The Generalized Core-EP Inverse for Anti-Triangular Matrices

1. Introduction

A Banach *-algebra $\mathcal{A}$ is a Banach algebra endowed with an involution $*:\mathcal{A}\to \mathcal{A}$ satisfying ${\left(x^{*}\right)}^{*}=x$, ${(x+y)}^{*}=x^{*}+y^{*}$, ${\left(\lambda x\right)}^{*}=\overline{\lambda }{x}^{*}$, ${\left(xy\right)}^{*}=y^{*}{x}^{*}$, and $\parallel x^{*}\parallel =\parallel x\parallel$ for all $x,y\in \mathcal{A}$ and $\lambda \in \mathbb{C}$. Let $a\in \mathcal{A}$. An element $x\in \mathcal{A}$ is the group inverse of a (denoted $a^{\#}$) if it satisfies: $a{x}^2=x,ax=xa,x{a}^2=a$ (see [17,18]. An element $x\in \mathcal{A}$ is the Drazin inverse of a (denoted $a^D$) if there exists a non-negative integer k (called the Drazin index) such that $a{x}^2=x,ax=xa,x{a}^{k+1}=a^k$ (see [12]). An element $x\in \mathcal{A}$ is the generalized Drazin inverse of a (denoted $a^d$) if $a{x}^2=x,ax=xa,a-x{a}^2\in {\mathcal{A}}^{qnil},$ where ${\mathcal{A}}^{qnil}$ denotes the set of quasinilpotent elements of $\mathcal{A}$ (elements with spectral radius equal to zero) (see [24]). Evidently, ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{{\displaystyle \frac{1}{n}}}=0\}.$

While the previous generalized inverses rely on spectral or nilpotent properties, the following inverses utilize the topological structure (specifically, closedness) of certain ranges or domains in Banach *-algebras. Let $\mathcal{A}$ be a Banach *-algebra. An element $x\in \mathcal{A}$ has a core inverse of a (denoted $a^{\#}$) if

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

(see [21,26]). An element $x\in \mathcal{A}$ is a core-EP inverse (i.e., pseudo-core inverse) of a (denoted by $a^{Ⓓ}$) if there exists a non-negative integer k such that

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

(see [13,19,20]).

Recently, the authors extended the generalized inverses mentioned above and introduced generalized core-EP inverse. An element $x\in \mathcal{A}$ is the generalized core-EP inverse of a (denoted $a^{ⓓ}$) if

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

This generalizes the core-EP inverse from complex matrices to infinite-dimensional Banach space operators. We list several characterizations of the generalized core-EP inverse.

Theorem 1.1. (see [4,7]) Let $\mathcal{A}$ be a Banach *-algebra, and let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{ⓓ}$.

(2)

There exist $x,y\in \mathcal{A}$ such that

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

(3)

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

$$a+p\in {\mathcal{A}}^{-1},pa=pap\in {\mathcal{A}}^{qnil}.$$

(4)

$xax=x,im\left(x\right)=im\left(x^{*}\right)=im\left(a^d\right)$.

(5)

$a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}^{\#}$. In this case, $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{\#}.$

(6)

$a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}^{(1,3)}$. In this case, $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{(1,3)}.$

The generalized inverses mentioned above are powerful tools in linear and operator algebra for handling matrices and operators lacking a traditional inverse. They provide a robust method for solving linear systems and have broad applications across numerous scientific and engineering disciplines. Recently, many authors have investigated these inverses from diverse perspectives, e.g., [1,3,5,6,16,25,26,27].

The core and core-EP inverses of triangular matrices have attracted considerable attention. Xu [22] considered the core inverse for a special form of triangular matrices, while Gao and Chen [14] provided further investigations for the core-EP inverse of the triangular matrices. It is worth noting that a triangular matrix may fail to have a core-EP inverse even when its diagonal entries do [14], highlighting the complexity of this problem. Recent effort have also explored the core-EP inverse of companion matrices [15].

In [9], Chen and Sheibani studied the generalized core-EP inverse for triangular matrices over a $C^{*}$-algebra. Overcoming the limitations of this specific algebraic and structural setting remains a non-trivial challenge. This paper aims to break this barrier by characterizing the generalized core-EP inverse for a novel class of anti-triangular matrices over a Banach algebra. Consequently, new representations of the core and core-EP inverses of such kind of anti-triangular matrices follow.

In this paper, we consider $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b\in {\mathcal{A}}^{ⓓ}$. This special class of matrices have been well studied, playing a key role in solving difference equations (see [3,6,12]). We establish novel criteria for the existence of the generalized core-EP inverse of the anti-triangular matrix x. Furthermore, explicit representations of its generalized core-EP inverse are presented. These newly established representations naturally induce novel formulations for the core and core-EP inverses, shedding new light on their algebraic structures.

Throughout the paper, all Banach *-algebras are assumed to be complex with an identity. We use $A^{-1}$, $A^{\#}$, $A^{Ⓓ}$, and $A^{ⓓ}$ to denote the sets of all invertible, core invertible, core-EP invertible, and generalized core-EP invertible elements in $\mathcal{A}$, respectively. Let ${\mathbb{C}}^{n\times n}$ denote the Banach algebra of all $n\times n$ complex matrices, equipped with the conjugate transpose as its involution. Let $\mathcal{A}$ be a Banach *-algebra. Then $M_2\left(\mathcal{A}\right)$ is a Banach *-algebra with *-transpose as the involution. The norm on $M_2\left(\mathcal{A}\right)$ is typically defined as the sum of the norms of its entries. Suppose that $a\in \mathcal{A}$ possesses the generalized Drazin inverse; we denote its spectral idempotent $1-a{a}^d$ by $a^{\pi }$.

2. Key Lemmas

To prove the main results, we present the necessary lemmas. Let $a^{\pi }=1-a{a}^d,a^{\sigma }=1-a{a}^{ⓓ}$ and $a^{\tau }=1-a{a}^{ⓓ}$. We begin with

Lemma 2.1.

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

(1)

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

(2)

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

(3)

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

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

$\left(2\right)\Rightarrow \left(3\right)$ Since $(1-a^{ⓓ}a)b=0$, we have $b=a^{ⓓ}ab$. Thus, $(1-a{a}^{ⓓ})b=(1-a{a}^{ⓓ})a^{ⓓ}ab=0.$

$\left(3\right)\Rightarrow \left(1\right)$ Since $(1-a{a}^{ⓓ})b=0$, we have $b=a{a}^{ⓓ}b$. Then

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

This implies that $a^{\pi }b=0$, as desired. □

Lemma 2.2.

Let $a\in {\mathcal{A}}^{ⓓ}$ and $b\in {\mathcal{A}}^{qnil}$. If $ba=0$, then $a+b\in {\mathcal{A}}^{ⓓ}$. In this case,

$${(a+b)}^{ⓓ}=a^{ⓓ}.$$

Proof.

We directly verify that

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

Moreover, we have

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

By using Cline’s formula, $a-a^2{a}^{ⓓ}\in {\mathcal{A}}^{qnil}$. By hypothesis, $b\in {\mathcal{A}}^{qnil}$. Since $b(a-a^2{a}^{ⓓ})=\left(ba\right)(1-a{a}^{ⓓ})=0$, it follows by [2] that $(a-a^2{a}^{ⓓ})+b\in {\mathcal{A}}^{qnil}$, i.e, $(a+b)-{(a+b)}^2{a}^{ⓓ}\in {\mathcal{A}}^{qnil}$. Therefore ${(a+b)}^{ⓓ}=a^{\#}$. □

Lemma 2.3.

Let $a\in {\mathcal{A}}^{Ⓓ}$ and $b\in {\mathcal{A}}^{nil}$. If $ba=0$, then $a+b\in {\mathcal{A}}^{Ⓓ}$. In this case,

$${(a+b)}^{Ⓓ}=a^{Ⓓ}.$$

Proof.

By virtue of Lemma 2.1, ${(a+b)}^{ⓓ}=a^{ⓓ}=a^{Ⓓ}$. Hence, $(a+b)-(a+b){\left[{(a+b)}^{ⓓ}\right]}^2=(a+b)-(a+b){\left(a^{Ⓓ}\right)}^2=[a-a{\left(a^{Ⓓ}\right)}^2]+b$. Since $a-a{\left(a^{Ⓓ}\right)}^2,b\in {\mathcal{A}}^{nil}$ and $b[a-a{\left(a^{Ⓓ}\right)}^2]=0$, we see that $(a+b)-(a+b){\left[{(a+b)}^{ⓓ}\right]}^2\in {\mathcal{A}}^{nil}$. Therefore $a+b\in {\mathcal{A}}^{Ⓓ}$ and

$${(a+b)}^{Ⓓ}=a^{Ⓓ}.$$

□

Lemma 2.4.

Let $a,e^2=e\in \mathcal{A}$ and $ea(1-e)=0$. If $a,a(1-e)\in {\mathcal{A}}^d$, then $ea\in \mathcal{A}$ and ${\left(ea\right)}^d=e{a}^d$.

Proof.

See [24]. □

Lemma 2.5.

Let $a\in {\mathcal{A}}^d$. Then $a\in {\mathcal{A}}^{\#}$ if and only if $\mathcal{A}a=\mathcal{A}{a}^2$.

Proof.

Write $a=z{x}^2$. Then $a=z{x}^2=z^n{x}^{n+1}=z^n(a^n-a^d{x}^{n+1})a+z^n{x}^d{x}^{n+2}$ and $a^2{a}^d=\left(z{x}^2\right)x{x}^d=z^n{x}^{n+2}{a}^d$. We verify that

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

Hence

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

We infer that $a=a^2{a}^d$. Then $a\in a^2\mathcal{A}\bigcap \mathcal{A}{a}^2$, and therefore $a\in {\mathcal{A}}^{\#}$. □

Lemma 2.6.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b\in {\mathcal{A}}^{nil}$. If $ba=0$, then x is nilpotent.

Proof.

Clearly, $x^2=\left(\begin{array}{cc}{a}^2+b& 1\\ 0& b\end{array}\right).$ Since $a\in {\mathcal{A}}^{nil}$, then so is $a^2$. As $b\in {\mathcal{A}}^{qnil}$ and $b{a}^2=0$, we see that $a^2+b\in {\mathcal{A}}^{nil}$. Therefore $x^2$ is nilpotent, and thus yielding the result. □

3. Main Results

Let $x\in R$ and $e\in R$ such that $e^2=e=p^{*}$ and $e^{\pi }=1-e$. Then x has the Peirce decomposition relative to e, i.e., $x=exe+ex{e}^{\pi }+e^{\pi }xe+e^{\pi }x{e}^{\pi }$. We denote it by a matrix form: ${\left(\begin{array}{cc}exe& ex{e}^{\pi }\\ e^{\pi }xe& e^{\pi }x{e}^{\pi }\end{array}\right)}_e$. We come now to investigate the generalized core-EP inverse for anti-triangular matrices over a Banach algebra, establishing novel existence criteria and deriving its explicit formula.

Theorem 3.1.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $b\in {\mathcal{A}}^{\#}$ and $b^{\sigma }a\in {\mathcal{A}}^{ⓓ}$. If $b^{\pi }ab=0$, then x has a generalized core-EP inverse. In this case,

$$x^{ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & (1-b^{\#}b){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b^{\#},\hfill \\ \hfill \gamma & =\hfill & b{b}^{\#}+(a{b}^{\#}b-b{b}^{\#}a){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a{b}^{\#}.\hfill \end{array}$$

Proof.

By hypothesis, $b^{\pi }ab=0$. It follows by Lemma 2.1 that $b^{\sigma }ab=0$. Set $e=\left(\begin{array}{cc}{b}^{\sigma }& 0\\ 0& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$. Then $M={\left(\begin{array}{cc}\theta & \zeta \\ \eta & \mu \end{array}\right)}_e$, where

$$\begin{array}{ccc}\hfill \theta & =\hfill & \left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}(1-b{b}^{\#})a& 0\\ 0& 0\end{array}\right),\hfill \\ \hfill \zeta & =\hfill & \left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}0& 1-b{b}^{\#}\\ 0& 0\end{array}\right),\hfill \\ \hfill \eta & =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}b{b}^{\#}a(1-b{b}^{\#})& 0\\ b-b^2{b}^{\#}& 0\end{array}\right),\hfill \\ \hfill \mu & =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}ab{b}^{\#}& b{b}^{\#}\\ b^2{b}^{\#}& 0\end{array}\right).\hfill \end{array}$$

Let $\xi =\left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)$. Then it is easy to check that

$$\begin{array}{ccc}\hfill \mu \xi & =\hfill & \left(\begin{array}{cc}ab{b}^{\#}& b{b}^{\#}\\ b^2{b}^{\#}& 0\end{array}\right)\left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& b{b}^{\#}\end{array}\right),\hfill \\ \hfill \xi \mu & =\hfill & \left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\left(\begin{array}{cc}ab{b}^{\#}& b{b}^{\#}\\ b^2{b}^{\#}& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& b{b}^{\#}\end{array}\right),\hfill \\ \hfill {\left(\mu \xi \right)}^{*}& =\hfill & \mu \xi ,\hfill \\ \hfill \mu {\xi }^2& =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& b{b}^{\#}\end{array}\right)\left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)=\xi ,\hfill \\ \hfill \xi {\mu }^2& =\hfill & \left(\begin{array}{cc}b{b}^{\#}& 0\\ 0& b{b}^{\#}\end{array}\right)\left(\begin{array}{cc}ab{b}^{\#}& b{b}^{\#}\\ b^2{b}^{\#}& 0\end{array}\right)=\left(\begin{array}{cc}ab{b}^{\#}& b{b}^{\#}\\ b^2{b}^{\#}& 0\end{array}\right)=\mu \hfill \end{array}$$

Thus

$$\begin{array}{ccc}\hfill {\mu }^{\#}& =\hfill & \xi =\left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right),\hfill \\ \hfill {\mu }^{\pi }& =\hfill & 1-\mu {\mu }^{\#}=\left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 1-b{b}^{\#}\end{array}\right),\hfill \\ \hfill {\mu }^{\pi }\eta & =\hfill & \left(\begin{array}{cc}1-b{b}^{\#}& 0\\ 0& 1-b{b}^{\#}\end{array}\right)\left(\begin{array}{cc}b{b}^{\#}a(1-b{b}^{\#})& 0\\ b-b^2{b}^{\#}& 0\end{array}\right)\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Moreover we verify that

$$\begin{array}{ccc}\hfill x& =\hfill & y+z,\hfill \\ \hfill y& =\hfill & \left(\begin{array}{cc}\theta & 0\\ \eta & \mu \end{array}\right),\hfill \\ \hfill z& =\hfill & \left(\begin{array}{cc}0& \zeta \\ 0& 0\end{array}\right).\hfill \end{array}$$

Set $w=\left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)$. Then we verify that

$$\begin{array}{ccc}\hfill yw& =\hfill & \left(\begin{array}{cc}\theta & 0\\ \eta & \mu \end{array}\right)\left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\theta {\theta }^{ⓓ}& 0\\ (1-\mu {\mu }^{\#})\eta {\theta }^{ⓓ}& \mu {\mu }^{\#}\end{array}\right)=\left(\begin{array}{cc}\theta {\theta }^{ⓓ}& 0\\ 0& \mu {\mu }^{\#}\end{array}\right),\hfill \\ \hfill {\left(yw\right)}^{*}& =\hfill & yw,\hfill \\ \hfill wy& =\hfill & \left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)\left(\begin{array}{cc}\theta & 0\\ \eta & \mu \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{\theta }^{ⓓ}\theta & 0\\ {\mu }^{\#}\eta (1-{\theta }^{ⓓ}\theta )& {\mu }^{\#}\mu \end{array}\right),\hfill \\ \hfill y{w}^2& =\hfill & \left(\begin{array}{cc}\theta {\theta }^{ⓓ}& 0\\ 0& \mu {\mu }^{\#}\end{array}\right)\left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)=w.\hfill \end{array}$$

Furthermore, we have

$$\begin{array}{ccc}\hfill (1-wy)y& =\hfill & \left(\begin{array}{cc}(1-{\theta }^{ⓓ}\theta )\theta & 0\\ -{\mu }^{\#}\eta (1-{\theta }^{ⓓ}\theta )\theta +(1-{\mu }^{ⓓ}\mu )\eta & (1-{\mu }^{\#}\mu )\mu \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}(1-{\theta }^{ⓓ}\theta )\theta & 0\\ -{\mu }^{\#}\eta (1-{\theta }^{ⓓ}\theta )\theta & 0\end{array}\right).\hfill \end{array}$$

$$(1-wy)y^n=\left(\begin{array}{cc}(1-{\theta }^{ⓓ}\theta ){\theta }^n& 0\\ -{\mu }^{\#}\eta (1-{\theta }^{ⓓ}\theta ){\theta }^n& 0\end{array}\right).$$

Thus,

$$\underset{n\to \infty }{lim}\left|\right|y^n-w{y}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Therefore $y\in {\mathcal{A}}^{ⓓ}$ and $y^{ⓓ}=w$.

Clearly, we have

$$\zeta \eta =\left(\begin{array}{cc}(1-b{b}^{\#})ab(1-b{b}^{\#})& 0\\ 0& 0\end{array}\right)=0,\zeta \mu =\left(\begin{array}{cc}(1-b{b}^{\#})b^2{b}^{\#}& 0\\ 0& 0\end{array}\right)=0.$$

Then $zy=\left(\begin{array}{cc}\zeta \eta & \zeta \mu \\ 0& 0\end{array}\right)=0$. By virtue of Lemma 2.2,

$$\begin{array}{ccc}\hfill x^{ⓓ}& =\hfill & y^{ⓓ}={\left(\begin{array}{cc}{\theta }^{ⓓ}& 0\\ -{\mu }^{\#}\eta {\theta }^{ⓓ}& {\mu }^{\#}\end{array}\right)}_e\hfill \\ & =\hfill & {\theta }^{ⓓ}+{\mu }^{\#}-{\mu }^{\#}\eta {\theta }^{ⓓ}\hfill \\ & =\hfill & \left(\begin{array}{cc}{\left((1-b{b}^{\#})a\right)}^{ⓓ}& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\hfill \\ & -\hfill & \left(\begin{array}{cc}0& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\left(\begin{array}{cc}b{b}^{\#}a(1-b{b}^{\#})& 0\\ b-b^2{b}^{\#}& 0\end{array}\right)\left(\begin{array}{cc}{\left((1-b{b}^{\#})a\right)}^{ⓓ}& 0\\ 0& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{\left((1-b{b}^{\#})a\right)}^{ⓓ}& b^{\#}\\ b{b}^{\#}& -a{b}^{\#}\end{array}\right)\hfill \\ & -\hfill & \left(\begin{array}{cc}\left(b^{\#}b-b{b}^{\#}\right){\left(b^{\sigma }a\right)}^{ⓓ}& 0\\ \left(b{b}^{\#}a(1-b{b}^{\#})-a(b^{\#}b-b{b}^{\#})\right){\left(b^{\sigma }a\right)}^{ⓓ}& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),\hfill \end{array},$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {\left(b^{\sigma }a\right)}^{ⓓ}-\left(b^{\#}b\right){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b^{\#},\hfill \\ \hfill \gamma & =\hfill & b{b}^{\#}-\left(b{b}^{\#}\right)a{\left(b^{\sigma }a\right)}^{ⓓ}+a\left(b^{\#}b\right){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a{b}^{\#},\hfill \end{array}$$

as required. □

Corollary 3.2.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $b\in {\mathcal{A}}^{\#}$ and $b^{\sigma }a\in {\mathcal{A}}^{ⓓ}$. If $b^{\pi }ab=0$, then x has a core inverse if and only if

$${\left(b^{\sigma }a\right)}^{ⓓ}b=0,{\left(b^{\sigma }a\right)}^{ⓓ}a=b^{\sigma }.$$

In this case,

$$x^{\#}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & (1-b^{\#}b){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b^{\#},\hfill \\ \hfill \gamma & =\hfill & b{b}^{\#}+(a{b}^{\#}b-b{b}^{\#}a){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a{b}^{\#}.\hfill \end{array}$$

Proof.

By virtue of Theorem 3.1, x has a generalized core-EP inverse. In this case,

$$x^{ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & (1-b^{\#}b){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b^{\#},\hfill \\ \hfill \gamma & =\hfill & b{b}^{\#}+(a{b}^{\#}b-b{b}^{\#}a){\left(b^{\sigma }a\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a{b}^{\#}.\hfill \end{array}$$

Then we see that

$$I_2-\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)=\left(\begin{array}{cc}1-\alpha a-\beta b& -\alpha \\ -\gamma a-\delta b& 1-\gamma \end{array}\right).$$

Hence $(I_2-x^{ⓓ}x)x=0$ if and only if

$$\begin{array}{cc}\hfill \alpha b& =0\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \alpha a+\beta b& =1\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill (1-\gamma )b& =0\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \gamma a+\delta b& =0\hfill \end{array}$$

(4)

⟹ By hypothesis, x has a core inverse. Then $x^{\#}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$. By $\left(1\right)$, we get

$${\left(b^{\sigma }a\right)}^{ⓓ}b=0.$$

From $\left(2\right)$, we see that

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

as desired.

⟸ By hypothesis,

$${\left(b^{\sigma }a\right)}^{ⓓ}b=0,{\left(b^{\sigma }a\right)}^{ⓓ}a=b^{\sigma }.$$

We check that

$$\begin{array}{ccc}\hfill \alpha b& =\hfill & (1-b^{\#}b){\left(b^{\sigma }a\right)}^{ⓓ}b=0,\hfill \\ \hfill \alpha a+\beta b& =\hfill & b^{\tau }{\left(b^{\sigma }a\right)}^{ⓓ}a+b^{\#}b\hfill \\ & =\hfill & b^{\tau }{b}^{\sigma }+b^{\#}b=(1-b^{\#}b)(1-b{b}^{\#})+b^{\#}b\hfill \\ & =\hfill & 1-b{b}^{\#}-b^{\#}b+b^{\#}{b}^2{b}^{\#}+b^{\#}b=1,\hfill \\ \hfill (1-\gamma )b& =\hfill & (1-b{b}^{\#})b-(a{b}^{\#}b-b{b}^{\#}a){\left(b^{\sigma }a\right)}^{ⓓ}b=0,\hfill \\ \hfill \gamma a+\delta b& =\hfill & b{b}^{\#}a+(a{b}^{\#}b-b{b}^{\#}a){\left(b^{\sigma }a\right)}^{ⓓ}a-a{b}^{\#}b\hfill \\ & =\hfill & b{b}^{\#}a+(a{b}^{\#}b-b{b}^{\#}a)b^{\sigma }-a{b}^{\#}b\hfill \\ & =\hfill & b{b}^{\#}a+a{b}^{\#}b{b}^{\sigma }-b{b}^{\#}a{b}^{\sigma }-a{b}^{\#}b\hfill \\ & =\hfill & b{b}^{\#}a+(a{b}^{\#}b-ab{b}^{\#})-b{b}^{\#}a+ab{b}^{\#}-a{b}^{\#}b=0.\hfill \end{array}$$

Then equality $\left(1\right)-\left(4\right)$ holds. Therefore $(I_2-x^{ⓓ}x)x=0$. Accordingly, x has a core inverse, as required. □

Corollary 3.3.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b\in {\mathcal{A}}^{\#}$. If $ab=ba$ and $a^{*}b=b{a}^{*}$, then x has a core inverse if and only if $a^{\pi }{b}^{\pi }=0.$ In this case,

$$x^{\#}=\left(\begin{array}{cc}{a}^{\#}(1-b^{\#}b)& b^{\#}\\ b^{\#}b& -a{b}^{\#}\end{array}\right).$$

Proof.

Since $ab=ba$ and $a^{*}b=b{a}^{*}$, it follows by [10] that $a{b}^{\#}=b^{\#}a$. Then we see that $b^{\sigma }a=a{b}^{\sigma }$. As ${\left(b^{\sigma }\right)}^{*}=b^{\sigma }$, we have $b^{\sigma }{a}^{*}=a^{*}{b}^{\sigma }$. In view of [10], $b^{\sigma }a\in {\mathcal{A}}^{\#}$ and ${\left(b^{\sigma }a\right)}^{\#}=b^{\sigma }{a}^{\#}$. Moreover, we have $b^{\pi }ab=\left(b^{\pi }b\right)a=0$. By virtue of x has a core inverse if and only if

$${\left(b^{\sigma }a\right)}^{ⓓ}b=0,{\left(b^{\sigma }a\right)}^{ⓓ}a=b^{\sigma }.$$

Obviously, ${\left(b^{\sigma }a\right)}^{ⓓ}b=\left(b^{\sigma }{a}^{\#}\right)b=\left(b^{\sigma }b\right)a^{\#}=0$. Moreover, we see that ,

$$b^{\sigma }-{\left(b^{\sigma }a\right)}^{ⓓ}a=b^{\sigma }-b^{\sigma }{a}^{\#}a=b^{\sigma }{a}^{\tau }.$$

If ${\left(b^{\sigma }a\right)}^{ⓓ}a=b^{\sigma }$, then $b^{\sigma }{a}^{\tau }=0$. By virtue of Lemma 2.1, $b^{\pi }{a}^{\tau }=0$. Hence, $a^{\tau }{b}^{\pi }=0$. By using Lemma 2.1 again, $a^{\pi }{b}^{\pi }=0$.

If $a^{\pi }{b}^{\pi }=0$, it follows by Lemma 2.1 that $a^{\tau }{b}^{\pi }=0$. Hence, $b^{\pi }{a}^{\tau }=0$. By Lemma 2.1 again, $b^{\sigma }{a}^{\tau }=0$. Thus $b^{\sigma }=b^{\sigma }{a}^{\#}a$. This implies that ${\left(b^{\sigma }a\right)}^{\#}a=b^{\sigma }$. Thus x has a core inverse if and only if $a^{\pi }{b}^{\pi }=0.$

In this case, we have $a^{\sigma }{b}^{\tau }=a^{\sigma }{b}^{\sigma }=0$ by Lemma 2.1. Then we directly verifies that

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

Therefore we complete the proof by Theorem 3.1. □

We are ready to prove:

Theorem 3.4.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b,a{a}^{ⓓ}{b}^{\sigma }{a}^2{a}^{ⓓ}\in {\mathcal{A}}^{ⓓ}$ and $a^{\sigma }b\in {\mathcal{A}}^{qnil}$. If $b^{\pi }ab=b^{\pi }ba=a^{\pi }ba=a^{\pi }{b}^{*}a=0$, then x has a generalized core-EP inverse. In this case,

$$x^{ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & a{a}^{ⓓ}{b}^{\pi }{\left(a{a}^{ⓓ}{b}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b{b}^{d}a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ},\hfill \\ \hfill \gamma & =\hfill & a{a}^{ⓓ}b{b}^{ⓓ}a{a}^{ⓓ}+\left(a^2{a}^{ⓓ}b{b}^{d}a{a}^{ⓓ}-a{a}^{ⓓ}b{b}^{ⓓ}{a}^2{a}^{ⓓ}\right){\left(a{a}^{ⓓ}{b}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a^2{a}^{ⓓ}b{b}^{d}a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ}.\hfill \end{array}$$

Proof.

As $a^{\pi }ba=0$, it follows by Lemma 2.1 that $a^{\sigma }ba=0$. Then $ba=a{a}^{ⓓ}ba$, and so $b=ba{a}^{ⓓ}+b(1-a{a}^{ⓓ})=a{a}^{ⓓ}ba{a}^{ⓓ}+b(1-a{a}^{ⓓ})$. Thus we get $x=y+z,$ where

$$y=\left(\begin{array}{cc}{a}^2{a}^{ⓓ}& a{a}^{ⓓ}\\ a{a}^{ⓓ}ba{a}^{ⓓ}& 0\end{array}\right),z=\left(\begin{array}{cc}a(1-a{a}^{ⓓ})& 1-a{a}^{ⓓ}\\ b(1-a{a}^{ⓓ})& 0\end{array}\right).$$

Claim 1. $zy=0$. One directly verifies that

$$\begin{array}{ccc}\hfill zy& =\hfill & \left(\begin{array}{cc}a(1-a{a}^{ⓓ})& 1-a{a}^{ⓓ}\\ b(1-a{a}^{ⓓ})& 0\end{array}\right)\left(\begin{array}{cc}{a}^2{a}^{ⓓ}& a{a}^{ⓓ}\\ a{a}^{ⓓ}ba{a}^{ⓓ}& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}a(1-a{a}^{ⓓ})a^2{a}^{ⓓ}& a(1-a{a}^{ⓓ})a{a}^{ⓓ}\\ b(1-a{a}^{ⓓ})a^2{a}^{ⓓ}& b(1-a{a}^{ⓓ})a{a}^{ⓓ}\end{array}\right)=0.\hfill \end{array}$$

Claim 2. $z\in M_2\left(\mathcal{A}\right)$ is quasinilpotent. Let $0\ne \lambda \in C$. Then

$$\begin{array}{ccc}\hfill I_2+\lambda z& =\hfill & \left(\begin{array}{cc}1+\lambda a(1-a{a}^{ⓓ})& \lambda (1-a{a}^{ⓓ})\\ \lambda b(1-a{a}^{ⓓ})& \lambda \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}1& 1-a{a}^{ⓓ}\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1+\lambda [a{a}^{\sigma }-a^{\sigma }b]& 0\\ \lambda b{a}^{\sigma }& \lambda \end{array}\right).\hfill \end{array}$$

Since $a{a}^{\sigma },a^{\sigma }b\in {\mathcal{A}}^{qnil}$ and $\left(a^{\sigma }b\right)\left(a{a}^{\sigma }\right)=\left(a^{\sigma }ba\right)a^{\sigma }=0$, it follows by [2] that $a{a}^{\sigma }-a^{\sigma }b\in {\mathcal{A}}^{qnil}$. Hence, $1+\lambda [a{a}^{\sigma }-a^{\sigma }b]\in {\mathcal{A}}^{-1}$. Therefore $I_2+\lambda z$ is invertible. This implies that $z\in M_2{\left(\mathcal{A}\right)}^{qnil}$.

Claim 3. Obviously, we have $a^2{a}^{ⓓ}=\left(a{a}^{ⓓ}\right)a\left(a{a}^{ⓓ}\right)\in \left(a{a}^{ⓓ}\right)\mathcal{A}\left(a{a}^{ⓓ}\right)$. Then

$$y=\left(\begin{array}{cc}{a}^2{a}^{ⓓ}& a{a}^{ⓓ}\\ a{a}^{ⓓ}ba{a}^{ⓓ}& 0\end{array}\right)\in M_2\left(\left(a{a}^{ⓓ}\right)\mathcal{A}\left(a{a}^{ⓓ}\right)\right).$$

We directly verify that

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

As $a^{\pi }{b}^{*}a=0$, by Lemma 2.1, we have $a^{\sigma }{b}^{*}a=0$, and so $a^{\sigma }{b}^{*}a{a}^{ⓓ}=0$. This implies that $a{a}^{ⓓ}b{a}^{\sigma }=0$. Hence $a{a}^{ⓓ}ba{a}^{ⓓ}=a{a}^{ⓓ}b.$

Since $b^{\pi }ba=0$, we have $ba=b{b}^{d}ba$. Thus

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

Thus we deduce that

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

Thus, $a{a}^{ⓓ}ba{a}^{ⓓ}\in \mathcal{A}$ has a $(1,3)$-inverse in $a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ}$.

Moreover, we have

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

This implies that $\mathcal{A}\left(ba{a}^{ⓓ}\right)=\mathcal{A}{\left(ba{a}^{ⓓ}\right)}^2$.

Since $(1-a{a}^{ⓓ})ba{a}^{ⓓ}=0$ and $a^{\sigma }b$ is quasinilpotent, it follows by Lemma 2.4 that ${\left(ba{a}^{ⓓ}\right)}^d=b^{d}a{a}^{ⓓ}$.

In light of Lemma 2.5, $ba{a}^{ⓓ}\in {\mathcal{A}}^{\#}$. By using [23], $ba{a}^{ⓓ}\in {\mathcal{A}}^{\#}$. In view of Lemma 2.1, we derive that

$$\begin{array}{ccc}\hfill {\left(a{a}^{ⓓ}ba{a}^{ⓓ}\right)}^{\#}& =\hfill & {\left(ba{a}^{ⓓ}\right)}^{\#}\hfill \\ & =\hfill & {\left(ba{a}^{ⓓ}\right)}^{\#}\left(ba{a}^{ⓓ}\right)a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ}\hfill \\ & =\hfill & {\left(ba{a}^{ⓓ}\right)}^{\#}ba{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ}\hfill \\ & \in \hfill & a{a}^{ⓓ}\mathcal{A}a{a}^{ⓓ},\hfill \end{array}$$

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

$$\begin{array}{cc}& {\left(a{a}^{ⓓ}ba{a}^{ⓓ}\right)}^{\pi }\left(a^2{a}^{ⓓ}\right)\left(a{a}^{ⓓ}ba{a}^{ⓓ}\right)\hfill \\ \hfill =& a{a}^{ⓓ}{b}^{\pi }a{a}^{ⓓ}a\left(a{a}^{ⓓ}ba\right)a^{ⓓ}\hfill \\ \hfill =& a{a}^{ⓓ}{b}^{\pi }a\left(a{a}^{ⓓ}ba\right)a^{ⓓ}\hfill \\ \hfill =& a{a}^{ⓓ}\left(b^{\pi }ab\right)a{a}^{ⓓ})=0.\hfill \end{array}$$

By virtue of Theorem 3.1, y has a generalized core-EP inverse in $\left(a{a}^{ⓓ}\right)\mathcal{A}\left(a{a}^{ⓓ}\right).$

Furthermore, we have

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

Accordingly, we have

$$y^{ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & \left(a{a}^{ⓓ}-{\left(ba{a}^{ⓓ}\right)}^{\#}\left(ba{a}^{ⓓ}\right)\right){\left({\left(ba{a}^{ⓓ}\right)}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & (ba{a}^{ⓓ}{)}^{\#},\hfill \\ \hfill \gamma & =\hfill & \left(ba{a}^{ⓓ}\right){\left(ba{a}^{ⓓ}\right)}^{\#}+(\left(a^2{a}^{ⓓ}\right){\left(ba{a}^{ⓓ}\right)}^{\#}\left(ba{a}^{ⓓ}\right)\hfill \\ & -\hfill & \left(ba{a}^{ⓓ}\right){\left(ba{a}^{ⓓ}\right)}^{\#}\left(a^2{a}^{ⓓ}\right)){\left({\left(ba{a}^{ⓓ}\right)}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a^2{a}^{ⓓ}{\left(ba{a}^{ⓓ}\right)}^{\#}.\hfill \end{array}$$

We compute that

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

Hence, we derive that

$$\begin{array}{ccc}\hfill \alpha & =\hfill & a{a}^{ⓓ}{\left(ba{a}^{ⓓ}\right)}^{\pi }{\left(a{a}^{ⓓ}{b}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \beta & =\hfill & b{b}^{d}a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ},\hfill \\ \hfill \gamma & =\hfill & a{a}^{ⓓ}b{b}^{ⓓ}a{a}^{ⓓ}+(a^2{a}^{ⓓ}ba{a}^{ⓓ}{\left(ba{a}^{ⓓ}\right)}^{\#}\hfill \\ & -\hfill & a{a}^{ⓓ}b{b}^{ⓓ}a{a}^{ⓓ}\left(a^2{a}^{ⓓ}\right)){\left({\left(ba{a}^{ⓓ}\right)}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ}\hfill \\ & =\hfill & a{a}^{ⓓ}b{b}^{ⓓ}a{a}^{ⓓ}+\left(a^2{a}^{ⓓ}b{b}^{d}a{a}^{ⓓ}-a{a}^{ⓓ}b{b}^{ⓓ}{a}^2{a}^{ⓓ}\right){\left(a{a}^{ⓓ}{b}^{\sigma }{a}^2{a}^{ⓓ}\right)}^{ⓓ},\hfill \\ \hfill \delta & =\hfill & -a^2{a}^{ⓓ}b{b}^{d}a{a}^{ⓓ}{b}^{ⓓ}a{a}^{ⓓ}.\hfill \end{array}$$

Therefore x has a generalized core-EP inverse by Lemma 2.2 and $x^{ⓓ}=y^{ⓓ}$, as desired. □

Corollary 3.5.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a\in {\mathcal{A}}^{ⓓ},b\in {\mathcal{A}}^{qnil}$. If $ab=ba=a^{*}b=0$, then x has a generalized core-EP inverse. In this case,

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

Proof.

Since $ab=ba=a^{*}b=0$, we have $a{a}^{ⓓ}b={\left(a^{ⓓ}\right)}^{*}\left(a^{*}b\right)=0,b{b}^{d}a=b^d\left(ba\right)=0$ and $b{b}^{ⓓ}a={\left(b^{ⓓ}\right)}^{*}{\left(a^{*}b\right)}^{*}=0$. Therefore we complete the proof by Theorem 3.4. □

Theorem 3.6.

Let $x=\left(\begin{array}{cc}a& 1\\ b& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b,a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\in {\mathcal{A}}^{Ⓓ}$ and $a^{\sigma }b\in {\mathcal{A}}^{qnil}$. If $b^{\pi }ab=b^{\pi }ba=a^{\pi }ba=a^{\pi }{b}^{*}a=0$, then x has a core-EP inverse. In this case, In this case,

$$x^{Ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & a{a}^{Ⓓ}{b}^{\pi }{\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ},\hfill \\ \hfill \beta & =\hfill & b{b}^{d}a{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ},\hfill \\ \hfill \gamma & =\hfill & a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}+\left(a^2{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}-a{a}^{Ⓓ}b{b}^{Ⓓ}{a}^2{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ},\hfill \\ \hfill \delta & =\hfill & -a^2{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}.\hfill \end{array}$$

Proof.

By virtue of Theorem 3.4, x has a generalized core-EP inverse. As in the proof of Theorem 3.4, $x=y+z,y\in M_2{\left(\mathcal{A}\right)}^{Ⓓ},z\in M_2{\left(\mathcal{A}\right)}^{nil}$ and $zy=0$, where

$$y=\left(\begin{array}{cc}{a}^2{a}^{Ⓓ}& a{a}^{Ⓓ}\\ a{a}^{Ⓓ}ba{a}^{Ⓓ}& 0\end{array}\right),z=\left(\begin{array}{cc}a(1-a{a}^{Ⓓ})& 1-a{a}^{Ⓓ}\\ b(1-a{a}^{Ⓓ})& 0\end{array}\right).$$

We see that $x{\left(x^{ⓓ}\right)}^2=x^{ⓓ}$ and ${\left(x{x}^{ⓓ}\right)}^{*}=x{x}^{ⓓ}$. Moreover, we have

$$y^{ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & a{a}^{Ⓓ}{\left(ba{a}^{Ⓓ}\right)}^{\pi }{\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ},\hfill \\ \hfill \beta & =\hfill & b{b}^{d}a{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ},\hfill \\ \hfill \gamma & =\hfill & a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}+\left(a^2{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}-a{a}^{Ⓓ}b{b}^{Ⓓ}{a}^2{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ},\hfill \\ \hfill \delta & =\hfill & -a^2{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}.\hfill \end{array}$$

We directly compute that

$$\begin{array}{ccc}\hfill y{y}^{ⓓ}& =\hfill & \left(\begin{array}{cc}{a}^2{a}^{Ⓓ}& a{a}^{Ⓓ}\\ a{a}^{Ⓓ}ba{a}^{Ⓓ}& 0\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}+(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}& 0\\ 0& a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}\end{array}\right).\hfill \end{array}$$

Then

$$a{a}^{Ⓓ}{I}_2-y{y}^{ⓓ}=\left(\begin{array}{cc}a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}-(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}& 0\\ 0& a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}\end{array}\right).$$

It is easy to verify that

$$\begin{array}{cc}& a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}\hfill \\ \hfill =& a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}\hfill \\ \hfill =& a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}-a{a}^{Ⓓ}b{b}^{Ⓓ}a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}\hfill \\ \hfill =& a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}-\left[a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}ba{a}^{Ⓓ}\right]b^{Ⓓ}a{a}^{Ⓓ}\hfill \\ \hfill =& a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}-a{a}^{Ⓓ}ba{a}^{Ⓓ}{b}^{Ⓓ}a{a}^{Ⓓ}=0.\hfill \end{array}$$

Therefore $a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}=a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}$.

Further, we derive that

$$\begin{array}{cc}& \left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}{a}^2{a}^{Ⓓ}\hfill \\ \hfill =& \left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right).\hfill \end{array}$$

Accordingly, we deduce that

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

Set

$$s=\left(\begin{array}{cc}{a}^2{a}^{Ⓓ}& a{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}\\ a{a}^{Ⓓ}ba{a}^{Ⓓ}& 0\end{array}\right),t=\left(\begin{array}{cc}0& a{a}^{Ⓓ}-\left(a{a}^{Ⓓ}ba{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}\right)}^{\#}\\ 0& 0\end{array}\right).$$

As in the proof of Theorem 3.4, we see that t is nilpotent, s has a generalized core-EP inverse and $ts=0$. In this case, we have $y^{ⓓ}=s^{ⓓ}$. Hence $I_2-y{y}^{ⓓ}=I_2-s{s}^{ⓓ}$. We check that

$$\begin{array}{cc}& (I_2-s{s}^{ⓓ})s=(I_2-y{y}^{ⓓ})s\hfill \\ \hfill =& \left(\begin{array}{cc}a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}-(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}& 0\\ 0& a{a}^{Ⓓ}{b}^{\sigma }a{a}^{Ⓓ}\end{array}\right)\hfill \\ & \left(\begin{array}{cc}{a}^2{a}^{Ⓓ}& a{a}^{Ⓓ}b{b}^{D}a{a}^{Ⓓ}\\ a{a}^{Ⓓ}ba{a}^{Ⓓ}& 0\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}-\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right){\left(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ}\right)}^{Ⓓ}(a{a}^{Ⓓ}{b}^{\sigma }{a}^2{a}^{Ⓓ})& 0\\ 0& 0\end{array}\right).\hfill \end{array}$$

Hence, $s-s^{ⓓ}{s}^2$ is nilpotent. This implies that s has a core-EP inverse. Since $y=s+t$, it follows by Lemma 2.3 that y has a core-EP inverse.

Since $(1-a{a}^{Ⓓ})b$ and $(1-a{a}^{Ⓓ})a$ is nilpotent, so are $(1-a{a}^{Ⓓ})b(1-a{a}^{Ⓓ})$ and $(1-a{a}^{Ⓓ})a(1-a{a}^{Ⓓ}).$ By hypothesis, $(1-a{a}^{Ⓓ})b(1-a{a}^{Ⓓ})=(1-a{a}^{Ⓓ})b$ is nilpotent and $\left[(1-a{a}^{Ⓓ})b(1-a{a}^{Ⓓ})\right]\left[(1-a{a}^{Ⓓ})a(1-a{a}^{Ⓓ})\right]=0$. It follows by Lemma 2.6 that $\left(\begin{array}{cc}(1-a{a}^{Ⓓ})a(1-a{a}^{Ⓓ})& (1-a{a}^{Ⓓ})\\ (1-a{a}^{Ⓓ})b(1-a{a}^{Ⓓ})& 0\end{array}\right)$ is nilpotent. This implies that $z=\left(\begin{array}{cc}a(1-a{a}^{Ⓓ})& 1-a{a}^{Ⓓ}\\ b(1-a{a}^{Ⓓ})& 0\end{array}\right)$ is nilpotent. By using Lemma 2.3 again, x has a core-EP inverse and $x^{Ⓓ}=x^{ⓓ}$, as required. □

We illustrate Theorem 3.6 by the following numerical example.

Example 3.7.

Let $A=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& -i\\ 0& 0& 0& 0& 0& 1\end{array}\right),B=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& i& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right)\in C^{6\times 6}$. We take the involution on $C^{6\times 6}$ as the conjugate transpose. We compute that

$$A^{Ⓓ}=A^D=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& i\\ 0& 0& 0& 0& 0& 1\end{array}\right),B^{Ⓓ}=B^D=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right).$$

Then

$$A^{\sigma }=A^{\pi }=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right),B^{\sigma }=B^{\pi }=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right).$$

We verify that $A^{\sigma }B=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& i& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right)$ is nilpotent. Moreover, we directly verify that

$$B^{\pi }AB=B^{\pi }BA=A^{\pi }BA=A^{\pi }{B}^{*}A=0.$$

Set $X=\left(\begin{array}{cc}A& I_6\\ B& 0\end{array}\right)$.

By virtue of Theorem 3.6, we have

$$X^{Ⓓ}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),$$

where

$$\begin{array}{cc}\hfill \alpha & =A{A}^{\mathring{D}}{B}^{\pi }{\left(A{A}^{\mathring{D}}{B}^{\sigma }{A}^2{A}^{\mathring{D}}\right)}^{\mathring{D}},\hfill \\ \hfill \beta & =B{B}^{D}A{A}^{\mathring{D}}{B}^{\mathring{D}}A{A}^{\mathring{D}},\hfill \\ \hfill \gamma & =A{A}^{\mathring{D}}B{B}^{\mathring{D}}A{A}^{\mathring{D}}\hfill \\ \hfill & +\left(A^2{A}^{\mathring{D}}B{B}^{D}A{A}^{\mathring{D}}-A{A}^{\mathring{D}}B{B}^{\mathring{D}}{A}^2{A}^{\mathring{D}}\right){\left(A{A}^{\mathring{D}}{B}^{\sigma }{A}^2{A}^{\mathring{D}}\right)}^{\mathring{D}},\hfill \\ \hfill \delta & =-A^2{A}^{\mathring{D}}B{B}^{D}A{A}^{\mathring{D}}{B}^{\mathring{D}}A{A}^{\mathring{D}}.\hfill \end{array}$$

Then

$$\begin{array}{c}\alpha =\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& i\\ 0& 0& 0& 0& 0& 1\end{array}\right),\beta =\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right),\\ \gamma =\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right),\delta =\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right).\end{array}$$

Evidently, we verify that $X^{Ⓓ}=X{\left(X^{Ⓓ}\right)}^2,{\left(X{X}^{Ⓓ}\right)}^{*}=X{X}^{Ⓓ}$ and $X-X^{Ⓓ}{X}^2$ is nilpotent, as desired.

Remark 3.8.

1. Analogously, to the preceding Theorems, the existence and representation of $x=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,b\in {\mathcal{A}}^{ⓓ}$ can be obtained by a similar route.

2. The generalized dual core-EP inverse can be discussed by the dual approach.