Perturbation for the Group Inverse in a Banach Algebra

Abstract

We present new additive results for the group inverse in a Banach algebra under certain perturbations. The upper bound of ${\parallel (a+b)}^{\#}-a^d\parallel$ is thereby given. These results extend the main results presented by Liu, Qin, and Wei. We then apply them to establish the representation of the group inverse of $a+b$ under a kind of commutative perturbation condition. Finally, some numerical examples are given to demonstrate the main results.

1. Introduction

Let $\mathcal{A}$ be a complex Banach algebra with an identity of 1 and where $a\in \mathcal{A}$. The Drazin inverse of a is the unique x in $\mathcal{A}$, satisfying

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

for some non-negative integers k. We denote x as $a^D$. The least non-negative integer k is called the index of a, denoted by $ind\left(a\right)=k$. If $ind\left(a\right)=1$, we say a is group-invertible; that is, $ax=xa,xax=x,axa=a.$ The element x will be denoted by $a^{\#}$. The symbol ${\mathcal{A}}^{\mathrm{qnil}}$ stands for the sets of all quasi-nilpotent elements in $\mathcal{A}$, i.e., $a\in {\mathcal{A}}^{qnil}\iff \underset{n\to \infty }{lim}{\parallel a^n\parallel }^{{\displaystyle \frac{1}{n}}}=0.$ We note that $ind\left(a\right)=0$ if and only if a is invertible in the usual sense. An element x is a generalized Drazin inverse of a in $\mathcal{A}$ if

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

The generalized Drazin inverse is unique if it exists. We denote x as $a^d$.

The preceding generalized inverses play important roles in matrix and operator theory (see [1,2,3,4,5,6,7,8,9]). In [10], Yang and Liu studied the Drazin inverse of the sum of complex matrices, P and Q, under the conditions $Q{P}^2=0$ and $QPQ=0$. The generalized Drazin inverse of $a+b$ in a Banach algebra in Theorem 2.1 was studied in [11] under the conditions $b{a}^2=0$ and $b^2=0$. In [12], Liu et al. investigated the Drazin inverse ${(P+Q)}^D$ of two complex matrices, P and Q, under the conditions $P^{2}Q=PQP$ and $Q^{2}P=QPQ$. In [13], Zhang et al. give a formal expression of the Drazin inverse of $P+Q$ under the conditions $P{Q}^2=0$, $PQ{P}^2=0$, ${\left(PQ\right)}^{2}P=0$, $ind\left(P\right)=s$, and $ind\left(Q\right)=t$. In [9], Zou et al. presented the generalized Drazin inverse of $a+b$ in a Banach algebra under $a^{2}b=aba$ and $b^{2}a=bab$. Furthermore, for $a,b\in \mathcal{A}$, Peng and Zhang [14] proved that either instance of generalized Drazin invertibility among $a,b$ and $a+b$ implies a third one under the conditions $ab(a+b)=(a+b)ab$ and $ab\in {\mathcal{A}}^{\mathrm{qnil}}$.

Perturbation bounds for the generalized inverse have been studied extensively (see [15,16]). The perturbation bounds for generalized inverse have important applications in various fields, for instance, in control theory and in signal processing. One of the key results is the perturbation bound for the group inverse. This bound provides a quantitative measure of how the group inverse changes under perturbations.

Liu et al. investigated the perturbation of a group inverse under the conditions $\parallel a^{d}ba{a}^d\parallel$$<1$ and $a^{\pi }ba=0$ (see Theorem 2.8 in [17]). They further studied the group invertibility under the condition $\parallel a^{d}ba{a}^d\parallel <1,a^{\pi }ba=ab{a}^{\pi }$ (see [17], Theorem 2.12). These studies inspired us to explore the perturbation for group invertibility in a more general setting. The previous results of X. Liu et al. are thereby generalized to wider cases.

In Section 2, we investigate the group invertibility of $a+b$ in a Banach algebra under the condition $a^{\pi }b{a}^2=0,a^{\pi }bab=0$. The upper bound of ${\parallel (a+b)}^{\#}-a^d\parallel$ is thereby given. These results extend the main results in [17]. In Section 3, we investigate the additive property for a group inverse under the condition $a^{2}b{a}^{\pi }=a^{\pi }aba,a^{\pi }{b}^{2}a=bab{a}^{\pi }$ and obtain the bound of ${\parallel (a+b)}^{\#}-a^d\parallel$.

Throughout the paper, all Banach algebras are complex with an identity. We use ${\mathcal{A}}^d$ to denote the set of all g-Drazin-invertible elements in $\mathcal{A}$. We use $a^{\pi }$ to denote the spectral idempotent $1-a{a}^d$ of an element where $a\in {\mathcal{A}}^d$.

2. Orthogonal Conditions

In this section we establish the additive property of the group inverse in a Banach algebra under certain orthogonal perturbation conditions. We start with some key lemmas for proving the main results in this paper.

Lemma 1.

Let $\mathcal{A}$ be a Banach algebra, $a,b\in {\mathcal{A}}^{\mathrm{d}}$, $c\in \mathcal{A}$, $p^2=p\in \mathcal{A}$,

$$x={\left(\begin{array}{cc}a& 0\\ c& b\end{array}\right)}_{p}or{\left(\begin{array}{cc}b& c\\ 0& a\end{array}\right)}_{1-p}.$$

Then

$$x^d={\left(\begin{array}{cc}{a}^d& 0\\ z& b^d\end{array}\right)}_{p}or{\left(\begin{array}{cc}{b}^d& z\\ 0& a^d\end{array}\right)}_{1-p},$$

where

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

Proof. 

See Lemma 1.1 in [8]. □

Lemma 2.

Let e be an idempotent in a Banach algebra $\mathcal{A}$, and let $a,ea\in {\mathcal{A}}^d$. If $ea(1-e)=0$, then $a(1-e)\in {\mathcal{A}}^d$, and

$${\left(ea\right)}^d=e{a}^d,{\left(a(1-e)\right)}^d=a^d(1-e).$$

Proof. 

See Lemma 2.2 in [8]. □

Thus, we can prove the following:

Theorem 1.

Let $\mathcal{A}$ be a Banach algebra, and let $a,b,a^{\pi }b\in {\mathcal{A}}^d$. If

$$\parallel a^{d}ba{a}^d\parallel <1,a^{\pi }b{a}^2=0,a^{\pi }bab=0,$$

then $a+b$ has a group inverse if and only if $a^{\pi }(a+b)$ has a group inverse. In this case,

$$\begin{array}{ccc}\hfill {\parallel (a+b)}^{\#}-a^d\parallel & \le & {\displaystyle {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}^2\parallel b{a}^{\pi }\parallel \left(\parallel a^{\pi }\parallel +\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+1}{\parallel a^{\pi }{b}^d\parallel }^{n+1}\right.}\hfill \\ & & {\displaystyle +\left.\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+2}\parallel a^{\pi }{b}^d{\parallel }^{n+2}+\parallel a^{\pi }{b}^d(a+b)\parallel \right)+\left(1+\frac{\parallel a^{d}b\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}\hfill \\ & & {\displaystyle ·\left(\parallel a^d\parallel +\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^n\parallel a^{\pi }{b}^d{\parallel }^{n+1}\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+1}{\parallel a^{\pi }{b}^d\parallel }^{n+2}\right)+\parallel a^d\parallel .}\hfill \end{array}$$

Proof. 

Let $p=a{a}^d$. Then we have

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}\right)}_p\mathrm{and}b={\left(\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right)}_p,$$

where $a_1\in {\left(p\mathcal{A}p\right)}^{-1}$, $a_2\in {\left((1-p)\mathcal{A}(1-p)\right)}^{\mathrm{qnil}}$.

Furthermore,

$$a^d={\left(\begin{array}{cc}{a}_1^{-1}& 0\\ 0& 0\end{array}\right)}_p\mathrm{and}{a}^{\pi }={\left(\begin{array}{cc}0& 0\\ 0& a^{\pi }\end{array}\right)}_p.$$

Since $a^{\pi }b{a}^2=0$, we see that $a^{\pi }ba{a}^d=a^{\pi }b{a}^2{\left(a^d\right)}^2=0$. This implies that $b_3=0$, and so

$$a+b={\left(\begin{array}{cc}{a}_1+b_1& b_2\\ 0& a_2+b_4\end{array}\right)}_p.$$

Since $\parallel a^{d}ba{a}^d\parallel <1$, we see that $\parallel a_1^{-1}{b}_1\parallel <1$, and so $a_1+b_1=a_1(1+a_1^{-1}{b}_1){\in \left(p\mathcal{A}p\right)}^{-1}$. One can easily check that $a_2=a-a^2{a}^d=a{a}^{\pi }\in {\left((1-p)\mathcal{A}(1-p)\right)}^{\mathrm{qnil}}$. On the other hand, we have $b_4=a^{\pi }b{a}^{\pi }=a^{\pi }b-a^{\pi }b{a}^2{\left(a^d\right)}^2=a^{\pi }b\in {\mathcal{A}}^d$, and so $b_4\in {\left((1-p)\mathcal{A}(1-p)\right)}^d$.

Moreover, we have

$$\begin{array}{c}{b}_4{a}_2^2=a^{\pi }b(1-a{a}^d)a^2{a}^{\pi }=a^{\pi }b{a}^2{a}^{\pi }=0,\\ b_4{a}_2{b}_4=a^{\pi }b(1-a{a}^d)ab{a}^{\pi }=a^{\pi }bab{a}^{\pi }=0.\end{array}$$

According to Theorem 2.2 in [18] and the quasi-nilpotency of $a_2$, we obtain the following formula for calculating the generalized Drazin inverse of $a_2+b_4$:

$$\begin{array}{ccc}\hfill {(a_2+b_4)}^d& =& {\displaystyle \sum _{n=0}^{\infty }{a}_2^n{\left(b_4^d\right)}^{n+1}+\sum _{n=0}^{\infty }{a}_2^n{\left(b_4^d\right)}^{n+2}{a}_2}\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a,}\hfill \end{array}$$

this formula is an extension of Theorem 2.1 in [10] for Drazin inverse. We note that the preceding infinite sums exist because $a^{\pi }a$ is quasi-nilpotent, and so $\underset{n\to \infty }{lim}\left|\right|{\left(a^{\pi }a\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. This implies that the preceding series absolutely converge. Therefore, according to Theorem 2.3 in [1],

$${(a+b)}^d={\left(\begin{array}{cc}{(a_1+b_1)}^{-1}& z\\ 0& {(a_2+b_4)}^d\end{array}\right)}_p,$$

where

$${\displaystyle z=\sum _{n=0}^{\infty }{(a_1+b_1)}^{-n-2}{b}_2{(a_2+b_4)}^n{(a_2+b_4)}^{\pi }-{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^d.}$$

We can check that

$$a_2+b_4=a^{\pi }a+a^{\pi }b=a^{\pi }(a+b).$$

If $a+b\in {\mathcal{A}}^{\#}$, we can easily see that $a_2+b_4\in {\mathcal{A}}^{\#}$, and so $a^{\pi }(a+b)\in {\mathcal{A}}^{\#}$. We now assume that $a^{\pi }(a+b)\in {\mathcal{A}}^{\#}.$ Then ${(a_2+b_4)}^d={(a_2+b_4)}^{\#}$. It follows from $(a_2+b_4){(a_2+b_4)}^{\pi }=0$ that

$${\displaystyle z={(a_1+b_1)}^{-2}{b}_2{(a_2+b_4)}^{\pi }-{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^{\#}.}$$

Obviously,

$$\begin{array}{c}{(a+b)}^d(a+b)=(a+b){(a+b)}^d,\\ {(a+b)}^d={(a+b)}^d(a+b){(a+b)}^d.\end{array}$$

It is easy to verify that

$$\begin{array}{ccc}\hfill \left[(a_1+b_1)z+b_2{(a_2+b_4)}^{\#}\right](a_2+b_4)& =& \left[{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^{\pi }-b_2{(a_2+b_4)}^{\#}\right](a_2+b_4)\hfill \\ & & +b_2{(a_2+b_4)}^{\#}(a_2+b_4)\hfill \\ & =& 0.\hfill \end{array}$$

Then we have

$$\begin{array}{ccc}\hfill (a+b){(a+b)}^d(a+b)& =& \left(\begin{array}{cc}{a}_1+b_1& b_2\\ 0& a_2+b_4\end{array}\right)\left(\begin{array}{cc}{(a_1+b_1)}^{-1}& z\\ 0& {(a_2+b_4)}^{\#}\end{array}\right)\left(\begin{array}{cc}{a}_1+b_1& b_2\\ 0& a_2+b_4\end{array}\right)\hfill \\ & =& a+b.\hfill \end{array}$$

The above facts show that $a+b$ is group-invertible and ${(a+b)}^{\#}={(a+b)}^d$. Accordingly,

$$\begin{array}{ccc}\hfill {(a_2+b_4)}^{\#}& =& {\displaystyle \sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a.}\hfill \end{array}$$

Since $\parallel a_1^{-1}{b}_1\parallel <1$, we can easily check that

$$\begin{array}{ccc}\hfill {(a_1+b_1)}^{-1}& =& {\displaystyle {\left(1+a_1^{-1}{b}_1\right)}^{-1}{a}_1^{-1}}\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(a_1^{-1}{b}_1\right)}^n{a}_1^{-1}}\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}ba{a}^d\right)}^n{a}^d}\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d.}\hfill \end{array}$$

On the other hand, $\parallel b_1{a}_1^{-1}\parallel =\parallel a_1^{-1}{b}_1\parallel <1$, so we get

$$\begin{array}{ccc}\hfill {(a_1+b_1)}^{-1}& =& {\displaystyle a_1^{-1}{\left(1+b_1{a}_1^{-1}\right)}^{-1}}\hfill \\ & =& {\displaystyle a_1^{-1}\sum _{n=0}^{\infty }{(-1)}^n{\left(b_1{a}_1^{-1}\right)}^n}\hfill \\ & =& {\displaystyle a^d\sum _{n=0}^{\infty }{(-1)}^n{\left(ba{a}^d·a^d\right)}^n}\hfill \\ & =& {\displaystyle a^d\sum _{n=0}^{\infty }{(-1)}^n{\left(b{a}^d\right)}^n.}\hfill \end{array}$$

Since $b_3=0$, it follows from Lemma 1 that $a^{\pi }{b}^d{a}^{\pi }=a^{\pi }{b}^d$. Moreover, we have

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

Then we can verify that

$$\begin{array}{ccc}\hfill a^{\pi }b{(a_2+b_4)}^{\#}& =& {\displaystyle a^{\pi }b\left(\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a\right)}\hfill \\ & =& {\displaystyle a^{\pi }b{a}^{\pi }{b}^d+a^{\pi }b{\left(a^{\pi }{b}^d\right)}^2{a}^{\pi }a+a^{\pi }b{a}^{\pi }a{a}^{\pi }{b}^d+a^{\pi }b{a}^{\pi }a{\left(a^{\pi }{b}^d\right)}^2{a}^{\pi }a}\hfill \\ & & {\displaystyle +\sum _{n=2}^{\infty }{a}^{\pi }b{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=2}^{\infty }{a}^{\pi }b{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a}\hfill \\ & =& {\displaystyle a^{\pi }{b}^d{a}^{\pi }b+a^{\pi }{b}^{d}a+a^{\pi }b{a}^{\pi }a{a}^{\pi }b{\left(b^d\right)}^2+a^{\pi }b{a}^{\pi }a{a}^{\pi }b{b}^d{a}^{\pi }{b}^d{a}^{\pi }a}\hfill \\ & & {\displaystyle +\sum _{n=2}^{\infty }{a}^{\pi }b{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=2}^{\infty }{a}^{\pi }b{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a}\hfill \\ & =& {\displaystyle a^{\pi }{b}^{d}a+a^{\pi }{b}^{d}b+b_4{a}_2{b}_4{\left(b^d\right)}^2+b_4{a}_2{b}_4{b}^d{a}^{\pi }{b}^d{a}^{\pi }a}\hfill \\ & & {\displaystyle +\sum _{n=2}^{\infty }{b}_4{a}_2^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=2}^{\infty }{b}_4{a}_2^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a}\hfill \\ & =& {\displaystyle a^{\pi }{b}^d(a+b).}\hfill \end{array}$$

Furthermore, we get

$$\begin{array}{ccc}\hfill z& =& {(a_1+b_1)}^{-2}{b}_2{(a_2+b_4)}^{\pi }-{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^{\#}\hfill \\ & =& {(a_1+b_1)}^{-2}{b}_2[a^{\pi }-a^{\pi }(a+b){(a_2+b_4)}^{\#}]-{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^{\#}\hfill \\ & =& {(a_1+b_1)}^{-2}{b}_2[a^{\pi }-a^{\pi }a{(a_2+b_4)}^{\#}-a^{\pi }b{(a_2+b_4)}^{\#}]-{(a_1+b_1)}^{-1}{b}_2{(a_2+b_4)}^{\#}\hfill \\ \hfill & =& {\displaystyle {\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^{2}a{a}^{d}b{a}^{\pi }}\hfill \\ \hfill & & {\displaystyle ·\left(a^{\pi }-\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^{n+1}{\left(a^{\pi }{b}^d\right)}^{n+1}-\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^{n+1}{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a-a^{\pi }{b}^d(a+b)\right)}\hfill \\ \hfill & & {\displaystyle -\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)a{a}^{d}b{a}^{\pi }\left(\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a\right)}\hfill \\ \hfill & =& {\displaystyle {\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^{2}b{a}^{\pi }}\hfill \\ \hfill & & {\displaystyle ·\left(a^{\pi }-\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^{n+1}{\left(a^{\pi }{b}^d\right)}^{n+1}-\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^{n+1}{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a-a^{\pi }{b}^d(a+b)\right)}\hfill \\ & & {\displaystyle -\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^{n+1}\right)\left(\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+1}+\sum _{n=0}^{\infty }{\left(a^{\pi }a\right)}^n{\left(a^{\pi }{b}^d\right)}^{n+2}{a}^{\pi }a\right).}\hfill \end{array}$$

Hence,

$${(a+b)}^{\#}={(a_1+b_1)}^{-1}+z+{(a_2+b_4)}^{\#}.$$

We compute that

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

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

Clearly, we have

$$\begin{array}{c}{\displaystyle ∥\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d∥\le \sum _{n=0}^{\infty }\parallel a^{d}ba{a}^d{\parallel }^n\parallel a^d\parallel =\frac{\parallel a^d\parallel }{1-\parallel a^{d}ba{a}^d\parallel },}\\ {\displaystyle ∥\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n∥\le 1+∥\sum _{n=0}^{\infty }{(-1)}^{n+1}{\left(a^{d}ba{a}^d\right)}^n∥\parallel a^{d}b\parallel \le 1+\frac{\parallel a^{d}b\parallel }{1-\parallel a^{d}ba{a}^d\parallel }.}\end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill {\parallel (a+b)}^{\#}-a^d\parallel & \le & {\displaystyle {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}^2\parallel b{a}^{\pi }\parallel \left(\parallel a^{\pi }\parallel +\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+1}{\parallel a^{\pi }{b}^d\parallel }^{n+1}\right.}\hfill \\ & & {\displaystyle +\left.\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+2}\parallel a^{\pi }{b}^d{\parallel }^{n+2}+\parallel a^{\pi }{b}^d(a+b)\parallel \right)+\left(1+\frac{\parallel a^{d}b\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}\hfill \\ & & {\displaystyle ·\left(\parallel a^d\parallel +\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^n\parallel a^{\pi }{b}^d{\parallel }^{n+1}\sum _{n=0}^{\infty }\parallel a^{\pi }{a\parallel }^{n+1}{\parallel a^{\pi }{b}^d\parallel }^{n+2}\right)+\parallel a^d\parallel ,}\hfill \end{array}$$

as asserted. □

Corollary 1.

Let $\mathcal{A}$ be a Banach algebra, and let $a,b,a^{\pi }b\in {\mathcal{A}}^d$. If

$$\begin{array}{c}{a}^{\pi }b{a}^2=0,a^{\pi }bab=0,\\ {\displaystyle max\left\{\parallel a^{d}ba{a}^d\parallel ,\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel \right\}<1,}\end{array}$$

then $a+b$ has a group inverse if and only if $a^{\pi }(a+b)$ has a group inverse. In this case,

$$\begin{array}{ccc}\hfill {\parallel (a+b)}^{\#}-a^d\parallel & \le & {\displaystyle {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}^2\parallel b{a}^{\pi }\parallel }\hfill \\ & & {\displaystyle ·\left(\parallel a^{\pi }\parallel +\frac{\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel }{1-\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel }+\frac{{\left(\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel \right)}^2}{1-\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel }+1+\parallel a^{\pi }b{b}^d\parallel \right)}\hfill \\ & & {\displaystyle +\left(1+\frac{\parallel a^{d}b\parallel }{1-\parallel a^{d}ba{a}^d\parallel }\right)}\hfill \\ & & {\displaystyle ·\left(\parallel a^d\parallel +\frac{\parallel a^{\pi }{b}^d\parallel }{1-\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel }+\frac{\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d{\parallel }^2}{1-\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel }\right)+\parallel a^d\parallel .}\hfill \end{array}$$

Proof. 

According to this hypothesis, $\parallel a^{\pi }a\parallel \parallel a^{\pi }{b}^d\parallel <1$. Therefore we complete the proof according to Theorem 1. □

The following example illustrates that Theorem 2.3 is a nontrivial generalization of Theorem 2.8 in [17].

Example 1.

Let

$$a=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right),b=\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 0& 0& 0\end{array}\right)\in {\mathbb{C}}^{3\times 3}.$$

Clearly, $a^d=b^d=0$ and $a^{\pi }=1$. Therefore, we have

$$\parallel a{a}^{d}b{a}^d\parallel =0<1,a^{\pi }b{a}^2=0and{a}^{\pi }bab=0.$$

But

$$a^{\pi }ba=\left(\begin{array}{ccc}0& 0& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\ne 0.$$

3. Commutative Conditions

In this section we shall establish the representation of the group inverse of $a+b$ under a kind of commutative perturbation condition that is weaker than $a^{\pi }ba=ab{a}^{\pi }$. The following lemma is crucial.

Lemma 3.

Let $\mathcal{A}$ be a Banach algebra, and let $a\in {\mathcal{A}}^{qnil},b\in \mathcal{A}$. If $a^{2}b=aba$, $b^{2}a=bab$, then $a+b\in {\mathcal{A}}^d$. In this case,

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

Proof. 

Since $a\in {\mathcal{A}}^{qnil}$, we have $a^d=0$. This completes the proof according to Theorem 3.3 in [9]. □

Thus, we can prove the following:

Theorem 2.

Let $\mathcal{A}$ be a Banach algebra, and let $a,b,a^{\pi }b\in {\mathcal{A}}^d$. If

$$\parallel a^{d}ba{a}^d\parallel <1,a^{2}b{a}^{\pi }=a^{\pi }aba,a^{\pi }{b}^{2}a=bab{a}^{\pi },$$

then $a+b$ has a group inverse if and only if $(a+b)a^{\pi }$ has a group inverse. In this case,

$$\begin{array}{ccc}\hfill {\parallel (a+b)}^{\#}-a^d\parallel & \le & {\displaystyle \frac{\parallel a^{d}b\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }+\parallel a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ & & {\displaystyle +\parallel (a+b)a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel \parallel (a+b)a^{\pi }ba{a}^d\parallel \sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n{\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ & & {\displaystyle +\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}\parallel a{a}^{\pi }{\parallel }^n+\frac{\parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel \sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\frac{\parallel b^{\pi }a{a}^{\pi }\parallel \parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n.}\hfill \end{array}$$

Proof. 

Let $p=a{a}^d$. Then we have

$$a={\left(\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}\right)}_p,b={\left(\begin{array}{cc}{b}_1& b_3\\ b_4& b_2\end{array}\right)}_p,$$

where $a_1$ is invertible, and $a_2$ is quasi-nilpotent. Then

$$a^d={\left(\begin{array}{cc}{a}_1^{-1}& 0\\ 0& 0\end{array}\right)}_p\mathrm{and}{a}^{\pi }={\left(\begin{array}{cc}0& 0\\ 0& a^{\pi }\end{array}\right)}_p.$$

Since $a^{2}b{a}^{\pi }=a^{\pi }aba$, we can check that

$$b_3=a{a}^{d}b{a}^{\pi }={\left(a^d\right)}^2{a}^{2}b{a}^{\pi }={\left(a^d\right)}^2{a}^{\pi }aba=0.$$

Therefore,

$$a+b=\left(\begin{array}{cc}{a}_1+b_1& 0\\ b_4& a_2+b_2\end{array}\right).$$

Obviously, $a^{d}b=a^{d}ba{a}^d=a_1^{-1}{b}_1$. Since $\parallel a^{d}ba{a}^d\parallel <1$, we have $\parallel a_1^{-1}{b}_1\parallel <1$. Thus, $a_1+b_1=a_1(1+a_1^{-1}{b}_1)$ is invertible in $p\mathcal{A}p$. Moreover, $a_1+b_1$ is group-invertible in $p\mathcal{A}p$. In view of $a^{2}b{a}^{\pi }=a^{\pi }aba$, we get $b_2=a^{\pi }b{a}^{\pi }=b{a}^{\pi }-{\left(a^d\right)}^2{a}^{2}b{a}^{\pi }=b{a}^{\pi }-{\left(a^d\right)}^2{a}^{\pi }aba=b{a}^{\pi }$. That is, $a_2+b_2=(a+b)a^{\pi }$. Therefore, according to Theorem 2.1 in [19] and the invertibility of $a_1+b_1$, we derive that $a+b$ is group-invertible if and only if $(a+b)a^{\pi }$ is group-invertible. In this case,

$${(a+b)}^{\#}=\left(\begin{array}{cc}{(a_1+b_1)}^{-1}& 0\\ {(a_2+b_2)}^{\pi }{b}_4{(a_1+b_1)}^{-2}-{(a_2+b_2)}^{\#}{b}_4{(a_1+b_1)}^{-1}& {(a_2+b_2)}^{\#}\end{array}\right).$$

As $a_2=a{a}^{\pi }\in {\mathcal{A}}^{\mathrm{qnil}}$ and $b_2=b{a}^{\pi }$, it follows from $a^{2}b{a}^{\pi }=a^{\pi }aba$ that

$$a{a}^{\pi }·a{a}^{\pi }·b{a}^{\pi }=a{a}^{\pi }·b{a}^{\pi }·a{a}^{\pi };$$

that is, $a_2^2{b}_2=a_2{b}_2{a}_2$. Likewise, since $a^{\pi }{b}^{2}a=bab{a}^{\pi }$, we obtain

$$b{a}^{\pi }·b{a}^{\pi }·a{a}^{\pi }=b{a}^{\pi }·a{a}^{\pi }·b{a}^{\pi };$$

that is, $b_2^2{a}_2=b_2{a}_2{b}_2$. According to this hypothesis, we calculate that $a^{\pi }b\in {\mathcal{A}}^d$ using Cline’s formula, $b_2\in {\mathcal{A}}^d$. In light of Lemma 3, $a_2+b_2\in {\left((1-p)\mathcal{A}(1-p)\right)}^d$. Since $b_3=pb(1-p)=0$, according to Lemma 1 and Lemma 2, we have $a^{\pi }{b}^d{a}^{\pi }=b^d{a}^{\pi }$ and $b_2^d=b^d(1-p)=b^d{a}^{\pi }$. These lead to $b_2^{\pi }=a^{\pi }-b_2{b}_2^d=a^{\pi }-b{a}^{\pi }{b}^d{a}^{\pi }=a^{\pi }-b{b}^d{a}^{\pi }=(1-b{b}^d)a^{\pi }=b^{\pi }{a}^{\pi }$ according to Theorem 1. So $b_2^{\pi }{a}_2=b^{\pi }{a}^{\pi }a{a}^{\pi }=b^{\pi }a{a}^{\pi }$. Therefore,

$$\begin{array}{ccc}\hfill {(a_2+b_2)}^{\#}& =& {(a_2+b_2)}^d\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(b_2^d\right)}^{n+1}{a}_2^n+b_2^{\pi }{a}_2\sum _{n=0}^{\infty }{(-1)}^n(n+1){\left(b_2^d\right)}^{n+2}{a}_2^n}\hfill \\ & =& {\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n+b^{\pi }a{a}^{\pi }\sum _{n=0}^{\infty }{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n.}\hfill \end{array}$$

We derive that

$${\displaystyle \begin{array}{cc}{(a+b)}^{\#}\hfill & ={(a_1+b_1)}^{-1}+{(a_2+b_2)}^{\pi }{b}_4{(a_1+b_1)}^{-2}-{(a_2+b_2)}^{\#}{b}_4{(a_1+b_1)}^{-1}+{(a_2+b_2)}^{\#}\hfill \\ & {\displaystyle =\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d+\left[a^{\pi }-\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n\right.\right.}\hfill \\ & {\displaystyle \left.\left.+b^{\pi }a{a}^{\pi }\sum _{n=0}^{\infty }{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n\right)(a+b)a^{\pi }\right]}\hfill \\ & {\displaystyle ·a^{\pi }ba{a}^d{\left(\sum _{n=0}^{\infty }{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^2}\hfill \\ & +\left({\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n+b^{\pi }a{a}^{\pi }{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n\right)\hfill \\ & ·\left(a^{\pi }-a^{\pi }ba{a}^d{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d+a^{\pi }ba{a}^d{\left({\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^2\hfill \\ & -{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n(a+b)a^{\pi }ba{a}^d{\left({\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^2\hfill \\ & -b^{\pi }a{a}^{\pi }{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n(a+b)a^{\pi }ba{a}^d{\left({\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\right)}^2\hfill \\ & +{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n\hfill \\ & -{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(b^d{a}^{\pi }\right)}^{n+1}{\left(a{a}^{\pi }\right)}^n{a}^{\pi }ba{a}^d{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d\hfill \\ & +b^{\pi }a{a}^{\pi }{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n\hfill \\ & -b^{\pi }a{a}^{\pi }{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n(n+1){\left(b^d{a}^{\pi }\right)}^{n+2}{\left(a{a}^{\pi }\right)}^n{a}^{\pi }ba{a}^d{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{\left(a^{d}b\right)}^n{a}^d.\hfill \end{array}}$$

From this, we obtain an estimation of the norm

$$\begin{array}{ccc}\hfill {\parallel (a+b)}^{\#}-a^d\parallel & \le & {\displaystyle \frac{\parallel a^{d}b\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }+\parallel a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ & & {\displaystyle +\parallel (a+b)a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel \parallel (a+b)a^{\pi }ba{a}^d\parallel \sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n{\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ & & {\displaystyle +\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}\parallel a{a}^{\pi }{\parallel }^n+\frac{\parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel \sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n}\hfill \\ & & {\displaystyle +\frac{\parallel b^{\pi }a{a}^{\pi }\parallel \parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\sum _{n=0}^{\infty }(n+1)\parallel b^d{a}^{\pi }{\parallel }^{n+2}{\parallel a{a}^{\pi }\parallel }^n.}\hfill \end{array}$$

The theorem is thus proven. □

As an immediate consequence of Theorem 2, we now derive

Corollary 2.

Let $\mathcal{A}$ be a Banach algebra, and let $a,b,a^{\pi }b\in {\mathcal{A}}^d$. If

$$a^{2}b{a}^{\pi }=a^{\pi }aba,a^{\pi }{b}^{2}a=bab{a}^{\pi },$$

$$max\left\{\parallel a^{d}ba{a}^d\parallel ,\parallel a{a}^{\pi }\parallel \parallel b^d{a}^{\pi }\parallel \right\}<1,$$

then $a+b$ has a group inverse if and only if $(a+b)a^{\pi }$ has a group inverse. In this case,

$$\begin{array}{c}{\displaystyle {\parallel (a+b)}^{\#}-a^d\parallel \le \frac{\parallel a^{d}b\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }+\parallel a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ {\displaystyle +\frac{\parallel (a+b)a^{\pi }ba{a}^d\parallel \parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }{\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel \parallel (a+b)a^{\pi }ba{a}^d\parallel {\left(\frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }\right)}^2{\left(\frac{\parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\right)}^2}\hfill \\ {\displaystyle +\frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }+\frac{\parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }\frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }}\hfill \\ {\displaystyle +\parallel b^{\pi }a{a}^{\pi }\parallel {\left(\frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }\right)}^2}\hfill \\ {\displaystyle +\frac{\parallel b^{\pi }a{a}^{\pi }\parallel \parallel a^{\pi }ba{a}^d\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }{\left(\frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }\right)}^2.}\hfill \end{array}$$

Proof. 

The first part follows from Theorem 2. For the second part, based on $\parallel a{a}^{\pi }\parallel \parallel b^d{a}^{\pi }\parallel <1$, we obtain the geometric series

$$\sum _{n=0}^{\infty }\parallel b^d{a}^{\pi }{\parallel }^{n+1}{\parallel a{a}^{\pi }\parallel }^n={\displaystyle \frac{\parallel b^d{a}^{\pi }\parallel }{1-\parallel b^d{a}^{\pi }\parallel \parallel a{a}^{\pi }\parallel }}.$$

For the same reason, applying term-by-term differentiation, we obtain the power series

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

So the upper bound of the norm ${\parallel (a+b)}^{\#}-a^d\parallel$ can be obtained using Theorem 2. □

Corollary 3

(see Theorem 2.12 in [17]). Let $\mathcal{A}$ be a Banach algebra, and let $a,b\in {\mathcal{A}}^d$. If

$$\parallel a^{d}ba{a}^d\parallel <1,a^{\pi }ba=ab{a}^{\pi },$$

then $a+b$ has a group inverse if and only if $a^{\pi }(a+b)$ has a group inverse. In this case,

$${\displaystyle {\parallel (a+b)}^{\#}-a^d\parallel \le \frac{\parallel a^{d}b\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }+\parallel a^{\pi }\parallel \sum _{n=0}^{\infty }\parallel b^d{\parallel }^{n+1}{\parallel a\parallel }^n.}$$

Proof. 

Since $a^{\pi }ba=ab{a}^{\pi }$, we have $a{a}^{d}b{a}^{\pi }=a^d\left(a^{\pi }ba\right)=0$. Hence, $a{a}^{d}b=a{a}^{d}ba{a}^d$. Also, we have $a^{\pi }ba{a}^d=ab{a}^{\pi }{a}^d=0$, and so $ba{a}^d=a{a}^{d}ba{a}^d$. Accordingly, $a^{d}ab=ba{a}^d$. Hence, $a^{\pi }b=b{a}^{\pi }$. Therefore, we check that

$$\begin{array}{c}{a}^{2}b{a}^{\pi }=a\left(ab{a}^{\pi }\right)=a\left(a^{\pi }ba\right)=a^{\pi }aba,\\ a^{\pi }{b}^{2}a=a^{\pi }b{a}^{\pi }ba=a^{\pi }bab{a}^{\pi }=a^{\pi }bab=b{a}^{\pi }ab=bab{a}^{\pi }.\end{array}$$

By virtue of Theorem 2, we obtain the upper bound of the norm

$${\displaystyle {\parallel (a+b)}^{\#}-a^d\parallel \le \frac{\parallel a^{d}b\parallel \parallel a^d\parallel }{1-\parallel a^{d}b\parallel }+\parallel a^{\pi }\parallel \sum _{n=0}^{\infty }\parallel b^d{\parallel }^{n+1}{\parallel a\parallel }^n.}$$

Therefore, we obtain the conclusion of the corollary. □

We present the following numerical example to demonstrate Theorem 2.

Example 2.

Let $a=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right),b=\left(\begin{array}{cccc}{\displaystyle \frac{1}{2}}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\end{array}\right)\in {\mathbb{C}}^{4\times 4}.$ Then

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

$$b^d=\left(\begin{array}{cccc}2& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& {\displaystyle \frac{1}{2}}\end{array}\right),b^{\pi }=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right).$$

We compute that

$$a^{d}ba{a}^d=\mathrm{diag}\left({\displaystyle \frac{1}{2}},0,0,0\right),$$

and so $\parallel a^{d}ba{a}^d\parallel <1$. Moreover, we have $a^{2}b{a}^{\pi }=a^{\pi }aba,a^{\pi }{b}^{2}a=bab{a}^{\pi }.$ But we have

$$ab{a}^{\pi }=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\\ 0& 0& 0& 0\end{array}\right),a^{\pi }ba=0,$$

and then $a^{\pi }ab\ne a^{\pi }ba$.

In this case,

$$a^{d}b=\mathrm{diag}\left({\displaystyle \frac{1}{2}},0,0,0\right),b^d{a}^{\pi }=\mathrm{diag}\left(0,0,0,{\displaystyle \frac{1}{2}}\right),a^{\pi }ba{a}^d=0,$$

$$a{a}^{\pi }=b^{\pi }a{a}^{\pi }=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right).$$

According to the upper bound in Theorem 2, we should have

$${\parallel (a+b)}^{\#}-a^d\parallel \le 2+0+0+0+1+0+1+0=4.$$

Indeed, since

$${(a+b)}^{\#}=\left(\begin{array}{cccc}{\displaystyle \frac{2}{3}}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& {\displaystyle \frac{1}{4}}\\ 0& 0& 0& {\displaystyle \frac{1}{2}}\end{array}\right),$$

we compute that

$${\parallel (a+b)}^{\#}-a^d\parallel ={\displaystyle \frac{13}{12}}<4.$$