Additive Results of Group Inverses in Banach Algebras

Abstract

In this paper, we present new presentations of group inverses for the sum of two group invertible elements in a Banach algebra. We then apply these results to block complex matrices. The group invertibility of certain block complex matrices is thereby obtained.

1. Introduction

A Banach algebra is an algebra $\mathcal{A}$ over $\mathbb{C}$, with an identity e, which has a norm $\parallel ·\parallel$, making it into a Banach space and satisfying $\parallel e\parallel =1$, $\parallel ab\parallel \le \parallel a\parallel \parallel b\parallel$, where $a,b\in \mathcal{A}$. Let X be a Banach space, then $L\left(X\right)$, the algebra of all bounded linear operators on X, is a Banach algebra with respect to the usual operator norm. The identity operator I is its unit elements. $L\left(X\right)$ is noncommutative when $\dim \left(X\right)>1$. An element a in a Banach algebra $\mathcal{A}$ has a group inverse, provided that there exists $b\in \mathcal{A}$ such that $a=aba,b=bab$ and $ab=ba$. Such a b is unique if exists, denoted by $a^{\#}$, and is called the group inverse of a. In view of ${\left(a{a}^{\#}\right)}^2=\left(a{a}^{\#}a\right)a^{\#}=a{a}^{\#}$, $a{a}^{\#}$ is an idempotent of $\mathcal{A}$. The symbol ${\mathcal{A}}^{\#}$ denotes the set of all group invertible elements in $\mathcal{A}$. For some examples related to the group inverse in Banach algebras see [1]. As is well known, a square complex matrix A has a group inverse if and only if $rank\left(A\right)=rank\left(A^2\right)$. The group invertibility in a ring is attractive. It has interesting applications of resistance distances to the bipartiteness of graphs (see [2,3]). Recently, the group inverse in a Banach algebra or a ring was extensively studied by many authors, e.g., [4,5,6,7,8,9,10]. In [11], Theorem 2.3, Liu et al. presented the group inverse of the combinations of two group invertible complex matrices P and Q under the condition $PQ{Q}^{\#}=QP{P}^{\#}$. In Theorem 3.1 of [12], Zhou et al. investigated the group inverse of $a+b$ under the condition $ab{b}^{\#}=ba{a}^{\#}$ in a Dedekind finite ring in which 2 is invertible. Group inverse is very useful, for example, in solving singular differential and difference equations formulated over a Banach space X [13]. In fact, since the structure of the Banach space is mainly considered, we can regard the operators on Banach space X as an element of the Banach algebra $L\left(X\right)$ of all bounded linear operators on a complex Banach space X. The motivation of this paper is to extend the preceding results to a general setting for Banach algebras.

In Section 2, we present the group inverse for the sum of two group invertible elements in a Banach algebra. Let $a,b\in {\mathcal{A}}^{\#}$. If $ab{b}^{\#}=\lambda ba{a}^{\#}$, then $a+b\in {\mathcal{A}}^{\#}$. The representation of its group inverse is also given. In Section 3, we apply our results and investigate the group inverse of a block complex matrix

$$M=\left(\begin{array}{cc}A& C\\ B& D\end{array}\right)$$

where $A\in {\mathbb{C}}^{m\times m},B\in {\mathbb{C}}^{n\times m},C\in {\mathbb{C}}^{m\times n},D\in {\mathbb{C}}^{n\times n}$. This problem is quite complicated, and was extensively studied by many authors. As applications, the group invertibility of certain block complex matrices M is thereby obtained. Additionally, this paper extends the results obtained in Theorem 2.3 of [11], and Theorem 3.1 of [12].

Throughout the paper, all Banach algebras are complex with an identity. Let $\mathcal{A}$ be the Banach algebra. We use ${\mathcal{A}}^{-1}$ to denote the set of all invertible elements in $\mathcal{A}$. $\lambda$ always stands for a complex number. ${\mathbb{C}}^{m\times n}$ stand for the set of all complex $m\times n$ matrices.

2. Main Results

Let $S=\{e_1,\cdots ,e_n\}$ be a complete set of idempotents in $\mathcal{A}$, i.e., $e_i{e}_j=0(i\ne j),e_i^2=e_i(1\le i\le n)$ and ${\displaystyle \sum _{i=1}^n}{e}_i=1$. Then, we have $a={\displaystyle \sum _{i,j=1}^n}{e}_{i}a{e}_j$. We write a as the matrix form $a={\left(a_{ij}\right)}_S$, where $a_{ij}=e_{i}a{e}_j\in e_i\mathcal{A}{e}_j$, and call it the Peirce matrix of a relatively to S. We shall use this new technique with relative Peirce matrices and generalize Theorem 2.3 of [11] and Theorem 3.1 of [12] as follows.

Theorem 1. 

Let $a,b\in {\mathcal{A}}^{\#},\lambda \in \mathbb{C}.$ If $ab{b}^{\#}=\lambda ba{a}^{\#}$; then, $a+b\in {\mathcal{A}}^{\#}$. In this case,

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

Proof. 

Let $p=a{a}^{\#}$. Write

$$b={\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right)}_p,b{b}^{\#}={\left(\begin{array}{cc}{x}_1& x_2\\ x_3& x_4\end{array}\right)}_p.$$

Then,

$$ab{b}^{\#}={\left(\begin{array}{cc}a{x}_1& a{x}_2\\ 0& 0\end{array}\right)}_p,ba{a}^{\#}={\left(\begin{array}{cc}{b}_1& 0\\ b_3& 0\end{array}\right)}_p.$$

Since $ab{b}^{\#}=\lambda ba{a}^{\#}$, we have

$$a{x}_1=\lambda b_1,x_2=0,b_3=0.$$

Then,

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

Moreover, we have

$$b^{\#}={\left(\begin{array}{cc}{b}_1^{\#}& z\\ 0& b_4^{\#}\end{array}\right)}_p$$

for some $z\in \mathcal{A}$. This implies that

$$x_1=b_1{b}_1^{\#},x_3=0,x_4=b_4{b}_4^{\#}.$$

Therefore,

$$b{b}^{\#}={\left(\begin{array}{cc}{b}_1{b}_1^{\#}& 0\\ 0& b_4{b}_4^{\#}\end{array}\right)}_p.$$

Since $b=\left(b{b}^{\#}\right)b=b\left(b{b}^{\#}\right)$, we see that

$$b_2=\left(b_1{b}_1^{\#}\right)b_2=b_2\left(b_4{b}_4^{\#}\right).$$

Then,

$$b_2=\left(b_1{b}_1^{\#}\right)b_2\left(b_4{b}_4^{\#}\right).$$

Let $e_1=b_1{b}_1^{\#},e_2=a{a}^{\#}-b_1{b}_1^{\#},e_3=b_4{b}_4^{\#},e_4=a^{\pi }-b_4{b}_4^{\#}$. Since $a,e_1\in a{a}^{\#}\mathcal{A}a{a}^{\#}$, we write

$$a={\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right)}_{e_1}\in a{a}^{\#}\mathcal{A}a{a}^{\#}.$$

Then,

$$\begin{array}{c}{a}_1=x_{1}a{x}_1=\lambda x_1{b}_1=\lambda b_1{b}_1^{\#}{b}_1=\lambda b_1,\hfill \\ a_3=(a{a}^{\#}-e_1)a{x}_1=a{x}_1-\lambda b_1=0.\hfill \end{array}$$

Moreover, we see that

$$a_1,a_4\in {\left(a{a}^{\#}\mathcal{A}a{a}^{\#}\right)}^{-1}.$$

Since $S=\{e_1,e_2,e_3,e_4\}$ is a complete set of idempotents in $\mathcal{A}$, we have two Peirce matrices of a and b relative to S:

$$a={\left(\begin{array}{cccc}{a}_1& a_2& 0& 0\\ 0& a_4& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S,$$

$$b={\left(\begin{array}{cccc}{b}_1& 0& b_2& 0\\ 0& 0& 0& 0\\ 0& 0& b_4& 0\\ 0& 0& 0& 0\end{array}\right)}_S.$$

Then,

$$a+b={\left(\begin{array}{cccc}(1+\lambda )b_1& a_2& b_2& 0\\ 0& a_4& 0& 0\\ 0& 0& b_4& 0\\ 0& 0& 0& 0\end{array}\right)}_S.$$

One directly checks that

$$a^{\#}={\left(\begin{array}{cccc}{a}_1^{-1}& -a_1^{-1}{a}_2{a}_4^{-1}& 0& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S,$$

$$b^{\#}={\left(\begin{array}{cccc}{b}_1^{-1}& 0& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& 0& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.$$

Then,

$$\begin{array}{c}\hfill a{a}^{\#}={\left(\begin{array}{cccc}{e}_1& 0& 0& 0\\ 0& e_2& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S,b{b}^{\#}={\left(\begin{array}{cccc}{e}_1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& e_3& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\end{array}$$

Case 1. $\lambda =-1$. Then,

$$\begin{array}{cc}\hfill a+b& ={\left(\begin{array}{cccc}0& a_2& b_2& 0\\ 0& a_4& 0& 0\\ 0& 0& b_4& 0\\ 0& 0& 0& 0\end{array}\right)}_S,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {(a+b)}^{\#}& ={\left(\begin{array}{cccc}0& a_2{\left(a_4\right)}^{-2}& b_2{\left(b_4\right)}^{-2}& 0\\ 0& {\left(a_4\right)}^{-1}& 0& 0\\ 0& 0& {\left(b_4\right)}^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\hfill \end{array}$$

Since $a_1^{-1}+b_1^{-1}=a^{-1}(b_1+a_1)b_1^{-1}=a^{-1}(1+\lambda )b_1{b}_1^{-1}=0$, we see that

$$\begin{array}{c}\hfill a^{\#}+b^{\#}={\left(\begin{array}{cccc}0& -a_1^{-1}{a}_2{a}_4^{-1}& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & (a+b){(a^{\#}+b^{\#})}^2\hfill \\ \hfill & ={\left(\begin{array}{cccc}0& a_2& b_2& 0\\ 0& a_4& 0& 0\\ 0& 0& b_4& 0\\ 0& 0& 0& 0\end{array}\right)}_S{\left(\begin{array}{cccc}0& -a_1^{-1}{a}_2{a}_4^{-1}& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S^2\hfill \\ \hfill & ={(a+b)}^{\#},\hfill \end{array}$$

as desired.

Case 2. $\lambda \ne -1$. Then,

$$\begin{array}{cc}\hfill & {(a+b)}^{\#}\hfill \\ \hfill & ={\left(\begin{array}{cccc}{(1+\lambda )}^{-1}{b}_1^{-1}& -{(1+\lambda )}^{-1}{b}_1^{-1}{a}_2{a}_4^{-1}& -{(1+\lambda )}^{-1}{b}_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S\hfill \\ \hfill & ={(1+\lambda )}^{-1}{\left(\begin{array}{cccc}{b}_1^{-1}& -b_1^{-1}{a}_2{a}_4^{-1}& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S\hfill \\ \hfill & +{\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S+{\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\hfill \end{array}$$

We compute that

$$\begin{array}{cc}\hfill & a^{\#}+b^{\#}-a^{\#}b{b}^{\#}\hfill \\ \hfill & ={\left(\begin{array}{cccc}{a}_1^{-1}& -a_1^{-1}{a}_2{a}_4^{-1}& 0& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S+{\left(\begin{array}{cccc}{b}_1^{-1}& 0& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& 0& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S\hfill \\ \hfill & -{\left(\begin{array}{cccc}{a}_1^{-1}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S\hfill \\ \hfill & ={\left(\begin{array}{cccc}{b}_1^{-1}& -a_1^{-1}{a}_2{a}_4^{-1}& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S,\hfill \end{array}$$

$$\begin{array}{c}\hfill b^{\pi }{a}^{\#}={\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S,a^{\pi }{b}^{\#}={\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{1+\lambda }[a^{\#}+b^{\#}-a^{\#}b{b}^{\#}]+\frac{\lambda }{1+\lambda }[b^{\pi }{a}^{\#}+a^{\pi }{b}^{\#}]}\hfill \\ \hfill & ={(1+\lambda )}^{-1}{\left(\begin{array}{cccc}{b}_1^{-1}& -b_1^{-1}{a}_2{a}_4^{-1}& -b_1^{-1}{b}_2{b}_4^{-1}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S\hfill \\ \hfill & +{\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& a_4^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)}_S+{\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& b_4^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)}_S.\hfill \end{array}$$

Therefore, we have

$${(a+b)}^{\#}=\frac{1}{1+\lambda }[a^{\#}+b^{\#}-a^{\#}b{b}^{\#}]+\frac{\lambda }{1+\lambda }[b^{\pi }{a}^{\#}+a^{\pi }{b}^{\#}],$$

as asserted. □

Corollary 1. 

Let $a,b\in {\mathcal{A}}^{\#},\lambda \in \mathbb{C}.$ If $a{a}^{\#}b=\lambda b{b}^{\#}a$, then $a+b\in {\mathcal{A}}^{\#}$. In this case,

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

Proof. 

Let $(R,\ast )$ be the opposite ring of R. That is, it is a ring with the multiplication $a\ast b=b·a$. Applying Theorem 1 to the opposite ring $(R,\ast )$ of R, we obtain the result. □

Corollary 2. 

Let $a,b\in \mathcal{A}$ be idempotents, and let $\lambda \in \mathbb{C}.$ If $ab=\lambda ba$, then $a+b\in {\mathcal{A}}^{\#}$. In this case,

$$\begin{array}{cc}\hfill {(a+b)}^{\#}& =\left\{\begin{array}{ccc}{(a+b)}^3\hfill & ,\hfill & \lambda =-1,\hfill \\ {\displaystyle a+b-\frac{2+\lambda }{1+\lambda }ab}\hfill & ,\hfill & \lambda \ne -1.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

Since a and b are idempotents, we have

$$a{a}^{\#}b=ab=\lambda ba=\lambda b{b}^{\#}a.$$

Therefore, we establish the result by Corollary 2.2 [8]. □

Theorem 2. 

Let $a,b\in {\mathcal{A}}^{\#},\lambda \in \mathbb{C}.$ If $ab{b}^{\#}=\lambda b(\lambda \ne -1)$, then $a+b\in {\mathcal{A}}^{\#}$. In this case,

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

Proof. 

Let $p=b{b}^{\#}$. Write $a={\left(\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right)}_p,b={\left(\begin{array}{cc}b& 0\\ 0& 0\end{array}\right)}_p$. Then, $a_1=b{b}^{\#}ab{b}^{\#}=\lambda b{b}^{\#}b=\lambda b$ and $a_3=(1-b{b}^{\#})ab{b}^{\#}=\lambda (1-b{b}^{\#})b=0$. Hence,

$$a={\left(\begin{array}{cc}\lambda b& a_2\\ 0& a_4\end{array}\right)}_p,a+b={\left(\begin{array}{cc}(1+\lambda )b& a_2\\ 0& a_4\end{array}\right)}_p.$$

Obviously, $a_4=(1-b{b}^{\#})a(1-b{b}^{\#})=b^{\pi }a$. We easily check that $a_4^{\#}=b^{\pi }{a}^{\#}{b}^{\pi }$ and $a_4^{\pi }=1-b^{\pi }a{b}^{\pi }{a}^{\#}{b}^{\pi }=1-b^{\pi }a{a}^{\#}{b}^{\pi }.$ Moreover, we have

$$\begin{array}{cc}\hfill {\left[(1+\lambda )b\right]}^{\pi }{a}_2{a}_4^{\pi }& =b^{\pi }b{b}^{\#}{a}_4^{\#}\hfill \\ \hfill & =b^{\pi }b{b}^{\#}{b}^{\pi }{a}^{\#}{b}^{\pi }\hfill \\ \hfill & =0.\hfill \end{array}$$

According to Theorem 2.1 of [8], $a+b\in {\mathcal{A}}^{\#}.$ Further, we have

$${(a+b)}^{\#}={\left(\begin{array}{cc}{\left[(1+\lambda )b\right]}^{\#}& z\\ 0& a_4^{\#}\end{array}\right)}_p,$$

where $z={\left[{(1+\lambda )}^{-1}{b}^{\#}\right]}^2{a}_2{a}_4^{\pi }-{(1+\lambda )}^{-1}{b}^{\#}{a}_2{a}_4^{\#}.$ We compute that

$$\begin{array}{cc}\hfill a_2{a}_4^{\pi }& =b{b}^{\#}a{b}^{\pi }[1-b^{\pi }a{a}^{\#}{b}^{\pi }]\hfill \\ \hfill & =b{b}^{\#}a{b}^{\pi }{a}^{\pi }{b}^{\pi }]\hfill \\ \hfill & =b{b}^{\#}ab{b}^{\#}{a}^{\pi }{b}^{\pi }]\hfill \\ \hfill & =\lambda ba{a}^{\#}{b}^{\pi }.\hfill \end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill {(a+b)}^{\#}& ={(1+\lambda )}^{-1}{b}^{\#}+b^{\pi }{a}^{\#}{b}^{\pi }\hfill \\ \hfill & +{\left[{(1+\lambda )}^{-1}{b}^{\#}\right]}^2{a}_2{a}_4^{\pi }-{(1+\lambda )}^{-1}{b}^{\#}{a}_2{a}_4^{\#}\hfill \\ \hfill & ={(1+\lambda )}^{-1}{b}^{\#}+b^{\pi }{a}^{\#}{b}^{\pi }\hfill \\ \hfill & +\lambda {(1+\lambda )}^{-2}{b}^{\#}a{a}^{\#}{b}^{\pi }-{(1+\lambda )}^{-1}{b}^{\#}a{b}^{\pi }{a}^{\#}{b}^{\pi },\hfill \end{array}$$

as asserted. □

Corollary 3. 

Let $a,b\in {\mathcal{A}}^{\#},\lambda \in \mathbb{C}.$ If $a{a}^{\#}b=\lambda a(\lambda \ne -1)$, then $a+b\in {\mathcal{A}}^{\#}$. In this case,

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

Proof. 

Let $(\mathcal{A},\ast )$ be the opposite algebra of $\mathcal{A}$. By applying Theorem 2 to elements $b,a$ in this opposite ring, we obtain the result. □

We demonstrate Theorem 2 by the following numerical example.

Example 1. 

Let $A=\left(\begin{array}{cc}-1& -1\\ 1& -3\end{array}\right),B=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)\in {\mathbb{C}}^{2\times 2}$. Then, A and B have group inverses and $AB{B}^{\#}=-2B$. Since $A^{\#}=\left(\begin{array}{cc}-\frac{3}{4}& \frac{1}{4}\\ -\frac{1}{4}& -\frac{1}{4}\end{array}\right)$ and $B^{\#}=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)$. By using Theorem 2, we obtain

$$\begin{array}{cc}\hfill {(A+B)}^{\#}& =-B^{\#}+B^{\pi }{A}^{\#}{B}^{\pi }-2B^{\#}A{A}^{\#}{B}^{\pi }+B^{\#}A{B}^{\pi }{A}^{\#}{B}^{\pi }\hfill \\ \hfill & =\left(\begin{array}{cc}-1& 0\\ -\frac{1}{2}& -\frac{1}{2}\end{array}\right).\hfill \end{array}$$

3. Applications

The aim of this section is to present the group invertibility of the block matrix M by using our main results. We are ready to prove the following.

Theorem 3. 

Let A and D have group inverses. If $A^{\pi }B=0,D^{\pi }C=0,AC{D}^{\#}=\lambda C$ and $BC{D}^{\#}=\lambda D,$ then M has a group inverse.

Proof. 

Write $M=P+Q$, where

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

Since $A^{\pi }B=0,D^{\pi }C=0$, it follows by Theorem 3.4 of [4] that P and Q have group inverses. Moreover, we obtain

$$Q^{\#}=\left(\begin{array}{cc}0& C{\left(D^{\#}\right)}^2\\ 0& D^{\#}\end{array}\right).$$

We easily check that

$$\begin{array}{cc}\hfill PQ{Q}^{\#}& =\left(\begin{array}{cc}A& 0\\ B& 0\end{array}\right)\left(\begin{array}{cc}0& C\\ 0& D\end{array}\right)\left(\begin{array}{cc}0& C{\left(D^{\#}\right)}^2\\ 0& D^{\#}\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}A& 0\\ B& 0\end{array}\right)\left(\begin{array}{cc}0& C{D}^{\#}\\ 0& D^{\#}\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}0& AC{D}^{\#}\\ 0& BC{D}^{\#}\end{array}\right)\hfill \\ \hfill & =\lambda \left(\begin{array}{cc}0& C\\ 0& D\end{array}\right)\hfill \\ \hfill & =\lambda Q.\hfill \end{array}$$

In light of Theorem 2, $M=P+Q$ has a group inverse, as desired. □

Corollary 4. 

Let A and D have group inverses. If $C{D}^{\pi }=0,B{A}^{\pi }=0,A^{\#}BD=\lambda B$ and $A^{\#}BC=\lambda A$, then M has a group inverse.

Proof. 

Applying Theorem 3 to the block matrix

$$M^T=\left(\begin{array}{cc}{D}^T& B^T\\ C^T& A^T\end{array}\right),$$

we prove that $M^T$ has a group inverse. Therefore, we easily check that $M={\left(M^T\right)}^T$ has a group inverse, as asserted. □

Theorem 4. 

Let A and D have group inverses. If $A^{\pi }C=0,D^{\pi }B=0,A^{\#}AB=\lambda A$ and $A^{\#}AD=\lambda C,$ then M has a group inverse.

Proof. 

Write $M=P+Q$, where

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

Since $A^{\pi }C=0,D^{\pi }B=0$, by using Theorem 3.4 of [4], we see that P and Q have group inverses. Moreover, we obtain

$$P^{\#}=\left(\begin{array}{cc}{A}^{\#}& {\left(A^{\#}\right)}^{2}A\\ 0& 0\end{array}\right).$$

Then, we have

$$\begin{array}{cc}\hfill P{P}^{\#}Q& =\left(\begin{array}{cc}A& C\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{A}^{\#}& {\left(A^{\#}\right)}^{2}A\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ B& D\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}A{A}^{\#}& A^{\#}A\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ B& D\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{A}^{\#}AB& A^{\#}AD\\ 0& 0\end{array}\right)\hfill \\ \hfill & =\lambda \left(\begin{array}{cc}A& C\\ 0& 0\end{array}\right)\hfill \\ \hfill & =\lambda P.\hfill \end{array}$$

In light of Corollary 3, $M=P+Q$ has a group inverse, as desired. □

Corollary 5. 

Let A and D have group inverses. If $B{D}^{\pi }=0,C{A}^{\pi }=0,CD{D}^{\#}=\lambda D$ and $AD{D}^{\#}=\lambda B,$ then M has a group inverse.

Proof. 

Applying Theorem 4 to the block matrix

$$M^T=\left(\begin{array}{cc}{D}^T& B^T\\ C^T& A^T\end{array}\right),$$

we easily obtain the result as in Corollary 4. □

It is convenient at this stage to prove the following.

Theorem 5. 

Let $A\in {\mathbb{C}}^{m\times m},D\in {\mathbb{C}}^{n\times n}$ be idempotents and $rank\left(B\right)=rank\left(C\right)=rank\left(BC\right)=rank\left(CB\right)$. If $AD=\lambda AC,A$$(I-CB)=0$ and $DB{A}^{\pi }C=0$, then M has a group inverse.

Proof. 

Since $r\left(B\right)=r\left(C\right)=r\left(BC\right)=r\left(CB\right)$, it follows by Lemma 2.3 of [14] that $BC$ and $CB$ have group inverses. Let $K=\left(\begin{array}{cc}0& C\\ B& 0\end{array}\right)$. Then, $K^2=\left(\begin{array}{cc}CB& 0\\ 0& BC\end{array}\right).$ By hypothesis, we have

$$\begin{array}{cc}\hfill rank\left(K^2\right)& =rank\left(CB\right)+rank\left(BC\right)\hfill \\ \hfill & =rank\left(C\right)+rank\left(B\right)\hfill \\ \hfill & =rank\left(K\right).\hfill \end{array}$$

Then, K has a group inverse.

Write $Q:=\left(\begin{array}{cc}0& A^{\pi }C\\ D^{\pi }B& 0\end{array}\right).$ Then, we have

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

By hypothesis, we see that

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

Therefore, N has a group inverse, and

$$Q^{\#}=\left(\begin{array}{cc}{A}^{\pi }& 0\\ 0& D^{\pi }\end{array}\right)\left(\begin{array}{cc}0& C{\left(BC\right)}^{\#}\\ B{\left(CB\right)}^{\#}& 0\end{array}\right).$$

Let $P=\left(\begin{array}{cc}A& AC\\ DB& D\end{array}\right)$. Then, $M=P+Q$. Clearly, $A^{\#}A\left(DB\right)=ADB=\lambda A,A^{\#}AD=AD=\lambda AC$, $A^{\pi }\left(AC\right)=0$ and $D^{\pi }\left(DB\right)=0$. In light of Theorem 4, P has a group inverse. Since $AC{D}^{\pi }B=0,DB{A}^{\pi }C=0$, we check that

$$\begin{array}{cc}\hfill PQ& =\left(\begin{array}{cc}A& AC\\ DB& D\end{array}\right)\left(\begin{array}{cc}0& A^{\pi }C\\ D^{\pi }B& 0\end{array}\right)\hfill \\ \hfill & =0.\hfill \end{array}$$

According to Theorem 2.1 of [4], M has a group inverse, as asserted. □

Corollary 6. 

Let $A\in {\mathbb{C}}^{m\times m},D\in {\mathbb{C}}^{n\times n}$ be idempotents and $rank\left(B\right)=rank\left(C\right)=rank\left(BC\right)=rank\left(CB\right)$. If $AD=\lambda BD,(I-CB)D=0$ and $B{D}^{\pi }CA=0$, then M has a group inverse.

Proof. 

Applying Theorem 5 to the block matrix

$$M^T=\left(\begin{array}{cc}{D}^T& B^T\\ C^T& A^T\end{array}\right),$$

we complete the proof as in Corollary 5. □