Generalized core-EP inverse for triangular operator matrices

1. Introduction

Let $\mathcal{A}$ be a Banach *-algebra. An element $a\in \mathcal{A}$ has Drazin inverse provided that there exists $x\in \mathcal{A}$ such that

$$a{x}^2=x,ax=xa,x{a}^{k+1}=a^k,$$

where k is the index of a (denoted by $ind\left(a\right)$), i.e., the smallest k such that the previous equations are satisfied. Such x is unique if exists, denoted by $a^D$, and called the Drazin inverse of a. We say that $a\in \mathcal{A}$ has group inverse x if $ind\left(a\right)=1$, i.e., there exists a unique $x\in \mathcal{A}$ such that

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

We denote the group inverse x by $a^{\#}$. Evidently, a square complex matrix A has group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$.

An element $a\in \mathcal{A}$ has core-EP inverse (i.e., pseudo core inverse) if there exist $x\in \mathcal{A}$ and $k\in N$ such that

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

where the smallest k is the index of a (denoted by $i\left(a\right)$). If such x exists, it is unique, and denote it by $a^D$. We say that $a\in \mathcal{A}$ has core inverse x if $i\left(a\right)=1$, i.e., there exists a unique $x\in \mathcal{A}$ such that

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

We denote the core inverse x by $a^{\#}$. As is well known, an element $a\in \mathcal{A}$ has core inverse x if and only if

$$a=axa,x\mathcal{A}=a\mathcal{A},\mathcal{A}x=\mathcal{A}{a}^{*}.$$

As a natural generalization of core-EP invertibility, the authors introduced the generalized core-EP inverse in a Banach algebra $\mathcal{A}$ with an involution *. An element $a\in \mathcal{A}$ has generalized core-EP inverse if there exists $x\in \mathcal{A}$ such that

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

If such x exists, it is unique, and denote it by $a^d$.

The generalized inverses mentioned above are powerful tools in linear algebra and operator algebra for dealing with matrices and operators that do not have a traditional inverse. They are used in various applications and provide a means to find solutions to linear systems and has applications across various scientific and engineering disciplines. Recently, many authors have studied them from many different views, e.g., [2,4,5,7,8,9,10,11,13,17,20,21,24,25].

Recall that $a\in \mathcal{A}$ has generalized Drazin inverse if there exists $x\in \mathcal{A}$ such that

$$a{x}^2=x,ax=xa,a-a^{2}x\in {\mathcal{A}}^{qnil}.$$

Here, ${\mathcal{A}}^{qnil}=\{a\in \mathcal{A}\mid 1+\lambda a\in {\mathcal{A}}^{-1}\}.$ Such x is unique, if exists, and denote it by $a^d$. We use ${\mathcal{A}}^d,{\mathcal{A}}^{\#}$ and ${\mathcal{A}}^d$ to denote the sets of all generalized Drazin inverse, core and generalized core-EP invertible elements in $\mathcal{A}$, respectively. If a and x satisfy the equations $a=axa$ and ${\left(ax\right)}^{*}=ax$, then x is called $(1,3)$-inverse of a and is denoted by $a^{(1,3)}$. We use ${\mathcal{A}}^{(1,3)}$ to stand for sets of all $(1,3)$-invertible elements in $\mathcal{A}$. We list several characterizations of generalized core-EP inverse.

Theorem 1.1.(see [3,6])Let $\mathcal{A}$ be a Banach *-algebra, and let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

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

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

(6)

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

(7)

$a\in {\mathcal{A}}^d$ and there exists a projection $q\in \mathcal{A}$ such that $a^d\mathcal{A}=q\mathcal{A}$. In this case, $a^d=a^{d}q.$

The motivation of this paper is to investigate the generalized core-EP inverse for the triangular matrices over a Banach *-algebra.

In Section 2, we establish necessary and sufficient conditions under which the block operator triangular matrix $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ over a Banach algebra has the generalized core-EP inverse with upper triangular form.

A $C^{*}$-algebra is a Banach algebra equipped with an involution operation * that satisfies satisfies the $C^{*}$-identity: $\left|\right|x^{*}x\left|\right|={\left|\right|x\left|\right|}^2$ for all $x\in \mathcal{A}$. In Section 3, we particularly investigate the generalized core-EP inverse of a triangular block operator matrices over a $C^{*}$-algebra. We prove that every triangular operator matrix $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ over a $C^{*}$-algebra with generalized core-EP invertible diagonal entries has the generalized core-EP inverse and its representation of generalized core-EP inverse is presented.

The set of all bounded linear operators on a Hilbert space H, denoted $\mathcal{B}\left(H\right)$, forms a $C^{*}$-algebra with the operator norm and the adjoint operation. Lex X and Y be Hilbert spaces. We use $\mathcal{B}(X,Y)$ to stand for the set of all bounded linear operators from X to Y. Finally, in Section 4, we apply our results and study the generalized core-EP inverse for the block operator matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right),$ where $A\in \mathcal{B}{\left(X\right)}^d,B\in \mathcal{B}(X,Y),C\in \mathcal{B}(Y,X),D\in \mathcal{B}{\left(Y\right)}^d$. Here, M is a linear operator on Hilbert space $X\oplus Y$.

Throughout the paper, all Banach *-algebras are complex with an identity. An element $p\in \mathcal{A}$ is a projection if $p^2=p=p^{*}$. ${\mathcal{A}}^D,{\mathcal{A}}^d$ and ${\mathcal{A}}^{nil}$ denote the sets of all Drazin, generalized core-EP invertible and nilpotent elements in $\mathcal{A}$ respectively. Let $a\in {\mathcal{A}}^d$. We use $a^{\pi }$ to stand for the spectral idempotent $a^{\pi }=1-a{a}^d$.

2. Triangular Operator Matrices over Banach *-Algebras

Let $\mathcal{A}$ be a Banach *-algebra. Then $M_2\left(\mathcal{A}\right)$ is a Banach *-algebra with *-transpose as the involution. We come now to generalized EP-inverse of a triangular matrix over $\mathcal{A}$. To prove the main results, some lemmas are needed. We begin with

Lemma 2.1. 

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

(1)

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

(2)

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

(3)

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

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

$\left(3\right)\Rightarrow \left(2\right)$ Since $a^d={\left(a^d\right)}^{2}a=a^d\left[a^d{\left(a^d\right)}^{\#}{a}^d\right]a=\left[{\left(a^d\right)}^2{\left(a^d\right)}^{\#}\right]a{a}^d=a^{d}a{a}^d$. Then $b=a{a}^{d}b=a^d{a}^2{a}^{d}b$; and so $(1-a{a}^d)b=(1-a{a}^d)a^d{a}^2{a}^{d}b=0$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Since $(1-a{a}^d)b=0$, we get $b=a{a}^{d}b$. Therefore $(1-a^{d}a)b=(1-a^d)a{a}^{d}b=(a-a^d{a}^2)a^{d}b=0$, as asserted. □

Let $\mathcal{A}$ be a Banach *-algebra. Then $M_2\left(\mathcal{A}\right)$ is a Banach *-algebra with *-transpose as the involution. We come now to generalized EP-inverse of a triangular matrix over $\mathcal{A}$.

Lemma 2.2. 

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

(1)

If $a,d\in {\mathcal{A}}^d$, then $x\in M_2{\left(\mathcal{A}\right)}^d$ and $x^d=\left(\begin{array}{cc}{a}^d& z\\ 0& d^d\end{array}\right)$, where

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

(2)

If $a,d\in {\mathcal{A}}^{\#}$ and $a^{\pi }b=0$, then $x\in M_2{\left(\mathcal{A}\right)}^{\#}$ and

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

Proof. 

See [26][Lemma 2.1] and [23][Theorem 2.5]. □

We are ready to prove:

Theorem 2.3. 

Let $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)\in M_2\left(\mathcal{A}\right)$ with $a,d\in {\mathcal{A}}^d$. Then the following are equivalent:

(1)

$x\in M_2\left(\mathcal{A}\right)$ has upper triangular generalized core-EP inverse.

(2)

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

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

In this case,

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

where $z=-a^{d}b{d}^d$.

Proof. 

By virtue of Lemma 2.2, we have

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

where

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

$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 1.1, $x^d$ has core inverse and that $x^d={\left(x^d\right)}^2{\left(x^d\right)}^{\#}.$ Hence,

$${\left(x^d\right)}^{\#}=x^2\left[{\left(x^d\right)}^2{\left(x^d\right)}^{\#}\right],$$

and so ${\left(x^d\right)}^{\#}$ is a upper triangular matrix. Write

$${\left(x^d\right)}^{\#}=\left(\begin{array}{cc}\alpha & \delta \\ 0& \beta \end{array}\right).$$

Then

$$x^d{\left({\left(x^d\right)}^{\#}\right)}^2={\left(x^d\right)}^{\#},{\left(x^d\right)}^{\#}{\left(x^d\right)}^2=x^d,{\left(x^d{\left(x^d\right)}^{\#}\right)}^{*}=x^d{\left(x^d\right)}^{\#}.$$

This implies that

$$a^d{\alpha }^2=\alpha ,\alpha {\left(a^d\right)}^2=a^d,{\left(a^d\alpha \right)}^{*}=a^d\alpha .$$

Hence, $a^d\in {\mathcal{A}}^{\#}.$ By using Theorem 1.1 again, $a\in {\mathcal{A}}^d$. Likewise, $d\in {\mathcal{A}}^d$. In view of Lemma 2.2, ${\left(a^d\right)}^{\pi }s=0$. This implies that

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

Therefore

$$\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}=0,$$

as asserted.

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

$${\left(a^d\right)}^{\pi }s=(1-a^d{a}^2{a}^d)s=a^{\pi }s=a^{\pi }[\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }-a^{d}b{d}^d]=0.$$

Then it follows by Lemma 2.2 that

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

where $t=-{\left(a^d\right)}^{\#}s{\left(d^d\right)}^{\#}.$ Hence, $t=-{\left(a^d\right)}^{\#}[\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }-a^{d}b{d}^d]{\left(d^d\right)}^{\#}={\left(a^d\right)}^{\#}{a}^{d}b{d}^d{\left(d^d\right)}^{\#}.$ Then we have

$${\left(x^d\right)}^2=\left(\begin{array}{cc}{\left(a^d\right)}^2& w\\ 0& {\left(d^d\right)}^2\end{array}\right),$$

where $w=\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+3}b{d}^i{d}^{\pi }-{\left(a^d\right)}^{2}b{d}^d-a^{d}b{\left(d^d\right)}^2.$ Therefore

$$\begin{array}{ccc}\hfill x^d& =\hfill & {\left(x^d\right)}^2{\left(x^d\right)}^{\#}\hfill \\ & =\hfill & \left(\begin{array}{cc}{\left(a^d\right)}^2& w\\ 0& {\left(d^d\right)}^2\end{array}\right)\left(\begin{array}{cc}{\left(a^d\right)}^{\#}& t\\ 0& {\left(d^d\right)}^{\#}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{a}^d& z\\ 0& d^d\end{array}\right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill z& =\hfill & {\left(a^d\right)}^{2}t+w{\left(d^d\right)}^{\#}\hfill \\ & =\hfill & {\left(a^d\right)}^2\left[{\left(a^d\right)}^{\#}{a}^{d}b{d}^d{\left(d^d\right)}^{\#}\right]-[{\left(a^d\right)}^{2}b{d}^d+a^{d}b{\left(d^d\right)}^2]{\left(d^d\right)}^{\#}\hfill \\ & =\hfill & {\left(a^d\right)}^{2}b{d}^d{\left(d^d\right)}^{\#}-a^d(a^{d}b+b{d}^d)d^d{\left(d^d\right)}^{\#}\hfill \\ & =\hfill & {\left(a^d\right)}^{2}b{d}^d{\left(d^d\right)}^{\#}-{\left(a^d\right)}^{2}b{d}^d{\left(d^d\right)}^{\#}-a^d\left[b{\left(d^d\right)}^2{\left(d^d\right)}^{\#}\right]\hfill \\ & =\hfill & -a^{d}b{d}^d\hfill \end{array}$$

This completes the proof. □

Corollary 2.4. 

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

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

Proof. 

Since $a^{\pi }b{d}^d=0,$ it follows by Theorem 1.1 that $a^{\pi }b{\left(d^d\right)}^2{\left(d^d\right)}^{\#}=0$; hence,

$$a^{\pi }b{d}^d=\left[a^{\pi }b{\left(d^d\right)}^2{\left(d^d\right)}^{\#}\right]b^{d}b=0.$$

By using Lemma 2.1, we have $(1-a{a}^d)b{d}^d=0$, and so $b{d}^d=a{a}^{d}b{d}^d$. Then

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

In light of Theorem 2.3,

$${\alpha }^d=\left(\begin{array}{cc}{a}^d& -a^{d}b{d}^d\\ 0& d^d\end{array}\right),$$

as asserted. □

It is very hard to determine the core-EP inverse of a triangular complex matrix (see [10]). As a consequence of Theorem 2.3, we now derive the following.

Corollary 2.5. 

Let $M=\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)$, $A,B,D\in C^{n\times n}$. If

$$\sum _{i=0}^{i\left(A\right)}{A}^i{A}^{\pi }B{\left(D^D\right)}^{i+2}=0,$$

then

$$M^D=\left(\begin{array}{cc}{A}^D& Z\\ 0& D^D\end{array}\right),$$

where $Z=-A^{D}B{D}^D.$

Proof. 

Since the generalized core-EP inverse and generalized core-EP inverse coincide with each other for a complex matrix, we obtain the result by Theorem 2.3. □

Corollary 2.6. 

Let $M=\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)$, $A,B,D\in C^{n\times n}$. If A is invertible, then

$$M^D=\left(\begin{array}{cc}{A}^{-1}& -A^{-1}B{D}^D\\ 0& D^D\end{array}\right).$$

Proof. 

Straightforward. □

The condition "$x^d\in M_2\left(\mathcal{A}\right)$ is upper triangular" in Theorem 2.3 is necessary as the following shows.

Example 2.7. 

Let σ and τ be linear operators, acting on separable Hilbert space $l_2\left(N\right)$ with the conjugate adjoint as an involution, defined as follows respectively:

$$\begin{array}{ccc}\sigma (x_1,x_2,x_3,x_4,\cdots )\hfill & =\hfill & (0,x_1,x_2,x_3,\cdots ),\hfill \\ \tau (x_1,x_2,x_3,x_4,\cdots )\hfill & =\hfill & (x_2,x_3,x_4,x_5,\cdots ).\hfill \end{array}$$

Then $\tau \sigma =1$. Take $M=\left(\begin{array}{cc}\sigma & 1-\sigma \tau \\ 0& \tau \end{array}\right)$. Then

$$M^d=M^{-1}=\left(\begin{array}{cc}\tau & 0\\ 1-\sigma \tau & \sigma \end{array}\right).$$

In this case, M is upper triangular matrix, but its generalized core-EP inverse is lower triangular.

3. Triangular Matrices with $C^{*}$-Algebra Entries

The aim of this section is to investigate the generalized core-EP inverse of triangular matrices over a $C^{*}$-algebra. Throughout this section, $\mathcal{A}$ is always a $C^{*}$-algebra. We start by

Lemma 3.1. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and let $a\in {\mathcal{A}}^{\#}\bigcap {\mathcal{A}}^{†}$. Then $a^{†}a{a}^{\#}=a^{†}.$

Proof. 

Since $(1-a{a}^d)a^d=0$, by virtue of [23][Lemma 2.4], we have $(1-a{a}^{\#})a^d=0$. This implies that $a^{†}{\left[(1-a{a}^{\#})a^d\left(a^2{a}^{†}\right)\right]}^{*}=0$, and so $a^{†}{\left[(1-a{a}^{\#})\left(a{a}^{†}\right)\right]}^{*}=0$. Hence, $a^{†}{\left(a{a}^{†}\right)}^{*}{(1-a{a}^{\#})}^{*}=0$. Therefore $a^{†}(1-a{a}^{\#})=0$, as required. □

Set $e_a=1-a{a}^{†}$ and $f_d=1-d^{†}d$. Then we derive

Lemma 3.2. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and let $a,d\in {\mathcal{A}}^{\#}\bigcap {\mathcal{A}}^{†}$. Then

$$d{d}^{\#}[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]=[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d{d}^{\#}.$$

Proof. 

It is easy to check that

$${\left(e_{a}b{d}^{†}d{d}^{\#}\right)}^{*}{e}_{a}b{d}^{†}={\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)={\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}d{d}^{\#}\right).$$

Then

$$d{d}^{\#}{\left(e_{a}b{d}^{†}\right)}^{*}{e}_{a}b{d}^{†}={\left(e_{a}b{d}^{†}\right)}^{*}{e}_{a}b{d}^{†}d{d}^{\#}.$$

Therefore

$$d{d}^{\#}[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]=[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d{d}^{\#},$$

as asserted. □

In [15], Li and Du investigate the core inverse of a triangular block complex matrix. We now extend Li and Du’s result to block operator matrices over a $C^{*}$-algebra by a new route.

Lemma 3.3. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and let $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$. If $a,d\in {\mathcal{A}}^{\#},a^{\pi }b{d}^{\pi }$$=0$ and $e_{a}b{f}_d=0$, then $x\in {\mathcal{A}}^{\#}$. In this case, $x^{\#}=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$, where

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

Proof. 

Since $a\in {\mathcal{A}}^{\#}$, by virtue of [23][Lemma 2.1], it has group inverse, and so a is regular. As $\mathcal{A}$ is a $C^{*}$-algebra, it follows by [14][Theorem 2.8] that $a\in {\mathcal{A}}^{†}$. Likewise, $d\in {\mathcal{A}}^{†}$. Since every $C^{*}$-algebra has the symmetry property, we have $1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\in {\mathcal{A}}^{-1}$.

Let $z=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$, where $\alpha ,\beta ,\gamma$ and $\delta$ as defined above.

Claim 1. ${\left(xz\right)}^{*}=xz$.

Since $e_{a}b{f}_d=0$, by virtue of Lemma 3.1, we have

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

Hence,

$$\begin{array}{cc}& {\left[(1-a{a}^{\#})b{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\right]}^{*}\hfill \\ \hfill =& \left[e_{a}b{d}^{†}\right]{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left[(1-a{a}^{\#})b{d}^{\#}\right]}^{*}\hfill \\ \hfill =& (1-a{a}^{\#})b{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}.\hfill \end{array}$$

Therefore

$$\begin{array}{cc}& a\alpha +b\gamma \hfill \\ \hfill =& a{a}^{\#}+(1-a{a}^{\#})b{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& a{a}^{\#}+e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}.\hfill \end{array}$$

Hence,

$${(a\alpha +b\gamma )}^{*}=a\alpha +b\gamma .$$

By virtue of Lemma 3.1, we have

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

Hence,

$$\begin{array}{ccc}\hfill d{d}^{\#}[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]& =\hfill & d{d}^{\#}+d{d}^{\#}{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\hfill \\ & =\hfill & d{d}^{\#}+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)d{d}^{\#}\hfill \\ & =\hfill & [1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d{d}^{\#}.\hfill \end{array}$$

Thus, we derive that

$$d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}={[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d{d}^{\#}.$$

Since $d\delta =d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1},$ we have ${\left(d\delta \right)}^{*}=d\delta .$

In view of Lemma 3.2, we verify that

$$\begin{array}{cc}& a\beta +b\delta \hfill \\ \hfill =& -a{a}^{\#}b{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}+b{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& [1-a{a}^{\#}]b{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& [1-a{a}^{\#}]b\left(d^{\#}d{d}^{\#}\right){[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& [1-a{a}^{\#}]b{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d{d}^{\#}\hfill \\ \hfill =& e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d{d}^{\#}\hfill \\ \hfill =& e_{a}b{d}^{†}\left(d{d}^{\#}\right){[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \end{array}$$

Therefore

$$\begin{array}{cc}& {(a\beta +b\delta )}^{*}\hfill \\ \hfill =& {\left[(1-a{a}^{\#})b{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}d{d}^{\#}\right]}^{*}\hfill \\ \hfill =& d{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left[(1-a{a}^{\#})b{d}^{\#}\right]}^{*}\hfill \\ \hfill =& d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& d\gamma .\hfill \end{array}$$

We compute that

$$\begin{array}{ccc}\hfill xz& =\hfill & \left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}a\alpha +b\gamma & a\beta +b\delta \\ d\gamma & d\delta \end{array}\right)\hfill \end{array}$$

Therefore ${\left(xz\right)}^{*}=xz$.

Claim 2. $x{z}^2=z$.

We compute that

$$\begin{array}{cc}& (1-xz)z\hfill \\ \hfill =& \left(\begin{array}{cc}1-(a\alpha +b\gamma )& -(a\beta +b\delta )\\ -d\gamma & 1-d\delta \end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}[1-(a\alpha +b\gamma \left)\right]\alpha -(a\beta +b\delta )\gamma & [1-(a\alpha +b\gamma \left)\right]\beta -(a\beta +b\delta )\delta \\ -d\gamma \alpha +(1-d\delta )\gamma & -d\gamma \beta +(1-d\delta )\delta \end{array}\right).\hfill \end{array}$$

Obviously, we have

$$a\alpha +b\gamma =a{a}^{\#}+(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}.$$

Then we compute that

$$\begin{array}{cc}& [1-(a\alpha +b\gamma \left)\right]\alpha \hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}][a^{\#}+(a^{\pi }b{d}^{†}-a^{\#}b)\gamma ]\hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}]\left[(a^{\pi }b{d}^{†}-a^{\#}b)\gamma \right]\hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}][\left(a^{\pi }b{d}^{†}\right]\gamma \hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}]\left(e_{a}b{d}^{†}\right)\gamma \hfill \\ \hfill =& e_{a}b{d}^{†}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\gamma \hfill \\ \hfill =& e_{a}b{d}^{†}-e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{\delta }^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\gamma \hfill \\ \hfill =& e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}[1+{\left(e_{a}b{d}^{†}\right)}^{*}{e}_{a}b{d}^{†}-{\left(e_{a}b{\delta }^{†}\right)}^{*}{e}_{a}b{d}^{†}]\gamma \hfill \\ \hfill =& e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\gamma \hfill \\ \hfill =& (a\beta +b\delta )\gamma ;\hfill \end{array}$$

hence,

$$[1-(a\alpha +b\gamma \left)\right]\alpha -(a\beta +b\delta )\gamma =0.$$

Analogously, we derive that

$$\begin{array}{cc}& [1-(a\alpha +b\gamma \left)\right]\beta \hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}][(a^{\pi }b{d}^{†}-a^{\#}b)d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& [1-a{a}^{\#}-(a\beta +b\delta ){\left(e_{a}b{\delta }^{†}\right)}^{*}][(a^{\pi }b{d}^{†}-a^{\#}b)\delta \hfill \\ \hfill =& (a\beta +b\delta )\delta ;\hfill \end{array}$$

hence,

$$[1-(a\alpha +b\gamma \left)\right]\beta -(a\beta +b\delta )\delta =0.$$

Also we check that

$$\begin{array}{cc}& -d\gamma \alpha +(1-d\delta )\gamma \hfill \\ \hfill =& -d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}[a^{\#}+[a^{\pi }b{d}^{†}-a^{\#}b]\hfill \\ & d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]\hfill \\ \hfill +& [1-d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& -{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}[a^{\pi }b{d}^{†}-a^{\#}b]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]\hfill \\ \hfill +& [1-d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& -{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{a}^{\pi }b{d}^{†}{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]\hfill \\ \hfill +& [1-{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}[-{\left(e_{a}b{d}^{†}\right)}^{*}{a}^{\pi }b{d}^{†}+{\left(e_{a}b{d}^{†}\right)}^{*}{e}_{a}b{d}^{†}]\hfill \\ & d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\left[{\left(e_{a}b{d}^{†}\right)}^{*}{e}_a(1-a^{\pi })b{d}^{†}\right]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]\hfill \\ \hfill =& 0.\hfill \end{array}$$

Furthermore, we verify that

$$\begin{array}{cc}& -d\gamma \beta +(1-d\delta )\delta \hfill \\ \hfill =& -d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\left[(a^{\pi }b{d}^{†}-a^{\#}b)d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\right]\hfill \\ \hfill +& [1-d{d}^{\#}{\left(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right)}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& -{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}(a^{\pi }b{d}^{†}-a^{\#}b)d^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}\hfill \\ \hfill +& [1-{\left(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right)}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& -{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{a}^{\pi }b{d}^{†}{d}^{\#}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}\hfill \\ \hfill +& [1-{\left(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right)}^{-1}]d^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}[-{\left(e_{a}b{d}^{†}\right)}^{*}{a}^{\pi }b{d}^{†}+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d^{\#}\hfill \\ & {(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}[-{\left(e_{a}b{d}^{†}\right)}^{*}{a}^{\pi }b{d}^{†}+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d^{\#}\hfill \\ & {(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}\hfill \\ \hfill =& {[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{e}_a(1-a^{\pi })b{d}^{†}+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]d^{\#}\hfill \\ & {(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}\hfill \\ \hfill =& 0.\hfill \end{array}$$

Therefore $x{z}^2=z$.

Claim 3. $xzx=x$.

$$\begin{array}{cc}& (1-xz)x\hfill \\ \hfill =& \left(\begin{array}{cc}1-(a\alpha +b\gamma )& -(a\beta +b\delta )\\ -d\gamma & 1-d\delta \end{array}\right)\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}[1-(a\alpha +b\gamma \left)\right]a& [1-(a\alpha +b\gamma \left)\right]b-(a\beta +b\delta )d\\ -d\gamma a\gamma & -d\gamma b+(1-d\delta )d\end{array}\right).\hfill \end{array}$$

Obviously, we have ${\left(e_a\right)}^{*}a=(1-a{a}^{†})a=0.$ Then

$$\begin{array}{cc}& [1-(a\alpha +b\gamma \left)\right]a\hfill \\ \hfill =& [1-a{a}^{\#}-e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]a\hfill \\ \hfill =& [(1-a{a}^{\#})a-e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{\left(e_a\right)}^{*}a)]\hfill \\ \hfill =& 0.\hfill \end{array}$$

Similarly, we have

$$\begin{array}{ccc}\hfill -d\gamma a& =\hfill & -d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}a\hfill \\ & =\hfill & -d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{\left(e_a\right)}^{*}a\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Clearly, $(1-a{a}^{\#})b=(1-a{a}^{\#}a{a}^{†})b=e_{a}b=e_{a}b(f_d+d^{†}d)=e_{a}b{d}^{dag}d$. Then we have

$$\begin{array}{cc}& [1-(a\alpha +b\gamma \left)\right]b\hfill \\ \hfill =& [1-a{a}^{\#}-e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}]b\hfill \\ \hfill =& e_{a}b-e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b\right)\hfill \\ \hfill =& e_{a}b{d}^{†}d-e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{dag}\right)d\hfill \\ \hfill =& e_{a}b{d}^{†}[1-{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{dag}\right)]d\hfill \\ \hfill =& e_{a}b{d}^{†}{(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right))}^{-1}[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)-{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{dag}\right)]d\hfill \\ \hfill =& e_{a}b{d}^{†}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d\hfill \\ \hfill =& (a\beta +b\delta )d.\hfill \end{array}$$

Finally, we verify that

$$\begin{array}{ccc}\hfill d\gamma b& =\hfill & d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}b\hfill \\ & =\hfill & d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{\left(e_a\right)}^{*}b\hfill \\ & =\hfill & d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}{\left(e_{a}b{d}^{†}\right)}^{*}{e}_{a}b(f_d+d^{†}d)\hfill \\ & =\hfill & d{d}^{\#}[1-{\left(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right)}^{-1}]\left[{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right]d\hfill \\ & =\hfill & d{d}^{\#}[1-{\left(1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)\right)}^{-1}]d\hfill \\ & =\hfill & d{d}^{\#}d-d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d\hfill \\ & =\hfill & d-d{d}^{\#}{[1+{\left(e_{a}b{d}^{†}\right)}^{*}\left(e_{a}b{d}^{†}\right)]}^{-1}d\hfill \\ & =\hfill & d(1-\delta d)\hfill \\ & =\hfill & (1-d\delta )d.\hfill \end{array}$$

Therefore $xzx=x$. In light of [24][Theorem 3.3], $x\in {\mathcal{A}}^{\#}$ and $x^{\#}=z$, as asserted. □

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

Theorem 3.4. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$. If $a,d\in {\mathcal{A}}^d$, then $x\in {\mathcal{A}}^d$ and

$$x^d=\left(\begin{array}{cc}{\left(a^d\right)}^2\alpha +w\gamma & {\left(a^d\right)}^2\beta +w\delta \\ {\left(d^d\right)}^2\gamma & {\left(d^d\right)}^2\delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill w& =\hfill & \sum _{i=0}^{\infty }{\left(a^d\right)}^{i+3}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+3}-{\left(a^d\right)}^{2}b{d}^d-a^{d}b{\left(d^d\right)}^2,\hfill \\ \hfill s& =\hfill & \sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}-a^{d}b{d}^d,\hfill \\ \hfill \alpha & =\hfill & a^2{a}^d+[a^{\pi }s{\left[d^d\right]}^{†}-a^2{a}^{d}s]d^2{d}^d{\left(1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)\right)}^{-1}{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \beta & =\hfill & (a^{\pi }s{\left[d^d\right]}^{†}-a^2{a}^{d}s)d^2{d}^d{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \gamma & =\hfill & d^2{d}^d{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1}{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \delta & =\hfill & d^2{d}^d{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1}.\hfill \end{array}$$

Proof. 

In view of Theorem 1.1, $a,d\in {\mathcal{A}}^d$ and $a^d,d^d\in {\mathcal{A}}^{\#}$. By virtue of Lemma 2.2, we have

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

where $s=\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}-a^{d}b{d}^d.$

Since $a^d=a^{d}a{a}^d$, i.e., $a^d$ is regular. As $\mathcal{A}$ is a $C^{*}$-algebra, it follows by [14][Theorem 2.8], $a^d\in {\mathcal{A}}^{†}$. Likewise, $d^d\in {\mathcal{A}}^{†}$. Let $e_{a^d}=1-a^d{\left(a^d\right)}^{†}$ and $f_{d^d}=1-{\left[d^d\right]}^{†}{d}^d$. Then we check that

$$\begin{array}{ccc}\hfill {\left(a^d\right)}^{\pi }s{\left(d^d\right)}^{\pi }& =\hfill & a^{\pi }[\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}-a^{d}b{d}^d]d^{\pi }\hfill \\ & =\hfill & a^{\pi }\left[\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}\right]d^{\pi }\hfill \\ & =\hfill & 0\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill e_{a^d}s{f}_{d^d}& =\hfill & [1-a^d{\left(a^d\right)}^{†}][\sum _{i=0}^{\infty }{\left(a^d\right)}^{i+2}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}\hfill \\ & -\hfill & a^{d}b{d}^d][1-{\left[d^d\right]}^{†}{d}^d]\hfill \\ & =\hfill & [1-a^d{\left(a^d\right)}^{†}]\left[\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left(d^d\right)}^{i+2}\right][1-{\left[d^d\right]}^{†}{d}^d]\hfill \\ & =\hfill & 0.\hfill \end{array}$$

It follows by Lemma 3.3 that

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {\left[a^d\right]}^{\#}+[a^{\pi }s{\left[d^d\right]}^{†}-{\left[a^d\right]}^{\#}s]{\left[d^d\right]}^{\#}{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1}{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \beta & =\hfill & (a^{\pi }s{\left[d^d\right]}^{†}-{\left[a^d\right]}^{\#}s){\left[d^d\right]}^{\#}{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \gamma & =\hfill & {\left[d^d\right]}^{\#}{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1}{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \delta & =\hfill & {\left[d^d\right]}^{\#}{[1+{\left(e_{a^d}s{\left[d^d\right]}^{†}\right)}^{*}\left(e_{a^d}s{\left[d^d\right]}^{†}\right)]}^{-1}.\hfill \end{array}$$

Set

$$\begin{array}{ccc}\hfill w& :=\hfill & a^{d}s+s{d}^d\hfill \\ & =\hfill & \sum _{i=0}^{\infty }{\left(a^d\right)}^{i+3}b{d}^i{d}^{\pi }+\sum _{i=0}^{\infty }{a}^i{a}^{\pi }b{\left[d^d\right]}^{i+3}-{\left(a^d\right)}^{2}b{d}^d-a^{d}b{\left(d^d\right)}^2.\hfill \end{array}$$

Accordingly,

$$\begin{array}{ccc}\hfill x^d& =\hfill & {\left(x^d\right)}^2{\left(x^d\right)}^{\#}\hfill \\ & =\hfill & \left(\begin{array}{cc}{\left(a^d\right)}^2& w\\ 0& {\left(d^d\right)}^2\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{\left(a^d\right)}^2\alpha +w\gamma & {\left(a^d\right)}^2\beta +w\delta \\ {\left(d^d\right)}^2\gamma & {\left(d^d\right)}^2\delta \end{array}\right).\hfill \end{array}$$

In view of Theorem 1.1, $a^d={\left(a^d\right)}^2{\left[a^d\right]}^{\#}$. Hence, ${\left(a^d\right)}^{\#}=a^2\left[{\left(a^d\right)}^2{\left[a^d\right]}^{\#}\right]=a^2{a}^d.$ Similarly, we have ${\left(d^d\right)}^{\#}=d^2{d}^d.$ Therefore we verify the formulas of $\alpha ,\beta ,\gamma$ and $\delta$ mentioned before. □

Corollary 3.5. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$. If $a\in {\mathcal{A}}^{qnil},d\in {\mathcal{A}}^d$, then $x\in {\mathcal{A}}^d$ and

$$x^d=\left(\begin{array}{cc}s{d}^d\gamma & s{d}^d\delta \\ {\left(d^d\right)}^2\gamma & {\left(d^d\right)}^2\delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill s& =\hfill & \sum _{i=0}^{\infty }{a}^{i}b{\left(d^d\right)}^{i+2},\hfill \\ \hfill \alpha & =\hfill & s{\left[d^d\right]}^{†}{d}^2{d}^d{\left(1+{\left(s{\left[d^d\right]}^{†}\right)}^{*}\left(s{\left[d^d\right]}^{†}\right)\right)}^{-1}{\left(s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \beta & =\hfill & \left(s{\left[d^d\right]}^{†}\right)d^2{d}^d{[1+{\left(s{\left[d^d\right]}^{†}\right)}^{*}\left(s{\left[d^d\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \gamma & =\hfill & d^2{d}^d{[1+{\left(s{\left[d^d\right]}^{†}\right)}^{*}\left(s{\left[d^d\right]}^{†}\right)]}^{-1}{\left(s{\left[d^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \delta & =\hfill & d^2{d}^d{[1+{\left(s{\left[d^d\right]}^{†}\right)}^{*}\left(s{\left[d^d\right]}^{†}\right)]}^{-1}.\hfill \end{array}$$

Proof. 

Since $a\in {\mathcal{A}}^{qnil}$, we see that $a^d=0$, and therefore we obtain the result by Theorem 3.4. □

Corollary 3.6. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$. If $a\in {\mathcal{A}}^d,d\in {\mathcal{A}}^{qnil}$, then $x\in {\mathcal{A}}^d$ and

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

Proof. 

Since ${\left(a^d\right)}^2{a}^2{a}^d=a{a}^d{a}^d=a^2{a}^d$ nd $d^d=0$, we complete the proof by Theorem 3.4. □

Since the algebra $C^{n\times n}$ of all $n\times n$ complex matrices is a $C^{*}$-algebra, the following result gives a simpler formula to compute the core-EP inverse of a block triangular complex matrices (see [10]).

Corollary 3.7. 

Let $\mathcal{A}$ be a $C^{*}$-algebra and $x=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$. If $a,d\in {\mathcal{A}}^D$, then $x\in {\mathcal{A}}^D$ and

$$x^D=\left(\begin{array}{cc}{\left(a^D\right)}^2\alpha +w\gamma & {\left(a^D\right)}^2\beta +w\delta \\ {\left(d^D\right)}^2\gamma & {\left(d^D\right)}^2\delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill w& =\hfill & \sum _{i=0}^m{\left(a^D\right)}^{i+3}b{d}^i{d}^{\pi }+\sum _{i=0}^m{a}^i{a}^{\pi }b{\left(d^D\right)}^{i+3}-{\left(a^D\right)}^{2}b{d}^D-a^{D}b{\left(d^D\right)}^2,\hfill \\ \hfill s& =\hfill & \sum _{i=0}^m{\left(a^D\right)}^{i+2}b{d}^i{d}^{\pi }+\sum _{i=0}^m{a}^i{a}^{\pi }b{\left(d^D\right)}^{i+2}-a^{d}b{d}^D,\hfill \\ \hfill \alpha & =\hfill & a^2{a}^D+[a^{\pi }s{\left[d^D\right]}^{†}-a^2{a}^{D}s]d^2{d}^D{\left(1+{\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*}\left(e_{a^D}s{\left[d^D\right]}^{†}\right)\right)}^{-1}\hfill \\ & & {\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*},\hfill \\ \hfill \beta & =\hfill & (a^{\pi }s{\left[d^D\right]}^{†}-a^2{a}^{D}s)d^2{d}^D{[1+{\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*}\left(e_{a^D}s{\left[d^D\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \gamma & =\hfill & d^2{d}^D{[1+{\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*}\left(e_{a^D}s{\left[d^D\right]}^{†}\right)]}^{-1}{\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*},\hfill \\ \hfill \delta & =\hfill & d^2{d}^D{[1+{\left(e_{a^D}s{\left[d^D\right]}^{†}\right)}^{*}\left(e_{a^D}s{\left[d^D\right]}^{†}\right)]}^{-1},\hfill \\ \hfill m& =\hfill & max\left\{ind\right(a),ind(d\left)\right\}.\hfill \end{array}$$

Proof. 

Since $a,d\in {\mathcal{A}}^D$, it follows by [3][Corollary 3.4] that $a,d\in {\mathcal{A}}^d\bigcap {\mathcal{A}}^D$. By virtue of Theorem 3.4, $x\in {\mathcal{A}}^d$. Clearly, $x\in {\mathcal{A}}^D$. By using [3][Corollary 3.4] again, $x\in {\mathcal{A}}^D$ and $x^D=x^d$, as required. □

4. Applications

Lex X and Y be Hilbert spaces, and let $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right),$ where $A\in \mathcal{B}{\left(X\right)}^d,B\in \mathcal{B}(X,Y),C\in \mathcal{B}(Y,X),D\in \mathcal{B}{\left(Y\right)}^d$. Choose $p=\left(\begin{array}{cc}{I}_X& 0\\ 0& I_Y\end{array}\right).$ Then M can be regarded as the Pierce matrix ${\left(\begin{array}{cc}pMp& pM{p}^{\pi }\\ p^{\pi }Mp& p^{\pi }M{p}^{\pi }\end{array}\right)}_p.$ Here, every subblock matrices can be seen as the bounded linear operators on Hilbert space $X\oplus Y$. Throughout this section, without loss the generality, we consider M as the block operator matrix in a specifical case $X=Y$. In this case, $\mathcal{B}(X\oplus X)$ is indeed a $C^{*}$-algebra. The following lemma is crucial.

Lemma 4.1. 

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

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

Proof. 

Since $a\in {\mathcal{A}}^d$, by virtue of Theorem 1.1, there exist $x\in {\mathcal{A}}^{\#}$ and $y\in {\mathcal{A}}^{qnil}$ such that $a=x+y,x^{*}y=0,yx=0$. As in the proof of [3], $x=a{a}^{d}a$ and $y=a-a{a}^{d}a$. Then $a=x+(y+b)$. Since $by=b(a-a{a}^{d}a)=0$, it follows by [26][Lemma 2.10] that $y+b\in {\mathcal{A}}^{qnil}$. We directly verify that

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

In light of Theorem 1.1, $a+b\in {\mathcal{A}}^d$. In this case,

$${(a+b)}^d=x^{\#}=a^d,$$

as asserted. □

Theorem 4.2. 

If $CA=0,CB=0$ and $D^{*}C=0$, then M has generalized core-EP inverse. In this case,

$$M^d=\left(\begin{array}{cc}{\left(A^d\right)}^2\Lambda +W\Gamma & {\left(A^d\right)}^2\Sigma +W\Delta \\ {\left(D^d\right)}^2\Gamma & {\left(D^d\right)}^2\Delta \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill W& =\hfill & \sum _{i=0}^{\infty }{\left(A^d\right)}^{i+3}B{D}^i{D}^{\pi }+\sum _{i=0}^{\infty }{A}^i{A}^{\pi }B{\left(D^d\right)}^{i+3}-{\left(A^d\right)}^{2}B{D}^d-A^{d}B{\left(D^d\right)}^2,\hfill \\ \hfill S& =\hfill & \sum _{i=0}^{\infty }{\left(A^d\right)}^{i+2}B{D}^i{D}^{\pi }+\sum _{i=0}^{\infty }{A}^i{A}^{\pi }B{\left(D^d\right)}^{i+2}-A^{d}B{D}^d,\hfill \\ \hfill \Lambda & =\hfill & A^2{A}^d+[A^{\pi }S{\left[D^d\right]}^{†}-A^2{A}^{d}S]D^2{D}^d{\left(I+{\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*}\left(e_{A^d}S{\left[D^d\right]}^{†}\right)\right)}^{-1}\hfill \\ & & {\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \Sigma & =\hfill & (A^{\pi }S{\left[D^d\right]}^{†}-A^2{A}^{d}S)D^2{D}^d{[I+{\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*}\left(e_{A^d}S{\left[D^d\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \Gamma & =\hfill & D^2{D}^d{[I+{\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*}\left(e_{A^d}S{\left[D^d\right]}^{†}\right)]}^{-1}{\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \Delta & =\hfill & D^2{D}^d{[I+{\left(e_{A^d}S{\left[D^d\right]}^{†}\right)}^{*}\left(e_{A^d}S{\left[D^d\right]}^{†}\right)]}^{-1}.\hfill \end{array}$$

Proof. 

Write $M=P+Q$, where

$$P=\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right),Q=\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right).$$

We easily check that

$$\begin{array}{ccc}\hfill P^{*}Q& =\hfill & \left(\begin{array}{cc}{A}^{*}& 0\\ B^{*}& D^{*}\end{array}\right)\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ D^{*}C& 0\end{array}\right)=0,\hfill \\ \hfill QP& =\hfill & \left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)=\left(\begin{array}{cc}0& 0\\ CA& CB\end{array}\right)=0\hfill \end{array}$$

Since A and D have generalized core-EP inverses, it follows by Theorem 3.4 that P has generalized core-EP inverse, and that

$$P^d=\left(\begin{array}{cc}{\left(A^d\right)}^2\Lambda +W\Gamma & {\left(A^d\right)}^2\Sigma +W\Delta \\ {\left(D^d\right)}^2\Gamma & {\left(D^d\right)}^2\Delta \end{array}\right),$$

where $W,S,\Lambda ,\Sigma ,\Gamma ,\Delta$ as defined before. Obviously, Q is nilpotent, and so it is quasinilpotent. According to Lemma 4.1, $M^d=P^d$, required. □

Corollary 4.3. 

If $BD=0,BC=0$ and $A^{*}B=0$, then M has generalized core-EP inverse. In this case,

$$M^d=\left(\begin{array}{cc}{\left(A^d\right)}^2\Delta & {\left(A^d\right)}^2\Gamma \\ {\left(D^d\right)}^2\Sigma +W\Delta & {\left(D^d\right)}^2\Lambda +W\Gamma \end{array}\right),$$

where

$$\begin{array}{ccc}\hfill W& =\hfill & \sum _{i=0}^{\infty }{\left(D^d\right)}^{i+3}C{A}^i{A}^{\pi }+\sum _{i=0}^{\infty }{D}^i{D}^{\pi }C{\left(A^d\right)}^{i+3}-{\left(D^d\right)}^{2}C{A}^d-D^{d}C{\left(A^d\right)}^2,\hfill \\ \hfill S& =\hfill & \sum _{i=0}^{\infty }{\left(D^d\right)}^{i+2}C{A}^i{A}^{\pi }+\sum _{i=0}^{\infty }{D}^i{D}^{\pi }C{\left(A^d\right)}^{i+2}-D^{d}C{A}^d,\hfill \\ \hfill \Lambda & =\hfill & D^2{D}^d+[D^{\pi }S{\left[A^d\right]}^{†}-D^2{D}^{d}S]A^2{A}^d{\left(I+{\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*}\left(e_{D^d}S{\left[A^d\right]}^{†}\right)\right)}^{-1}\hfill \\ & & {\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \Sigma & =\hfill & (D^{\pi }S{\left[A^d\right]}^{†}-D^2{D}^{d}S)A^2{A}^d{[I+{\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*}\left(e_{D^d}S{\left[A^d\right]}^{†}\right)]}^{-1},\hfill \\ \hfill \Gamma & =\hfill & A^2{A}^d{[I+{\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*}\left(e_{D^d}S{\left[A^d\right]}^{†}\right)]}^{-1}{\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*},\hfill \\ \hfill \Delta & =\hfill & A^2{A}^d{[I+{\left(e_{D^d}S{\left[A^d\right]}^{†}\right)}^{*}\left(e_{D^d}S{\left[A^d\right]}^{†}\right)]}^{-1}.\hfill \end{array}$$

Proof. 

Obviously, $\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)M\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)=\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)$. Applying Theorem 4.2 to the matrix $\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)$, we see that $\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)$ has generalized core-EP inverse. This implies that M has generalized core-EP inverse. Additionally,

$$M^d=\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right){\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)}^d\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right).$$

Therefore we complete the proof by Theorem ???. □

Lemma 4.4. 

Let $a\in {\mathcal{A}}^{qnil},b\in {\mathcal{A}}^d$. If $ab=ba,a^{*}b=b{a}^{*}$ and $b^{*}\left(ab\right)=\left(ba\right)b^{*}$, then $a+b\in {\mathcal{A}}^d$. In this case,

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

Proof. 

Since $ab=ba$, it follows by [26][Theorem 2.3] that $a{b}^d=b^{d}a$. Likewise, we have $a{\left(b^d\right)}^{*}=a{\left(b^{*}\right)}^d={\left(b^{*}\right)}^{d}a={\left(b^d\right)}^{*}a$. Since $\left(a{b}^{*}\right)b=\left(b^{*}a\right)b=b^{*}\left(ab\right)=\left(ba\right)b^{*}=b\left(a{b}^{*}\right)$, by using [26][Theorem 2.3] again, $\left(a{b}^{*}\right)b^d=b^d\left(a{b}^{*}\right)$. Then $b^{*}\left(a{b}^d\right)=\left(b^{*}a\right)b^d=\left(a{b}^{*}\right)b^d=b^d\left(a{b}^{*}\right)=\left(a{b}^d\right)b^{*}$. Hence, ${\left(b^d\right)}^{*}\left(a{b}^d\right)=\left(a{b}^d\right){\left(b^d\right)}^{*}$. In view of Theorem 1.1 and [7], $b^d\left(a{b}^d\right)=\left(a{b}^d\right)b^d$. We directly verify that

$$\begin{array}{ccc}\hfill (a+b)b^d{(1+a{b}^d)}^{-1}& =\hfill & (a{b}^d+b{b}^d){(1+a{b}^d)}^{-1}\hfill \\ & =\hfill & b{b}^d(1+a{b}^d){(1+a{b}^d)}^{-1}=b{b}^d\hfill \\ & =\hfill & {(1+a{b}^d)}^{-1}(1+a{b}^d)b{b}^d\hfill \\ & =\hfill & {(1+a{b}^d)}^{-1}{b}^d(a+b)\hfill \\ & =\hfill & b^d{(1+a{b}^d)}^{-1}(a+b),\hfill \\ \hfill (a+b){\left[b^d{(1+a{b}^d)}^{-1}\right]}^2& =\hfill & \left(b{b}^d\right)b^d{(1+a{b}^d)}^{-1}=b^d{(1+a{b}^d)}^{-1},\hfill \\ \hfill (a+b)-\left[b^d{(1+a{b}^d)}^{-1}\right]{(a+b)}^2& =\hfill & (a+b)-b{b}^d(a+b)\hfill \\ & =\hfill & (b-b^d{b}^2)+(1-b{b}^d)a\hfill \\ & \in \hfill & {\mathcal{A}}^{qnil}.\hfill \end{array}$$

This implies that ${(a+b)}^d=b^d{(1+a{b}^d)}^{-1}$. We easily verify that

$$b^d(1+a{b}^d)=(1+a{b}^d)b^d,{\left(b^d\right)}^{*}(1+a{b}^d)=(1+a{b}^d){\left(b^d\right)}^{*}.$$

Then we derive that

$$b^d{(1+a{b}^d)}^{-1}={(1+a{b}^d)}^{-1}{b}^d,{\left(b^d\right)}^{*}{(1+a{b}^d)}^{-1}={(1+a{b}^d)}^{-1}{\left(b^d\right)}^{*}.$$

According to [7][Theorem 3.5], ${(a+b)}^d\in {\mathcal{A}}^{\#}$ and

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

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

□

Theorem 4.5. 

If $BC=0,CB=0,CA=DC,C{A}^{*}=D^{*}C$ and $D^{*}CA=DC{A}^{*}$, then M has generalized core-EP inverse. In this case,

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {\left(A^d\right)}^2\Lambda +W\Gamma ,\hfill \\ \hfill \beta & =\hfill & {\left(A^d\right)}^2\Sigma +W\Delta ,\hfill \\ \hfill \gamma & =\hfill & -C{A}^d[{\left(A^d\right)}^2\Lambda +W\Gamma ]+(I-CS){\left(D^d\right)}^2\Gamma ,\hfill \\ \hfill \delta & =\hfill & -C{A}^d[{\left(A^d\right)}^2\Sigma +W\Delta ]+(I-CS){\left(D^d\right)}^2\Delta \hfill \end{array}$$

and $W,S,\Lambda ,\Sigma ,\Gamma$ and Δ constructed as in Theorem 4.2.

Proof. 

Write $M=P+Q$, where

$$P=\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right),Q=\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right).$$

Then

$$P^d=\left(\begin{array}{cc}{A}^d& S\\ 0& D^d\end{array}\right),$$

where $S=\sum _{i=0}^{\infty }{\left(A^d\right)}^{i+2}B{D}^i{D}^{\pi }+\sum _{i=0}^{\infty }{A}^i{A}^{\pi }B{\left(D^d\right)}^{i+2}-A^{d}B{D}^d.$ Hence, we have

$$\begin{array}{ccc}\hfill I-Q{P}^d& =\hfill & I-\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)\left(\begin{array}{cc}{A}^d& S\\ 0& D^d\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}I& 0\\ -C{A}^d& I-CS\end{array}\right).\hfill \end{array}$$

We easily check that

$$\begin{array}{ccc}\hfill PQ& =\hfill & \left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)=\left(\begin{array}{cc}BC& 0\\ DC& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}0& 0\\ CA& CB\end{array}\right)=\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)\left(\begin{array}{cc}A& B\\ 0& D\end{array}\right)=QP,\hfill \\ \hfill P^{*}Q& =\hfill & \left(\begin{array}{cc}{A}^{*}& 0\\ B^{*}& D^{*}\end{array}\right)\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ D^{*}C& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}0& 0\\ C{A}^{*}& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ C& 0\end{array}\right)\left(\begin{array}{cc}{A}^{*}& 0\\ B^{*}& D^{*}\end{array}\right)=QP,\hfill \\ \hfill P^{*}\left(QP\right)& =\hfill & \left(\begin{array}{cc}{A}^{*}& 0\\ B^{*}& D^{*}\end{array}\right)\left(\begin{array}{cc}0& 0\\ CA& CB\end{array}\right)=\left(\begin{array}{cc}0& 0\\ D^{*}CA& D^{*}CB\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}BC{A}^{*}& 0\\ DC{A}^{*}& 0\end{array}\right)=\left(\begin{array}{cc}BC& 0\\ DC& 0\end{array}\right)\left(\begin{array}{cc}{A}^{*}& 0\\ B^{*}& D^{*}\end{array}\right)=\left(PQ\right)P^{*}.\hfill \end{array}$$

Since A and D have generalized core-EP inverses, it follows by Theorem 3.4 that P has generalized core-EP inverse, and that

$$P^d=\left(\begin{array}{cc}{\left(A^d\right)}^2\Lambda +W\Gamma & {\left(A^d\right)}^2\Sigma +W\Delta \\ {\left(D^d\right)}^2\Gamma & {\left(D^d\right)}^2\Delta \end{array}\right),$$

where $W,S,\Lambda ,\Sigma ,\Gamma ,\Delta$ as defined before. Obviously, $Q^2=0$, and so it is quasinilpotent. According to Lemma 4.4,

$$\begin{array}{ccc}\hfill M^d& =\hfill & {(I+Q{P}^d)}^{-1}{P}^d\hfill \\ & =\hfill & (I-Q{P}^d)P^d\hfill \\ & =\hfill & \left(\begin{array}{cc}I& 0\\ -C{A}^d& I-CS\end{array}\right)\left(\begin{array}{cc}{\left(A^d\right)}^2\Lambda +W\Gamma & {\left(A^d\right)}^2\Sigma +W\Delta \\ {\left(D^d\right)}^2\Gamma & {\left(D^d\right)}^2\Delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {\left(A^d\right)}^2\Lambda +W\Gamma ,\hfill \\ \hfill \beta & =\hfill & {\left(A^d\right)}^2\Sigma +W\Delta ,\hfill \\ \hfill \gamma & =\hfill & -C{A}^d[{\left(A^d\right)}^2\Lambda +W\Gamma ]+(I-CS){\left(D^d\right)}^2\Gamma ,\hfill \\ \hfill \delta & =\hfill & -C{A}^d[{\left(A^d\right)}^2\Sigma +W\Delta ]+(I-CS){\left(D^d\right)}^2\Delta .\hfill \end{array}$$

as required. □

Corollary 4.6. 

If $BC=0,CB=0,AB=BD,A^{*}B=B{D}^{*}$ and $A^{*}BD=AB{D}^{*}$, then M has generalized core-EP inverse. In this case,

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & -B{D}^d[{\left(D^d\right)}^2\Sigma +W\Delta ]+(I-BS){\left(A^d\right)}^2\Delta ,\hfill \\ \hfill \beta & =\hfill & -B{D}^d[{\left(D^d\right)}^2\Lambda +W\Gamma ]+(I-BS){\left(A^d\right)}^2\Gamma ,\hfill \\ \hfill \gamma & =\hfill & {\left(D^d\right)}^2\Sigma +W\Delta ,\hfill \\ \hfill \delta & =\hfill & {\left(D^d\right)}^2\Lambda +W\Gamma \hfill \end{array}$$

and $W,S,\Lambda ,\Sigma ,\Gamma$ and Δ constructed as in Corollary 4.3.

Proof. 

Applying Theorem 4.5 to the matrix $\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)$, we see that it has generalized core-EP inverse. Analogously to Corollary 4.3, we have

$$M^d=\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right){\left(\begin{array}{cc}D& C\\ B& A\end{array}\right)}^d\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right).$$

Therefore we obtain the result by Theorem 4.5. □