Some Results on the Symmetric Representation of the Generalized Drazin Inverse in a Banach Algebra

Abstract

Based on the conditions $a{b}^2=0$ and $b^{\pi }(ab)\in {\mathcal{A}}^{\mathrm{d}}$, we derive that ${(ab)}^n$, ${(ba)}^n$, and $ab+ba$ are all generalized Drazin invertible in a Banach algebra $\mathcal{A}$, where $n\in \mathbb{N}$ and a and b are elements of $\mathcal{A}$. By using these results, some results on the symmetry representations for the generalized Drazin inverse of $ab+ba$ are given. We also consider that additive properties for the generalized Drazin inverse of the sum $a+b$.

1. Introduction

Let $\mathcal{A}$ be a complex unital Banach algebra with unit 1. The sets of all invertible elements and quasinilpotent elements of $\mathcal{A}$ are denoted by ${\mathcal{A}}^{-1}$ and ${\mathcal{A}}^{\mathrm{qnil}}$, respectively, where ${\mathcal{A}}^{-1}:=\{a\in \mathcal{A}:\exists x\in \mathcal{A}:ax=xa=1\}$ and ${\mathcal{A}}^{\mathsf{qnil}}:=\{a\in \mathcal{A}:\underset{n\to +\infty }{lim}\parallel a^n{\parallel }^{1/n}=0\}.$ Let $a\in \mathcal{A}$ and, if there is a element $b\in \mathcal{A}$ such that

$$bab=b,ab=ba,\mathrm{and}a(1-ab)\mathrm{is}\mathrm{quasinilpotent},$$

(1)

then b is the generalized Drazin inverse of a, denoted by $a^{\mathsf{d}}$, and it is unique. The set of generalized Drazin-invertible elements is denoted by ${\mathcal{A}}^{\mathsf{d}}=\{a\in \mathcal{A}:\exists a^{\mathsf{d}}\}.$ In particular, if $a(1-ab)=0$ (or $aba=a$), then b is called the group inverse of a. Note that $a{a}^{\mathsf{d}}$ is an idempotent element and let $a^{\pi }=1-a{a}^{\mathsf{d}}$. It was given, in [1] (Lemma 2.4), that $a^{\mathsf{d}}$ exists if and only if there is an idempotent $q\in \mathcal{A}$, such that $aq=qa$, $aq$ is quasinilpotent, and $a+q$ is invertible.

The generalized inverse in a matrix or operator theory is very useful in scientific calculation and in various engineering technologies [2,3,4]. It is well known that the Drazin inverse has been applied in a few fields, such as statistics and probability [5], ordinary differential equations [6], Markov chains [7], operator matrices [8], neural network models [9,10], and the references therein. In [11], a study of the Drazin inverse for bounded linear operators in a Banach space X is given, when 0 is an isolated spectral point of the operator. In [12], some additive results on the Drazin inverse, under the condition $ab=0$, are obtained. However, as in [12,13], this condition was not enough to derive a formula for the generalized Drazin inverse for $a+b$. In [14], authors investigated how to express the Drazin inverse of sums, differences, and products of two matrices P and Q, under the conditions $P^{3}Q=QP$ and $Q^{3}P=PQ$. The representations of the Drazin inverse for $(P+Q)$, such that $PQP=0$ and $P{Q}^2=0$, is given in [15]. The generalized inverses in $C^{*}$-algebras has been investigated in [16] and a symmetry of the generalized Drazin inverse in a $C^{*}$-algebra has been considered in [17].

Some additive properties of the generalized Drazin inverse in a Banach algebra were investigated in [18]. Recently, the expression for the generalized Drazin inverse of the sum $a+b$ on Banach algebra has been studied, such as in the representations of the generalized Drazin inverse for $a+b$ in a Banach algebra, obtained in [19]; some new additive results for the generalized Drazin inverse in a Banach algebra, given in [20]; and additive results on the generalized Drazin inverse of a sum of two elements in a Banach algebra are derived in [21] and the references therein. In this paper, we consider the representations of the generalized Drazin inverse of the sum of two elements in a Banach algebra. By using the assumed conditions $a{b}^2=0$ and $b^{\pi }(ab)\in {\mathcal{A}}^{\mathrm{d}}$, it is implied that ${(ab)}^n$, ${(ba)}^n$, and $ab+ba\in {\mathcal{A}}^{\mathrm{d}}$, and a symmetry representation for the generalized Drazin inverse of $ab+ba$ is obtained, where $n\in \mathbb{N}$ and $a,b\in {\mathcal{A}}^{\mathrm{d}}$. We also consider the additive properties for the generalized Drazin inverse of the sum $a+b$.

In Section 2, some notation is introduced and lemmas are given. In Section 3, a symmetric representation of the generalized Drazin inverse for $ab+ba$ in a Banach algebra is derived. The additive properties of the generalized Drazin inverse of $a+b$ are investigated in Section 4.

2. Preliminaries

Let $\mathcal{B}$ be a subalgebra of the unital algebra $\mathcal{A}$. For an element $b\in {\mathcal{B}}^{-1}$, the inverse of b in $\mathcal{B}$ is denoted by ${\left[b^{-1}\right]}_{\mathcal{B}}$. As in [19], it is given that ${\mathcal{B}}^{-1}\not\subset {\mathcal{A}}^{-1}$. Let $\mathcal{P}=\{p_1,p_2,\dots ,p_n\}$ be a total system of idempotents in $\mathcal{A}$ if $p_i^2=p_i$, for all i, $p_i{p}_j=0$ if $i\ne j$, and $p_1+\cdots +p_n=1$, as in [22]. If $a\in {\mathcal{A}}^{\mathsf{d}}$, then

$$a={\left[\begin{array}{cc}{a}_1& 0\\ 0& a_2\end{array}\right]}_{\mathcal{P}},a^{\mathsf{d}}={\left[\begin{array}{cc}{\left[{a_1}^{-1}\right]}_{p\mathcal{A}p}& 0\\ 0& 0\end{array}\right]}_{\mathcal{P}},a^{\pi }={\left[\begin{array}{cc}0& 0\\ 0& 1-p\end{array}\right]}_{\mathcal{P}},$$

(2)

where $p=a{a}^{\mathsf{d}}$, $\mathcal{P}=\{p,1-p\}$, $a_1\in {\left[p\mathcal{A}p\right]}^{-1}$, and $a_2\in {\left[(1-p)\mathcal{A}(1-p)\right]}^{\mathsf{qnil}}$. If a has the representation given as in (2), then $a^{\mathsf{d}}={\left[{a_1}^{-1}\right]}_{p\mathcal{A}p}=a_1^{\mathsf{d}}$.

The following lemmas are required in what follows.

Lemma 1

([19]). Let $\mathcal{P}=\{p,1-p\}$ be a total system of idempotents in $\mathcal{A}$, and let $a,b\in \mathcal{A}$ have the following representation

$$a={\left[\begin{array}{cc}x& 0\\ z& y\end{array}\right]}_{\mathcal{P}},b={\left[\begin{array}{cc}x& t\\ 0& y\end{array}\right]}_{\mathcal{P}}.$$

Then there exist ${(z_n)}_{n=0}^{\infty }\subset (1-p)\mathcal{A}p$ and ${(t_n)}_{n=0}^{\infty }\subset p\mathcal{A}(1-p)$, such that

$$a^n={\left[\begin{array}{cc}{x}^n& 0\\ z_n& y^n\end{array}\right]}_{\mathcal{P}},b^n={\left[\begin{array}{cc}{x}^n& t_n\\ 0& y^n\end{array}\right]}_{\mathcal{P}},\forall n\in \mathbb{N}.$$

Lemma 2

([22]). Let $a,b\in \mathcal{A}$ be generalized Drazin invertible and $ab=0$. Then, $a+b$ is generalized Drazin invertible and

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

Lemma 3

([22]). Let $x,y\in \mathcal{A}$, p be an idempotent of $\mathcal{A}$, and let x and y have the representation

$$x={\left[\begin{array}{cc}a& 0\\ c& b\end{array}\right]}_{\{p,1-p\}},y={\left[\begin{array}{cc}b& c\\ 0& a\end{array}\right]}_{\{1-p,p\}}.$$

(3)

(i)

If $a\in {\left[p\mathcal{A}p\right]}^{\mathsf{d}}$ and $b\in {\left[(1-p)\mathcal{A}(1-p)\right]}^{\mathsf{d}}$, then $x,y\in {\mathcal{A}}^{\mathsf{d}}$ and

$$x^{\mathsf{d}}={\left[\begin{array}{cc}{a}^{\mathsf{d}}& 0\\ u& b^{\mathsf{d}}\end{array}\right]}_{\{p,1-p\}},y^{\mathsf{d}}={\left[\begin{array}{cc}{b}^{\mathsf{d}}& u\\ 0& a^{\mathsf{d}}\end{array}\right]}_{\{1-p,p\}},$$

(4)

where

$$u=\sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{n+2}c{a}^n{a}^{\pi }+\sum _{n=0}^{\infty }{b}^{\pi }{b}^{n}c{(a^{\mathsf{d}})}^{n+2}-b^{\mathsf{d}}c{a}^{\mathsf{d}}.$$

(5)

(ii)

If $x\in {\mathcal{A}}^{\mathsf{d}}$ and $a\in {\left[p\mathcal{A}p\right]}^{\mathsf{d}}$, then $b\in {\left[(1-p)\mathcal{A}(1-p)\right]}^{\mathsf{d}}$, and $x^{\mathsf{d}}$ and $y^{\mathsf{d}}$ are given by (4) and (5).

Lemma 4

([11]). Let $a\in {\mathcal{A}}^{\mathsf{d}}$. Then ${\left[{(a)}^n\right]}^{\mathsf{d}}={\left[a^{\mathsf{d}}\right]}^n$, for all $n=1,2,\cdots$.

Lemma 5

([11]). If $a,b\in {\mathcal{A}}^{\mathsf{d}}$ and $ab=ba=0$. Then, ${(a+b)}^{\mathsf{d}}$ also exists and ${(a+b)}^{\mathsf{d}}=a^{\mathsf{d}}+b^{\mathsf{d}}$.

Lemma 6

([23]). Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$. Then ${(ab)}^{n+1}$ is generalized Drazin invertible, for some $n\in \mathbb{N}$, if and only if $ab$ is generalized Drazin invertible.

Lemma 7

([23]). Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ and ${(ab)}^{n+1}$ be generalized Drazin invertible for some $n\in \mathbb{N}$. Then, ${(ba)}^n$ is generalized Drazin invertible and ${\left[{(ba)}^n\right]}^{\mathsf{d}}=b{\left[{(ab)}^{n+1}\right]}^{\mathsf{d}}a$.

3. The Symmetric Representation for the Generalized Drazin Inverse of $\mathit{ab}+\mathit{ba}$

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$. A symmetric expression of ${(ab+ba)}^{\mathsf{d}}$ is given, by using $ab$, $ba$, ${(ab)}^{\mathsf{d}}$, and ${(ba)}^{\mathsf{d}}$, with the following assumed conditions

$$\begin{array}{c}\hfill a{b}^2=0,b^{\pi }(ab)\in {\mathcal{A}}^{\mathsf{d}}.\end{array}$$

(6)

Theorem 1.

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ satisfy (6). Then, ${(ab)}^n,{(ba)}^n,ab+ba\in {\mathcal{A}}^{\mathsf{d}}$ $(n=1,2,\cdots )$, and a representation of ${(ab+ba)}^{\mathrm{d}}$ is given as

$$\begin{array}{c}\hfill {(ab+ba)}^{\mathsf{d}}={(ba)}^{\pi }\sum _{n=1}^{\infty }{(ba)}^{n-1}{\left[{(ab)}^n\right]}^{\mathsf{d}}+\sum _{n=1}^{\infty }{\left[{(ba)}^n\right]}^{\mathsf{d}}{(ab)}^{n-1}{(ab)}^{\pi }.\end{array}$$

(7)

Proof. 

Let $b={\left[\begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\right]}_{\mathcal{P}}$, where $\mathcal{P}=\{b{b}^{\mathsf{d}},b^{\pi }\}$, $b_1$ is invertible in the subalgebra $b{b}^{\mathsf{d}}\mathcal{A}b{b}^{\mathsf{d}}$, and $b_2$ is quasinilpotent. Let us write $a={\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]}_{\mathcal{P}}$. From $a{b}^2=0$, we have

$$\begin{array}{c}\hfill a_{11}=0,a_{21}=0,a_{12}{b}_2^2=0,\mathrm{and}{a}_{22}{b}_2^2=0.\end{array}$$

(8)

Thus, we have $ab=\left[\begin{array}{cc}0& a_{12}{b}_2\\ 0& a_{22}{b}_2\end{array}\right]$. By Lemma 3, we obtain that $ab\in {\mathcal{A}}^{\mathsf{d}}$ if and only if $a_{22}{b}_2$ is generalized Drazin invertible. Thus, ${(b^{\pi }ab)}^{\mathsf{d}}$ exists. By using Cline’s formula, it proves that ${(ab)}^{\mathsf{d}}$ also is. Therefore, we obtain ${(ab)}^n,{(ba)}^n\in {\mathcal{A}}^{\mathsf{d}}$ by using Lemma 6 and 7. Since $a{b}^2=0$, by Lemma 2 we can prove that $ab+ba$ is generalized Drazin invertible and that (7) holds. If $n=1$, then ${(ab+ba)}^{\mathsf{d}}={(ba)}^{\pi }{(ab)}^{\mathsf{d}}+{(ba)}^{\mathsf{d}}{(ab)}^{\pi }.$ By using mathematical induction, we derive that the representation can be given, as in (7). □

Remark 1.

Note that the expression given in Theorem 1 is symmetric.

Theorem 2.

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ satisfy (6) and $a^2=0$. Then $ab+ba\in {\mathcal{A}}^{\mathrm{d}}$ and ${\left[{(ab+ba)}^{\mathsf{d}}\right]}^n={\left[{(ab)}^{\mathsf{d}}\right]}^n+{\left[{(ba)}^{\mathsf{d}}\right]}^n$, for all $n=1,2,\cdots$.

Proof. 

Let $a,b$ be written as in the proof of Theorem 1, and, by $a{b}^2=0$, we derive $ab=\left[\begin{array}{cc}0& a_{12}{b}_2\\ 0& a_{22}{b}_2\end{array}\right]$ and $ab,ba,{(ab)}^n,{(ba)}^n\in {\mathcal{A}}^{\mathsf{d}}$. Since $a{b}^2=0$ and $a^2=0$, we have

$$\begin{array}{c}\hfill {(ab)}^n{(ba)}^n={(ba)}^n{(ab)}^n=0,{(ab+ba)}^n={(ab)}^n+{(ba)}^n,\end{array}$$

(9)

for all $n=1,2,\cdots$. By Lemma 4, Lemma 5, and the first equality of (9), we derive

$$\begin{array}{c}\hfill {\left[{(ab+ba)}^{\mathsf{d}}\right]}^n={\left[{(ab+ba)}^n\right]}^{\mathsf{d}}={[{(ab)}^n+{(ba)}^n]}^{\mathsf{d}}={\left[{(ab)}^n\right]}^{\mathsf{d}}+{\left[{(ba)}^n\right]}^{\mathsf{d}}={\left[{(ab)}^{\mathsf{d}}\right]}^n+{\left[{(ba)}^{\mathsf{d}}\right]}^n.\end{array}$$

 □

At the end of Section 3, let $\mathcal{A}$ be a $C^{*}$-algebra, as in [17]. Then, a simple application of the generalized Drazin inverse in a $C^{*}$-algebra can be given, as follows.

Theorem 3.

Let $a,b\in \mathcal{A}$ be group invertible. If (6) is satisfied, then ${(ab+ba)}^{†}$ exists.

Proof. 

By using Theorem 1, we derive that $ab+ba$ is group invertible. As pointed out in [16], $ab+ba$ is generalized invertible. Thus, ${(ab+ba)}^{†}$ exists. □

Theorem 4.

Let $a,b\in {\mathcal{A}}^{\mathrm{d}}$. If (6) is satisfied, then ${(ab+ba)}^{\mathrm{d}}$ is self-adjoint in a $C^{*}$-algebra.

Proof. 

Note that $ab+ba$ is self-adjoint in a $C^{*}$-algebra. By Theorem 1 and using [17] (Theorem 3.2), we obtain that ${(ab+ba)}^{\mathrm{d}}$ is self-adjoint in a $C^{*}$-algebra. □

4. The Representation for the Generalized Drazin Inverse of $\mathit{a}+\mathit{b}$

In this section, we consider some results on the expression of ${(a+b)}^{\mathsf{d}}$, by using a, b, $a^{\mathsf{d}}$, and $b^{\mathsf{d}}$, where $a,b\in {\mathcal{A}}^{\mathsf{d}}$.

Lemma 8.

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ satisfy $a{b}^2=0$. Then, ${(a+b)}^{\mathsf{d}}$ exists if and only if $b^{\pi }(a+b)\in {\mathcal{A}}^{\mathsf{d}}$.

Proof. 

Similarly, we rewrite $a,b$ as in the proof of Theorem 1. Since $a{b}^2=0$, we derive

$$\begin{array}{c}\hfill a+b={\left[\begin{array}{cc}{b}_1& a_{12}\\ 0& b_2+a_{22}\end{array}\right]}_{\mathcal{P}}.\end{array}$$

(10)

By Lemma 3, note that ${(a+b)}^{\mathsf{d}}$ exists if and only if ${(a_{22}+b_2)}^{\mathsf{d}}$ exists; that is, ${(a+b)}^{\mathsf{d}}$ exists if and only if $b^{\pi }(a+b)$ is generalized Drazin invertible. □

Theorem 5.

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ satisfy the conditions of Theorem 2. Then

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

Proof. 

By Lemma 8, it also leads to (10). By Lemma 3, we can prove that ${(a+b)}^{\mathsf{d}}$ exists if and only if ${(a_{22}+b_2)}^{\mathsf{d}}$ exists; that is, ${(a+b)}^{\mathsf{d}}$ exists if and only if $b^{\pi }(a+b)$ is generalized Drazin invertible. If $b^{\pi }ab\in {\mathcal{A}}^{\mathsf{d}}$, then ${(a_{22}{b}_2)}^{\mathsf{d}}$ exists. By Cline’s formula, we have that ${(b_2{a}_{22})}^{\mathsf{d}}$ exists. As in the proof of Theorem 1, by Lemma 6 and 7, we also obtain that ${(ab)}^n,{(ba)}^n\in {\mathcal{A}}^{\mathsf{d}}$, for all $n=1,2,\cdots$.

By $a^2=0$, we get

$$a_{12}{a}_{22}=0\mathrm{and}{a}_{22}^2=0.$$

(11)

By (8) and (11), we have $(b_2{a}_{22})(a_{22}{b}_2)=0,(a_{22}{b}_2)(b_2{a}_{22})=0.$ Using Lemma 5, and by Cline’s formula, we derive

$$\begin{array}{c}\hfill {(a_{22}{b}_2+b_2{a}_{22})}^{\mathsf{d}}={(a_{22}{b}_2)}^{\mathsf{d}}+{(b_2{a}_{22})}^{\mathsf{d}}.\end{array}$$

(12)

By induction, let ${[{(a_{22}{b}_2)}^{\mathsf{d}}+{(b_2{a}_{22})}^{\mathsf{d}}]}^n={\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^n+{\left[{(b_2{a}_{22})}^{\mathsf{d}}\right]}^n$ for all $n\ge 1$. Therefore, we can prove that

$$\begin{array}{c}\left[{(a_{22}{b}_2+b_2{a}_{22})}^{\mathsf{d}}\right]{[{(a_{22}{b}_2)}^{\mathsf{d}}+{(b_2{a}_{22})}^{\mathsf{d}}]}^n={\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}+{\left[{(b_2{a}_{22})}^{\mathsf{d}}\right]}^{n+1}.\hfill \end{array}$$

Since $(a_{22}{b}_2+b_2{a}_{22})b_2^2=0$ and $b_2$ are quasinilpotent, by Lemma 5 and (12), we obtain

$$\begin{array}{ccc}\hfill {\left[{(a_{22}+b_2)}^2\right]}^{\mathsf{d}}& =& {(a_{22}{b}_2+b_2{a}_{22}+b_2^2)}^{\mathsf{d}}\hfill \\ & =& \sum _{n=0}^{\infty }{b}_2^{2n}{\left[{(a_{22}{b}_2+b_2{a}_{22})}^{\mathsf{d}}\right]}^{n+1}\hfill \\ & =& \sum _{n=0}^{\infty }{b}_2^{2n}{\left[{(a_{22}{b}_2)}^{\mathsf{d}}+{(b_2{a}_{22})}^{\mathsf{d}}\right]}^{n+1}.\hfill \end{array}$$

(13)

Then, $b^{\pi }(a+b)\in {\mathcal{A}}^{\mathsf{d}}$ implies that ${(a_{22}+b_2)}^{\mathsf{d}}$ exists and ${(a_{22}+b_2)}^{\mathsf{d}}={\left[{(a_{22}+b_2)}^2\right]}^{\mathsf{d}}(a_{22}+b_2)$. Finally, by (13), and ${(b_2{a}_{22})}^{\mathsf{d}}=b_2{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^2{a}_{22}$, we obtain

$$\begin{array}{ccc}\hfill {(a_{22}+b_2)}^{\mathsf{d}}& =& {\left[{(a_{22}+b_2)}^{\mathsf{d}}\right]}^2(a_{22}+b_2)\hfill \\ & =& \sum _{n=0}^{\infty }{b}_2^{2n}\left\{{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}+{\left(b_2{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^2{a}_{22}\right)}^{n+1}\right\}(a_{22}+b_2)\hfill \\ & =& \sum _{n=0}^{\infty }{b}_2^{2n}{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}{a}_{22}+\sum _{n=0}^{\infty }{b}_2^{2n}{\left[b_2{\left({(a_{22}{b}_2)}^{\mathsf{d}}\right)}^2{a}_{22}\right]}^{n+1}{b}_2\hfill \\ & =& \sum _{n=0}^{\infty }{b}_2^{2n}\left\{{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}{a}_{22}+b_2{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}\right\}\hfill \end{array}$$

(14)

and

$$\begin{array}{c}\hfill {(a_{22}+b_2)}^{\pi }={(a_{22}{b}_2)}^{\pi }-\sum _{n=0}^{\infty }{b}_2^{2n+1}\left\{{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}{a}_{22}+b_2{\left[{(a_{22}{b}_2)}^{\mathsf{d}}\right]}^{n+1}\right\}.\end{array}$$

(15)

By Lemma 3, we get that $a+b\in {\mathcal{A}}^{\mathsf{d}}$ and

$${(a+b)}^{\mathsf{d}}={\left[\begin{array}{cc}{b}_1^{-\mathsf{1}}& u\\ 0& {(a_{22}+b_2)}^{\mathsf{d}}\end{array}\right]}_{\mathcal{P}},$$

(16)

and

$$u=\sum _{n=0}^{\infty }{(b_1^{-\mathsf{1}})}^{n+2}{a}_{12}{(b_2+a_{22})}^n{(a_{22}+b_2)}^{\pi }-{(a_{22}+b_2)}^{\mathsf{d}}{a}_{12}{b}_1^{-1}.$$

(17)

Evidently, we have ${\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}}=b^{\mathsf{d}}$ and

$$b^{\mathsf{d}}ba={\left[\begin{array}{cc}{b}_1^{-\mathsf{1}}{b}_1& 0\\ 0& 0\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}0& a_{12}\\ 0& a_{22}\end{array}\right]}_{\mathcal{P}}={\left[\begin{array}{cc}0& a_{12}\\ 0& 0\end{array}\right]}_{\mathcal{P}}=a_{12}.$$

One easily has (by induction and by using (8) and (11)) that, if $n\ge 1$, then

$$a_{12}{(a_{22}+b_2)}^n=\left\{\begin{array}{cc}{a}_{12}{(b_2{a}_{22})}^{n/2}\hfill & \mathrm{if}n\mathrm{is}\mathrm{even},\\ a_{12}{(b_2{a}_{22})}^{(n-1)/2}{b}_2\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}.\end{array}\right.$$

(18)

By Lemma 1, we obtain that, for any $n\ge 1$,

$$b^{\pi }{(ba)}^n={\left[\begin{array}{cc}0& 0\\ 0& b^{\pi }\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}0& x_n\\ 0& {(b_2{a}_{22})}^n\end{array}\right]}_{\mathcal{P}}={\left[\begin{array}{cc}0& 0\\ 0& {(b_2{a}_{22})}^n\end{array}\right]}_{\mathcal{P}}={(b_2{a}_{22})}^n,$$

where ${(x_n)}_{n=0}^{\infty }$ is a sequence in $\mathcal{A}$. Furthermore, one has $b_2=b^{\pi }b=b{b}^{\pi }$ and $a{b}^{\pi }=a(1-b{b}^{\mathsf{d}})=a(1-b^2{(b^{\mathsf{d}})}^2)=a$. Hence, if $n\ge 1$ is even, then

$$a_{12}{(a_{22}+b_2)}^n=a_{12}{(b_2{a}_{22})}^{n/2}=b^{\mathsf{d}}ba{b}^{\pi }{(ba)}^{n/2}=b^{\mathsf{d}}ba{(ba)}^{n/2}=b^{\mathsf{d}}{(ba)}^{(n+2)/2},$$

and if $n\ge 1$ is odd, then

$$a_{12}{(a_{22}+b_2)}^n=a_{12}{(b_2{a}_{22})}^{(n-1)/2}{b}_2=b^{\mathsf{d}}ba{b}^{\pi }{(ba)}^{(n-1)/2}{b}^{\pi }b=b^{\mathsf{d}}{(ba)}^{(n+1)/2}b.$$

From (15), we have

$$\begin{array}{c}{a}_{12}{(a_{22}+b_2)}^{\pi }=a_{12}(1-b_2{(a_{22}{b}_2)}^{\mathsf{d}}{a}_{22}),\hfill \\ a_{22}{(a_{22}+b_2)}^{\pi }={(a_{22}{b}_2)}^{\pi }{a}_{22},\hfill \\ a_{12}{b}_2{(a_{22}+b_2)}^{\pi }=a_{12}{b}_2{(a_{22}{b}_2)}^{\pi },\hfill \\ a_{22}{b}_2{(a_{22}+b_2)}^{\pi }=a_{22}{b}_2{(a_{22}{b}_2)}^{\pi }.\hfill \end{array}$$

Thus, by using the obvious equality ${(ba)}^{k}b=b{(ab)}^k$, and by (14)–(16) and (18), we have

$$\begin{array}{ccc}\hfill {(a+b)}^{\mathsf{d}}& =& b_1^{\mathsf{d}}+u={\left[b_1\right]}_{\mathcal{P}}^{-\mathsf{1}}+\sum _{n=0}^{\infty }{({\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}})}^{n+2}{a}_{12}{(b_2+a_{22})}^n{(a_{22}+b_2)}^{\pi }\hfill \\ & & -{(a_{22}+b_2)}^{\mathsf{d}}{a}_{12}{b}_1^{-1}+{(a_{22}+b_2)}^{\mathsf{d}}\hfill \\ & =& \sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{2n+2}{b}^{\pi }{(ab)}^{n}a+\sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{2n+1}{b}^{\pi }{(ab)}^n\hfill \\ & & -\sum _{n=0}^{\infty }{b}^{\pi }{b}^{2n}\left\{{\left[{(ab)}^{\mathsf{d}}\right]}^{n+1}a+b{\left[{(ab)}^{\mathsf{d}}\right]}^{n+1}\right\}\hfill \\ & =& \sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{2n+1}\left[b^{\mathsf{d}}{(ab)}^{\pi }{(ab)}^{n}a+{(ab)}^{\pi }{(ab)}^n\right]\hfill \\ & & -\sum _{n=0}^{\infty }{b}^{\pi }{b}^{2n}\left\{{\left[{(ab)}^{\mathsf{d}}\right]}^{n+1}a+b{\left[{(ab)}^{\mathsf{d}}\right]}^{n+1}\right\}.\hfill \end{array}$$

The proof is completed. □

Theorem 6.

Let $a,b\in {\mathcal{A}}^{\mathsf{d}}$ satisfy (6) and $b^{\pi }{a}^2=0$. Then,

$$\begin{array}{c}\hfill {(a+b)}^{\mathsf{d}}=b^{\mathsf{d}}+u+v,\end{array}$$

where

$$\begin{array}{cc}\hfill v& =-\left\{b^{\mathrm{d}}a{(ba)}^{\mathsf{d}}+\sum _{n=0}^{\infty }{b}^{\mathsf{d}}{b}^{2n+1}\left[{({(ab)}^{\mathsf{d}})}^{n+1}+{({(ba)}^{\mathsf{d}})}^{n+1}\right]\right\},\hfill \\ \hfill u& =\sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{n+2}a{(a+b)}^n+\sum _{n=0}^{\infty }(1-b^{\pi })b^{n}a{v}^{n+2}-b^{\mathsf{d}}av.\hfill \end{array}$$

Proof. 

Let $p=b{b}^{\mathsf{d}}$ and $\mathcal{P}=\{p,1-p\}$. Let a and b have the following representation

$$b={\left[\begin{array}{cc}{b}_1& 0\\ 0& b_2\end{array}\right]}_{\mathcal{P}},a={\left[\begin{array}{cc}{a}_3& a_1\\ a_4& a_2\end{array}\right]}_{\mathcal{P}},$$

(19)

where $b_1$ is invertible in $p\mathcal{A}p$ and $b_2$ is quasinilpotent in $(1-p)\mathcal{A}(1-p)$. Let us find the expression of $b^{\pi }{a}^2$ in the system of idempotents $\mathcal{P}$:

$$b^{\pi }{a}^2={\left[\begin{array}{cc}0& 0\\ 0& 1-p\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}0& a_1\\ 0& a_2\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}0& a_1\\ 0& a_2\end{array}\right]}_{\mathcal{P}}={\left[\begin{array}{cc}0& 0\\ 0& a_2^2\end{array}\right]}_{\mathcal{P}}=a_2^2.$$

Thus, $a_2^2=0$. On the other hand,

$$a{b}^2={\left[\begin{array}{cc}0& a_1\\ 0& a_2\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}{b}_1^2& 0\\ 0& b_2^2\end{array}\right]}_{\mathcal{P}}={\left[\begin{array}{cc}0& a_1{b}_2^2\\ 0& a_2{b}_2^2\end{array}\right]}_{\mathcal{P}}.$$

Therefore, $a_2{b}_2^2=0$. By $b^{\pi }ab,b^{\pi }ba\in {\mathcal{A}}^d$, we obtain $(a_2{b}_2),(b_2{a}_2)\in {\mathcal{A}}^d$. We can appeal to Theorem 5, obtaining (recall that $b_2$ is quasinilpotent and $b_2^{\mathsf{d}}=0$) that

$$\begin{array}{c}\hfill {(a_2+b_2)}^{\mathsf{d}}=-a_2{(b_2{a}_2)}^{\mathsf{d}}-\sum _{n=0}^{\infty }{b}_2^{2n+1}\left[{({(a_2{b}_2)}^{\mathsf{d}})}^{n+1}+{({(b_2{a}_2)}^{\mathsf{d}})}^{n+1}\right].\end{array}$$

From Lemma 3 and the representation of $a+b$ in (16), we have

$$\begin{array}{ccc}\hfill {(a+b)}^{\mathsf{d}}& =& {\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}}+{(a_2+b_2)}^{\mathsf{d}}+u\hfill \\ & =& {\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}}+u-\left\{a_2{(b_2{a}_2)}^d+\sum _{n=0}^{\infty }{b}_2^{2n+1}\left[{({(a_2{b}_2)}^d)}^{n+1}+{({(b_2{a}_2)}^d)}^{n+1}\right]\right\},\hfill \end{array}$$

(20)

where

$$\begin{array}{ccc}\hfill u& =& \sum _{n=0}^{\infty }{({\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}})}^{n+2}{a}_1{(a_2+b_2)}^n{(a_2+b_2)}^{\pi }\hfill \\ & +& \sum _{n=0}^{\infty }{b}_1^{\pi }{b}_1^n{a}_1{({(a_2+b_2)}^{\mathsf{d}})}^{n+2}-{\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}}{a}_1{(a_2+b_2)}^{\mathsf{d}}\hfill \\ & =& \sum _{n=0}^{\infty }{(b_1^{\mathsf{d}})}^{n+2}{a}_1{(a_2+b_2)}^n.\hfill \end{array}$$

Observe that ${\left[b_1^{-\mathsf{1}}\right]}_{\mathcal{P}}=b^{\mathsf{d}}$, and

$$\begin{array}{ccc}\hfill {(b^{\mathsf{d}})}^{n+2}a{(a+b)}^n& =& {\left[\begin{array}{cc}{(b_1^{\mathsf{d}})}^{n+2}& 0\\ 0& 0\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}0& a_1\\ 0& a_2\end{array}\right]}_{\mathcal{P}}{\left[\begin{array}{cc}{b}_1^n& x_n\\ 0& {(a_2+b_2)}^n\end{array}\right]}_{\mathcal{P}}\hfill \\ & =& {\left[\begin{array}{cc}0& 0\\ 0& {(b_1^{\mathsf{d}})}^{n+2}{a}_1{(a_2+b_2)}^n\end{array}\right]}_{\mathcal{P}}={(b_1^{\mathsf{d}})}^{n+2}{a}_1{(a_2+b_2)}^n,\hfill \\ \hfill v=b^{\pi }{(a+b)}^{\mathsf{d}}& =& {(a_2+b_2)}^{\mathsf{d}}=-\left\{b^{\mathsf{d}}a{(ba)}^{\mathsf{d}}+\sum _{n=0}^{\infty }{b}^{\mathsf{d}}{b}^{2n+1}\left[{({(ab)}^{\mathsf{d}})}^{n+1}+{({(ba)}^{\mathsf{d}})}^{n+1}\right]\right\}.\hfill \end{array}$$

Thus, the above expression of u reduces to

$$u=\sum _{n=0}^{\infty }{(b^{\mathsf{d}})}^{n+2}a{(a+b)}^n+\sum _{n=0}^{\infty }(1-b^{\pi })b^{n}a{(v)}^{n+2}-b^{\mathsf{d}}av.$$

(21)

Expressions (20) and (21) finish the proof. □

5. Conclusions

In this paper, we have proved that the multiplications ${(ab)}^n$ and ${(ba)}^n$ of elements $a,b\in {\mathcal{A}}^{\mathrm{d}}$ in a Banach algebra are both generalized Drazin invertible with the conditions (6). A symmetry representation of the generalized Drazin inverse for $ab+ba$ has been derived. The expression given in Theorem 1 is symmetric, as in Remark 1. In the other words, if the result is applied in the computation of ${(ab+ba)}^{\mathrm{d}}$, maybe it will improve the corresponding computational effectiveness and reduce its complexity. The additive properties of ${(a+b)}^{\mathrm{d}}$ have been investigated under the conditions $a{b}^2=0$, $b^{\pi }ab\in {\mathcal{A}}^{\mathrm{d}}$, and $a^2=0$. With similar conditions, but $a^2=0$ being replaced by $b^{\pi }{a}^2=0$, we have also given a resulting expression of ${(a+b)}^{\mathrm{d}}$.

In fact, as pointed out as in [19], it is still an interesting and open problem to express the generalized Drazin inverse of $a+b$ as a function of a, b, and their respective generalized Drazin inverses. In the future, we plan to consider the representations of the generalized Drazin inverse for $a\pm b$ by using a, b, and their generalized Drazin inverses, without side conditions.