse if and only if a has Drazin inverse and it has $(b,c)$-inverse, where $(b,c)=(a^D,{\left(a^D\right)}^{\ast })$.

Corollary 14. 

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

(1)

$a\in R_{b,b}^{Ⓓ}$.

(2)

$a\in R^{Ⓓ},b\in R^{\#⃝}$ and $a{a}^{Ⓓ}=b^{Ⓓ}b.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 4.3, $a\in R^D$ and $a^D\in R_{b,c}^{\#⃝}$. By virtue of Theorem 2.6, $a^D,b\in R^{\#⃝}$ and $\left(a^D\right){\left(a^D\right)}^{\#⃝}=b^{\#⃝}b$. In view of Theorem 4.3, $a\in R^{Ⓓ}$ and $a^{Ⓓ}={\left(a^D\right)}^2{\left(a^D\right)}^{\#⃝}$. Therefore $a{a}^{Ⓓ}=\left[a{\left(a^D\right)}^2\right]{\left(a^D\right)}^{\#⃝}=a^D{\left(a^D\right)}^{\#⃝}=b^{\#⃝}b$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By virtue of [12], Theorem 2.3 and [3], Theorem 4.1, $a\in R^D$ and $a^{Ⓓ}={\left(a^D\right)}^2{\left(a^D\right)}^{\#⃝}$. Then $a^D{\left(a^D\right)}^{\#⃝}=a{a}^{Ⓓ}=b^{\#⃝}b$. In light of Theorem 2.6, $a^D\in R_{b,b}^{\#⃝}$. According to Theorem 4.3, $a\in R_{b,b}^{Ⓓ}$. □

Corollary 15. 

Let $a\in R_{b,b}^{Ⓓ}$. Then $1-a{a}^{Ⓓ}=b^{\pi },1-a^{Ⓓ}a={\left(ba\right)}^{\pi }$.

Proof. 

In light of Corollary 4.4, $a{a}^{Ⓓ}=b^{\#⃝}b.$ Hence, $1-a{a}^{Ⓓ}=1-b^{\#⃝}b=1-\left[b^{\#}b{b}^{(1,3)}\right]b=1-b{b}^{\#}=b^{\pi }$.

By hypothesis, we have $a^{Ⓓ}=a^{(b,b)}$. By virtue of [14], Theorem 2.13, $a^{(b,b)}={\left(ba\right)}^{\#}b$. Hence $a^{Ⓓ}={\left(ba\right)}^{\#}b$, and then $a^{Ⓓ}a={\left(ba\right)}^{\#}ba$. Therefore $1-a^{Ⓓ}a=1-ba{\left(ba\right)}^{\#}={\left(ba\right)}^{\pi }$. □

Corollary 16. 

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

(1)

A has core-EP-$(B,C)$-inverse.

(2)

$rank\left(A^D\right)=rank\left(B\right)=rank\left(C\right)=rank\left(C^{\ast }\right)=rank\left(C{A}^{D}B\right)$.

In this case, $A^{Ⓓ}={\left(A^D\right)}^{2}B{\left(C{A}^{D}B\right)}^{-}C.$

Proof. 

This is obvious by Corollary 2.4 and Theorem 4.3. □

Theorem 12. 

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

(1)

$a\in R_{b,b}^{Ⓓ}$.

(2)

$a\in R^{Ⓓ},b,ba\in R^{\#⃝},a^{D}R=bR=b^{\ast }R$.

In this case, ${\left(ba\right)}^{Ⓓ}=a^{Ⓓ}{b}^{Ⓓ}.$

Proof. $\left(1\right)\Rightarrow \left(2\right)$ Since $a\in R_{b,b}^{Ⓓ}$, by virtue of Corollary 4.4, $a,b\in R^{Ⓓ}$. In view of Theorem 4.3, $a^D\in R_{b,b}^{\#⃝}$. By virtue of Theorem 2.6, $a^{D}R=bR=b^{\ast }R$. According to [14], Theorem 2.13, we have

$$\begin{array}{ccc}\hfill a^{Ⓓ}{b}^{Ⓓ}& =\hfill & \left[{\left(a^D\right)}^2{\left(a^D\right)}^{\#⃝}\right]b^{\#⃝}\hfill \\ & =\hfill & {\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}b{b}^{\#⃝}\hfill \\ & =\hfill & {\left(a^D\right)}^2\left[{\left(a^D\right)}^{\#⃝}{b}^{\#⃝}\right]b{b}^{\#⃝}\hfill \\ & =\hfill & {\left(a^D\right)}^2[{\left(a^D\right)}^{\#⃝}\left[b^{\#⃝}b{b}^{\#⃝}\right]\hfill \\ & =\hfill & {\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}.\hfill \end{array}$$

Then $ba\left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]=\left(b{a}^D\right){\left(b{a}^D\right)}^{\#⃝}$, and so ${\left(ba\left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]\right)}^{\ast }$$=\left(b{a}^D\right){\left(b{a}^D\right)}^{\#⃝}=ba\left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]$. By using [14], Theorem 2.13, we check that

$$\begin{array}{ccc}\hfill ba{\left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]}^2& =\hfill & \left(b{a}^D\right){\left(b{a}^D\right)}^{\#⃝}\left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]\hfill \\ & =\hfill & \left(b{a}^D\right){\left(b{a}^D\right)}^{\#⃝}\left[{\left(a^D\right)}^{\#}\right]{\left(a^D\right)}^3{\left(b{a}^D\right)}^{\#⃝}]\hfill \\ & =\hfill & \left(b{a}^D\right){\left(b{a}^D\right)}^{\#⃝}\left[{\left(b{a}^D\right)}^{\#⃝}b\right]{\left(a^D\right)}^3{\left(b{a}^D\right)}^{\#⃝}]\hfill \\ & =\hfill & \left[{\left(b{a}^D\right)}^{\#⃝}b\right]{\left(a^D\right)}^3{\left(b{a}^D\right)}^{\#⃝}]\hfill \\ & =\hfill & {\left[a^D\right)}^{\#}\left]{\left(a^D\right)}^3{\left(b{a}^D\right)}^{\#⃝}\right]\hfill \\ & =\hfill & {\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}.\hfill \end{array}$$

Obviously, ${\left(a^D\right)}^{\#}{a}^{D}b=b$, and so $a{a}^{D}b=b$. This implies that $a{a}^{Ⓓ}b=a{a}^{Ⓓ}a{a}^{D}ba{a}^{D}b=b$. In view of Theorem 2.6, we have $a^D{\left(a^D\right)}^{\#⃝}=b^{\#⃝}b$, and so

$$\begin{array}{ccc}\hfill \left[{\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}\right]{\left(ba\right)}^2& =\hfill & \left[{\left(a^D\right)}^2{\left(a^D\right)}^{\#⃝}{b}^{\#⃝}\right]{\left(ba\right)}^2\hfill \\ & =\hfill & {\left(a^D\right)}^2{\left(a^D\right)}^{\#⃝}\left[a^D{\left(b{a}^D\right)}^{\#⃝}\right]{\left(ba\right)}^2\hfill \\ & =\hfill & a^D\left[b^{\#⃝}b\right]aba\hfill \\ & =\hfill & a^D\left[a^D{\left(a^D\right)}^{\#⃝}\right]aba\hfill \\ & =\hfill & a^D\left[a^D{\left(a^D\right)}^{\#⃝}\right]a\left[{\left(a^D\right)}^{\#}{a}^{d}b\right]a\hfill \\ & =\hfill & a{a}^{d}ba=a{a}^D\left[{\left(a^D\right)}^{\#}{a}^{d}b\right]a=ba.\hfill \end{array}$$

This implies that ${\left(ba\right)}^{\#⃝}={\left(a^D\right)}^2{\left(b{a}^D\right)}^{\#⃝}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Since $(1-a{a}^{Ⓓ})a^D=0$, we have $(1-a{a}^{Ⓓ})b^{\ast }=0$. Hence $b(1-a{a}^{Ⓓ})=0$. This implies that $b^{\#⃝}b=b^{\#⃝}ba{a}^{Ⓓ}=b^{\#}ba{a}^{Ⓓ}.$ On the other hand, $(1-b^{\#}b)b=0$, and then $(1-b^{\#}b)a^D=0$. This implies that $(1-b^{\#}b)a{a}^{Ⓓ}=0$. Hence $a{a}^{Ⓓ}=b^{\#}ba{a}^{Ⓓ}$, and so $a{a}^{Ⓓ}=b^{\#⃝}b.$ According to Corollary 4.4, $a\in R_{b,b}^{Ⓓ}$. □

Corollary 17. 

Let $a\in R^{Ⓓ},b\in R^{\#⃝}$. If $ab=ba,a^{\ast }b=b{a}^{\ast }$ and $a^{D}R=bR$, then $a\in R_{b,b}^{Ⓓ}$.

Proof. 

Since $a\in R^{Ⓓ}$, we see that ${\left(a^D\right)}^{\#⃝}=a^2{a}^{Ⓓ}$. As $ab=ba$ and $a^{\ast }b=b{a}^{\ast }$, by using [9], Theorem 2.2, we have $a^{D}b=b{a}^D$ and $a^D{b}^{\ast }=b^{\ast }{a}^D$. Hence $b{a}^D\in R^{\#⃝}$ and ${\left(b{a}^D\right)}^{\#⃝}=b^{\#⃝}{\left(a^D\right)}^{\#⃝}$. Since $a^{D}R=bR$, we get ${\left(a^D\right)}^{D}R=a^2{a}^{D}R=a^{D}R=bR$.

Since $(1-a{a}^{Ⓓ})a^D=0$, we have $(1-a{a}^{Ⓓ})b=0$; whence, $b=a{a}^{Ⓓ}b=ba{a}^{Ⓓ}$. This implies that $b^{\ast }=a{a}^{Ⓓ}{b}^{\ast }$, and so $b^{\ast }=a^D{a}^2{a}^{Ⓓ}{b}^{\ast }$. Then $b^{\ast }R\subseteq a^{D}R$. On the other hand, we have $(1-b{b}^{\#})a{a}^{Ⓓ}=0$, and then $a{a}^{Ⓓ}=b{b}^{\#}a{a}^{Ⓓ}=a{a}^{Ⓓ}{b}^{\#}b$. This implies that $a{a}^{Ⓓ}={\left(a{a}^{Ⓓ}\right)}^{\ast }={\left(a{a}^{Ⓓ}{b}^{\#}b\right)}^{\ast }$. We infer that $a^{D}R\subseteq a{a}^{Ⓓ}R\subseteq b^{\ast }R$, and then $a^{D}R=b^{\ast }R$. According to Theorem 4.7, $a\in R_{b,b}^{Ⓓ}$. □

Theorem 13. 

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

(1)

$a\in R_{b,b}^{Ⓓ}$.

(2)

$ba\in R^{\#⃝},b\in \left(bab\right)R\left(bab\right)$ and $a^{Ⓓ}=b{\left(bab\right)}^{-}b$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 4.7, $a\in R^{Ⓓ},b,ba\in R^{\#⃝}$. Since $a^{Ⓓ}=a^{(b,b)}$, we have $a^{Ⓓ}ab=b=ba{a}^{Ⓓ}$. Thus, $b=\left(ba\right)a^{Ⓓ}={\left(ba\right)}^2{\left[{\left(ba\right)}^{\#⃝}\right]}^2{a}^{Ⓓ}\in babR$. Moreover, we have

$$\begin{array}{ccc}\hfill b& =\hfill & b^{\#⃝}{b}^2=b^{\#⃝}b\left(ba\right)a^{Ⓓ}\hfill \\ & =\hfill & b^{\#⃝}{\left(ba\right)}^{\#⃝}\left(ba\right)\left(\left(ba\right)a^{Ⓓ}\right)\hfill \\ & =\hfill & b^{\#⃝}{\left(ba\right)}^{\#⃝}\left(bab\right)\in Rbab.\hfill \end{array}$$

Write $a^{Ⓓ}=bx=yb$ for some $x,y\in R$. Then we verify that

$$\begin{array}{ccc}\hfill bab& =\hfill & ba{a}^{Ⓓ}ab=babxab=babxa{a}^{Ⓓ}ab\hfill \\ & =\hfill & babxa\left(yb\right)ab=\left(bab\right)\left(xay\right)\left(bab\right),\hfill \end{array}$$

and so $bab\in R^{-}$. Write $b=\left(bab\right)s=t\left(bab\right)$ for some $s,t\in R$. We verify that

$$\begin{array}{ccc}\hfill \left[b{\left(bab\right)}^{-}b\right]a\left[b{\left(bab\right)}^{-}b\right]& =\hfill & b{\left(bab\right)}^{-}\left(bab\right){\left(bab\right)}^{-}b\hfill \\ & =\hfill & t\left[\left(bab\right){\left(bab\right)}^{-}\left(bab\right)\right]{\left(bab\right)}^{-}b\hfill \\ & =\hfill & t\left(bab\right){\left(bab\right)}^{-}b=b{\left(bab\right)}^{-}b,\hfill \\ \hfill \left(b{\left(bab\right)}^{-}b\right)ab& =\hfill & t\left(\left(bab\right){\left(bab\right)}^{-}\left(bab\right)\right)=tbab=b,\hfill \\ \hfill bab{\left(bab\right)}^{-}b& =\hfill & \left(\left(bab\right){\left(bab\right)}^{-}\left(bab\right)\right)s=\left(bab\right)s=b.\hfill \end{array}$$

Therefore $b{\left(bab\right)}^{-}b=a^{(b,b)}$, and then $a^{Ⓓ}=b{\left(bab\right)}^{-}b$.

$\left(2\right)\Rightarrow \left(1\right)$ By the argument above, we have $b{\left(bab\right)}^{-}b=a^{(b,b)}$. Accordingly, $a^{Ⓓ}=b{\left(bab\right)}^{-}b=a^{(b,b)}$, as asserted. □

Corollary 18. 

Let $A,B\in {\mathbb{C}}^{n\times n}$. Then A has core-EP-$(B,B)$-inverse if and only if

(1)

$BA$ has group inverse;

(2)

$rank\left(B\right)=rank\left(BAB\right)$ and $A^{Ⓓ}=B{\left(BAB\right)}^{-}B$.

Proof. 

Since every complex matrix has Moore-Penrose inverse, it follows by [25], Remark 2.16 that $BA$ has group inverse if and only if it has core inverse. Therefore we complete the proof by Theorem 4.9. □