The Core Inverse of Block Anti-Triangular Matrices with Product Subblocks

1. Introduction

Let R be an associative ring equipped with a unit element. An $n\times n$ matrix X is termed the group inverse of $n\times n$ matrix A over R if it satisfies the system of equations

$$A{X}^2=X,X{A}^2=A,\mathrm{and}AX=XA.$$

(1.1)

When such an X exists, it is uniquely determined and is denoted by $A^{\#}$. While the genesis of the group inverse lies in the theory of matrices and linear operators, the concept has since permeated various algebraic and analytical domains [2,7,11].

In the context of a ring R endowed with an involution *, we define the core inverse. An $n\times n$ matrix A over R is said to possess such a generalized inverse if there exists an $n\times n$ matrix such that

$$A{X}^2=X,{\left(AX\right)}^{*}=AX,\mathrm{and}X{A}^2=A.$$

(1.2)

The existence of this matrix implies its uniqueness; we denote it by $A^{\overline{)\#}}$ (see [4,8,13,15,16]). An $n\times n$ matrix A is classified as EP (or an EP matrix) if it admits a core inverse and the condition $A^{\overline{)\#}}=A^{\#}$ holds (see [9,10,14]).

Considering the specific case of ${\mathbb{C}}^{n\times n}$ the algebra of $n\times n$ complex matrices under conjugate transpose the following characterizations are standard. For a matrix $A\in {\mathbb{C}}^{n\times n}$, the existence of the group inverse is equivalent to the existence of the core inverse, a property that holds if and only if $rank\left(A\right)=rank\left(A^2\right)$ (see [1]). Furthermore, A is an EP matrix precisely when $rank\left(A\right)=rank\left(A^{*}\right)$ (see [14]).

The literature on the core inverse of matrices is extensive. Xu [13] explored the core inverses of a particular class of $2\times 2$ upper triangular matrices, demonstrating that the core inverse preserves this triangular structure. Later, Zhu [17] focused on the core inverse of $2\times 2$ lower triangular matrices featuring an invertible diagonal entry. Additionally, the scope of inquiry was extended by Chen and Sheibani [6], who analyzed core inverses of triangular matrices within the context of $C^{*}$-algebras. Most recently, attention has also turned to companion matrices [8].

The study of the generalized inverse of block anti-triangular matrices is of significant interest, particularly due to its application in finding closed-form solutions to systems of second-order linear differential equations [3,5,7]. The aim of this paper is to investigate the core inverse of the block anti-triangular matrix $M=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$ defined over a ring. We derive novel explicit representations for the core inverse of M with core invertible product subblocks. Specifically, we examine the EP property of M in each of these cases.

Throughout the paper, all rings are assumed to be associative and equipped with an involution *. The ring of $n\times n$ matrices under the *-transpose is denoted by $R^{n\times n}$, and ${\mathbb{C}}^{n\times n}$ refers to the algebra of $n\times n$ complex matrices with the conjugate transpose. For an $n\times n$ matrix A with a group inverse $A^{\#}$, we define the spectral idempotent matrix as $A^{\pi }=I-A{A}^{\#}$.

2. Main Results

This paper presents the core inverses of certain anti-triangular block matrices with product subblocks over a ring. In the following, we derive

Theorem 2.1.

Let $M=\left(\begin{array}{cc}ACB& A\\ B& 0\end{array}\right)\in R^{(m+n)\times (m+n)},A\in R^{m\times n},C\in R^{n\times n},B\in R^{n\times m}$. If $AB,BA$ has core inverses, ${\left(AB\right)}^{\pi }A=0$ and $B{\left(AB\right)}^{\pi }=0$, Then M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right).$$

Proof. 

Let $X=\left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right).$ Obviously, we have

$$\begin{array}{cc}& ACBA{\left(BA\right)}^{\overline{)\#}}-AB{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\hfill \\ \hfill =& \left[(I-\left(AB\right){\left(AB\right)}^{\overline{)\#}})A\right]CBA{\left(BA\right)}^{\overline{)\#}}\hfill \\ \hfill =& 0.\hfill \end{array}$$

Then we verify that

$$\begin{array}{ccc}\hfill MX& =\hfill & \left(\begin{array}{cc}AB{\left(AB\right)}^{\overline{)\#}}& ACBA{\left(BA\right)}^{\overline{)\#}}-AB{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\\ 0& BA{\left(BA\right)}^{\overline{)\#}}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}AB{\left(AB\right)}^{\overline{)\#}}& 0\\ 0& BA{\left(BA\right)}^{\overline{)\#}}\end{array}\right).\hfill \end{array}$$

Hence, ${\left(MX\right)}^{*}=MX$. Since ${\left(AB\right)}^{\pi }A=0$, we get $[I-AB{\left(AB\right)}^{\overline{)\#}}]A{\left(BA\right)}^{\overline{)\#}}=[I-AB{\left(AB\right)}^{\overline{)\#}}]A\left(BA\right){\left[{\left(BA\right)}^{\overline{)\#}}\right]}^2=0$. Thus we have $AB{\left(AB\right)}^{\overline{)\#}}A{\left(BA\right)}^{\overline{)\#}}=A{\left(BA\right)}^{\overline{)\#}}$.

Since ${\left(BA\right)}^{\pi }B=0$, analogously, we deduce that $BA{\left(BA\right)}^{\overline{)\#}}B{\left(AB\right)}^{\overline{)\#}}=B{\left(AB\right)}^{\overline{)\#}}.$

Accordingly, we have

$$\begin{array}{cc}& M{X}^2=\left(MX\right)X\hfill \\ \hfill =& \left(\begin{array}{cc}AB{\left(AB\right)}^{\overline{)\#}}& 0\\ 0& BA{\left(BA\right)}^{\overline{)\#}}\end{array}\right)\left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}0& AB{\left(AB\right)}^{\overline{)\#}}A{\left(BA\right)}^{\overline{)\#}}\\ BA{\left(BA\right)}^{\overline{)\#}}B{\left(AB\right)}^{\overline{)\#}}& -BA{\left(BA\right)}^{\overline{)\#}}B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right)=X;\hfill \\ & X{M}^2\hfill \\ \hfill =& \left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right){\left(\begin{array}{cc}ACB& A\\ B& 0\end{array}\right)}^2\hfill \\ \hfill =& \left(\begin{array}{cc}A{\left(BA\right)}^{\overline{)\#}}B& 0\\ B{\left(AB\right)}^{\overline{)\#}}ACB-B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}B& B{\left(AB\right)}^{\overline{)\#}}A\end{array}\right)\left(\begin{array}{cc}ACB& A\\ B& 0\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}A{\left(BA\right)}^{\overline{)\#}}BACB& A{\left(BA\right)}^{\overline{)\#}}BA\\ B{\left(AB\right)}^{\overline{)\#}}AB& B{\left(AB\right)}^{\overline{)\#}}AC[I-BA{\left(BA\right)}^{\overline{)\#}}]BA\end{array}\right)=M.\hfill \end{array}$$

Therefore $M^{\overline{)\#}}=X$, as asserted. □

Corollary 2.2.

Let $M=\left(\begin{array}{cc}0& A\\ B& 0\end{array}\right)\in R^{(m+n)\times (m+n)}$. If $AB,BA$ have generalized core inverses, ${\left(AB\right)}^{\pi }A=0$ and $B{\left(AB\right)}^{\pi }=0$, then M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& 0\end{array}\right).$$

Proof. 

This is obvious by choosing $C=0$ in Theorem 2.1. □

Corollary 2.3.

Let $M=\left(\begin{array}{cc}C& A\\ B& 0\end{array}\right)\in R^{(m+n)\times (m+n)}$. If $AB,BA,C$ have core inverses, $CA=0,C^{*}A=0,BC=0,{\left(AB\right)}^{\pi }A=0$ and $B{\left(AB\right)}^{\pi }=0$, then M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}{A}^{\overline{)\#}}& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& 0\end{array}\right).$$

Proof. 

we have $M=P+Q,$ where

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

In view of Theorem 2.1, Q has core inverse. By hypothesis,

$$PQ=0,P^{*}Q=0,QP=0.$$

By virtue of ???, M has core inverse and

$$M^{\overline{)\#}}=P^{\overline{)\#}}+Q^{\overline{)\#}}=\left(\begin{array}{cc}{A}^{\overline{)\#}}& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& 0\end{array}\right),$$

as asserted. □

Corollary 2.4.

Let $M=\left(\begin{array}{cc}ACB& A\\ B& 0\end{array}\right)\in R^{(m+n)\times (m+n)}$. If $AB,BA$ are EP, ${\left(AB\right)}^{\pi }A=0$ and $B{\left(AB\right)}^{\pi }=0$, then M is EP.

Proof. 

Since $AB,BA$ are EP, it follows by [14] Lemma 3.5, that $AB,BA$ have core inverses and ${\left(AB\right)}^{\overline{)\#}}={\left(AB\right)}^{\#},{\left(BA\right)}^{\overline{)\#}}={\left(BA\right)}^{\#}$. By virtue of Theorem ??, M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}0& A{\left(BA\right)}^{\overline{)\#}}\\ B{\left(AB\right)}^{\overline{)\#}}& -B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}\end{array}\right).$$

As in the proof of Theorem 2.1, we have

$$M{M}^{\overline{)\#}}=\left(\begin{array}{cc}\left(AB\right){\left(AB\right)}^{\overline{)\#}}& 0\\ 0& \left(BA\right){\left(BA\right)}^{\overline{)\#}}\end{array}\right).$$

Moreover, we have

$$\begin{array}{ccc}\hfill M^{\overline{)\#}}M& =\hfill & \left(\begin{array}{cc}A{\left(BA\right)}^{\overline{)\#}}B& 0\\ B{\left(AB\right)}^{\overline{)\#}}ACB-B{\left(AB\right)}^{\overline{)\#}}ACBA{\left(BA\right)}^{\overline{)\#}}B& B{\left(AB\right)}^{\overline{)\#}}A\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A{\left(BA\right)}^{\#}B& 0\\ B{\left(AB\right)}^{\#}ACB[I-A{\left(BA\right)}^{\#}B]& B{\left(AB\right)}^{\#}A\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\left(AB\right){\left(AB\right)}^{\#}& 0\\ 0& \left(BA\right){\left(BA\right)}^{\#}\end{array}\right).\hfill \end{array}$$

Thus, $M{M}^{\overline{)\#}}=M^{\overline{)\#}}M$. Therefore M is EP by [14], Lemma 3.5. □

Next, we investigate the core inverse of anti-triangular block matrices with product anti-diagonal entries.

Theorem 2.5.

Let R be a ring, let $M=\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)$, and let $A,CAB$ have core inverses. If $A^{\pi }B=0,{\left(CAB\right)}^{\pi }C=0$ and $CA{A}^{\overline{)\#}}=C$, then M has core inverse and

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & A^{\overline{)\#}}-B{\left(CAB\right)}^{\overline{)\#}}C,\hfill \\ \hfill \beta & =\hfill & B{\left(CAB\right)}^{\overline{)\#}},\hfill \\ \hfill \gamma & =\hfill & {\left(CAB\right)}^{\overline{)\#}}C,\hfill \\ \hfill \delta & =\hfill & -{\left(CAB\right)}^{\overline{)\#}}.\hfill \end{array}$$

Proof. 

Let $X=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$, where $\alpha ,\beta ,\gamma$ and $\delta$ are constructed as above. We directly verify that

$$\begin{array}{ccc}\hfill A\alpha +AB\gamma & =\hfill & A{A}^{\overline{)\#}}-AB{\left(CAB\right)}^{\overline{)\#}}C+AB{\left(CAB\right)}^{\overline{)\#}}C=A{A}^{\overline{)\#}},\hfill \\ \hfill A\beta +AB\delta & =\hfill & AB{\left(CAB\right)}^{\overline{)\#}}-AB{\left(CAB\right)}^{\overline{)\#}}=0,\hfill \\ \hfill CA\alpha & =\hfill & CA[A^{\overline{)\#}}-B{\left(CAB\right)}^{\overline{)\#}}C]\hfill \\ & =\hfill & CA{A}^{\overline{)\#}}-CAB{\left(CAB\right)}^{\overline{)\#}}C=C-[I-{\left(CAB\right)}^{\pi }]C=0,\hfill \\ \hfill CA\beta & =\hfill & CA\left[B{\left(CAB\right)}^{\overline{)\#}}\right]=CAB{\left(CAB\right)}^{\overline{)\#}}.\hfill \end{array}$$

Moreover, we deduce that

$$\begin{array}{ccc}\hfill \alpha A+\beta CA& =\hfill & A^{\overline{)\#}}A-B{\left(CAB\right)}^{\overline{)\#}}CA+B{\left(CAB\right)}^{\overline{)\#}}CA\hfill \\ & =\hfill & A^{\overline{)\#}}A-B{\left(CAB\right)}^{\overline{)\#}}CA+B{\left(CAB\right)}^{\overline{)\#}}CA=A^{\overline{)\#}}A,\hfill \\ \hfill \alpha AB& =\hfill & A^{\overline{)\#}}AB-B{\left(CAB\right)}^{\overline{)\#}}CAB=B{\left(CAB\right)}^{\tau },\hfill \\ \hfill \gamma A+\delta CA& =\hfill & {\left(CAB\right)}^{\overline{)\#}}CA-{\left(CAB\right)}^{\overline{)\#}}CA=0,\hfill \\ \hfill \gamma AB& =\hfill & {\left(CAB\right)}^{\overline{)\#}}CAB.\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\hfill MX& =\hfill & \left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A\alpha +AB\gamma & A\beta +AB\delta \\ CA\alpha & CA\beta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}& 0\\ 0& CAB{\left(CAB\right)}^{\overline{)\#}}\end{array}\right),\hfill \\ \hfill XM& =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha A+\beta CA& \alpha AB\\ \gamma A+\delta CA& \gamma AB\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{A}^{\overline{)\#}}A& B{\left(CAB\right)}^{\tau }\\ 0& {\left(CAB\right)}^{\overline{)\#}}CAB\end{array}\right).\hfill \end{array}$$

This implies that ${\left(MX\right)}^{*}=MX$.

Moreover, we see that

$$\begin{array}{ccc}\hfill M{X}^2& =\hfill & \left(MX\right)X\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}& 0\\ 0& CAB{\left(CAB\right)}^{\overline{)\#}}\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}\alpha & A{A}^{\overline{)\#}}\beta \\ CAB{\left(CAB\right)}^{\overline{)\#}}\gamma & CAB{\left(CAB\right)}^{\overline{)\#}}\delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)=X;\hfill \\ \hfill X{M}^2& =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right){\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)}^2\hfill \\ & =\hfill & \left(\begin{array}{cc}{A}^{\overline{)\#}}A& B{\left(CAB\right)}^{\tau }\\ 0& {\left(CAB\right)}^{\overline{)\#}}CAB\end{array}\right)\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)=M.\hfill \end{array}$$

Therefore M has core inverse and $M^{\overline{)\#}}=X$, as required. □

Corollary 2.6.

Let R be a ring, let $M=\left(\begin{array}{cc}I& B\\ C& 0\end{array}\right)$, and let $CB$ have core inverse. If ${\left(CB\right)}^{\pi }c=0$, then M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}I-B{\left(CB\right)}^{\overline{)\#}}C& B{\left(CB\right)}^{\overline{)\#}}\\ {\left(CB\right)}^{\overline{)\#}}C& -{\left(CB\right)}^{\overline{)\#}}.\end{array}\right).$$

Proof. 

This is obvious by choosing $A=I$ in Theorem 2.5. □

Corollary 2.7.

Let R be a ring, let $M=\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)$, and let $A,CAB$ be EP. If $A^{\pi }B=0,{\left(CAB\right)}^{\pi }C=0$ and $CA{A}^{\overline{)\#}}=C$, then M is EP.

Proof. 

By virtue of Theorem 2.5, we have

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & A^{\#}-B{\left(CAB\right)}^{\#}C,\hfill \\ \hfill \beta & =\hfill & B{\left(CAB\right)}^{\#},\hfill \\ \hfill \gamma & =\hfill & {\left(CAB\right)}^{\#}C,\hfill \\ \hfill \delta & =\hfill & -{\left(CAB\right)}^{\#}.\hfill \end{array}$$

Similarly to Corollary 2.4, we verify that $M{M}^{\overline{)\#}}=M^{\overline{)\#}}M$. In light of [14], Lemma 3.5, M is EP, as asserted. □

We now provide another related result.

Theorem 2.8.

Let R be a ring, let $A\in R^{n\times n},B\in R^{n\times m},C\in R^{m\times n}$ and $M=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$, and let $A,CB$ have core inverses. If $CA=C,A^{\pi }B=0,B{\left(CB\right)}^{\pi }=0$ and ${\left(CB\right)}^{\pi }C=0$, then M has core inverse and

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & A^{\overline{)\#}}-A^{\overline{)\#}}B{\left(CB\right)}^{\overline{)\#}}C,\hfill \\ \hfill \beta & =\hfill & A^{\overline{)\#}}B{\left(CB\right)}^{\overline{)\#}},\hfill \\ \hfill \gamma & =\hfill & {\left(CB\right)}^{\overline{)\#}}C,\hfill \\ \hfill \delta & =\hfill & -{\left(CB\right)}^{\overline{)\#}}.\hfill \end{array}$$

Proof. 

Let $X=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$, where $\alpha ,\beta ,\gamma$ and $\delta$ are constructed as above. We directly verify that

$$\begin{array}{ccc}\hfill A\alpha +B\gamma & =\hfill & A{A}^{\overline{)\#}}+(I-A{A}^{\overline{)\#}})B{\left(CB\right)}^{\overline{)\#}}C=A{A}^{\overline{)\#}},\hfill \\ \hfill A\beta +B\delta & =\hfill & A{A}^{\overline{)\#}}B{\left(CB\right)}^{\overline{)\#}}-B{\left(CB\right)}^{\overline{)\#}}=-(I-A{A}^{\overline{)\#}})B{\left(CB\right)}^{\overline{)\#}}=0,\hfill \\ \hfill C\alpha & =\hfill & (I-CB{\left(CB\right)}^{\overline{)\#}})C-C{A}^{\pi }B{\left[{\left(CB\right)}^{\overline{)\#}}\right]}^{2}C\hfill \\ & =\hfill & (I-CB{\left(CB\right)}^{\overline{)\#}})C=0,\hfill \\ \hfill C\beta & =\hfill & CB{\left(CB\right)}^{\overline{)\#}}=CB{\left(CB\right)}^{\overline{)\#}}.\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\hfill MX& =\hfill & \left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A\alpha +B\gamma & A\beta +B\delta \\ C\alpha & C\beta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}& 0\\ 0& CB{\left(CB\right)}^{\overline{)\#}}\end{array}\right).\hfill \end{array}$$

This implies that ${\left(MX\right)}^{*}=MX$.

Moreover, we see that

$$\begin{array}{ccc}\hfill M{X}^2& =\hfill & \left(MX\right)X\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}& 0\\ 0& CB{\left(CB\right)}^{\overline{)\#}}\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A{A}^{\overline{)\#}}\alpha & A{A}^{\overline{)\#}}\beta \\ CB{\left(CB\right)}^{\overline{)\#}}\gamma & CB{\left(CB\right)}^{\overline{)\#}}\delta \end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)=X;\hfill \\ \hfill X{M}^2& =\hfill & \left(\begin{array}{cc}\alpha A+\beta C& \alpha B\\ \gamma A+\delta c& \gamma B\end{array}\right)\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{A}^{\overline{)\#}}A& 0\\ 0& {\left(CB\right)}^{\overline{)\#}}\left(CB\right)\end{array}\right)\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)=M,\hfill \end{array}$$

thus yielding the result. □

Corollary 2.9.

Let R be a ring, let $M=\left(\begin{array}{cc}I& B\\ C& 0\end{array}\right)$, and let $CB$ have core inverse. If $B{\left(CB\right)}^{\pi }=0$ and ${\left(CB\right)}^{\pi }C=0$, then M has core inverse and

$$M^{\overline{)\#}}=\left(\begin{array}{cc}I-B{\left(CB\right)}^{\overline{)\#}}C& B{\left(CB\right)}^{\overline{)\#}}\\ {\left(CB\right)}^{\overline{)\#}}C& -{\left(CB\right)}^{\overline{)\#}}\end{array}\right).$$

Proof. 

This is obvious by Theorem 2.8. □

Corollary 2.10.

Let R be a ring, let $A\in R^{n\times n},B\in R^{n\times m},C\in R^{m\times n}$ and $M=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$, and let $A,CB$ be EP. If $CA=C,A^{\pi }B=0,B{\left(CB\right)}^{\pi }=0$ and ${\left(CB\right)}^{\pi }C=0$, then M is EP.

Proof. 

In view of Theorem 2.8,

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

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & A^{\#}-A^{\#}B{\left(CB\right)}^{\#}C,\hfill \\ \hfill \beta & =\hfill & A^{\#}B{\left(CB\right)}^{\#},\hfill \\ \hfill \gamma & =\hfill & {\left(CB\right)}^{\#}C,\hfill \\ \hfill \delta & =\hfill & -{\left(CB\right)}^{\#}.\hfill \end{array}$$

Then we directly verify that $M{M}^{\overline{)\#}}=M^{\overline{)\#}}M$. This completes the proof by [14], Lemma 3.5. □

3. Numerical Examples

We now illustrate the main results of the previous section through numerical examples in which the involution on ${\mathbb{C}}^{n\times n}$ is taken to be the conjugate transpose.

Example 3.1.

Let

$$A=\left(\begin{array}{c}1+i\\ -2+2i\end{array}\right),B=\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& 0\end{array}\right),C=\left[\begin{array}{c}3-2i\end{array}\right]$$

and let $M=\left(\begin{array}{cc}ACB& A\\ B& 0\end{array}\right).$ Then

$$AB=\left(\begin{array}{c}1+i\\ -2+2i\end{array}\right)\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ -2i& 0\end{array}\right).$$

$AB$ is an idempotent matrix, so its group inverse satisfies ${\left(AB\right)}^{\#}=AB$. We verify that

$$\left(AB\right){\left(AB\right)}^{\#}A={\left(AB\right)}^{2}A=ABA=\left(\begin{array}{cc}1& 0\\ -2i& 0\end{array}\right)\left(\begin{array}{c}1+i\\ -2+2i\end{array}\right)=\left(\begin{array}{c}1+i\\ -2+2i\end{array}\right)=A.$$

Moreover, we see that

$$BA=\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& 0\end{array}\right)\left(\begin{array}{c}1+i\\ -2+2i\end{array}\right)=\left[\begin{array}{c}1\end{array}\right]$$

As a non-zero $1\times 1$ matrix, its group inverse satisfies ${\left(BA\right)}^{\#}=BA$. We check that

$$B\left(AB\right){\left(AB\right)}^{\#}=B{\left(AB\right)}^2=BAB=\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ -2i& 0\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& 0\end{array}\right)=B.$$

Then we compute that

$${\left(AB\right)}^{\overline{)\#}}={\displaystyle \frac{1}{5}}\left(\begin{array}{cc}1& 2i\\ -2i& 4\end{array}\right)$$

Obviously,

$${\left(BA\right)}^{\overline{)\#}}=\left[\begin{array}{c}1\end{array}\right].$$

By virtue of Theorem 2.1, we deduce that

$$M^{\overline{)\#}}=\left(\begin{array}{cc}0& \left(\begin{array}{c}1+i\\ -2+2i\end{array}\right)\\ {\displaystyle \frac{1}{5}}\left(\begin{array}{cc}{\displaystyle \frac{1-i}{2}}& i(1-i)\end{array}\right)& {\displaystyle \frac{9-6i}{5}}\end{array}\right).$$

Evidently, we check that $M{\left(M^{\overline{)\#}}\right)}^2=M^{\overline{)\#}},{\left(M{M}^{\overline{)\#}}\right)}^{*}=M{M}^{\overline{)\#}},M^{\overline{)\#}}{M}^2=M.$

Example 3.2.

Let

$$C=\left(\begin{array}{cc}i& 1\end{array}\right),A={\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& -i\\ i& 1\end{array}\right)andB=\left(\begin{array}{c}1\\ i\end{array}\right)$$

Then $A^2=A=A^{*}$ and $A^{\pi }B=0,{\left(CAB\right)}^{\pi }C=0$ and $CA{A}^{\overline{)\#}}=C$. In view of Theorem 2.5, $M=\left(\begin{array}{cc}A& AB\\ CA& 0\end{array}\right)$ has core inverse and

$$\begin{array}{ccc}\hfill M^{\overline{)\#}}& =\hfill & \left(\begin{array}{cc}A-B{\left(CAB\right)}^{-1}C& B{\left(CAB\right)}^{-1}\\ {\left(CAB\right)}^{-1}C& -{\left(CAB\right)}^{-1}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{ccc}0& 0& -{\displaystyle \frac{i}{2}}\\ 0& 0& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{i}{2}}& {\displaystyle \frac{i}{2}}\end{array}\right).\hfill \end{array}$$

We directly verify that all three conditions for the core inverse are satisfied: $M{\left(M^{\overline{)\#}}\right)}^2=M^{\overline{)\#}},{\left(M{M}^{\overline{)\#}}\right)}^{*}=M{M}^{\overline{)\#}},M^{\overline{)\#}}{M}^2=M.$

A matrix $A\in C^{n\times n}$ is EP if and only if $\mathcal{R}\left(A\right)=\mathcal{R}\left(A^{*}\right)$, if and only if A commutes with its Moore-Penrose inverse $A^{†},$ i.e., $A{A}^{†}=A^{†}A$. To illustrate Corollary 2.10, we present an block operator matrix that is EP.

Example 3.3.

Let

$$A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right),B=\left(\begin{array}{cc}2& 1\\ 1& 1\\ 0& 0\end{array}\right),C={\displaystyle \frac{1}{5}}\left(\begin{array}{ccc}1& -1& 0\\ -1& 2& 0\end{array}\right).$$

Then $CB={\displaystyle \frac{1}{5}}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).$ Moreover, we check that

$$CA=C,A^{\pi }B=0,B{\left(CB\right)}^{\pi }=0,{\left(CB\right)}^{\pi }C=0.$$

Since A and $CB$ are symmetric matrices, we have $rank\left(A\right)=rank\left(A^{*}\right)$ and $rank\left(CB\right)=rank{\left(CB\right)}^{*}$. Hence A and $CB$ are EP (see [14]). By virtue of Corollary 2.10, the $2\times 2$ block matrix $M=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$ is EP.