Generalized EP elements in Banach *-algebras

1. Introduction

Let $\mathcal{A}$ be a Banach algebra with involution *. An element a in $\mathcal{A}$ has core inverse if there exists some $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=ax,a=x{a}^2.$ Such x is unique if it exists and is denoted by $a^{\#}.$ The core inverse was extensively considered in the context of Banach algebras, e.g., [1,6,17,18,19,26].

An element a in $\mathcal{A}$ is EP (i.e., an EP element) if there exists some $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=xa,a=x{a}^2.$ Evidently, $a\in \mathcal{A}$ is EP if and only if there exists $x\in \mathcal{A}$ such that $a^{2}x=a,ax=xa,{\left(ax\right)}^{*}=ax$ if and only if there exists $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(xa\right)}^{*}=xa,x{a}^2=a$ if and only if $a\in {\mathcal{A}}^{\#}$ and ${\left(a{a}^{\#}\right)}^{*}=a{a}^{\#}$ ([2,16,23,24,25,27]). Here, $a\in \mathcal{A}$ has group inverse provided that there exists $x\in \mathcal{A}$ such that $a{x}^2=x,ax=xa,a=x{a}^2.$ Such x is unique if exists, denoted by $a^{\#}$, and called the group inverse of a.

An element a in a Banach *-algebra $\mathcal{A}$ is *-DMP (i.e., *-DMP element) if there exist $m\in N$ and $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=xa=ax,a^m=x{a}^{m+1}.$ As is well known, $a\in \mathcal{A}$ is *-DMP if and only if $a^m\in \mathcal{A}$ is EP for some $m\in N$ (see [6,11,13]). In [21], Mosic and Djordjevic introduced and studied the gDMP inverse for a Hilbert space operator using its generalized Drazin inverse and its Moore-Penrose inverse.

The motivation of this paper is to introduce and study a new kind of generalized inverse as a natural generalization of EP and *-DMP elements mentioned above. Let ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{\frac{1}{n}}=0\}.$ As is well known, $x\in {\mathcal{A}}^{qnil}$ if and only if $1+\lambda x\in \mathcal{A}$ is invertible for any $\lambda \in C$.

Definition 1.1. 

An element $a\in \mathcal{A}$ is generalized EP (i.e., generalized EP element) if there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in \mathcal{A}isEP,y\in {\mathcal{A}}^{qnil}.$$

Recall that $a\in \mathcal{A}$ has g-Drazin inverse (i.e., generalized Drazin inverse) if there exists $x\in \mathcal{A}$ such that $a{x}^2=x,ax=xa,a-a^{2}x\in {\mathcal{A}}^{qnil}.$ Such x is unique, if exists, and denote it by $a^d$. AS it is well known, a has g-Drazin inverse if and only if a has quasi-polar property, i.e., there exists an idempotent $p\in \mathcal{A}$ such that $a+p\in {\mathcal{A}}^{-1}$ and $ap\in {\mathcal{A}}^{qnil}$ (see [3]). In Section 2, we investigate polar-like characterizations of generalized EP elements. We prove that $a\in \mathcal{A}$ is generalized EP if and only if there exists a projection $p\in \mathcal{A}$ (i.e., $p^2=p=p^{*}$)such that $a+p\in {\mathcal{A}}^{-1},ap=pa$ and $ap\in {\mathcal{A}}^{qnil}$.

In Section 3, we are concerned with additive properties of generalized EP elements. The necessary and sufficient conditions under which the sum of two generalized EP elements is a generalized EP element are investigated by using orthogonal and commuting perturbations.

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

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

The preceding x is unique if it exists, and denoted by $a^d$. Let $a,b\in \mathcal{A}$ have generalized core-EP inverses. Recall that $a{\le }^{d}b$ if $a{a}^d=b{a}^d$ and $a^{d}a=a^{d}b$. We refer the reader to [4] for properties of generalized core-EP inverses in a Banach *-algebra. Finally, in Section 4, the generalized core EP-orders for generalized EP elements in a Banach *-algebra are characterized. The properties of core-EP orders are thereby extended to wider cases.

Throughout the paper, all Banach *-algebras are complex with an identity. An element p in $\mathcal{A}$ is a projection provided that $p^2=p=p^{*}$. We use ${\mathcal{A}}^d$ and ${\mathcal{A}}^e$ to denote the sets of all generalized core-EP invertible and generalized EP elements in $\mathcal{A}$. The commutant of $a\in R$ is defined by $comm\left(a\right)=\{x\in R|xa=ax\}$. The double commutant of $a\in R$ is defined by $com{m}^2\left(a\right)=\{x\in R|xy=yxforally\in comm\left(a\right)\}$.

2. Polar-like Characterizations

In this section, we present a polar-like property for EP elements in a Banach *-algebra. The related characterize of EP elements are thereby derived. We begin with

Lemma 2.1. 

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

(1)

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

(2)

There exists $x\in comm\left(a\right)$ such that

$$a{x}^2=x,{\left(ax\right)}^{*}=ax,a-x{a}^2\in {\mathcal{A}}^{qnil}.$$

(3)

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

$$a{x}^2=x,{\left(ax\right)}^{*}=xa,a-x{a}^2\in {\mathcal{A}}^{qnil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that $a=z+y,z^{*}y=yz=0,z\in \mathcal{A}isEP,y\in {\mathcal{A}}^{qnil}.$ Set $x=z^{\#}$. Then $z{z}^{\#}=z^{\#}z$ by [27]. Hence $z^{\#}y=z^{\#}z{z}^{\#}y=z^{\#}{\left(z{z}^{\#}\right)}^{*}y=z^{\#}{\left(z^{\#}\right)}^{*}\left(z^{*}y\right)=0$. We check that $ax=(z+y)z^{\#}=z{z}^{\#},xa=z^{\#}(z+y)=z^{\#}z;a{x}^2=(z+y){\left(z^{\#}\right)}^2=z{\left(z^{\#}\right)}^2=z^{\#}=x,x{a}^2=z^{\#}{(z+y)}^2=z^{\#}{z}^2=z.$ Therefore ${\left(ax\right)}^{*}={\left(z{z}^{\#}\right)}^{*}=z{z}^{\#}=ax,ax=z{z}^{\#}=z^{\#}z=xa,x{a}^2-a=z-a=-y\in {\mathcal{A}}^{qnil}.$

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

$\left(3\right)\Rightarrow \left(1\right)$ By hypotheses, we have $z\in \mathcal{A}$ such that $a{z}^2=z,{\left(az\right)}^{*}=za,a-z{a}^2\in {\mathcal{A}}^{qnil}.$ Then ${\left(za\right)}^{*}={\left({\left(az\right)}^{*}\right)}^{*}=az$. In view of [25], $az=za$. Set $x=aza$ and $y=a-aza.$ We claim that x is EP. Evidently, we verify that $z{x}^2=za\left(z{a}^{2}z\right)a=z{a}^{2}za=aza=x,x{z}^2=aza{z}^2=a{z}^2=z,xz=azaz=zaza=zx,{\left(xz\right)}^{*}={\left(azaz\right)}^{*}={\left(az\right)}^{*}=az=\left(aza\right)z=xz.$ Therefore $x\in \mathcal{A}$ is EP.

By hypothesis, $a(1-za)\in {\mathcal{A}}^{qnil}$. By virtue of Cline’s formula (see [3]), $y=a-z{a}^2\in {\mathcal{A}}^{qnil}$. Moreover, we see that $x^{*}y={\left(aza\right)}^{*}(1-az)a=a^{*}{\left(az\right)}^{*}(1-az)a=a^{*}\left(az\right)(1-az)a=0,yx=(a-aza)aza=a(a-z{a}^2)za=0.$ This completes the proof. □

We are ready to prove:

Theorem 2.2. 

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

(1)

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

(2)

There exists a projection $p\in comm\left(a\right)$ such that $a+p\in {\mathcal{A}}^{-1}$ and $ap\in {\mathcal{A}}^{qnil}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Lemma 2.1, there exists $x\in comm\left(a\right)$ such that $a{x}^2=x,{\left(ax\right)}^{*}=ax,a-x{a}^2\in {\mathcal{A}}^{qnil}.$ Let $p=1-ax$. Then $p=p^2\in comm\left(a\right)$ and $a+p=a+1-ax$. We check that $(a+1-ax)(x+1-ax)=ax+a(1-ax)+(1-ax)x+(1-ax)=1$. Likewise, $(x+1-ax)(a+1-ax)=1$. Then $a+p\in {\mathcal{A}}^{-1}$. Additionally, $ap=a-a^{2}x\in {\mathcal{A}}^{qnil}$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists an idempotent $p\in comm\left(a\right)$ such that $a+p\in {\mathcal{A}}^{-1}$ and $ap\in {\mathcal{A}}^{qnil}$. Set $x=(1-p){(a+p)}^{-1}$. Then $x\in comm\left(a\right)$ and $a{x}^2=a(1-p){(a+p)}^{-2}=(1-p)(a+p){(a+p)}^{-2}=(1-p){(a+p)}^{-1}=x$. Moreover, we verify that $a-a^{2}x=a-a^2(1-p){(a+p)}^{-1}=a-{(a+p)}^2(1-p){(a-p)}^{-1}=a-(a+p)(1-p)=ap\in {\mathcal{A}}^{qnil},$ as desired. □

Corollary 2.3. 

Every generalized EP element in a Banach *-algebra is the sum of three invertible elements.

Proof. 

Let $a\in {\mathcal{A}}^e$. In view of Theorem 2.2, we have $p^2=p=p^{*}\in \mathcal{A}$ such that $u:=a+p\in {\mathcal{A}}^{-1}$. Then $a=u-p$. Obviously, $-p=\frac{1-2p}{2}-\frac{1}{2}$. It is easy to verify that

$${\left(\frac{1-2p}{2}\right)}^2=\frac{1}{4},$$

and so

$${\left(\frac{1-2p}{2}\right)}^{-1}=2(1-2p).$$

Therefore $a=u+\frac{1-2p}{2}-\frac{1}{2}$, as desired. □

Theorem 2.4. 

Every Hermitan periodic element (i.e., $a^{*}=a$ and $a^k=a^k$ for some distinct $k,l\in N$) in a Banach *-algebra is generalized EP.

Proof. 

Assume that $a^{*}=a$ and $a^k=a^k$ for some positive integers $k,l(k>l)$. Then $a^l=a^k=a^{(k-l)+l}=\cdots =a^{l(k-l)+l}$, and so $a^l={\left(a^l\right)}^{k-l+1}$. Choose $m=l(k-l)$. Then $a-a^{m+1}\in {\mathcal{A}}^{nil}$. According to the Dirichlet Theorem, there exists a prime k such that $k=sm+1$ for some $s\in N$. One easily checks that

$$\begin{array}{c}a-a{a}^m\in {\mathcal{A}}^{nil},\\ a{a}^m-a{a}^{2m}=(a-a^{m+1})a^m\in {\mathcal{A}}^{nil},\\ a{a}^{2m}-a{a}^{3m}=(a-a^{m+1})a^{2m}\in {\mathcal{A}}^{nil},\\ ⋮\\ a{a}^{(s-1)m}-a{a}^{sm}=(a-a^{m+1})a^{(s-1)m}\in {\mathcal{A}}^{nil}.\end{array}$$

Therefore $a-a^k=a-a^{sm+1}={\displaystyle \sum _{i=0}^{s-1}}(a{a}^{im}-a{a}^{(i+1)m}\in {\mathcal{A}}^{nil}$. Set $n=k-1$. Then $a-a^{n+1}\in {\mathcal{A}}^{nil}$.

In view of [9, Theorem 3.5], there exists some $e^{n+1}=e\in Z[a,\frac{1}{n}]$ such that

$$w:=a-e\in Z\left[\frac{1}{n}\right](a^{n+1}-a)\subseteq {\mathcal{A}}^{nil}.$$

Thus, $ea=ae$. Let $q=e^{n-1}$. Then $q^2=e^{2n-1}=e^{n+1}{e}^{n-2}=e^{n-1}=q\in comm\left(a\right)$. Since $a=a^{*}$, we see that $q=q^{*}$. Since $a-e\in {\mathcal{A}}^{nil}$ and $ea=ae$, we see that $a^{n-1}-q\in {\mathcal{A}}^{nil}$. Set $p=1-q$. Then $p^2=p^{*}=p\in comm\left(a\right)$. One easily checks that

$$a^{n-1}+p^{n-1}=a^{n-1}+p=(a^{n-1}-q)+1\in {\mathcal{A}}^{-1}.$$

Since $n-1=k$ is prime, we see that $a+p\in {\mathcal{A}}^{-1}$. Moreover, we have that

$$\begin{array}{ccc}\hfill a^{n-1}p& =\hfill & a^{n-1}(1-q)=[a^{n-1}-q](1-q)\hfill \\ & \in \hfill & {\mathcal{A}}^{nil}.\hfill \end{array}$$

Hence ${\left(ap\right)}^{n-1}\in {\mathcal{A}}^{nil}$, and so $ap\in {\mathcal{A}}^{nil}$. Therefore $a\in {\mathcal{A}}^e$ by Theorem 2.2. □

We are ready to prove:

Theorem 2.5. 

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

(1)

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

(2)

There exists $x\in comm\left(a\right)$ such that

$${\left(ax\right)}^{*}=ax,a-a^{2}x\in {\mathcal{A}}^{qnil}.$$

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

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists some $x\in comm\left(a\right)$ such that ${\left(ax\right)}^{*}=ax,a-a^{2}x\in {\mathcal{A}}^{qnil}$. Set $s=xax$. Then $z\in comm\left(a\right)$. We check that

$$\begin{array}{ccc}a-a^{2}s\hfill & =\hfill & a-axaxa\hfill \\ & =\hfill & (1+ax)(a-a^{2}x)\hfill \\ & \in \hfill & {\mathcal{A}}^{qnil},\hfill \\ s-s^{2}a\hfill & =\hfill & xax-xaxaxax\hfill \\ & =\hfill & x(a-a^{2}x)x+xax(a-a^{2}x)x\hfill \\ & \in \hfill & {\mathcal{A}}^{qnil}.\hfill \end{array}$$

$$as-{\left(as\right)}^2=(a-a^{2}s)s\in {\mathcal{A}}^{qnil}.$$

Let

$$q=as-{\left(as\right)}^2,z=-\frac{1}{2}{\displaystyle \sum _{k=1}^{\infty }}\left(\begin{array}{c}\frac{1}{2}\\ k\end{array}\right){\left(4q{(4q-1)}^{-1}\right)}^k,e=as-(2as-1)z.$$

As in the proof of [3], we have an idempotent $e\in com{m}^2\left(az\right)$ such that $az-e\in {\mathcal{A}}^{qnil}$. We easily check that

$$(a+1-as)((s+1-as)=1+(a-a^{2}s)(1-s)+(s-s^{2}a).$$

Hence,

$$\begin{array}{ccc}a+1-e\hfill & =\hfill & (a+1-as)+(as-e)\in {\mathcal{A}}^{-1},\hfill \\ a(1-e)\hfill & =\hfill & (a-a^{2}s)+a(as-e)\in {\mathcal{A}}^{qnil}.\hfill \end{array}$$

Since $a\in comm\left(as\right)$, we have $ea=ae$. As ${\left(ax\right)}^{*}=ax$, we see that ${\left(as\right)}^{*}=as$. This implies that $q^{*}=q$, and then $z^{*}=z$. Therefore $e^{*}=e$. In light of Theorem 2.2, $a\in {\mathcal{A}}^e$. □

Theorem 2.6. 

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

(1)

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

(2)

There exists a projection $p\in comm\left(a\right)$ such that

$$1-p\in a\mathcal{A}aandpap\in {\mathcal{A}}^{qnil}.$$

(3)

There exists a projection $p\in comm\left(a\right)$ such that

$$pap\in {\left(p\mathcal{A}p\right)}^{qnil}and(1-p)a(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 2.2, there exists a projection $p\in comm\left(a\right)$ such that $u:=a+p\in {\mathcal{A}}^{-1}$ and $ap\in {\mathcal{A}}^{qnil}$. By using Cline’s formula, we have $pap\in {\mathcal{A}}^{qnil}$. Moreover, $(1-p)a=(1-p)u$, and then $u^{-1}(1-p)a=1-p$. Similarly, $a(1-p)u^{-1}=1-p$. Accordingly, $1-p=a(1-p)u^{-2}(1-p)a\in a\mathcal{A}a$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ As $pap\in {\mathcal{A}}^{qnil}$, we see that $pap\in {\left(e\mathcal{A}e\right)}^{qnil}$. Write $1-e=ara$ for some $r\in \mathcal{A}$. Hence, $(1-p)a(1-p)(1-p)ra(1-p)=1-p=(1-p)ar(1-p)a(1-p)$. Therefore $(1-p)a(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}$, as desired.

$\left(3\right)\Rightarrow \left(1\right)$ By using Cline’s formula, we have $pa,ap\in {\mathcal{A}}^{qnil}$. Since $(1-p)a(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}$, we have some $b\in \mathcal{A}$ such that $(1-p)a(1-p)b(1-p)=(1-p)b(1-p)a(1-p)=1-p$. This implies that

$$(a+p)\left((1-p)b(1-p)+p\right)=a(1-p)b(1-p)+ap+p=1+ap\in {\mathcal{A}}^{-1}.$$

Analogously, $\left((1-p)b(1-p)+p\right)(a+p)=1+pa\in {\mathcal{A}}^{qnil}$. Accordingly, $a+p\in {\mathcal{A}}^{qnil}$ and $ap\in {\mathcal{A}}^{qnil}$, thus yielding the result. □

Corollary 2.7. 

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

(1)

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

(2)

There exists a projection $p\in comm\left(a\right)$ such that

$$1-p\in a\mathcal{A}\bigcap \mathcal{A}aandpap\in {\mathcal{A}}^{qnil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is clear by Theorem 2.6 as $a\mathcal{A}a\subseteq a\mathcal{A}\bigcap \mathcal{A}a$.

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

$$1-p\in a\mathcal{A}\bigcap \mathcal{A}aandpap\in {\mathcal{A}}^{qnil}.$$

Write $1-p=ax=ya$ for some $x,y\in \mathcal{A}$. Then $1-p=ax=ax(1-p)=axyz\in a\mathcal{A}a$. This completes the proof by Theorem 2.6. □

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

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

the preceding x is unique if it exists, and denoted by $a^d$. We use ${\mathcal{A}}^d$ to stand for the set of all generalized core-EP invertible element a in $\mathcal{A}$. We refer the reader to [4] for more properties of the generalized core-EP inverse in Banach *-algebra. We say that a has dual generalized core-EP inverse if $a^{*}\in {\mathcal{A}}^d$ denote $a_d={\left({\left(a^{*}\right)}^d\right)}^{*}.$ We now derive

Theorem 2.8. 

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

(1)

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

(2)

$a\in {\mathcal{A}}^d$ and $a^{\pi }$ is a projection.

(3)

$a\in {\mathcal{A}}^d$ and $a^d=a^d$.

(4)

$a\in {\mathcal{A}}^d\bigcap {\mathcal{A}}_d$ and $a^d=a_d$.

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

$\left(2\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}^d$, there exists $x\in comm\left(a\right)$ such that $a{x}^2=x,a-a^{2}x\in {\mathcal{A}}^{qnil}$. By hypothesis, $a^{\pi }=1-ax$ is a projection. Hence, ${\left(ax\right)}^{*}=ax$. In view of Lemma 2.1, $a\in {\mathcal{A}}^e$.

$\left(1\right)\Rightarrow \left(3\right)$ Since every EP element has core inverse, it follows by [4] that $a\in {\mathcal{A}}^d$; hence, $a\in {\mathcal{A}}^d$. By the uniqueness of the g-Drazin inverse of a, we have $a^d=a^d$.

$\left(3\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}^d$, there exists $x\in \mathcal{A}$ such that

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

By hypothesis, $x=a^d$, nd then $x\in comm\left(a\right)$. This implies that ${lim}_{n\to \infty }\left|\right|{(a-x{a}^2)}^n{\left|\right|}^{\frac{1}{n}}=0,$ and so $a-x{a}^2\in {\mathcal{A}}^{qnil}$. Therefore $a\in {\mathcal{A}}^e$ by Theorem 2.5.

$\left(1\right)\Rightarrow \left(4\right)$ By the discussion above, $a\in {\mathcal{A}}^d$ and $a^d=a^d$. Dually, we have $a_d={\left({\left(a^{*}\right)}^d\right)}^{*}=a^d=a^d$, as desired.

$\left(4\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}^d$, it follows by [4, Theorem 1.2] that there exist $x,y\in \mathcal{A}$ such that

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

By hypothesis, $x^{\#}=a^d=a_d{\left({\left(a^{*}\right)}^d\right)}^{*}$. Hence, ${\left[a^{*}{\left(x^{\#}\right)}^{*}\right]}^{*}=a^{*}{\left(x^{\#}\right)}^{*}$. This implies that ${\left(x^{\#}a\right)}^{*}=x^{\#}a$. We easily check that

$$\begin{array}{ccc}\hfill x^{\#}a& =\hfill & x^{\#}(x+y)\hfill \\ & =\hfill & x^{\#}x+x^{\#}x{x}^{\#}y\hfill \\ & =\hfill & x^{\#}x+x^{\#}{\left(x^{\#}\right)}^{*}\left(x^{*}y\right)\hfill \\ & =\hfill & x^{\#}x.\hfill \end{array}$$

Then ${\left(x^{\#}x\right)}^{*}=x^{\#}x$. In view of [27, Theorem 2.2], $x\in {\mathcal{A}}^{\#}$. Accordingly, $a\in {\mathcal{A}}^e$, as asserted. □

Corollary 2.9. 

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

(1)

$A\in \mathcal{B}{\left(X\right)}^e$.

(2)

$A\in \mathcal{B}{\left(X\right)}^d$ and $A{A}^{d}A\in \mathcal{B}\left(X\right)$ is EP.

(3)

$A\in \mathcal{B}{\left(X\right)}^d$ and $A^d{A}^{*}A{A}^{\pi }=0$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.8, $A\in \mathcal{B}{\left(X\right)}^d$. By hypothesis, there exist $X,Y\in \mathcal{B}\left(X\right)$ such that

$$A=X+Y,X^{*}Y=YX=0,X\in \mathcal{B}\left(X\right)isEP,Y\in \mathcal{B}{\left(X\right)}^{qnil}.$$

In view of [24, Corollary 5], $X=A{A}^{d}A$. Hence, $A{A}^{d}A\in \mathcal{B}\left(X\right)$ is EP.

$\left(2\right)\Rightarrow \left(1\right)$ In view of [24, Theorem 4], there exist unique $X,Y\in \mathcal{B}\left(X\right)$ such that

$$A=X+Y,X^{*}Y=0,YX=0,X\in \mathcal{B}{\left(X\right)}^{\#},Y\in \mathcal{B}{\left(X\right)}^{qnil}.$$

Explicitly, $X=A{A}^{d}A$. By hypothesis, X is $EP$. Therefore $A\in \mathcal{B}{\left(X\right)}^e$.

$\left(1\right)\Rightarrow \left(3\right)$ Clearly, $A\in \mathcal{B}{\left(X\right)}^d$. By hypothesis, there exist $X,Y\in \mathcal{B}\left(X\right)$ such that

$$A=X+Y,X^{*}Y=YX=0,X\in \mathcal{B}\left(X\right)isEP,Y\in \mathcal{B}{\left(X\right)}^{qnil}.$$

By virtue of [24, Corollary 1], $X=A^2{A}^d$. Therefore

$$X^{*}Y={\left(A^2{A}^d\right)}^{*}\left(A-A^2{A}^d\right)=0.$$

Hence,

$$A{A}^d{A}^{*}A\left(I-A{A}^d\right)=0.$$

$$A^d{A}^{*}A{A}^{\pi }=0.$$

$\left(3\right)\Rightarrow \left(1\right)$ In view of [24, Theorem 1], there exist unique $X,Y\in \mathcal{B}\left(X\right)$ such that

$$A=X+Y,YX=0,X\in \mathcal{B}\left(X\right)isEP,Y\in \mathcal{B}{\left(X\right)}^{qnil}.$$

Explicitly, $X=A^2{A}^d$. By hypothesis, $X^{*}Y=0$. Therefore $A\in \mathcal{B}{\left(X\right)}^e$. □

3. Additive Properties

In this section, we are concerned with additive properties of generalized EP elements. Let $a,p^2=p\in \mathcal{A}$. Then a has the Pierce decomposition relative to p, and we denote it by ${\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)}_p$. We now derive

Lemma 3.1. 

Let p be a projection, $a\in {\left(p\mathcal{A}p\right)}^d$ and $x={\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)}_p.$ Then $x\in \mathcal{A})$ is generalized EP if and only if $a,d\in \mathcal{A}$ are generalized EP and

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{a}^n{a}^{\pi }b{\left(d^d\right)}^{n+2}=0,\\ {\displaystyle \sum _{n=0}^{\infty }}{\left(a^d\right)}^{n+2}b{d}^n{d}^{\pi }=0.\end{array}$$

Proof. 

⟹ Since $x\in {\mathcal{A}}^e$, it follows by [4, Theorem 1.2] that $x^d\in {\mathcal{A}}^{\#}$. In this case, ${\left(x^d\right)}^{\#}={\left(x^e\right)}^d$ and $x^e={\left(x^d\right)}^2{\left(x^d\right)}^{\#}$. In view of Theorem 2.8, $x^e=x^d$. By virtue of [8, Theorem 2.1], we can write $x^e=\left(\begin{array}{cc}*& *\\ 0& *\end{array}\right),$ and so ${\left(x^e\right)}^d=\left(\begin{array}{cc}*& *\\ 0& *\end{array}\right).$ This implies that ${\left(x^d\right)}^{\#}=\left(\begin{array}{cc}*& *\\ 0& *\end{array}\right).$ Obviously, we have $x^d={\left(\begin{array}{cc}{a}^d& z\\ 0& d^d\end{array}\right)}_p,$ where

$$z={\displaystyle \sum _{n=0}^{\infty }}{\left(a^d\right)}^{n+2}b{d}^n{d}^{\pi }+{\displaystyle \sum _{n=0}^{\infty }}{a}^n{a}^{\pi }b{\left(d^d\right)}^{n+2}-a^{d}b{d}^d.$$

In light of [26, Theorem 2.5], $a^d,d^d\in {\mathcal{A}}^{\#}$ and ${\left(a^d\right)}^{\pi }z=0$. This implies that $a^{\pi }z=0.$ Since $x^d\in {\mathcal{A}}_{\#}$, then ${\left(x^d\right)}^{*}{\in \mathcal{A})}^{\#}$. It follows by [26, Theorem 2.5] that ${\left({\left(d^{*}\right)}^d\right)}^{\pi }{z}^{*}=0$, and so $z{\left(d^d\right)}^{\pi }=0$, i.e., $z{d}^{\pi }=0$. We easily check that

$$\begin{array}{ccc}\hfill a^{\pi }z& =\hfill & {\displaystyle \sum _{n=0}^{\infty }}{a}^n{a}^{\pi }b{\left(d^d\right)}^{n+2},\hfill \\ \hfill z{d}^{\pi }& =\hfill & {\displaystyle \sum _{n=0}^{\infty }}{\left(a^d\right)}^{n+2}b{d}^n{d}^{\pi }.\hfill \end{array}$$

Thus, we have

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{a}^n{a}^{\pi }b{\left(d^d\right)}^{n+2}=0,\\ {\displaystyle \sum _{n=0}^{\infty }}{\left(a^d\right)}^{n+2}b{d}^n{d}^{\pi }=0.\end{array}$$

In view of Theorem 2.8, $a,d\in {\mathcal{A}}^e$, as desired.

⟸ Since $a,d\in \mathcal{A}$ are generalized EP, it follows by Theorem 2.8 that $a^d,d^d\in {\mathcal{A}}^{\#}\bigcap {\mathcal{A}}_{\#}$. In view of [8], $x\in {\mathcal{A}}^d$ and $x^d=\left(\begin{array}{cc}{a}^d& z\\ 0& d^d\end{array}\right),$ where

$$\begin{array}{ccc}\hfill z& =\hfill & {\displaystyle \sum _{n=0}^{\infty }}{\left(a^d\right)}^{n+2}b{d}^n{d}^{\pi }+{\displaystyle \sum _{n=0}^{\infty }}{a}^n{a}^{\pi }b{\left(d^d\right)}^{n+2}-a^{d}b{d}^d\hfill \\ & =\hfill & -a^{d}b{d}^d.\hfill \end{array}$$

We easily check that $a^{\pi }z=0$ and $z{d}^{\pi }=0$. In light of [26, Theorem 2.5], $x^d\in {\mathcal{A}}^{\#}$. In this case,

$${\left(x^d\right)}^{\#}=\left(\begin{array}{cc}{\left(a^d\right)}^{\#}& -{\left(a^d\right)}^{\#}z{\left(d^d\right)}^{\#}\\ 0& {\left(d^d\right)}^{\#}\end{array}\right).$$

Thus, we have

$$\begin{array}{cc}& x^d={\left(x^d\right)}^2{\left(x^d\right)}^{\#}\hfill \\ \hfill =& {\left(\begin{array}{cc}{a}^d& -a^{d}b{d}^d\\ 0& d^d\end{array}\right)}^2\left(\begin{array}{cc}{\left(a^d\right)}^{\#}& {\left(a^d\right)}^{\#}{a}^{d}b{d}^d{\left(d^d\right)}^{\#}\\ 0& {\left(d^d\right)}^{\#}\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}{\left(a^d\right)}^2& -{\left(a^d\right)}^{2}b{d}^d-a^{d}b{\left(d^d\right)}^2\\ 0& {\left(d^d\right)}^2\end{array}\right)\left(\begin{array}{cc}{\left(a^d\right)}^{\#}& {\left(a^d\right)}^{\#}{a}^{d}b{d}^d{\left(d^d\right)}^{\#}\\ 0& {\left(d^d\right)}^{\#}\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}{a}^d& -a^{d}b{d}^d\\ 0& d^d\end{array}\right).\hfill \end{array}$$

This implies that $x^d=x^d$. According to Theorem 2.8, $x\in \mathcal{A}$ is generalized EP. □

Lemma 3.2. 

Let $a,b\in \mathcal{A}$ be generalized EP. If $ab=ba=a^{*}b=0$, then $a+b$ is generalized EP.

Proof. 

In view of Theorem 2.8 , $a,b\in {\mathcal{A}}^d$, $a^d=a^d$ and $b^d=b^d$. Since $a^{*}b=0$, we have $b^{*}a=0$. By virtue of [5], we have ${(a+b)}^d=a^d+b^d$. Clearly, ${(a+b)}^d=a^d+b^d$, and then ${(a+b)}^d={(a+b)}^d$. By using Theorem 2.8 again, $a+b$ is generalized EP. □

We come now to the demonstration for which this section has been developed.

Theorem 3.3. 

Let $a,b,a^{\pi }b\in \mathcal{A}$ be generalized EP. If $a^{\pi }ab=a^{\pi }ba=a^{\pi }{a}^{*}b=0$, then the following are equivalent:

(1)

$a+b\in \mathcal{A}$ is generalized EP.

(2)

$(a+b)a{a}^d\in \mathcal{A}$ is generalized EP

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{(a+b)}^{n}a{a}^d{(a+b)}^{\pi }a{a}^{d}b{\left(a^{\pi }{b}^d\right)}^{n+2}=0,\\ \hfill {\displaystyle \sum _{n=0}^{\infty }}{\left({(a+b)}^{d}a{a}^d\right)}^{n+2}b{a}^{\pi }(a^n+b^n)a^{\pi }{b}^{\pi }=0.\end{array}$$

Proof. 

Let $p=a{a}^d$. By hypothesis, $p^{\pi }bp=(1-a{a}^d)ba{a}^d=\left(a^{\pi }ba\right)a^d=0$. So we get

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_4\end{array}\right)}_p,b={\left(\begin{array}{cc}{b}_1& b_2\\ 0& b_4\end{array}\right)}_p.$$

Hence

$$a+b={\left(\begin{array}{cc}{a}_1+b_1& b_2\\ 0& a_4+b_4\end{array}\right)}_p.$$

Here, $a_1=a^2{a}^d$ and $b_1=a{a}^{d}ba{a}^d=ba{a}^d$. Then

$$a_1+b_1=(a+b)a{a}^d.$$

Moreover, we see that

$$\begin{array}{ccc}\hfill {(a_1+b_1)}^i& =\hfill & {(a+b)}^{i}a{a}^d,\hfill \\ \hfill {(a_1+b_1)}^d& =\hfill & {(a+b)}^{d}a{a}^d,\hfill \\ \hfill {(a_1+b_1)}^{\pi }& =\hfill & 1-(a+b){(a+b)}^{d}a{a}^d.\hfill \end{array}$$

Also we have $a_4=a{a}^{\pi }$ and $b_4=a^{\pi }b{a}^{\pi }=a^{\pi }b$, and so

$$a_4+b_4=a^{\pi }a+a^{\pi }b.$$

Then we check that

$$\begin{array}{ccc}\hfill {(a_4+b_4)}^i& =\hfill & a^{\pi }(a^i+b^i),\hfill \\ \hfill {(a_4+b_4)}^d& =\hfill & a^{\pi }{b}^d,\hfill \\ \hfill {(a_4+b_4)}^{\pi }& =\hfill & 1-a^{\pi }b{b}^d.\hfill \end{array}$$

Clearly, $a^{\pi }a$ is generalized EP. By hypothesis, $a^{\pi }b$ is generalized EP. Further, we see that

$$\begin{array}{ccc}\hfill \left(a^{\pi }a\right)\left(a^{\pi }b\right)& =\hfill & 0,\hfill \\ \hfill \left(a^{\pi }b\right)\left(a^{\pi }a\right)& =\hfill & 0,\hfill \\ \hfill {\left(a^{\pi }a\right)}^{*}\left(a^{\pi }b\right)& =\hfill & 0.\hfill \end{array}$$

In view of Lemma 3.2, $a_4+b_4$ is generalized EP.

In light of Lemma 3.1, $a+b$ is generalized EP if and only if $a_1+b_1$ is generalized EP and

$$\begin{array}{c}{\displaystyle {\displaystyle \sum _{n=0}^{\infty }}}{(a_1+b_1)}^n{(a_1+b_1)}^{\pi }{b}_2{\left({(a_4+b_4)}^d\right)}^{n+2}=0,\\ {\displaystyle \sum _{n=0}^{\infty }}{\left({(a_1+b_1)}^d\right)}^{n+2}{b}_2{(a_4+b_4)}^n{(a_4+b_4)}^{\pi }=0.\end{array}$$

Therefore $a+b$ is generalized EP if and only if $(a+b)a{a}^d$ is generalized EP and

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{(a+b)}^{n}a{a}^d{(a+b)}^{\pi }a{a}^{d}b{\left(a^{\pi }{b}^d\right)}^{n+2}=0,\\ {\displaystyle \sum _{n=0}^{\infty }}{\left({(a+b)}^{d}a{a}^d\right)}^{n+2}b{a}^{\pi }(a^n+b^n)a^{\pi }{b}^{\pi }=0.\end{array}$$

This completes the proof. □

Corollary 3.4. 

Let $a,b,a^{\pi }b\in \mathcal{A}$ be EP. If $a^{\pi }ba=0$, then the following are equivalent:

(1)

$a+b\in \mathcal{A}$ is generalized EP.

(2)

$(a+b)a{a}^{\#}\in \mathcal{A}$ is generalized EP and

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{(a+b)}^{n}a{a}^{\#}{(a+b)}^{\pi }a{a}^{\#}b{\left(a^{\pi }{b}^{\#}\right)}^{n+2}=0,\\ \hfill {\displaystyle \sum _{n=0}^{\infty }}{\left({(a+b)}^{d}a{a}^{\#}\right)}^{n+2}b{a}^{\pi }(a^n+b^n)a^{\pi }{b}^{\pi }=0.\end{array}$$

Proof. 

Since $a\in \mathcal{A}$ is EP, we see that $a^{\pi }a=a^{\pi }{a}^{*}=0$. Therefore we complete the proof by Theorem 3.3. □

Lemma 3.5. 

Let $a,b\in \mathcal{A}$ be generalized EP. If $ab=ba$ and $a^{*}b=b{a}^{*}$, then the following are equivalent:

(1)

$a+b\in \mathcal{A}$ is generalized EP.

(2)

$1+b{a}^d\in \mathcal{A}$ is generalized EP.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of [5, Theorem 3.4], $1+a^{d}b\in {\mathcal{A}}^d$ and

$${(1+a^{d}b)}^d=a^{\pi }+a^2{a}^d{(a+b)}^d.$$

Then

$$\begin{array}{ccc}\hfill (1+a^{d}b){(1+a^{d}b)}^d& =\hfill & a^{\pi }+(1+a^{d}b)a^2{a}^d{(a+b)}^d\hfill \\ & =\hfill & a^{\pi }+a{a}^d(a+b){(a+b)}^d\hfill \\ & =\hfill & 1-a{a}^d{(a+b)}^{\pi }.\hfill \end{array}$$

In view of Theorem 2.8, $a{a}^d$ are ${(a+b)}^{\pi }$ are projections. Then $(1+a^{d}b){(1+a^{d}b)}^d$ is a projection, and so is ${(1+a^{d}b)}^{\pi }$. Therefore $1+a^{d}b\in \mathcal{A}$ is generalized EP by Theorem 2.8. Since $ab=ba$, it follows by [3] that $a^{d}b=b{a}^d$, and so $1+b{a}^d$ is generalized EP.

$\left(2\right)\Rightarrow \left(1\right)$ Since $1+a^{d}b=1+b{a}^d\in \mathcal{A}$ is generalized EP, it follows by Theorem 2.8 that $1+a^{d}b\in {\mathcal{A}}^d$ and ${(1+a^{d}b)}^{\pi }$ is a projection. In view of [28, Theorem 3.3], $a+b\in {\mathcal{A}}^d$ and

$${(a+b)}^d={(1+a^{d}b)}^d{a}^d+b^d{(1+a{a}^{\pi }{b}^d)}^{-1}{a}^{\pi }.$$

Since $(1-a^{\pi }b{b}^d)(1+a{a}^{\pi }{b}^d)=1-a^{\pi }b{b}^d$, we have

$$(1-a^{\pi }b{b}^d){(1+a{a}^{\pi }{b}^d)}^{-1}=1-a^{\pi }b{b}^d.$$

Then we check that

$$\begin{array}{cc}& (a+b){(1+b)}^d\hfill \\ \hfill =& a^d(a+b){(1+a^{d}b)}^d+(a+b)a^{\pi }{b}^d{(1+a{a}^{\pi }{b}^d)}^{-1}\hfill \\ \hfill =& a{a}^d(1+a^{d}b){(1+a^{d}b)}^d+(a{a}^{\pi }{b}^d+a^{\pi }b{b}^d){(1+a{a}^{\pi }{b}^d)}^{-1}\hfill \\ \hfill =& a{a}^d(1+a^{d}b){(1+a^{d}b)}^d+1-(1-a^{\pi }b{b}^d){(1+a{a}^{\pi }{b}^d)}^{-1}\hfill \\ \hfill =& a{a}^d(1+a^{d}b){(1+a^{d}b)}^d+1-[1-a^{\pi }b{b}^d]\hfill \\ \hfill =& a{a}^d(1+a^{d}b){(1+a^{d}b)}^d+a^{\pi }b{b}^d.\hfill \end{array}$$

Therefore

$$\begin{array}{ccc}\hfill {(a+b)}^{\pi }& =\hfill & 1-a{a}^d(1+a^{d}b){(1+a^{d}b)}^d-a^{\pi }b{b}^d\hfill \\ & =\hfill & a{a}^d-a{a}^d(1+a^{d}b){(1+a^{d}b)}^d+a^{\pi }{b}^{\pi }\hfill \\ & =\hfill & a{a}^d{(1+a^{d}b)}^{\pi }+a^{\pi }{b}^{\pi }.\hfill \end{array}$$

Hence, ${(a+b)}^{\pi }$ is a projection. Accordingly, $a+b\in \mathcal{A}$ is generalized EP by Theorem 2.8. □

We are ready to prove:

Theorem 3.6. 

Let $a,b,a^{\pi }b\in \mathcal{A}$ be generalized EP. If $a^{\pi }ab=a^{\pi }ba$ and $a^{\pi }{a}^{*}b=a^{\pi }b{a}^{*}$, then the following are equivalent:

(1)

$a+b\in \mathcal{A}$ is generalized EP.

(2)

$1+b{a}^d\in \mathcal{A}$ is generalized EP and

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{(a+b)}^{n}a{a}^d{(a+b)}^{\pi }a{a}^{d}b{\left(a^{\pi }{b}^{\pi }\right)}^{n+2}=0,\\ \hfill {\displaystyle \sum _{n=0}^{\infty }}{\left({(a+b)}^{d}a{a}^d\right)}^{n+2}b{a}^{\pi }(a^n+b^n)a^{\pi }{b}^{\pi }=0.\end{array}$$

Proof. 

Since $a^{\pi }ab=a^{\pi }ba$, we have $a\left(a^{\pi }b\right)=\left(a^{\pi }b\right)a$. In view of [3, Theorem 15.2.12], $a^d\left(a^{\pi }b\right)=\left(a^{\pi }b\right)a^d$. Hence, $a^{\pi }b{a}^d=0$. Let $p=a{a}^d$. Then $p^{\pi }bp=\left(a^{\pi }b{a}^d\right)a=0$, and then we have

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_4\end{array}\right)}_p,b={\left(\begin{array}{cc}{b}_1& b_2\\ 0& b_4\end{array}\right)}_p.$$

Thus,

$$a+b={\left(\begin{array}{cc}{a}_1+b_1& b_2\\ 0& a_4+b_4\end{array}\right)}_p.$$

Here, $a_1=a^2{a}^d$ and $b_1=a{a}^{d}ba{a}^d=(1-a^{\pi })ba{a}^d=ba{a}^d$. Then

$$a_1+b_1=(a+b)a{a}^d=(1+b{a}^d)a^2{a}^d.$$

We verify that

$$\begin{array}{ccc}\hfill {(a_1+b_1)}^i& =\hfill & {(a+b)}^{i}a{a}^d,\hfill \\ \hfill {(a_1+b_1)}^d& =\hfill & {(a+b)}^{d}a{a}^d,\hfill \\ \hfill {(a_1+b_1)}^{\pi }& =\hfill & 1-(a+b){(a+b)}^{d}a{a}^d.\hfill \end{array}$$

Further, we have $a_4=a{a}^{\pi }$ and $b_4=a^{\pi }b{a}^{\pi }=a^{\pi }b$; hence,

$$a_4+b_4=a^{\pi }a+a^{\pi }b.$$

Then we check that

$$\begin{array}{ccc}\hfill {(a_4+b_4)}^i& =\hfill & a^{\pi }(a^i+b^i),\hfill \\ \hfill {(a_4+b_4)}^d& =\hfill & a^{\pi }{b}^d,\hfill \\ \hfill {(a_4+b_4)}^{\pi }& =\hfill & 1-a^{\pi }b{b}^d.\hfill \end{array}$$

Obviously, $a^{\pi }a$ and $a^{\pi }b$ are generalized EP. Furthermore, we have

$$\begin{array}{ccc}\hfill \left(a^{\pi }a\right)\left(a^{\pi }b\right)& =\hfill & a^{\pi }ab=a^{\pi }ba=\left(a^{\pi }b\right)\left(a^{\pi }a\right),\hfill \\ \hfill {\left(a^{\pi }a\right)}^{*}\left(a^{\pi }b\right)& =\hfill & a^{\pi }{a}^{*}b=a^{\pi }b{a}^{*}=a^{\pi }b{a}^{\pi }{a}^{*}=\left(a^{\pi }b\right){\left(a^{\pi }a\right)}^{*},\hfill \\ \hfill 1+{\left(a^{\pi }a\right)}^d\left(a^{\pi }b\right)& =\hfill & 1hasgeneralizedcore-EPinverse.\hfill \end{array}$$

By virtue Lemma 3.5, $a_4+b_4$ is generalized EP.

By virtue of Lemma 3.1, $a+b$ is generalized EP if and only if $a_1+b_1$ is generalized EP and

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{(a_1+b_1)}^n{(a_1+b_1)}^{\pi }{b}_2{\left({(a_4+b_4)}^d\right)}^{n+2}=0,\\ \hfill {\displaystyle \sum _{n=0}^{\infty }}{\left({(a_1+b_1)}^d\right)}^{n+2}{b}_2{(a_4+b_4)}^n{(a_4+b_4)}^{\pi }=0.\end{array}$$

Claim 1. Assume that $1+b{a}^d$ is generalized EP. Then we see that

$$\begin{array}{ccc}\hfill \left(b{a}^d\right)\left(a{a}^d\right)& =\hfill & b{a}^d=(1-a^{\pi })b{a}^d=\left(a{a}^d\right)\left(b{a}^d\right),\hfill \\ \hfill \left(b{a}^d\right){\left(a{a}^d\right)}^{*}& =\hfill & \left(b{a}^d\right)\left(a{a}^d\right)=\left(a{a}^d\right)\left(b{a}^d\right)={\left(a{a}^d\right)}^{*}\left(b{a}^d\right),\hfill \end{array}$$

and then

$$\begin{array}{ccc}\hfill (1+b{a}^d)\left(a{a}^d\right)& =\hfill & \left(a{a}^d\right)(1+b{a}^d),\hfill \\ \hfill (1+b{a}^d){\left(a{a}^d\right)}^{*}& =\hfill & {\left(a{a}^d\right)}^{*}(1+b{a}^d).\hfill \end{array}$$

By virtue of Lemma 3.5, $a_1+b_1=(1+b{a}^d)\left(a{a}^d\right)$ is generalized EP.

Claim 2. Assume that $a_1+b_1=(1+b{a}^d)a{a}^d$ is generalized EP. Obviously, we have $a^{\pi }(1+b{a}^d)a{a}^d={\left(a^{\pi }\right)}^{*}(1+b{a}^d)a{a}^d=(1+b{a}^d)a{a}^d{a}^{\pi }=0$. It follows by Lemma 3.2 that $1+b{a}^d=a^{\pi }+(1+b{a}^d)a{a}^d$ is generalized EP.

Thus, we conclude that $a_1+b_1$ is generalized EP if and only if so is $1+b{a}^d$. Therefore $a+b$ is generalized EP if and only if $1+b{a}^d$ is generalized EP and

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{(a+b)}^{n}a{a}^d{(a+b)}^{\pi }a{a}^{d}b{\left(a^{\pi }{b}^{\pi }\right)}^{n+2}=0,\\ \hfill {\displaystyle \sum _{n=0}^{\infty }}{\left({(a+b)}^{d}a{a}^d\right)}^{n+2}b{a}^{\pi }(a^n+b^n)a^{\pi }{b}^{\pi }=0.\end{array}$$

□

Corollary 3.7. 

Let $a,b\in \mathcal{A}$ be EP. If $ab=ba$ and $a^{*}b=b{a}^{*}$, then the following are equivalent:

(1)

$a+b\in \mathcal{A}$ is generalized EP.

(2)

$1+b{a}^{\#}\in \mathcal{A}$ is generalized EP.

Proof. 

This is obvious by Theorem 3.6. □

4. Generalized Core-EP Orders

This section is devoted to the generalized core-EP orders involved in generalized EP elements. We now extend [13, Theorem 4.4] as follows.

Theorem 4.1. 

$a\in {\mathcal{A}}^e,b\in {\mathcal{A}}^d$. Then the following are equivalent:

(1)

$a{\le }^{d}b$.

(2)

$a^d{\le }^d{b}^d$ and $a^{d}a=a^{d}b$.

(3)

$a^d{b}^d=b^d{a}^d$ and $a^{d}a=a^{d}b$.

Proof.$\left(1\right)\Rightarrow \left(3\right)$ By hypothesis, we have $a^{d}a=a^{d}b,a{a}^d=b{a}^d.$ Since $a\in {\mathcal{A}}^e$, it follows by Theorem 2.8 that $a^d=a^d$, and then

$$a{a}^{d}b=a{a}^{d}a=b{a}^{d}a=b{a}^{d}a=ba{a}^d=ba{a}^d.$$

Moreover, ${\left(a{a}^d\right)}^{*}b=b{\left(a{a}^d\right)}^{*}$, and so $a{a}^d{b}^{*}=b^{*}a{a}^d.$ In light of [5, Lemma 3.2],

$$a{a}^d{b}^d=b^{d}a{a}^d.$$

Therefore

$$\begin{array}{cc}& \left|\right|a^d-a{a}^d{b}^d\left|\right|\hfill \\ \hfill =& \left|\right|a^k{\left(a^d\right)}^{k+1}-b^d{b}^{k+1}{\left(a^d\right)}^{k+1}\left|\right|\hfill \\ \hfill \le & \left|\right|a^k{\left(a^d\right)}^{k+1}-b^k{\left(a^d\right)}^{k+1}\left|\right|+\left|\right|b^k-b^d{b}^{k+1}\left|\right|\left|\right|a^d{)}^{k+1}\left|\right|\hfill \\ \hfill =& \left|\right|b^k-b^d{b}^{k+1}\left|\right|\left|\right|a^d{)}^{k+1}\left|\right|.\hfill \end{array}$$

Since

$$\underset{k\to \infty }{lim}\left|\right|b^k-b^d{b}^{k+1}{\left|\right|}^{\frac{1}{k}}=0,$$

we deduce that

$$\underset{k\to \infty }{lim}\left|\right|a^d-a{a}^d{b}^d{\left|\right|}^{\frac{1}{k}}=0.$$

Then $a{a}^d{b}^d=a^d;$ hence,

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

Accordingly,

$$a^d{b}^d={\left(a^d\right)}^2=a^d{a}^d=b^d{a}^d=b^d{a}^d.$$

$\left(3\right)\Rightarrow \left(1\right)$ Since $a,b\in {\mathcal{A}}^e$, then $a^d=a^d$ and $b^d=b^d$. We verify that

$$\begin{array}{ccc}\hfill b{a}^d& =\hfill & b{\left(a^d\right)}^{2}a=b{\left(a^d\right)}^{2}b=b{\left(a^d\right)}^{k+1}{b}^k,\hfill \\ \hfill b{b}^{d}a{a}^d& =\hfill & b{b}^d{\left(a^d\right)}^{k+1}{b}^{k+1}=b{\left(a^d\right)}^{k+1}{b}^d{b}^{k+1},\hfill \\ \hfill a{a}^d& =\hfill & a^{d}a=a^{d}b={\left(a^d\right)}^k{b}^k,\hfill \\ \hfill b{b}^{d}a{a}^d& =\hfill & b^d{\left(a^d\right)}^k{b}^{k+1}={\left(a^d\right)}^k{b}^d{b}^{k+1}.\hfill \end{array}$$

Then

$$\left|\right|b{a}^d-b{b}^{d}a{a}^d{\left|\right|}^{\frac{1}{k}}\le {\left|\right|b\left|\right|}^{\frac{1}{k}}\left|\right|a^d{\left|\right|}^{1+\frac{1}{k}}\left|\right|b^k-b^d{b}^{k+1}{\left|\right|}^{\frac{1}{k}}.$$

Since ${lim}_{k\to \infty }\left|\right|b^k-b^d{b}^{k+1}{\left|\right|}^{\frac{1}{k}}=0$, we have

$$\underset{k\to \infty }{lim}\left|\right|b{a}^d-b{b}^{d}a{a}^d{\left|\right|}^{\frac{1}{k}}=0.$$

This implies that $b{a}^d=b{b}^{d}a{a}^d$. Likewise, $a{a}^d=b{b}^{d}a{a}^d$. Therefore $b{a}^d=b{b}^{d}a{a}^d=a{a}^d$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, we have ${\left(a^d\right)}^d{a}^d={\left(a^d\right)}^d{b}^d$. In view of [5, Theorem 3.5], ${\left(a^d\right)}^d=a^2{a}^d$. Then

$$a^2{a}^d{a}^d=a^2{a}^d{b}^d.$$

Hence, $a{a}^d=a^2{a}^d{b}^d.$

On the other hand, $a^d{\left(a^d\right)}^d=b^d{\left(a^d\right)}^d$. Then $a^d{a}^2{a}^d=b^d{a}^2{a}^d.$ This implies that $a{a}^d=b^d{a}^2{a}^d.$ Therefore

$$\begin{array}{ccc}\hfill a^d{b}^d& =\hfill & {\left(a^d\right)}^2\left(a^2{a}^d{b}^d\right)\hfill \\ & =\hfill & {\left(a^d\right)}^2\left(a{a}^d\right)={\left(a^d\right)}^2=a{a}^d{\left(a^d\right)}^2\hfill \\ & =\hfill & \left(b^d{a}^2{a}^d\right){\left(a^d\right)}^2=b^d{a}^d,\hfill \end{array}$$

as desired.

$\left(3\right)\Rightarrow \left(2\right)$ In view of [5, Theorem 3.5], ${\left(a^d\right)}^d=a^2{a}^d$. Then we check that

$$\begin{array}{ccc}\hfill {\left(a^d\right)}^d{a}^d& =\hfill & a^2{\left(a^d\right)}^2=a^2\left(a{a}^d\right){\left(a^d\right)}^2=a^{2}b{\left(a^d\right)}^3=a^{2}b\left(a{a}^d\right){\left(a^d\right)}^3\hfill \\ & =\hfill & a^2{b}^2{\left(a^d\right)}^4=\cdots =a^2{b}^k{\left(a^d\right)}^{k+2},\hfill \\ \hfill {\left(a^d\right)}^d{b}^d& =\hfill & a^2{a}^d{b}^d=a^2{b}^d{a}^d=a^2{b}^{d}a{\left(a^d\right)}^2=a^2{b}^d{b}^{k+1}{\left(a^d\right)}^{k+2}\hfill \end{array}$$

Hence,

$$\begin{array}{cc}& \left|\right|{\left(a^d\right)}^d{a}^d-a^d{)}^d{b}^d{\left|\right|}^{\frac{1}{k}}\hfill \\ \hfill \le & \left|\right|a^2{\left|\right|}^{\frac{1}{k}}\left|\right|b^k-b^d{b}^{k+1}{\left|\right|}^{\frac{1}{k}}\left|\right|a^d{\left|\right|}^{1+\frac{2}{k}}.\hfill \end{array}$$

Since ${lim}_{k\to \infty }\left|\right|b^k-b^d{b}^{k+1}{\left|\right|}^{\frac{1}{k}}=0$, we deduce that

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

This implies that

$${\left(a^d\right)}^d{a}^d={\left(a^d\right)}^d{b}^d.$$

On the other hand, we have

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

Thus, $a^d{\left(a^d\right)}^d=b^d{\left(a^d\right)}^d.$ This completes the proof. □

The core-EP order for core-EP inverse of complex matrices was studied in [25, Theorem 4.2]. As an immediate consequence of Theorem 4.1, we give an alternative characterization of core-EP order for core-EP inverses as follows.

Corollary 4.2. 

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

(1)

$A{\le }^{D}B$.

(2)

$A^D{\le }^D{B}^D$ and $A^{D}A=A^{D}B$.

(3)

$A^D{B}^D=B^D{A}^D$ and $A^{D}A=A^{D}B$.

Theorem 4.3. 

Let $a\in {\mathcal{A}}^e,b\in {\mathcal{A}}^d$. If $a{\le }^{d}b$, then the following are equivalent:

(1)

$b\in {\mathcal{A}}^e$.

(2)

$b(1-a{a}^d)\in {\mathcal{A}}^e$.

Proof. 

Since $a{\le }^{d}b$, we have that $a^{d}a=a^{d}b$ and $a{a}^d=b{a}^d$. Then

$$a{a}^{d}b=a{a}^{d}a=b{a}^{d}a=ba{a}^d.$$

$\left(1\right)\Rightarrow \left(2\right)$ Since $b(1-a{a}^d)=(1-a{a}^d)b$, it follows by [28, Theorem 3.1] that

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

We verify that

$$\begin{array}{c}\left[b(1-a{a}^d)\right][b^d-a^d]=b{b}^d-a{a}^d;\\ {\left[\left(b(1-a{a}^d)\right)(b^d-a^d)\right]}^{*}=\left[b(1-a{a}^d)\right][b^d-a^d],\end{array}$$

Moreover, we check that

$$\begin{array}{cc}& {\left(b(1-a{a}^d)\right)}^n-\left(b^d-a^d\right){\left(b(1-a{a}^d)\right)}^{n+1}\hfill \\ \hfill =& b^n(1-a{a}^d)-\left(b^d-a^d\right)b^{n+1}(1-a{a}^d)\hfill \\ \hfill =& b^n(1-a{a}^d)-b^d{b}^{n+1}(1-a{a}^d)\hfill \\ \hfill =& (b^n-b^d{b}^{n+1})(1-a{a}^d).\hfill \end{array}$$

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

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

Therefore

$${\left[b(1-a{a}^d)\right]}^d=b^d-a^d={\left[b(1-a{a}^d)\right]}^d.$$

By virtue of Theorem 2.8, $b(1-a{a}^d)$ is generalized EP.

$\left(2\right)\Rightarrow \left(1\right)$ Obviously, $b=x+y$, where $x=ba{a}^d$ and $y=b(1-a{a}^d)$.

Claim 1. $x\in \mathcal{A}$ are generalized EP. We directly verify that $x^d=a^d=a^d=x^d$. In view of Theorem 2.8, $x\in {\mathcal{A}}^e$.

Claim 2. $xy=yx=x^{*}y=0$. Since $a{a}^{d}a=ba{a}^d$, we have that $a^{*}a{a}^d=a{a}^d{b}^{*}$. Then we verify that

$$\begin{array}{ccc}\hfill xy& =\hfill & ba{a}^{d}b(1-a{a}^d)=ba{a}^{d}a(1-a{a}^d)=0,\hfill \\ \hfill yx& =\hfill & b(1-a{a}^d)b{a}^{d}a=b(1-a{a}^d)a{a}^{d}a=0,\hfill \\ \hfill x^{*}y& =\hfill & a{a}^d{b}^{*}b(1-a{a}^d)=a^{*}a{a}^{d}b(1-a{a}^d)\hfill \\ & =\hfill & a^{*}a{a}^{d}a(1-a{a}^d)=0.\hfill \end{array}$$

In light of Lemma 3.2, $b=x+y\in \mathcal{A}$ is generalized EP, as asserted. □

As an immediate consequence, we now improve [13, Theorem 4.5] as follows.

Corollary 4.4. 

Let $a\in {\mathcal{A}}^e,b\in {\mathcal{A}}^D$. If $a{\le }^{d}b$, then the following are equivalent:

(1)

$b\in \mathcal{A}$ is *-DMP.

(2)

$b(1-a{a}^d)\in \mathcal{A}$ is *-DMP.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 2.8, $b(1-a{a}^d)\in {\mathcal{A}}^e$ and ${\left[b(1-a{a}^d)\right]}^d={\left[b(1-a{a}^d)\right]}^d$. Since $b\in \mathcal{A}$ is *-DMP, it follows by Theorem 2.8 that $b\in {\mathcal{A}}^D$. Since $a{\le }^{d}b$, we have

$$a{a}^{d}b=a{a}^{d}b=a{a}^{d}a=b{a}^{d}a=ba{a}^d.$$

In view of [28, Theorem 3.1], $a{a}^{d}b\in {\mathcal{A}}^D$. Obviously, $a{a}^{d}b{a}^{\pi }=ba{a}^d{a}^{\pi }=0$. Set $p=a{a}^d$. Then

$$b={\left(\begin{array}{cc}a{a}^{d}b& 0\\ a^{\pi }ba{a}^d& b{a}^{\pi }\end{array}\right)}_p.$$

By virtue of [8, Theorem 2.1], $b{a}^{\pi }\in {\mathcal{A}}^D$. Hence,

$${\left[b(1-a{a}^d)\right]}^d={\left[b(1-a{a}^d)\right]}^D.$$

Therefore

$${\left[\left(b(1-a{a}^d)\right){\left(b(1-a{a}^d)\right)}^D\right]}^{*}=\left(b(1-a{a}^d)\right){\left(b(1-a{a}^d)\right)}^D.$$

Accordingly, $b(1-a{a}^d)\in \mathcal{A}$ is *-DMP.

$\left(2\right)\Rightarrow \left(1\right)$ In view of Theorem 4.3, $b\in {\mathcal{A}}^e$. Then $b\in {\mathcal{A}}^d$ and $b^d=b^d$. Since $b\in {\mathcal{A}}^D$, it follows by [10, Theorem 2.3] that $b\in {\mathcal{A}}^D$. Therefore $b^d=b^D$. Then ${\left(b{b}^D\right)}^{*}=b{b}^D$. This implies that $b\in \mathcal{A}$ is *-DMP by [13, Lemma 2.2]. □