Representations of the G-drazin Inverse for Certain Anti-triangular Matrices

1. Introduction

Let $\mathcal{A}$ be an is a Banach algebra with identity 1. An element $a\in \mathcal{A}$ has generalized Drazin inverse (g-Drazin inverse) if there exists $x\in \mathcal{A}$ such that

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

If such an x exists, it is unique and is denoted by $a^d$. Here, ${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid 1+\lambda x\in \mathcal{A}\mathrm{is} \mathrm{invertible} \mathrm{for} \mathrm{all}\lambda \in C\}.$ It is well known that $x\in {\mathcal{A}}^{qnil}$ if and only if $\underset{n\to \infty }{lim}\left|\right|x^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ If we replace the quasinilpotent set ${\mathcal{A}}^{qnil}$ with the set of all nilpotent elements in $\mathcal{A}$, we refer to the unique x as the Drazin inverse of a, and denote it by $a^D$. Both the Drazin and g-Drazin inverses play significant roles in ring and matrix theory (see [5]).

It is intriguing to investigate the Drazin and g-Drazin inverses of the anti-triangular matrix $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\in M_2\left(\mathcal{A}\right)$. One motivation for exploring this problem is the quest for a closed-form solution to systems of second-order linear differential equations, which can be expressed in the following vector-valued form: $Ax\left(t\right)+Bx\left(t\right)+Cx\left(t\right)=0$ where $A,B,C\in C^{n\times n}$ (with A being potentially singular) and x is an $C^n$-valued function. Clearly, the solutions to singular systems of differential equations are determined by the Drazin inverse of the aforementioned anti-triangular matrix M (see [2,3]). Although the Drazin and g-Drazin inverses of anti-triangular matrices are valuable tools in the context of differential equations, finding representations for such generalized inverses remains a challenging task.

In 2005, Castro-González and Dopazo gave the representations of the Drazin inverse for a class of complex matrix $\left(\begin{array}{cc}I& F\\ I& 0\end{array}\right)$ (see [9] [Theorem 3.3]).

In 2011, Bu et al. investigated the Drazin inverse of the complex matrix $\left(\begin{array}{cc}E& F\\ I& 0\end{array}\right)$ under the condition $EF=FE$ (see [2] [Theorem 3.3]).

In 2013, Xu, Song and Zhang studied an expression of the Drazin inverse of the operator matrix $\left(\begin{array}{cc}E& F\\ I& 0\end{array}\right)\in M_2\left(\mathcal{B}\left(X\right)\right)$ under the same condition, where $\mathcal{B}\left(X\right)$ is the Banach algebra of bounded linear operators on a complex Banach space X (see [16] [Theorem 3.8]).

In 2016, Yu, Wang and Deng characterized the Drazin invertibility of the anti-triangular operator matrix $\left(\begin{array}{cc}E& F\\ I& 0\end{array}\right)\in M_2\left(\mathcal{B}\left(\mathcal{H}\right)\right)$ under the conditions $F^{\pi }E{F}^D=0$, $F^{\pi }EF=F^{\pi }FE$, where $\mathcal{B}\left(\mathcal{H}\right)$ is the Banach algebra of bounded linear operators on a complex Hilbert space $\mathcal{H}$ (see [18] [Theorem 4.1]).

Recently, many authors have explored various conditions under which representations of the Drazin (g-Drazin) inverse of such anti-triangular matrices can be established. For additional references, we direct the reader to [10,11,19,20,21,23].

The motivation of this paper is to further investigate the representation of the g-Drazin inverse of the anti-triangular matrix $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ in a Banach algebra $\mathcal{A}$. We begin by examining the solvability of a quadratic equation in the Banach algebra $\mathcal{A}$ using Catalan numbers $C_n$. Next, we study the representation of M under the conditions $ab=ba,a\in \mathcal{A}isinvertible,b\in {\mathcal{A}}^{qnil}$. We then employ the ring of Morita context and the Pierce representation of a Banach algebra element as tools to extend the previous special case to the more general condition $ab=ba,a,b\in {\mathcal{A}}^d$. Consequently, the known results are extended to a broader context within a Banach algebra.

Throughout this paper, all Banach algebras are considered to be complex and possess an identity element. Let $M_2\left(\mathcal{A}\right)$ be the Banach algebra of all $2\times 2$ matrices over the Banach algebra $\mathcal{A}$. We use ${\mathcal{A}}^{-1},{\mathcal{A}}^D$ and ${\mathcal{A}}^d$ to stand for the sets of all invertible, Drazin invertible and g-Drazin invertible elements in $\mathcal{A}$, respectively. For $a\in {\mathcal{A}}^d$, we define $a^{\pi }=1-a{a}^d$. Let $a,p^2=p\in \mathcal{A}$. Then a has the Pierce decomposition given by $pap+pa{p}^{\pi }+p^{\pi }ap+p^{\pi }a{p}^{\pi }$, which we denote in matrix form as ${\left(\begin{array}{cc}pap& pa{p}^{\pi }\\ p^{\pi }ap& p^{\pi }a{p}^{\pi }\end{array}\right)}_p$.

2. key Lemmas

In this section, we present some necessary lemmas which will be used in the sequel. We start by

Lemma 2.1. 

Let $a,b\in {\mathcal{A}}^d$. If $ab=0$, then $a+b\in {\mathcal{A}}^d$ and

$${(a+b)}^d={\displaystyle \sum _{i=0}^{\infty }}{\left(a^d\right)}^{i+1}{b}^i{b}^{\pi }+{\displaystyle \sum _{i=0}^{\infty }}{a}^i{a}^{\pi }{\left(b^d\right)}^{i+1}.$$

Proof. 

See [5] [Lemma 15.2.2]. □

Lemma 2.2. 

Let $a,b\in {\mathcal{A}}^d$. If $a{b}^2=0$ and $aba=0$, then $a+b\in {\mathcal{A}}^d$ and

$$\begin{array}{ccc}\hfill {(a+b)}^d& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(b^d\right)}^{i+1}{a}^i{a}^{\pi }+{\displaystyle \sum _{i=0}^{\infty }}{b}^i{b}^{\pi }{\left(a^d\right)}^{i+1}+{\displaystyle \sum _{i=0}^{\infty }}{b}^i{b}^{\pi }{\left(a^d\right)}^{i+2}b\hfill \\ & +\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(b^d\right)}^{i+3}{a}^{i+1}{a}^{\pi }b-b^d{a}^{d}b-{\left(b^d\right)}^{2}a{a}^{d}b.\hfill \end{array}$$

Proof. 

See [17] [Theorem 2.1] and [5] [Corollary 15.2.4]. □

Lemma 2.3. 

Let

$$x=\left(\begin{array}{cc}a& 0\\ c& b\end{array}\right)or\left(\begin{array}{cc}b& c\\ 0& a\end{array}\right)$$

Then

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

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

Proof. 

See [5] [Lemma 15.2.1]. □

Lemma 2.4. 

Let $\mathcal{A}$ be a Banach algebra and $a\in {\mathcal{A}}^{-1},b\in {\mathcal{A}}^{qnil}$. If $ab=ba$, then the equation $ax+x^2=b$ has a solution x such that $a+bx\in {\mathcal{A}}^{-1},x\in {\mathcal{A}}^{qnil}.$

Proof. 

Let $x={\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{{\alpha }_i}{b}^{i+1}$, where $c_i\in C,{\alpha }_i\in Z.$ Choose ${\alpha }_i=-(2i+1)$, Since $ab=ba$, we have

$$\begin{array}{ccc}\hfill ax+x^2& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{{\alpha }_i+1}{b}^{i+1}+\left[{\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{{\alpha }_i}{b}^{i+1}\right]\left[{\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{{\alpha }_i}{b}^{i+1}\right]\hfill \\ & =\hfill & c_0{a}^{{\alpha }_0+1}b+[c_1{a}^{{\alpha }_1+1}+c_0^2{a}^{2{\alpha }_0}]b^2\hfill \\ & +\hfill & [c_2{a}^{{\alpha }_2+1}+c_0{c}_1{a}^{{\alpha }_0+{\alpha }_1}+c_1{c}_0{a}^{{\alpha }_1+{\alpha }_0}]b^3\hfill \\ & +\hfill & [c_3{a}^{{\alpha }_3+1}+c_0{c}_2{a}^{{\alpha }_0+{\alpha }_2}+c_1{c}_1{a}^{{\alpha }_1+{\alpha }_1}+c_2{c}_0{a}^{{\alpha }_2+{\alpha }_0}]b^4\hfill \\ & +\hfill & \cdots \hfill \\ & =\hfill & c_{0}b+[c_1+c_0^2]a^{-2}{b}^2+[c_2+c_0{c}_1+c_1{c}_0]{\alpha }^{-4}{b}^3\hfill \\ & +\hfill & [c_3+c_0{c}_2+c_1{c}_1+c_2{c}_0]a^{-6}{b}^4+\cdots \hfill \\ & =\hfill & b,\hfill \end{array}$$

hence, we choose

$$\begin{array}{c}{c}_0=1,c_1=-1,c_2=2,c_3=-5,c_4=14,c_5=-42,\cdots \\ c_i=-(c_0{c}_{i-1}+c_1{c}_{i-2}+\cdots +c_{i-1}{c}_0)(i\in \mathbb{N}).\end{array}$$

Let $\left\{C_n\right\}$ be the series of Catalan numbers, i.e.,

$$\begin{array}{c}{C}_0=1,C_1=1,c_2=2,c_3=5,c_4=14,c_5=42,\cdots ,\\ C_n=C_0{C}_{n-1}+\cdots +C_{n-1}{C}_0(n\in \mathbb{N}).\end{array}$$

Then $c_0=C_0,c_1=-C_1$. By induction, we claim that $c_{2n}=C_{2n},c_{2n+1}=-C_{2n+1}(n\ge 0)$. Hence, $|c_n|=C_n(n\ge 1)$. By using the asymptotic expression of the Catalan numbers $C_n$, we have

$$\underset{n\to \infty }{lim}{C}_n/\left({\displaystyle \frac{4^n}{\sqrt{\pi }{\left(n\right)}^{\frac{3}{2}}}}\right)=1.$$

Therefore

$$\underset{n\to \infty }{lim}\sqrt[n]{|c_n|}=\underset{n\to \infty }{lim}{\displaystyle \frac{4}{{\pi }^{\frac{1}{2n}}{\left(\sqrt[n]{n}\right)}^{\frac{3}{2}}}}=4.$$

Since $b\in {\mathcal{A}}^{qnil}$, we have $\underset{n\to \infty }{lim}{\Vert b^n\Vert }^{{\displaystyle \frac{1}{n}}}=0$. Since

$$\sqrt[n]{\Vert c_n{a}^{-(2n+1)}{b}^{n+1}\Vert }\le \sqrt[n]{|c_n|}\Vert a^{-1}{\Vert }^{2+{\displaystyle \frac{1}{n}}}\sqrt[n]{\Vert b\Vert }{\Vert b^n\Vert }^{{\displaystyle \frac{1}{n}}},$$

we deduce that

$$\underset{n\to \infty }{lim}\sqrt[n]{\Vert c_n{a}^{-(2n+1)}{b}^{n+1}\Vert }=0.$$

This implies that ${\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{-(2i+1)}{b}^{i+1}$ absolutely converges.

Accordingly, the equation $ax+x^2=b$ has a solution $x={\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{-(2i+1)}{b}^{i+1},$ where $c_0=1,c_{k+1}=-{\displaystyle \sum _{i=0}^k}{c}_i{c}_{k-i}(k\ge 0).$ Moreover, we verify that

$$\begin{array}{c}{c}_n={(-1)}^n{C}_n={(-1)}^n{\displaystyle \frac{\left(2n\right)!}{n!(n+1)!}},\\ x=\left[{\displaystyle \sum _{i=0}^{\infty }}{c}_i{a}^{-(2i+1)}{b}^i\right]b\in {\mathcal{A}}^{qnil},\\ a+x=a[1-a^{-1}x]\in {\mathcal{A}}^{-1}.\end{array}$$

This completes the proof. □

Lemma 2.5. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a\in {\mathcal{A}}^{-1},b\in {\mathcal{A}}^{qnil}$. If $ab=ba$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

$$M^d=\left(\begin{array}{cc}{(a+x)}^{-1}-xy& {(a+x)}^{-1}x-xyx\\ y& yx\end{array}\right),$$

where $x={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-(2i+1)}{b}^{i+1},y={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a+x)}^{-i-2}{x}^i.$

Proof. 

In view of Lemma 2.4, the equation $ax+x^2=b$ has a solution x such that $a+x\in {\mathcal{A}}^{-1},x\in {\mathcal{A}}^{qnil}.$ Here,

$$x={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-(2i+1)}{b}^{i+1}.$$

It is easy to verify that

$$M=\left(\begin{array}{cc}1& -x\\ 0& 1\end{array}\right)\left(\begin{array}{cc}a+x& 0\\ 1& -x\end{array}\right)\left(\begin{array}{cc}1& x\\ 0& 1\end{array}\right).$$

Since $x\in {\mathcal{A}}^{qnil}$ and $a+x\in {\mathcal{A}}^{-1}$. Then $\left(\begin{array}{cc}a+x& 0\\ 1& x\end{array}\right)$ has g-Drazin inverse. Therefore M has g-Drazin inverse. Exactly, we have

$$\begin{array}{ccc}\hfill M^d& =\hfill & \left(\begin{array}{cc}1& -x\\ 0& 1\end{array}\right){\left(\begin{array}{cc}a+x& 0\\ 1& -x\end{array}\right)}^d\left(\begin{array}{cc}1& x\\ 0& 1\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}1& -x\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{(a+x)}^{-1}& 0\\ y& 0\end{array}\right)\left(\begin{array}{cc}1& x\\ 0& 1\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{(a+x)}^{-1}-xy& {(a+x)}^{-1}x-xyx\\ y& yx\end{array}\right),\hfill \end{array}$$

where $y={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a+x)}^{-i-2}{x}^i.$ □

Lemma 2.6. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a\in \mathcal{A},b\in {\mathcal{A}}^{-1}$. Then $M\in M_2{\left(\mathcal{A}\right)}^{-1}$ and

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

Proof. 

Straightforward. □

Let $p^2=p\in \mathcal{A}$ and let ${\mathcal{A}}_1=p\mathcal{A}p,{\mathcal{A}}_2=p^{\pi }\mathcal{A}{p}^{\pi }$. Let T be the ring of Morita context $({\mathcal{A}}_1,{\mathcal{A}}_2,\phi ,\psi )$, i.e.,

$$T={\left(\begin{array}{cc}{\mathcal{A}}_1& M_2\left(p\mathcal{A}{p}^{\pi }\right)\\ M_2\left(p^{\pi }\mathcal{A}p\right)& {\mathcal{A}}_2\end{array}\right)}_{(\phi ,\psi )}$$

with the bimodule homomorphisms of the form

$$\begin{array}{c}\phi :M_2\left(p\mathcal{A}{p}^{\pi }\right)\times M_2\left(p^{\pi }\mathcal{A}p\right)\to {\mathcal{A}}_1,\\ \psi :M_2\left(p^{\pi }\mathcal{A}p\right)\times M_2\left(p\mathcal{A}{p}^{\pi }\right)\to {\mathcal{A}}_2.\end{array}$$

Then we have a natural isomorphism of rings given by

$$\begin{array}{c}\rho :M_2\left(\mathcal{A}\right)\cong T,\\ \left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\mapsto {\left(\begin{array}{cc}\begin{array}{cc}p{a}_{11}p& p{a}_{12}p\\ p{a}_{21}p& p{a}_{22}p\end{array}& \begin{array}{cc}p{a}_{11}{p}^{\pi }& p{a}_{12}{p}^{\pi }\\ p{a}_{21}{p}^{\pi }& p{a}_{22}{p}^{\pi }\end{array}\\ \begin{array}{cc}{p}^{\pi }{a}_{11}p& p^{\pi }{a}_{12}p\\ p^{\pi }{a}_{21}p& p^{\pi }{a}_{22}p\end{array}& \begin{array}{cc}{p}^{\pi }{a}_{11}{p}^{\pi }& p^{\pi }{a}_{12}{p}^{\pi }\\ p^{\pi }{a}_{21}{p}^{\pi }& p^{\pi }{a}_{22}{p}^{\pi }\end{array}\end{array}\right)}_{(\phi ,\psi )}.\end{array}$$

Lemma 2.7. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a\in {\mathcal{A}}^{-1},b\in {\mathcal{A}}^d$. If $ab=ba$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

$$M^d=\left(\begin{array}{cc}{z}_{11}& z_{12}\\ z_{21}& z_{22}\end{array}\right)$$

with $z_{ij}$ are formulated by

$$\begin{array}{ccc}\hfill z_{11}& =\hfill & {(a{b}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill z_{12}& =\hfill & b{b}^d+{(a{b}^{\pi }+x)}^{-1}x-xyx,\hfill \\ \hfill z_{21}& =\hfill & b^d,\hfill \\ \hfill z_{22}& =\hfill & -a{b}^d+y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-(2i+1)}{b}^{i+1}{b}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a{b}^{\pi }+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

Let $p=b{b}^d$. Since $ab=ba$, we have

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}\right)}_p,b={\left(\begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\right)}_p\in \mathcal{A}.$$

Then

$$M=\left(\begin{array}{cc}\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}& \begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\\ \begin{array}{cc}p& 0\\ 0& p^{\pi }\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\end{array}\right)\in M_2\left(\mathcal{A}\right).$$

Hence, we have

$$\rho \left(M\right)={\left(\begin{array}{cc}{M}_1& 0\\ 0& M_2\end{array}\right)}_{(\phi ,\psi )},$$

where

$$M_1=\left(\begin{array}{cc}{a}_1& b_1\\ p& 0\end{array}\right)\in M_2\left({\mathcal{A}}_1\right),M_2=\left(\begin{array}{cc}{a}_2& b_2\\ p^{\pi }& 0\end{array}\right)\in M_2\left({\mathcal{A}}_2\right).$$

Claim 1. $M_1\in M_2{\left({\mathcal{A}}_1\right)}^d$. Clearly, $a_1=ab{b}^d,b_1=b^2{b}^d\in {\mathcal{A}}_1^{-1}$. By Lemma 2.6, we have

$$M_1^d=M_1^{-1}=\left(\begin{array}{cc}0& b{b}^d\\ b_1^{-1}& -b_1^{-1}{a}_1\end{array}\right).$$

Claim 2. $M_2\in M_2{\left({\mathcal{A}}_2\right)}^d$. Clearly, $a_2=a{b}^{\pi }\in {\mathcal{A}}_1^{-1},b_2=b{b}^{\pi }\in {\mathcal{A}}_2^{qnil}$. By virtue of Lemma 2.5, we have

$$M_2^d=\left(\begin{array}{cc}{(a_2+x)}^{-1}-xy& {(a_2+x)}^{-1}x-xyx\\ y& yx\end{array}\right),$$

where $x={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}_2^{-(2i+1)}{b}_2^{i+1},y={\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a_2+x)}^{-i-2}{x}^i.$ Therefore $\rho \left(M\right)\in T^d$ and

$${\left[\rho \left(M\right)\right]}^d={\left(\begin{array}{cc}\begin{array}{cc}0& b{b}^d\\ b_1^{-1}& -b_1^{-1}{a}_1\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\\ \begin{array}{cc}0& 0\\ 0& 0\end{array}& \begin{array}{cc}{(a_2+x)}^{-1}-xy& {(a_2+x)}^{-1}x-xyx\\ y& yx\end{array}\end{array}\right)}_{(\phi ,\psi )}.$$

Therefore $M\in M_2{\left(\mathcal{A}\right)}^d$. Furthermore, we have

$$\begin{array}{ccc}\hfill M^d& =\hfill & \left(\begin{array}{cc}\begin{array}{cc}0& 0\\ 0& {(a_2+x)}^{-1}-xy\end{array}& \begin{array}{cc}b{b}^d& 0\\ 0& {(a_2+x)}^{-1}x-xyx\end{array}\\ \begin{array}{cc}{b}_1^{-1}& 0\\ 0& 0\end{array}& \begin{array}{cc}-b_1^{-1}{a}_1& 0\\ y& yx\end{array}\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{z}_{11}& z_{12}\\ z_{21}& z_{22}\end{array}\right)\hfill \end{array}$$

with $z_{ij}$ are formulated by

$$\begin{array}{ccc}\hfill z_{11}& =\hfill & {(a{b}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill z_{12}& =\hfill & b{b}^d+{(a{b}^{\pi }+x)}^{-1}x-xyx,\hfill \\ \hfill z_{21}& =\hfill & b^d,\hfill \\ \hfill z_{22}& =\hfill & -a{b}^d+y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-(2i+1)}{b}^{i+1}{b}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a{b}^{\pi }+x)}^{-i-2}{x}^i.\hfill \end{array}$$

This completes the proof. □

Lemma 2.8. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a\in {\mathcal{A}}^{qnil},b\in {\mathcal{A}}^d$. If $ab=ba$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

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

Proof. 

Let $X=\left(\begin{array}{cc}0& b{b}^d\\ b^d& -a{b}^d\end{array}\right)$. One directly verify that

$$\begin{array}{ccc}\hfill MX& =\hfill & \left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& b{b}^d\\ b^d& -a{b}^d\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}b{b}^d& 0\\ 0& b{b}^d\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}0& b{b}^d\\ b^d& -a{b}^d\end{array}\right)\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\hfill \\ & =\hfill & XM,\hfill \\ \hfill M{X}^2& =\hfill & \left(MX\right)X=X,\hfill \\ \hfill M-\left(MX\right)M& =\hfill & \left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)-\left(\begin{array}{cc}b{b}^d& 0\\ 0& b{b}^d\end{array}\right)\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}a(1-b{b}^d)& b-b^2{b}^d\\ 1-b{b}^d& 0\end{array}\right)\in M_2{\left(\mathcal{A}\right)}^{qnil}.\hfill \end{array}$$

Therefore M has g-Drazin inverse and $M^d=X$, as desired. □

3. Main Results

We now present the main results of this paper, which extend [16] [Theorem 3.8] and [18] [Theorem 4.1] to anti-triangular matrices in Banach algebras.

Theorem 3.1. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a,b\in {\mathcal{A}}^d$. If $ab=ba$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

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

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(a^2{a}^d{b}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & a{a}^{d}b{b}^d+{(a^2{a}^d{b}^{\pi }+x)}^{-1}x-xyx+a^{\pi }b{b}^d,\hfill \\ \hfill \gamma & =\hfill & a{a}^d{b}^d+a^{\pi }{b}^d,\hfill \\ \hfill \delta & =\hfill & -a^2{a}^d{b}^d+y+yx-a{a}^{\pi }{b}^d,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{a}^d{b}^{i+1}{b}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a^2{a}^d{b}^{\pi }+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

Let $q=a{a}^d$. Since $ab=ba$, we have

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}\right)}_q,b={\left(\begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\right)}_q.$$

Then

$$M=\left(\begin{array}{cc}\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}& \begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\\ \begin{array}{cc}p& 0\\ 0& p^{\pi }\end{array}& \begin{array}{cc}0& 0\\ 0& 0\end{array}\end{array}\right)\in M_2\left(\mathcal{A}\right).$$

By using the isomorphism $\rho$ between the matrix ring $M_2\left(\mathcal{A}\right)$ and the the ring of Morita context $({\mathcal{A}}_1,{\mathcal{A}}_2,\phi ,\psi )$ mentioned above, we have

$$\rho \left(M\right)={\left(\begin{array}{cc}{M}_1& 0\\ 0& M_2\end{array}\right)}_{(\phi ,\psi )},$$

where

$$M_1=\left(\begin{array}{cc}{a}_1& b_1\\ q& 0\end{array}\right)\in M_2\left({\mathcal{A}}_1\right),M_2=\left(\begin{array}{cc}{a}_2& b_2\\ q^{\pi }& 0\end{array}\right)\in M_2\left({\mathcal{A}}_2\right).$$

Claim 1. $M_1\in M_2{\left({\mathcal{A}}_1\right)}^d$. Obviously, $a_1\in {\mathcal{A}}_1^{-1},b_1\in {\mathcal{A}}_1^d$. In view of Lemma 2.7, we have

$$M_1^d=\left(\begin{array}{cc}{z}_{11}& z_{12}\\ z_{21}& z_{22}\end{array}\right)$$

with $z_{ij}$ are formulated by

$$\begin{array}{ccc}\hfill z_{11}& =\hfill & {(a^2{a}^d{b}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill z_{12}& =\hfill & a{a}^{d}b{b}^d+{(a^2{a}^d{b}^{\pi }+x)}^{-1}x-xyx,\hfill \\ \hfill z_{21}& =\hfill & a{a}^d{b}^d,\hfill \\ \hfill z_{22}& =\hfill & -a^2{a}^d{b}^d+y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{a}^d{b}^{i+1}{b}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(a^2{a}^d{b}^{\pi }+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Claim 2. $M_2\in M_2{\left({\mathcal{A}}_2\right)}^d$. Obviously, $a_2\in {\mathcal{A}}_2^{qnil},b_2\in {\mathcal{A}}_2^d$. By virtue of Lemma 2.8, we derive that

$$M_2^d=\left(\begin{array}{cc}0& b_2{b}_2^d\\ b_2^d& -a_2{b}_2^d\end{array}\right).$$

Therefore $\rho \left(M\right)\in T^d$ and

$${\left[\rho \left(M\right)\right]}^d={\left(\begin{array}{cc}{M}_1^d& 0\\ 0& M_2^d\end{array}\right)}_{(\phi ,\psi )}.$$

Therefore

$$M^d=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\in M_2\left(\mathcal{A}\right),$$

where

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {\left(\begin{array}{cc}{z}_{11}& 0\\ 0& 0\end{array}\right)}_p=z_{11},\hfill \\ \hfill \beta & =\hfill & {\left(\begin{array}{cc}{z}_{12}& 0\\ 0& b_2{b}_2^d\end{array}\right)}_p=z_{12}+a^{\pi }b{b}^d,\hfill \\ \hfill \gamma & =\hfill & {\left(\begin{array}{cc}{z}_{21}& 0\\ 0& b_2^d\end{array}\right)}_p=z_{21}+a^{\pi }{b}^d,\hfill \\ \hfill \delta & =\hfill & {\left(\begin{array}{cc}{z}_{22}& 0\\ 0& -a_2{b}_2^d\end{array}\right)}_p=z_{22}-a{a}^{\pi }{b}^d.\hfill \end{array}$$

This completes the proof. □

Corollary 3.2. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a,b\in {\mathcal{A}}^D$. If $ab=ba$, then $M\in M_2{\left(\mathcal{A}\right)}^D$ and

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

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(a^2{a}^D{b}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & a{a}^{d}b{b}^D+{(a^2{a}^D{b}^{\pi }+x)}^{-1}x-xyx+a^{\pi }b{b}^D,\hfill \\ \hfill \gamma & =\hfill & a{a}^D{b}^D+a^{\pi }{b}^D,\hfill \\ \hfill \delta & =\hfill & -a^2{a}^D{b}^D+y+yx-a{a}^{\pi }{b}^D,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{ind\left(b\right)-1}}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{a}^D{b}^{i+1}{b}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{ind\left(b\right)-1}}{(-1)}^i{(a^2{a}^D{b}^{\pi }+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

Evidently, $z\in {\mathcal{A}}^D$ if and only if $z\in {\mathcal{A}}^d$ and $a-a^2{a}^d\in \mathcal{A}$ is nilpotent. In this case, $z^D=z^d$. Therefore we complete the proof by Theorem 3.1. □

We are now ready to prove:

Theorem 3.3. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a,b,b^{\pi }a\in {\mathcal{A}}^d$. If $b^{\pi }a{b}^d=0$ and $b^{\pi }\left(ab\right)=b^{\pi }\left(ba\right)$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

$$M^d={\displaystyle \sum _{i=0}^{\infty }}{P}^i[I-P{P}^d]{\left(Q^d\right)}^{i+1},$$

where

$$\begin{array}{ccc}\hfill P& =\hfill & \left(\begin{array}{cc}{b}^{\pi }a& b^{\pi }b\\ b^{\pi }& 0\end{array}\right),P^d=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),\hfill \\ \hfill Q& =\hfill & \left(\begin{array}{cc}b{b}^{d}a& b^2{b}^d\\ b{b}^d& 0\end{array}\right),Q^d=\left(\begin{array}{cc}0& b{b}^d\\ b^d& -b^{d}a\end{array}\right)\hfill \end{array}$$

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(b^{\pi }{a}^2{a}^d+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & {(b^{\pi }{a}^2{a}^d+x)}^{-1}x-xyx,\hfill \\ \hfill \gamma & =\hfill & 0,\hfill \\ \hfill \delta & =\hfill & y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{b}^{\pi }{a}^d{b}^{i+1},\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(b^{\pi }{a}^2{a}^d+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

Write $M=P+Q$, where

$$P=\left(\begin{array}{cc}{b}^{\pi }a& b^{\pi }b\\ b^{\pi }& 0\end{array}\right),Q=\left(\begin{array}{cc}b{b}^{d}a& b^2{b}^d\\ b{b}^d& 0\end{array}\right).$$

Step 1. P has g-Drazin inverse. By hypothesis, we verify that

$$\left(b^{\pi }a\right)\left(b^{\pi }b\right)=b^{\pi }\left(ab\right)=b^{\pi }\left(ba\right)=\left(b^{\pi }b\right)\left(b^{\pi }a\right).$$

In light of Theorem 3.1, we have

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

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(b^{\pi }{a}^2{a}^d+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & {(b^{\pi }{a}^2{a}^d+x)}^{-1}x-xyx,\hfill \\ \hfill \gamma & =\hfill & 0,\hfill \\ \hfill \delta & =\hfill & y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{b}^{\pi }{a}^d{b}^{i+1},\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{(b^{\pi }{a}^2{a}^d+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Step 2. Q has g-Drazin inverse. By virtue of Lemma 2.6,

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

Step 3. Since $PQ=0$, it follows by Lemma 2.1 that

$$\begin{array}{ccc}\hfill M^d& =\hfill & {(P+Q)}^d\hfill \\ & =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(P^d\right)}^{i+1}{Q}^i{Q}^{\pi }+{\displaystyle \sum _{i=0}^{\infty }}{P}^i{P}^{\pi }{\left(Q^d\right)}^{i+1}\hfill \\ & =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{P}^i{P}^{\pi }{\left(Q^d\right)}^{i+1}.\hfill \end{array}$$

This completes the proof. □

Corollary 3.4. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)$ with $a,b,b^{\pi }a\in {\mathcal{A}}^D$. If $b^{\pi }a{b}^D=0$ and $b^{\pi }\left(ab\right)=b^{\pi }\left(ba\right)$, then $M\in M_2{\left(\mathcal{A}\right)}^D$ and

$$M^D={\displaystyle \sum _{i=0}^{ind\left(P\right)}}{P}^i[I-P{P}^D]{\left(Q^D\right)}^{i+1},$$

where

$$\begin{array}{ccc}\hfill P& =\hfill & \left(\begin{array}{cc}{b}^{\pi }a& b^{\pi }b\\ b^{\pi }& 0\end{array}\right),P^D=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right),\hfill \\ \hfill Q& =\hfill & \left(\begin{array}{cc}b{b}^{D}a& b^2{b}^D\\ b{b}^D& 0\end{array}\right),Q^D=\left(\begin{array}{cc}0& b{b}^D\\ b^D& -b^{D}a\end{array}\right)\hfill \end{array}$$

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(b^{\pi }{a}^2{a}^D+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & {(b^{\pi }{a}^2{a}^D+x)}^{-1}x-xyx,\hfill \\ \hfill \gamma & =\hfill & 0,\hfill \\ \hfill \delta & =\hfill & y+yx,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{ind\left(b\right)-1}}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{b}^{\pi }{a}^D{b}^{i+1},\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{ind\left(b\right)-1}}{(-1)}^i{(b^{\pi }{a}^2{a}^D+x)}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

It is immediate from Theorem 3.3. □

It is convenient at this stage to derive the following:

Theorem 3.5. 

Let $\mathcal{A}$ be a Banach algebra and $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $a,d,bc\in {\mathcal{A}}^d$. If $abc=bca,bdc=0$ and $b{d}^2=0$, then $M\in M_2{\left(\mathcal{A}\right)}^d$ and

$$\begin{array}{ccc}\hfill M^d& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(Q^d\right)}^{i+1}{P}^i(I-P{P}^d)+{\displaystyle \sum _{i=0}^{\infty }}{Q}^i{Q}^{\pi }{\left(P^d\right)}^{i+1}+{\displaystyle \sum _{i=0}^{\infty }}{Q}^i(I-Q{Q}^d){\left(P^d\right)}^{i+2}Q\hfill \\ & +\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(Q^d\right)}^{i+3}{P}^{i+1}(I-P{P}^d)Q-Q^d{P}^{d}Q-{\left(Q^d\right)}^{2}P{P}^{d}Q,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill P& =\hfill & \left(\begin{array}{cc}a& b\\ c& 0\end{array}\right),P^d=\left(\begin{array}{cc}{\alpha }^{2}a+\alpha \beta +\beta \gamma a+\beta \delta & {\alpha }^{2}b+\beta \gamma b\\ c\gamma \alpha a+c\gamma \beta +c\delta \gamma a+c\delta \delta & c\gamma \alpha b+c\delta \gamma b\end{array}\right);\hfill \\ \hfill Q& =\hfill & \left(\begin{array}{cc}0& 0\\ 0& d\end{array}\right),Q^d=\left(\begin{array}{cc}0& 0\\ 0& d^d\end{array}\right)\hfill \end{array}$$

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(a^2{a}^d{\left(bc\right)}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & a{a}^{d}bc{\left(bc\right)}^d+{(a^2{a}^d{\left(bc\right)}^{\pi }+x)}^{-1}x-xyx+a^{\pi }bc{\left(bc\right)}^d,\hfill \\ \hfill \gamma & =\hfill & a{a}^d{\left(bc\right)}^d+a^{\pi }{\left(bc\right)}^d,\hfill \\ \hfill \delta & =\hfill & -a^2{a}^d{\left(bc\right)}^d+y+yx-a{a}^{\pi }{\left(bc\right)}^d,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{a}^d{\left(bc\right)}^{i+1}{\left(bc\right)}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{[a^2{a}^d{\left(bc\right)}^{\pi }+x]}^{-i-2}{x}^i.\hfill \end{array}$$

Proof. 

Let $P=\left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)$ and $Q=\left(\begin{array}{cc}0& 0\\ 0& d\end{array}\right).$ In view of Theorem 3.1, we have

$${\left(\begin{array}{cc}a& bc\\ 1& 0\end{array}\right)}^d=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$$

with $\alpha ,\beta ,\gamma ,\delta$ are formulated by

$$\begin{array}{ccc}\hfill \alpha & =\hfill & {(a^2{a}^d{\left(bc\right)}^{\pi }+x)}^{-1}-xy,\hfill \\ \hfill \beta & =\hfill & a{a}^{d}bc{\left(bc\right)}^d+{(a^2{a}^d{\left(bc\right)}^{\pi }+x)}^{-1}x-xyx+a^{\pi }bc{\left(bc\right)}^d,\hfill \\ \hfill \gamma & =\hfill & a{a}^d{\left(bc\right)}^d+a^{\pi }{\left(bc\right)}^d,\hfill \\ \hfill \delta & =\hfill & -a^2{a}^d{\left(bc\right)}^d+y+yx-a{a}^{\pi }{\left(bc\right)}^d,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill x& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{\displaystyle \frac{\left(2i\right)!}{i!(i+1)!}}{a}^{-2i}{a}^d{\left(bc\right)}^{i+1}{\left(bc\right)}^{\pi },\hfill \\ \hfill y& =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{(-1)}^i{[a^2{a}^d{\left(bc\right)}^{\pi }+x]}^{-i-2}{x}^i.\hfill \end{array}$$

One easily verifies that

$$\begin{array}{ccc}\hfill \left(\begin{array}{cc}a& bc\\ 1& 0\end{array}\right)& =\hfill & \left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& c\end{array}\right),\hfill \\ \hfill \left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)& =\hfill & \left(\begin{array}{cc}1& 0\\ 0& c\end{array}\right)\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right).\hfill \end{array}$$

By using Cline’s formula (see [14] [Theorem 2.2]), P has g-Drazin inverse and

$$\begin{array}{ccc}\hfill P^d& =\hfill & \left(\begin{array}{cc}1& 0\\ 0& c\end{array}\right){\left[{\left(\begin{array}{cc}a& bc\\ 1& 0\end{array}\right)}^d\right]}^2\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}1& 0\\ 0& c\end{array}\right){\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)}^2\left(\begin{array}{cc}a& b\\ 1& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}\alpha & \beta \\ c\gamma & c\delta \end{array}\right)\left(\begin{array}{cc}\alpha a+\beta & \alpha b\\ \gamma a+\delta & \gamma b\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}{\alpha }^{2}a+\alpha \beta +\beta \gamma a+\beta \delta & {\alpha }^{2}b+\beta \gamma b\\ c\gamma \alpha a+c\gamma \beta +c\delta \gamma a+c\delta \delta & c\gamma \alpha b+c\delta \gamma b\end{array}\right).\hfill \end{array}$$

Obviously, we have

$$Q^d=\left(\begin{array}{cc}0& 0\\ 0& d^d\end{array}\right),Q^{\pi }=\left(\begin{array}{cc}1& 0\\ 0& d^{\pi }\end{array}\right).$$

One easily checks that

$$\begin{array}{ccc}\hfill P{Q}^2& =\hfill & \left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& d^2\end{array}\right)=\left(\begin{array}{cc}0& b{d}^2\\ 0& 0\end{array}\right)=0,\hfill \\ \hfill PQP& =\hfill & \left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& d\end{array}\right)\left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)\hfill \\ & =\hfill & \left(\begin{array}{cc}bdc& 0\\ 0& 0\end{array}\right)=0.\hfill \end{array}$$

According to Lemma 2.2, we derive that

$$\begin{array}{ccc}\hfill M^d& =\hfill & {(P+Q)}^d\hfill \\ & =\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(Q^d\right)}^{i+1}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{\infty }}{Q}^i{Q}^{\pi }{\left(P^d\right)}^{i+1}+{\displaystyle \sum _{i=0}^{\infty }}{Q}^i{Q}^{\pi }{\left(P^d\right)}^{i+2}Q\hfill \\ & +\hfill & {\displaystyle \sum _{i=0}^{\infty }}{\left(Q^d\right)}^{i+3}{P}^{i+1}{P}^{\pi }Q-Q^d{P}^{d}Q-{\left(Q^d\right)}^{2}P{P}^{d}Q,\hfill \end{array}$$

as asserted. □