Some converse problems on the g-Drazin invertibility in Banach algebras

Abstract

The main purpose of this paper is to investigate the converse problems of some well-known results related to the generalized Drazin (g-Drazin for short) inverse in Banach algebras. Let ${\mathcal {A}}$ be a Banach algebra and $a,b\in {\mathcal {A}}$. First, we give the relationship between the Drazin (g-Drazin, group) invertibility of a, b and that of the sum $a+b$ under certain conditions. Then, for a given polynomial $f(x)\in {\mathbb {C}}[x]$, the g-Drazin invertibility of f(a), $f(a^{d})$, f(ab), $f(1-ab)$ and $f(a+b)$ are investigated.

1 Introduction

Let ${\mathcal {A}}$ be a complex Banach algebra with unity 1. For $a \in {\mathcal {A}}$, denote the spectrum and the spectral radius of a by $\sigma (a)$ and r(a), respectively. In addition, ${\mathcal {A}}^{nil}$ and ${\mathcal {A}}^{qnil}$ stand for the sets of all nilpotent and quasinilpotent elements ($\sigma (a)$ =$\{0\}$) in ${\mathcal {A}}$, respectively.

Generalized inverses have wider applications in many mathematical areas: matrix theory, operator theory, differential equations, numerical analysis and so on [1, 4, 5, 10, 22, 23]. For the readers’ convenience, we first recall the definitions of some generalized inverses. The Drazin inverse [12] of $a\in {\mathcal {A}}$ is the element $x\in {\mathcal {A}}$ which satisfies

$$\begin{aligned} xax=x,\ \ ax=xa,\ \ \textrm{and} \ \ a-a^{2}x\in {\mathcal {A}}^{nil}. \end{aligned}$$

The element x above is unique if it exists and is denoted by $a^{D}$. The power of nilpotency of $a-a^{2}a^{D}$ is called the index of a, and will be denoted by ind(a). In particular, if ind$(a)=1$, then the Drazin inverse of a is called the group inverse of a, and will be denoted by $a^{\#}$. The symbols ${\mathcal {A}}^{D}$ and ${\mathcal {A}}^{\#}$ denote the sets of all Drazin and group invertible elements in ${\mathcal {A}}$, respectively.

The generalized Drazin inverse (or g-Drazin inverse) of $a\in {\mathcal {A}}$ [16] is the element $x\in {\mathcal {A}}$ which satisfies

$$\begin{aligned} xax=x, \ \ ax=xa \ \ \text{ and } \ \ a-a^{2}x \in {\mathcal {A}}^{qnil}. \end{aligned}$$

Such x, if it exists, is unique and will be denoted by $a^{d}$. It is well known that $a\in {\mathcal {A}}$ has the g-Drazin inverse if and only if 0 is not an accumulation point of $\sigma (a)$. Let ${\mathcal {A}}^{d}$ denote the set of all g-Drazin invertible elements in ${\mathcal {A}}$. If $a\in {\mathcal {A}}^{d}$, the spectral idempotent $a^{\pi }$ of a corresponding to the set {0} is given by $a^{\pi }=1-aa^{d}$.

As is well known that if a, b is Drazin invertible in a ring and $ab=0$, then $a+b$ is Drazin invertible and $(a+b)^{D}=a^{D}+b^{D}$ [12]. Since that time many researchers have been interested in this problem, the generalized invertibility of the sum. And, there were lots of results related to this topic in different settings, such as complex matrices [9, 15, 21], operators on a Hilbert space [11], Banach algebras [3, 7, 10, 20], morphisms [6] and so on. The most general result was Theorem 2.1 in [8] as follows.

Throughout this paper, for $k\in {\mathbb {N}}$, $U_{k}$ is defined as

$$\begin{aligned} U_{k}= & {} \{(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})|\sum \limits _{i=1}^{k}(p_{i}+q_{i})\\= & {} k-1,p_{i},q_{i}\in \{0,1,\dots ,k-1\},i=\overline{1,k}\}. \end{aligned}$$

Lemma 1.1

Let $a,b\in {\mathcal {A}}^{D}$ and let ind$(a)=r$, ind$(b)=s$ and $k\in {\mathbb {N}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$, then $a+b\in {\mathcal {A}}^{D}$ and

$$\begin{aligned} (a+b)^{D}= & {} y_{1}+y_{2}+\sum \limits _{i=1}^{k-1}\left( y_{1}(a^{D})^{i+1}+(b^{D})^{i+1}y_{2}-\sum \limits _{j=1}^{i+1}(b^{D})^{j}(a^{D})^{i+2-j}\right) \\{} & {} ab(a+b)^{i-1}, \end{aligned}$$

where $y_{1}=\sum \nolimits _{i=0}^{s-1}b^{\pi }b^{i}(a^{D})^{i+1}$ and $y_{2}=\sum \nolimits _{i=0}^{r-1}(b^{D})^{i+1}a^{i}a^{\pi }$.

However, there were no references related to the converse implications of all the results mentioned above. Motivated by them, in this paper we will give further consideration about the converse proposition of Lemma 1.1 for some different types of generalized inverses (for example, Drazin, g-Drazin, group inverse) under the one-sided and the two-sided conditions, respectively. In addition, for a given polynomial $f(x)\in {\mathbb {C}}[x]$ and $a,b\in {\mathcal {A}}$, the g-Drazin invertibility of f(a), $f(a^{d})$, f(ab), $f(1-ab)$ and $f(a+b)$ are investigated.

2 Some lemmas

In this section, we present some useful lemmas, which play an important role in the sequel. First, Jacobson’s lemma and Cline’s formula for the g-Draizn inverse are introduced as follows.

Lemma 2.1

[24, Theorem 2.3] Let $a,b\in {\mathcal {A}}$. Then $1-ab\in {\mathcal {A}}^{d}$ if and only if $1-ba\in {\mathcal {A}}^{d}$.

Lemma 2.2

[17, Theorem 2.2] Let $a,b\in {\mathcal {A}}$. Then $ab\in {\mathcal {A}}^{d}$ if and only if $ba\in {\mathcal {A}}^{d}$.

Next, several known characterizations and properties of some types of generalized inverses are given.

Lemma 2.3

Let $a\in {\mathcal {A}}$. Then

• (i) [13, Theorem 1] $a\in {\mathcal {A}}^{\#}$ if and only if $a\in a^{2}{\mathcal {A}}\cap {\mathcal {A}}a^{2}$. Moreover, if $a=a^{2}x=ya^{2}$, where $x,y\in {\mathcal {A}}$, then $a^{\#}=ax^{2}=yax=y^{2}a$.

• (ii) [12, Theorem 4] $a\in {\mathcal {A}}^{D}$ if and only if $a^{k}\in a^{k+1}{\mathcal {A}}\cap {\mathcal {A}}a^{k+1}$, for some $k\in {\mathbb {N}}$. Moreover, if $a^{k}=a^{k+1}x=ya^{k+1}$, where $x,y\in {\mathcal {A}}$, then $a^{D}=a^{k}x^{k+1}=y^{k+1}a^{k}$.

Lemma 2.4

[19, Corollary 2.2 and Remark 2.1] Let $a\in {\mathcal {A}}$ and $n\in {\mathbb {N}}$. Then

• (i) $a\in {\mathcal {A}}^{D}$ if and only if $a^{n}\in {\mathcal {A}}^{D}$.

• (ii) $a\in {\mathcal {A}}^{d}$ if and only if $a^{n}\in {\mathcal {A}}^{d}$.

Lemma 2.5

[16, Theorems 5.4 and 5.5] Let $a,b\in {\mathcal {A}}$.

• (i) If $a\in {\mathcal {A}}^{d}$, then $a^{d}\in {\mathcal {A}}^{d}$ and $(a^{d})^{d}=a^{2}a^{d}$.

• (ii) If $a,b\in {\mathcal {A}}^{d}$ and $ab=ba$, then $ab\in {\mathcal {A}}^{d}$ and $(ab)^{d}=a^{d}b^{d}$.

Lemma 2.6

[2, Example 4.5] Let $a, b\in {\mathcal {A}}^{d}$. If $ab=0$, then $a+b\in {\mathcal {A}}^{d}$ and

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

Let $p \in {\mathcal {A}}$ be an idempotent ($p^{2}=p$). Then we can represent element $a \in {\mathcal {A}}$ as

$$\begin{aligned} a=\left( \begin{matrix} a_{1}&{}a_{2} \\ a_{3}&{}a_{4} \\ \end{matrix}\right) _{p}, \end{aligned}$$

where $a_{1}=pap$, $a_{2}=pa(1-p)$, $a_{3}=(1-p)ap$ and $a_{4}=(1-p)a(1-p)$. In what follows, by ${\mathcal {A}}_{1}$ and ${\mathcal {A}}_{2}$ we denote the algebra $p{\mathcal {A}}p$ and $(1-p){\mathcal {A}}(1-p)$, respectively.

Lemma 2.7

Let $p^{2}=p$, $x,y\in {\mathcal {A}}$ and let x and y have the representations

$$\begin{aligned} x= \left( \begin{matrix} a&{}c \\ 0&{}b\\ \end{matrix}\right) _{p},\ \ \ y= \left( \begin{matrix} b&{}0 \\ c&{}a \\ \end{matrix}\right) _{1-p}. \end{aligned}$$

• (i) [2, Theorem 2.3] If $a\in {\mathcal {A}}_{1}^{d}$, then $x\in {\mathcal {A}}^{d}$ $(resp.\ y \in {\mathcal {A}}^{d})$ if and only if $b\in {\mathcal {A}}_{2}^{d}$. In this case,

  $$\begin{aligned} x^{d}= \left( \begin{matrix} a^{d}&{}u \\ 0&{}b^{d} \\ \end{matrix}\right) _{p},\ \ \ y^{d}= \left( \begin{matrix} b^{d}&{}0 \\ u&{}a^{d} \\ \end{matrix}\right) _{1-p}, \end{aligned}$$

  where

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

• (ii) [14, Theorem 1] If $a\in {\mathcal {A}}_{1}^{D}$, then

  • (1) $x\in {\mathcal {A}}^{D}$ $(resp.\ y \in {\mathcal {A}}^{D})$ if and only if $b\in {\mathcal {A}}_{2}^{D}$. In this case,

    $$\begin{aligned} x^{D}= \left( \begin{matrix} a^{D}&{}u \\ 0&{}b^{D} \\ \end{matrix}\right) _{p},\ \ \ y^{D}= \left( \begin{matrix} b^{D}&{}0 \\ u&{}a^{D} \\ \end{matrix}\right) _{1-p}, \end{aligned}$$

    where

    $$\begin{aligned} u=\sum _{n=0}^{ind (b)-1}(a^{D})^{n+2}cb^{n}b^{\pi }+\sum _{n=0}^{ind (a)-1}a^{\pi }a^{n}c(b^{D})^{n+2}-a^{D}cb^{D}. \end{aligned}$$

  • (2) $x\in {\mathcal {A}}^{\#}$ $(resp.\ y \in {\mathcal {A}}^{\#})$ if and only if $a\in {\mathcal {A}}_{1}^{\#}$, $b\in {\mathcal {A}}_{2}^{\#}$ and $a^{\pi }cb^{\pi }=0$. In this case,

    $$\begin{aligned} x^{\#}= \left( \begin{matrix} a^{\#}&{}u \\ 0&{}b^{\#} \\ \end{matrix}\right) _{p},\ \ \ y^{\#}= \left( \begin{matrix} b^{\#}&{}0 \\ u&{}a^{\#} \\ \end{matrix}\right) _{1-p}, \end{aligned}$$

    where

    $$\begin{aligned} u=(a^{\#})^{2}cb^{\pi }+a^{\pi }c(b^{\#})^{2}-a^{\#}cb^{\#}. \end{aligned}$$

3 The relationship of the g-Drazin invertibility among the elements a, b and $a+b$

In this section, for $a,b\in {\mathcal {A}}$ we are mainly concerned with the relationship between the g-Drazin (resp. Drazin, group) invertibility of a, b and that of the sum $a+b$ under certain conditions. First, this topic related to the Drazin invertibility is studied as follows.

Theorem 3.1

Let $k,m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$ and $b\in {\mathcal {A}}^{D}$, then

$$\begin{aligned} (i) \ a\in {\mathcal {A}}^{D} \Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{D} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{D}. \end{aligned}$$

Proof

(i) $\Rightarrow$ (ii) follows directly from Lemma 1.1.

(ii) $\Rightarrow$ (i). We consider the following two cases.

Case 1: Assume that $b \in {\mathcal {A}}^{nil}$ and $b^{l}=0$, where $l\in {\mathbb {N}}$. Since $a+b \in {\mathcal {A}}^{D}$, there exists a positive integer $n_{0}>l$ such that

$$\begin{aligned} (a+b)^{n_{0}}=(a+b)^{n_{0}+1}(a+b)^{D}=(a+b)^{D}(a+b)^{n_{0}+1}. \end{aligned}$$

Thus, on one hand, by $ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$, we get

$$\begin{aligned} a^{n_{0}+k}= & {} a(a+b)^{D}(a+b)^{n_{0}+1}a^{k-1}\\= & {} a(a+b)^{D}\sum \limits _{i=0}^{l-1}b^{i}a^{n_{0}+k-i}\\= & {} a\left( (a+b)^{D}\right) ^{2}(a+b)\sum \limits _{i=0}^{l-1}b^{i}a^{n_{0}+k-i}\\= & {} a\left( (a+b)^{D}\right) ^{2}\sum \limits _{i=0}^{l-1}b^{i}a^{n_{0}+k+1-i}\\{} & {} \ \ \vdots \\= & {} a\left( (a+b)^{D}\right) ^{l+1}\sum \limits _{i=0}^{l-1}b^{i}a^{n_{0}+k+l-i}\\= & {} \left( a\left( (a+b)^{D}\right) ^{l+1}\sum \limits _{i=0}^{l-1}b^{i}a^{l-1-i}\right) a^{n_{0}+k+1}\in {\mathcal {A}}a^{n_{0}+k+1}. \end{aligned}$$

On the other hand, by computation we deduce that

$$\begin{aligned} a^{n_{0}+k}= & {} a(a+b)^{n_{0}}a^{k-1}\\= & {} a(a+b)^{n_{0}+1}(a+b)^{D}a^{k-1}\\= & {} a(a+b)^{n_{0}+1}(a+b)^{k-1}\left( (a+b)^{D}\right) ^{k}a^{k-1}\\= & {} a^{n_{0}+2}(a+b)^{k-1}\left( (a+b)^{D}\right) ^{k}a^{k-1}\\= & {} a^{n_{0}+k+1}\left( (a+b)^{D}\right) ^{k}a^{k-1}\in a^{n_{0}+k+1}{\mathcal {A}}. \end{aligned}$$

Therefore, applying Lemma 2.3(ii) we have $a\in {\mathcal {A}}^{D}$.

Case 2: Assume that $b \notin {\mathcal {A}}^{nil}$. Let $p=bb^{D}$. We consider the matrix representations of a and b relative to the idempotent p:

$$\begin{aligned} b= \left( \begin{matrix} b_{1}&{}0\\ 0&{}b_{2}\\ \end{matrix}\right) _{p} \ \text{ and }\ \ a= \left( \begin{matrix} a_{1}&{}a_{3}\\ a_{4}&{}a_{2}\\ \end{matrix}\right) _{p}, \end{aligned}$$

where $b_{1}\in {\mathcal {A}}_{1}^{-1}$ and $b_{2}\in {\mathcal {A}}_{2}^{nil}$.

Note that

$$\begin{aligned} ab^{k}=\left( \begin{matrix} a_{1}b_{1}^{k}&{}a_{3}b_{2}^{k}\\ a_{4}b_{1}^{k}&{}a_{2}b_{2}^{k}\\ \end{matrix}\right) _{p}=0. \end{aligned}$$

So, we obtain $a_{1}b_{1}^{k}=0$ and $a_{4}b_{1}^{k}=0$, which infer that $a_{1}=a_{4}=0$. Thus, we derive

$$\begin{aligned} a= \left( \begin{matrix} 0&{}a_{3}\\ 0&{}a_{2}\\ \end{matrix}\right) _{p} \ \text{ and }\ \ a+b= \left( \begin{matrix} b_{1}&{}a_{3}\\ 0&{}a_{2}+b_{2}\\ \end{matrix}\right) _{p}. \end{aligned}$$

Since $a+b\in {\mathcal {A}}^{D}$ and $b_{1}\in {\mathcal {A}}_{1}^{-1}$, from Lemma 2.7(ii)(1) we see $a_{2}+b_{2}\in {\mathcal {A}}_{2}^{D}$. According to $ab\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$, we deduce $a_{2}b_{2}\prod \nolimits _{i=1}^{k}(a_{2}^{p_{i}}b_{2}^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$, Applying Case 1 to elements $a_{2}$ and $b_{2}$, we claim that $a_{2}\in {\mathcal {A}}_{2}^{D}$, which yields $a\in {\mathcal {A}}^{D}$.

(i) $\Leftrightarrow$ (iii). In view of Lemma 2.4(i), we obtain $b\in {\mathcal {A}}^{D}$ if and only if $b^{n}\in {\mathcal {A}}^{D}$. Observe that $a^{m}b^{n}\prod \nolimits _{i=1}^{k}\left( (a^{m})^{p_{i}}(b^{n})^{q_{i}}\right) =0$, for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, Then, by the equivalence of (i) and (ii) it follows that $a^{m}\in {\mathcal {A}}^{D}$ is equivalent to $a^{m}+b^{n}\in {\mathcal {A}}^{D}$, as required. $\square$

Similarly, we have the following result.

Remark 3.2

Let $k,m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$ and $a\in {\mathcal {A}}^{D}$, then

$$\begin{aligned} (i) \ b\in {\mathcal {A}}^{D} \Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{D} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{D}. \end{aligned}$$

Based on Theorem 3.1 and Remark 3.2, we get

Corollary 3.3

Let $k\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$ and $a+b\in {\mathcal {A}}^{D}$, then $a\in {\mathcal {A}}^{D}$ if and only if $b \in {\mathcal {A}}^{D}$.

Furthermore, under the two-sided condition we have the following result for the Drazin invertibility.

Theorem 3.4

Let $k,m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=ba\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\dots ,p_{k},q_{k})\in U_{k}$, then

$$\begin{aligned} (i) \ a, b\in {\mathcal {A}}^{D}\Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{D} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{D}. \end{aligned}$$

In this case, $a^{D}=\left( (a+b)^{D}\right) ^{k+1}a^{k}$.

Proof

(i) $\Rightarrow$ (ii) can be obtained by Lemma 1.1.

(ii) $\Rightarrow$ (i). Suppose that $a+b \in {\mathcal {A}}^{D}$. Then, we derive

$$\begin{aligned} (a+b)^{n_{0}}=(a+b)^{n_{0}+1}(a+b)^{D}=(a+b)^{D}(a+b)^{n_{0}+1}, \end{aligned}$$

where $n_{0}\in {\mathbb {N}}$. By $ba\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, one can check that

$$\begin{aligned} a^{n_{0}+k}= & {} (a+b)^{n_{0}}a^{k}\\= & {} (a+b)^{D}(a+b)^{n_{0}+1}a^{k}\\= & {} (a+b)^{D}a^{n_{0}+k+1}\in {\mathcal {A}}a^{n_{0}+k+1}. \end{aligned}$$

In addition, from the proof of (ii) $\Rightarrow$ (i) of Theorem 3.1 we know that $a^{n_{0}+k}\in a^{n_{0}+k+1}{\mathcal {A}}$. Therefore, by Lemma 2.3(ii) we have that $a\in {\mathcal {A}}^{D}$ and

$$\begin{aligned} a^{D}= & {} \left( (a+b)^{D}\right) ^{n_{0}+k+1}a^{n_{0}+k}\\= & {} \left( (a+b)^{D}\right) ^{n_{0}+k+1}(a+b)^{n_{0}}a^{k}\\= & {} \left( (a+b)^{D}\right) ^{k+1}a^{k}. \end{aligned}$$

(i) $\Leftrightarrow$ (iii) can be deduced using the equivalence (i) $\Leftrightarrow$ (ii). $\square$

Corresponding to the Drazin invertibility above, we will consider the case of g-Drazin invertibility. The proof of the following theorem is similar to that of an analogous result for matrices [10, Theorem 2.1]. For the completeness, we give the proof.

Theorem 3.5

Let $k,m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$ and $a\in {\mathcal {A}}^{d}$, then

$$\begin{aligned} (i) \ b\in {\mathcal {A}}^{d} \Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{d} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{d}. \end{aligned}$$

In this case,

$$\begin{aligned} (a+b)^{d}=y_{1}+y_{2}+\sum \limits _{i=1}^{k-1}\left( y_{1}(a^{d})^{i+1}+(b^{d})^{i+1}y_{2}-\sum \limits _{j=1}^{i+1}(b^{d})^{j}(a^{d})^{i+2-j}\right) ab(a+b)^{i-1}, \end{aligned}$$

(1)

where $y_{1}=\sum \nolimits _{i=0}^{\infty }b^{\pi }b^{i}(a^{d})^{i+1}$ and $y_{2}=\sum \nolimits _{i=0}^{\infty }(b^{d})^{i+1}a^{i}a^{\pi }$.

Proof

In view of [10, Theorem 2.1] and Lemma 2.4(ii), it is sufficient to prove $b\in {\mathcal {A}}^{d}$ if and only if $a+b \in {\mathcal {A}}^{d}$. From Lemma 2.2, one easily checks

$$\begin{aligned} a+b\in {\mathcal {A}}^{d}\Longleftrightarrow & {} \left( \begin{matrix} a+b&{}0\\ 0&{}0\\ \end{matrix}\right) =\left( \begin{matrix} 1&{}b\\ 0&{}0\\ \end{matrix}\right) \left( \begin{matrix} a&{}0\\ 1&{}0\\ \end{matrix}\right) \in M_{2}({\mathcal {A}})^{d}\\\Longleftrightarrow & {} N:=\left( \begin{matrix} a&{}0\\ 1&{}0\\ \end{matrix}\right) \left( \begin{matrix} 1&{}b\\ 0&{}0\\ \end{matrix}\right) =\left( \begin{matrix} a&{}ab\\ 1&{}b\\ \end{matrix}\right) \in M_{2}({\mathcal {A}})^{d}.\\ \end{aligned}$$

We use mathematical induction on k to prove this result. When $k=1$, as $ab=0$ and $a\in {\mathcal {A}}^{d}$, in light of Lemma 2.7(i) and above equivalence we have that $b\in {\mathcal {A}}^{d}$ if and only if $a+b \in {\mathcal {A}}^{d}$. Now, assume that this result holds for $k-1$. Next, we prove it is true for k. Let

$$\begin{aligned} M:=N^{2}=\left( \begin{matrix} a^{2}+ab&{}a^{2}b+ab^{2}\\ a+b&{}ab+b^{2}\\ \end{matrix}\right) =\left( \begin{matrix} ab&{}a^{2}b+ab^{2}\\ 0&{}ab\\ \end{matrix}\right) + \left( \begin{matrix} a^{2}&{}0\\ a+b&{}b^{2}\\ \end{matrix}\right) . \end{aligned}$$

Setting $M_{1}=\left( \begin{matrix} ab&{}a^{2}b+ab^{2}\\ 0&{}ab\\ \end{matrix}\right)$ and $M_{2}=\left( \begin{matrix} a^{2}&{}0\\ a+b&{}b^{2}\\ \end{matrix}\right)$. Clearly, $M_{1}^{n}=0$ for any $n\ge \frac{k+1}{2}$. So, $M_{1}\in M_{2}({\mathcal {A}})^{d}$. In addition, we have

$$\begin{aligned} M_{1}M_{2}\prod \limits _{i=1}^{k-1}(M_{1}^{p_{i}}M_{2}^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k-1},q_{k-1})\in U_{k-1}$. Therefore, by the induction hypothesis we conclude that $M\in M_{2}({\mathcal {A}})^{d}$ if and only if $M_{2}\in M_{2}({\mathcal {A}})^{d}$, which yields $b\in {\mathcal {A}}^{d}$ if and only if $a+b\in {\mathcal {A}}^{d}$ by Lemma 2.4(ii) and Lemma 2.7(i). $\square$

Now, we present the main result of this section.

Theorem 3.6

Let $k,m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=ba\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, then

$$\begin{aligned} (i) \ a, b\in {\mathcal {A}}^{d} \Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{d} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{d}. \end{aligned}$$

In this case,

$$\begin{aligned} a^{d}= & {} a\left( (a+b)^{d}\right) ^{2}+\sum \limits _{n=0}^{k-1}\left( a^{n+2}\left( (a+b)^{d}\right) ^{2n+4}a^{n+1} -a^{n+3}\left( (a+b)^{d}\right) ^{2n+4}a^{n}\right) \\= & {} \left( (a+b)^{d}\right) ^{k+1}a^{k}. \end{aligned}$$

Proof

(i) $\Rightarrow$ (ii) is obvious by Theorem 3.5.

(ii) $\Rightarrow$ (i). Suppose that $a+b \in {\mathcal {A}}^{d}$ and let $p=(a+b)(a+b)^{d}$. Then, consider the matrix representations of $a+b$, a and b relative to the idempotent p:

$$\begin{aligned} a+b= \left( \begin{matrix} x_{1}&{}0 \\ 0&{}x_{2}\\ \end{matrix}\right) _{p}, \ a= \left( \begin{matrix} a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\\ \end{matrix}\right) _{p} \ \text{ and }\ \ b= \left( \begin{matrix} x_{1}-a_{1}&{}-a_{2}\\ -a_{3}&{}x_{2}-a_{4}\\ \end{matrix}\right) _{p}, \end{aligned}$$

where $x_{1}\in {\mathcal {A}}_{1}^{-1}$ and $x_{2}\in {\mathcal {A}}_{2}^{qnil}$.

Since $ab\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$ for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, we have $ab(a+b)^{k-1}=0$, which can be expressed in the following matrix form:

$$\begin{aligned} ab(a+b)^{k-1}=\left( \begin{matrix} \left( a_{1}x_{1}-(a_{1}^{2}+a_{2}a_{3})\right) x_{1}^{k-1}&{}a_{2}x_{2}^{k}-(a_{1}a_{2}+a_{2}a_{4})x_{2}^{k-1}\\ \left( a_{3}x_{1}-(a_{3}a_{1}+a_{4}a_{3})\right) x_{1}^{k-1}&{}a_{4}x_{2}^{k}-(a_{3}a_{2}+a_{4}^{2})x_{2}^{k-1}\\ \end{matrix}\right) _{p}=0. \end{aligned}$$

Thus, we get

$$\begin{aligned} a_{1}x_{1}= & {} a_{1}^{2}+a_{2}a_{3}, \end{aligned}$$

(2)

$$\begin{aligned} a_{2}x_{2}^{k}= & {} (a_{1}a_{2}+a_{2}a_{4})x_{2}^{k-1}, \end{aligned}$$

(3)

$$\begin{aligned} a_{3}x_{1}= & {} a_{3}a_{1}+a_{4}a_{3}, \end{aligned}$$

(4)

$$\begin{aligned} a_{4}x_{2}^{k}= & {} (a_{3}a_{2}+a_{4}^{2})x_{2}^{k-1}. \end{aligned}$$

(5)

Similarly, from $ba\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, it follows that $ba(a+b)^{k-1}=0$, which yields

$$\begin{aligned} x_{1}a_{1}= & {} a_{1}^{2}+a_{2}a_{3}, \end{aligned}$$

(6)

$$\begin{aligned} x_{1}a_{2}x_{2}^{k-1}= & {} (a_{1}a_{2}+a_{2}a_{4})x_{2}^{k-1}, \end{aligned}$$

(7)

$$\begin{aligned} x_{2}a_{3}= & {} a_{3}a_{1}+a_{4}a_{3}, \end{aligned}$$

(8)

$$\begin{aligned} x_{2}a_{4}x_{2}^{k-1}= & {} (a_{3}a_{2}+a_{4}^{2})x_{2}^{k-1}. \end{aligned}$$

(9)

Then, by (2)–(9) we can verify that

$$\begin{aligned} a_{1}x_{1}=x_{1}a_{1},\ \ a_{2}x_{2}^{k}=x_{1}a_{2}x_{2}^{k-1},\ a_{3}x_{1}=x_{2}a_{3}\hbox { and }a_{4}x_{2}^{k}=x_{2}a_{4}x_{2}^{k-1}. \end{aligned}$$

According to $a_{3}x_{1}=x_{2}a_{3}$, we obtain

$$\begin{aligned} a_{3}=x_{2}a_{3}x_{1}^{-1}=x_{2}^{2}a_{3}x_{1}^{-2}=\cdots =x_{2}^{n}a_{3}x_{1}^{-n}, \end{aligned}$$

for any $n\in {\mathbb {N}}$. Since $x_{2}\in {\mathcal {A}}_{2}^{qnil}$, we have

$$\begin{aligned} \Vert a_{3}\Vert ^{\frac{1}{n}}=\Vert x_{2}^{n}a_{3}x_{1}^{-n}\Vert ^{\frac{1}{n}}\leqslant \Vert x_{2}^{n}\Vert ^{\frac{1}{n}}\Vert a_{3}\Vert ^{\frac{1}{n}}\Vert x_{1}^{-1}\Vert \xrightarrow {n\rightarrow \infty }0. \end{aligned}$$

Hence, $a_{3}=0$, i.e. $pap=ap$. Likewise, from $a_{2}x_{2}^{k}=x_{1}a_{2}x_{2}^{k-1}$ it follows that $a_{2}x_{2}^{k-1}=0$.

Note that

$$\begin{aligned} a= \left( \begin{matrix} a_{1}&{}a_{2}\\ 0&{}a_{4}\\ \end{matrix}\right) _{p}. \end{aligned}$$

To prove $a\in {\mathcal {A}}^{d}$, by Lemma 2.7(i) we only need to show that $a_{1}\in {\mathcal {A}}_{1}^{d}$ and $a_{4}\in {\mathcal {A}}_{2}^{d}$. Applying $a_{3}=0$ and (2), we have $a_{1}x_{1}=a_{1}^{2}=x_{1}a_{1}$. So, $a_{1}=x_{1}^{-1}a_{1}^{2}=a_{1}^{2}x_{1}^{-1}$, which shows that $a_{1}\in {\mathcal {A}}_{1}^{\#}$ and $a_{1}^{\#}=x_{1}^{-1}a_{1}x_{1}^{-1}=a_{1}x_{1}^{-2}$ by Lemma 2.3(i). Note that $x_{1}^{-1}=(a+b)^{d}$ and $a_{1}=pap=ap$. Accordingly, we have $a_{1}^{\#}=(a+b)^{d}a(a+b)^{d}=a\left( (a+b)^{d}\right) ^{2}$. Since

$$\begin{aligned} ba^{k}= & {} \left( \begin{matrix} x_{1}-a_{1}&{}-a_{2}\\ 0&{}x_{2}-a_{4}\\ \end{matrix}\right) _{p} \left( \begin{matrix} a_{1}^{k}&{}\sum \limits _{i=0}^{k-1}a_{1}^{i}a_{2}a_{4}^{k-1-i}\\ 0&{}a_{4}^{k}\\ \end{matrix}\right) _{p}\\= & {} \left( \begin{matrix} 0&{}x_{1}a_{2}a_{4}^{k-1}-(a_{1}a_{2}a_{4}^{k-1}+a_{2}a_{4}^{k})\\ 0&{}x_{2}a_{4}^{k}-a_{4}^{k+1}\\ \end{matrix}\right) _{p}=0, \end{aligned}$$

we have $a_{4}^{k+1}=x_{2}a_{4}^{k}$ and the following equation holds

$$\begin{aligned} x_{1}a_{2}a_{4}^{k-1}-(a_{1}a_{2}a_{4}^{k-1}+a_{2}a_{4}^{k})=0. \end{aligned}$$

(10)

From $a_{4}^{k+1}=x_{2}a_{4}^{k}$, by induction we get $a_{4}^{n+k}=x_{2}^{n}a_{4}^{k}$ for any $n\in {\mathbb {N}}$. So,

$$\begin{aligned} \Vert a_{4}^{n+k}\Vert ^{\frac{1}{n}}=\Vert x_{2}^{n}a_{4}^{k}\Vert ^{\frac{1}{n}}\leqslant \Vert x_{2}^{n}\Vert ^{\frac{1}{n}}\Vert a_{4}^{k}\Vert ^{\frac{1}{n}}\xrightarrow {n\rightarrow \infty }0, \end{aligned}$$

which gives that $r(a_{4})=0$, i.e., $a_{4}\in {\mathcal {A}}_{2}^{qnil}$. Then, $a_{4}\in {\mathcal {A}}_{2}^{d}$ with $a_{4}^{d}=0$. Hence, in view of Lemma 2.7(i), we claim that $a\in {\mathcal {A}}^{d}$ and

$$\begin{aligned} a^{d}=a_{1}^{\#}+\sum \limits _{n=0}^{\infty }(a_{1}^{\#})^{n+2}a_{2}a_{4}^{n}a_{4}^{\pi }. \end{aligned}$$

According to the symmetry of a, b, we conclude that $b\in {\mathcal {A}}^{d}$.

Next, we show that $a_{2}a_{4}^{k}=0$. By $aba^{k-1}=0$, we deduce that

$$\begin{aligned} a_{2}x_{2}a_{4}^{k-1}-(a_{1}a_{2}a_{4}^{k-1}+a_{2}a_{4}^{k})=0. \end{aligned}$$

(11)

Hence, using (10) and (11) we get $x_{1}a_{2}a_{4}^{k-1}=a_{2}x_{2}a_{4}^{k-1}$, which yields $a_{2}a_{4}^{k-1}=x_{1}^{-1}a_{2}x_{2}a_{4}^{k-1}$. Since $a_{4}^{2k-1}=x_{2}^{k-1}a_{4}^{k}$, we conclude that

$$\begin{aligned} a_{2}a_{4}^{k}= & {} x_{1}^{-1}a_{2}(x_{2}a_{4}^{k})=x_{1}^{-1}a_{2}a_{4}^{k+1}=x_{1}^{-1}(a_{2}a_{4}^{k})a_{4}\\= & {} x_{1}^{-2}a_{2}a_{4}^{k+2}=\cdots =x_{1}^{1-k}a_{2}a_{4}^{2k-1}\\= & {} x_{1}^{1-k}(a_{2}x_{2}^{k-1})a_{4}^{k}=0. \end{aligned}$$

So, by computation we obtain

$$\begin{aligned} a^{d}= & {} a_{1}^{\#}+\sum \limits _{n=0}^{k-1}(a_{1}^{\#})^{n+2}a_{2}a_{4}^{n}\\= & {} a\left( (a+b)^{d}\right) ^{2}+\sum \limits _{n=0}^{k-1}\left( a^{n+2}\left( (a+b)^{d}\right) ^{2n+4}a^{n+1} -a^{n+3}\left( (a+b)^{d}\right) ^{2n+4}a^{n}\right) . \end{aligned}$$

Since $ba^{k}=0$, we get $ba^{d}=ba^{k}(a^{d})^{k+1}=0$ and $b^{d}a^{k}=0$. Then, post-multiplying by $a^{k}$ on the equality (1) of Theorem 3.5 we announce that $(a+b)^{d}a^{k}=a^{d}a^{k}$, which implies $a^{d}=\left( a^{d}\right) ^{k+1}a^{k}=\left( (a+b)^{d}\right) ^{k+1}a^{k}$.

(i) $\Leftrightarrow$ (iii). By Lemma 2.4(2) and (i) $\Leftrightarrow$ (ii) it follows that $a, b\in {\mathcal {A}}^{d}$ is equivalent to $a^{m}+b^{n}\in {\mathcal {A}}^{d}$. $\square$

In Theorem 3.6, let $k=1$, then it is reduced as the following.

Corollary 3.7

Let $m,n\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If $ab=ba=0$, then

$$\begin{aligned} (i) \ a, b\in {\mathcal {A}}^{d} \Longleftrightarrow (ii) \ a+b \in {\mathcal {A}}^{d} \Longleftrightarrow (iii) \ a^{m}+b^{n}\in {\mathcal {A}}^{d}. \end{aligned}$$

Next, we continue to investigate the group invertibility of a, b and $a+b$, which is somewhat different from the g-Draizn inverse case.

Theorem 3.8

Let $k\in {\mathbb {N}}$ and $a,b\in {\mathcal {A}}$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=ba\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$ and $b\in {\mathcal {A}}^{\#}$, then

$$\begin{aligned} (i) \ a\in {\mathcal {A}}^{\#}\Longleftrightarrow (ii) \ a+b\in {\mathcal {A}}^{\#}\ and\ \left( 1-(a+b)^{\pi }\right) a(a+b)^{\pi }=0. \end{aligned}$$

Proof

Since $ab^{k}=0$ and $b\in {\mathcal {A}}^{\#}$, we obtain $ab=ab^{k}(b^{\#})^{k-1}=0$.

(i) $\Rightarrow$ (ii). Note that $ba^{k}=0$ and $a\in {\mathcal {A}}^{\#}$. Then, $ba=0$. Thus, $a+b\in {\mathcal {A}}^{\#}$ and $(a+b)^{\#}=a^{\#}+b^{\#}$. Also, $\left( 1-(a+b)^{\pi }\right) a(a+b)^{\pi }=0$.

(ii) $\Rightarrow$ (i). Let us consider the matrix representations of $a+b$, a and b relative to the idempotent $p=(a+b)(a+b)^{\#}$:

$$\begin{aligned} a+b= \left( \begin{matrix} x_{1}&{}0 \\ 0&{}0\\ \end{matrix}\right) _{p}, \ a= \left( \begin{matrix} a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\\ \end{matrix}\right) _{p} \ \text{ and }\ \ b= \left( \begin{matrix} x_{1}-a_{1}&{}-a_{2}\\ -a_{3}&{}-a_{4}\\ \end{matrix}\right) _{p}, \end{aligned}$$

where $x_{1}\in {\mathcal {A}}_{1}^{-1}$. Then, according to the proof of Theorem 3.6 we see that $a_{3}=0$ and $a_{1}x_{1}=a_{1}^{2}=x_{1}a_{1}$. So, $a_{1}\in {\mathcal {A}}_{1}^{\#}$. Moreover, from

$$\begin{aligned} ab=\left( \begin{matrix} 0&{}-a_{1}a_{2}-a_{2}a_{4}\\ 0&{}-a_{4}^{2}\\ \end{matrix}\right) _{p}=0, \end{aligned}$$

it follows that $a_{1}a_{2}+a_{2}a_{4}=0$ and $a_{4}^{2}=0$. Since $b\in {\mathcal {A}}^{\#}$, by virtue of Lemma 2.7(ii)(2) we obtain $a_{4}\in {\mathcal {A}}^{\#}$, which gives $a_{4}=a_{4}^{2}a_{4}^{\#}=0$. Hence, $a_{1}a_{2}=-a_{2}a_{4}=0$ and

$$\begin{aligned} a= \left( \begin{matrix} a_{1}&{}a_{2}\\ 0&{}0\\ \end{matrix}\right) _{p}. \end{aligned}$$

Applying Lemma 2.7(ii)(2) again, we get

$$\begin{aligned} a\in {\mathcal {A}}^{\#} \Longleftrightarrow a_{1}^{\pi }a_{2}=0 \Longleftrightarrow a_{2}=\left( 1-(a+b)^{\pi }\right) a(a+b)^{\pi }=0. \end{aligned}$$

$\square$

Remark 3.9

When $k\ge 2$ in Theorem 3.8, the condition $ab\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=ba\prod \nolimits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0$ for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$, $a+b\in {\mathcal {A}}^{\#}$ and $b\in {\mathcal {A}}^{\#}$ do not imply $a\in {\mathcal {A}}^{\#}$ in general, which can be seen from the following example.

Example 3.10

Let ${\mathcal {A}}=M_{2}({\mathbb {C}})$. Setting $a=\left( {\begin{matrix} 0&{}1\\ 0&{}0\\ \end{matrix}}\right)$, $b=\left( {\begin{matrix} 1&{}0\\ 0&{}0\\ \end{matrix}}\right)$. Obviously, for any $k\ge 2$, such a, b satisfy the condition of Remark 3.9. But, $a\notin {\mathcal {A}}^{\#}$.

However, when $k=1$ in Theorem 3.8, using Theorem 3.8 and the double commuting property of the group inverse, we directly get the following result.

Corollary 3.11

Let $a,b\in {\mathcal {A}}$ be such that $ab=ba=0$. If $b\in {\mathcal {A}}^{\#}$ $(resp.\ a\in {\mathcal {A}}^{\#})$, then $a+b\in {\mathcal {A}}^{\#}$ if and only if $a\in {\mathcal {A}}^{\#}$ $(resp.\ b\in {\mathcal {A}}^{\#})$.

Remark 3.12

In Corollary 3.11, the two-sided condition $ab=ba=0$ in general cannot be replaced by the one-sided condition $ab=0$, which can also be illustrated by Example 3.10. In addition, the condition $b\in {\mathcal {A}}^{\#}$ cannot be dropped. In Example 3.10, choose $a=\left( {\begin{matrix} 0&{}1\\ 0&{}0\\ \end{matrix}}\right)$ and $b=-a$. Evidently, $ab=ba=0$ and $a+b\in {\mathcal {A}}^{\#}$. However, $a\notin {\mathcal {A}}^{\#}$.

4 The g-Drazin invertibility involving the polynomial

A polynomial over the complex field ${\mathbb {C}}$ is an equation of the formal

$$\begin{aligned} f(x)=\lambda _{n}x^{n}+\lambda _{n-1}x^{n-1}+\cdots +\lambda _{0} \end{aligned}$$

where $n\in {\mathbb {N}}\cup \{0\}$ and the coefficients $\lambda _{n},\lambda _{n-1},\ldots ,\lambda _{0}\in {\mathbb {C}}$. The set of all polynomials over the complex field ${\mathbb {C}}$ is denoted by ${\mathbb {C}}[x]$.

In this section, for $a\in {\mathcal {A}}$ and $f(x)\in {\mathbb {C}}[x]$, the g-Drazin invertibility of the polynomial f(a) and its factors are investigated. And, we give the relationship of g-Drazin invertibility of several pairs of elements, such as (f(ab), f(ba)), ($f(1-ab)$, $f(1-ba)$) and (f(a), $f(a^{d})$).

Theorem 4.1

Let $a\in {\mathcal {A}}$ and $0 \ne f(x)=\lambda (x-\varepsilon _{1})(x-\varepsilon _{2})\cdots (x-\varepsilon _{n})\in {\mathbb {C}}[x]$, where $n\in {\mathbb {N}}$ and $\lambda , \varepsilon _{1},\varepsilon _{2},\ldots ,\varepsilon _{n}\in {\mathbb {C}}$. Then, $f(a)\in {\mathcal {A}}^{d}$ if and only if $a-\varepsilon _{i}\in {\mathcal {A}}^{d}$, for $i=\overline{1,n}$.

Proof

$``\Leftarrow ''$. It is obvious by Lemma 2.5.

$``\Rightarrow ''$. Suppose that $f(a)=\lambda (a-\varepsilon _{1})(a-\varepsilon _{2})\cdots (a-\varepsilon _{n})\in {\mathcal {A}}^{d}$. Let $u=a-\varepsilon _{1}$. Next, we only need to prove that $u\in {\mathcal {A}}^{d}$. Note that there exist suitable $\lambda _{1}, \ldots , \lambda _{n-1}\in {\mathbb {C}}$ satisfying $f(a)=\lambda u^{n}+\lambda _{n-1}u^{n-1}+\cdots +\lambda _{1}u$. Let $g(x)=\lambda x^{n}+\lambda _{n-1}x^{n-1}+\cdots +\lambda _{1}x\in {\mathbb {C}}[x]$. So, $g(u)=f(a)\in {\mathcal {A}}^{d}$. Now, we can consider the following two cases.

Case 1: Suppose that $\lambda _{1}=\cdots =\lambda _{n-1}=0$. Obviously, by the hypothesis we have $\lambda u^{n}\in {\mathcal {A}}^{d}$. Thus, in view of Lemma 2.4(ii) we get $u\in {\mathcal {A}}^{d}$.

Case 2: Suppose that $\lambda _{i}$ is the first non-zero in $\lambda _{1}$, $\ldots$, $\lambda _{n-1}$. Now, we assume that u is not generalized Drazin invertible, i.e. $0\in$ acc $(\sigma (u))$. Then, there is a sequence $t_{m}\in \sigma (u)\backslash \{0\}$ such that $\lim \limits _{m\rightarrow \infty }t_{m}=0$. Therefore, for any $m\in {\mathbb {N}}$, we can suppose that $0\ne |t_{m}|<$min$\{\frac{|\lambda _{i}|}{|\lambda |+|\lambda _{n-1}|+\cdots +|\lambda _{i+1}|},\frac{1}{2}\}$, which yields

$$\begin{aligned} \left| \frac{\lambda }{\lambda _{i}}t_{m}^{n-i}+\frac{\lambda _{n-1}}{\lambda _{i}}t_{m}^{n-1-i}+\cdots +\frac{\lambda _{i+1}}{\lambda _{i}}t_{m}\right|< \frac{\left| \lambda \right| +|\lambda _{n-1}|+\cdots +|\lambda _{i+1}|}{|\lambda _{i}|}\left| t_{m}\right| <1. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\lambda }{\lambda _{i}}t_{m}^{n-i}+\frac{\lambda _{n-1}}{\lambda _{i}}t_{m}^{n-1-i}+\cdots +\frac{\lambda _{i+1}}{\lambda _{i}}t_{m}\ne -1, \end{aligned}$$

i.e.

$$\begin{aligned} g(t_{m})=\lambda t_{m}^{n}+\lambda _{n-1}t_{m}^{n-1}+\cdots +\lambda _{i}t_{m}^{i}\ne 0. \end{aligned}$$

Note that $g(t_{m}) \in \sigma (g(u))$ and $\lim \limits _{m\rightarrow \infty }g(t_{m})=0$. Hence, we obtain that $0\in$ acc $(\sigma (g(u))$, which contradicts with $g(u)\in {\mathcal {A}}^{d}$, as required. $\square$

Applying Theorem 4.1, we get the following results.

Corollary 4.2

Let $a,b\in {\mathcal {A}}$ and $0\ne f(x), p(x), q(x)\in {\mathbb {C}}[x]$. The following hold.

• (i) If $f(x)=p(x)q(x)$, then $f(a)\in {\mathcal {A}}^{d}$ if and only if $p(a), q(a)\in {\mathcal {A}}^{d}$.

• (ii) $f(ab) \in {\mathcal {A}}^{d}$ if and only if $f(ba) \in {\mathcal {A}}^{d}$.

• (iii) $f(1-ab) \in {\mathcal {A}}^{d}$ if and only if $f(1-ba) \in {\mathcal {A}}^{d}$.

Proof

 

• (i) Let $p(x)=\lambda (x-\xi _{1})(x-\xi _{2})\cdots (x-\xi _{n})$ and $q(x)=\mu (x-\eta _{1})(x-\eta _{2})\cdots (x-\eta _{m})$, where $n,m\in {\mathbb {N}}$ and $\lambda ,\mu ,\xi _{1},\ldots ,\xi _{n},\eta _{1},\ldots ,\eta _{m}\in {\mathbb {C}}$. Then,

  $$\begin{aligned} f(x)=\lambda \mu (x-\xi _{1})(x-\xi _{2})\cdots (x-\xi _{n})(x-\eta _{1})(x-\eta _{2})\cdots (x-\eta _{m}). \end{aligned}$$

  Since $f(a)\in {\mathcal {A}}^{d}$, then using Theorem 4.1 we have $a-\xi _{i}\in {\mathcal {A}}^{d}$ and $a-\eta _{j}\in {\mathcal {A}}^{d}$, where $i=\overline{1,n}$ and $j=\overline{1,m}$. Hence, $p(a), q(a)\in {\mathcal {A}}^{d}$. The converse is trivial.

• (ii) Let $f(x)=\lambda (x-\varepsilon _{1})(x-\varepsilon _{2})\cdots (x-\varepsilon _{n})$, where $n\in {\mathbb {N}}$ and $\lambda ,\varepsilon _{1},\ldots ,\varepsilon _{n}\in {\mathbb {C}}$. Suppose that $f(ab) \in {\mathcal {A}}^{d}$. By Theorem 4.1 it follows that $ab-\varepsilon _{i}\in {\mathcal {A}}^{d}$, for $i=\overline{1,n}$. In terms of Lemmas 2.1 and 2.2, we obtain $ba-\varepsilon _{i}\cdot 1\in {\mathcal {A}}^{d}$, for $i=\overline{1,n}$. Therefore, $f(ba) \in {\mathcal {A}}^{d}$. On the contrary, it also holds by the symmetry of a and b.

• (iii) This can be proved in the same way as shown in the proof of item (ii). $\square$

In what follows, for a given polynomial $f(x)\in {\mathbb {C}}[x]$, we discuss the relation between the g-Drazin invertibility of f(a) and that of $f(a^{d})$.

Theorem 4.3

Let $a\in {\mathcal {A}}$, $n\in {\mathbb {N}}$ and $f(x)=x^{n}+x^{n-1}+\cdots +x \in {\mathbb {C}}[x]$.

• (i) If $f(a)\in {\mathcal {A}}^{d}$, then $a\in {\mathcal {A}}^{d}$ and $f(a^{d})\in {\mathcal {A}}^{d}$.

• (ii) Let $h(x)=(1-x^{n})f(x)$. If $h(a)\in {\mathcal {A}}^{d}$, then $a\in {\mathcal {A}}^{d}$ and $h(a^{d})\in {\mathcal {A}}^{d}$.

• (iii) Let $g(x)=1+f(x)$. If $a\in {\mathcal {A}}^{d}$ and $g(a)\in {\mathcal {A}}^{d}$, then $g(a^{d})\in {\mathcal {A}}^{d}$.

Proof

 

• (i) Since $f(a)=a(a^{n-1}+a^{n-2}+\cdots +1)\in {\mathcal {A}}^{d}$, using Corollary 4.2(i) we obtain $a\in {\mathcal {A}}^{d}$ and $a^{n-1}+a^{n-2}+\cdots +1\in {\mathcal {A}}^{d}$, which by Lemma 2.5 lead to

  $$\begin{aligned} (a^{d})^{n-1}+(a^{d})^{n-2}+\cdots +a^{d}+a^{d}a=(a^{d})^{n-1}(a^{n-1}+a^{n-2}+\cdots +1)\in {\mathcal {A}}^{d}. \end{aligned}$$

  Thus, in view of Lemma 2.6 we deduce

  $$\begin{aligned}{} & {} (a^{d})^{n-1}+(a^{d})^{n-2}+\cdots +1\\{} & {} \quad =(1-aa^{d})+\left( (a^{d})^{n-1}+(a^{d})^{n-2}+\cdots +a^{d}+a^{d}a\right) \in {\mathcal {A}}^{d}. \end{aligned}$$

  So

  $$\begin{aligned} f(a^{d})=a^{d}\left( (a^{d})^{n-1}+(a^{d})^{n-2}+\cdots +1\right) \in {\mathcal {A}}^{d}. \end{aligned}$$

• (ii) By the hypothesis $h(a)=(1-a^{n})f(a)\in {\mathcal {A}}^{d}$, we get $1-a^{n}\in {\mathcal {A}}^{d}$ and $f(a)\in {\mathcal {A}}^{d}$. Applying item (i), we deduce that $a\in {\mathcal {A}}^{d}$ and $f(a^{d})\in {\mathcal {A}}^{d}$. Note that

  $$\begin{aligned} 1-\left( a^{d}\right) ^{n}=\left( 1-aa^{d}\right) -\left( a^{d}\right) ^{n}\left( 1-a^{n}\right) \in {\mathcal {A}}^{d}. \end{aligned}$$

  Thus, we claim that $h(a^{d})=\left( 1-(a^{d})^{n}\right) f(a^{d})\in {\mathcal {A}}^{d}$.

• (iii) can be obtained by the proof of item (i).

$\square$

Let $f(x)=x$ in Theorem 4.3, we directly get

Corollary 4.4

Let $a\in {\mathcal {A}}$ and $n\in {\mathbb {N}}$.

• (i) If $a-a^{n+1}\in {\mathcal {A}}^{d}$, then $a\in {\mathcal {A}}^{d}$ and $a^{d}-(a^{d})^{n+1}\in {\mathcal {A}}^{d}$.

• (ii) If $1+a\in {\mathcal {A}}^{d}$ and $a\in {\mathcal {A}}^{d}$, then $1+a^{d}\in {\mathcal {A}}^{d}$.

Next, for a general polynomial $f(x)\in {\mathbb {C}}[x]$, the g-Drazin invertibility of f(a), f(b) and $f(a+b)$ are investigated under certain conditions.

Theorem 4.5

Let $a,b\in {\mathcal {A}}^{d}$, $k,n\in {\mathbb {N}}$ and $f(x)=\lambda _{n}x^{n}+\lambda _{n-1}x^{n-1}+\cdots +\lambda _{0}\in {\mathbb {C}}[x]$. If

$$\begin{aligned} ab\prod \limits _{i=1}^{k}(a^{p_{i}}b^{q_{i}})=0, \end{aligned}$$

for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$ and $f(a)\in {\mathcal {A}}^{d}$, then

$$\begin{aligned} f(b)\in {\mathcal {A}}^{d}\ \Longleftrightarrow \ f(a+b) \in {\mathcal {A}}^{d}. \end{aligned}$$

Proof

“$\Rightarrow$”. Suppose that $f(a)\in {\mathcal {A}}^{d}$. Let $g(x)=xf(x)$. Then, we have

$$\begin{aligned} g(a+b)=\left( \sum \limits _{i}s_{i}abt_{i}+g(a)\right) + \left( \sum \limits _{j}u_{j}bav_{j}+g(b)\right) , \end{aligned}$$

for some suitable $s_{i}, t_{i}, u_{j}, v_{j}\in {\mathcal {A}}$. So, Setting $x_{1}=\sum \limits _{i}s_{i}abt_{i}$, $x_{2}=g(a)$, $y_{1}=\sum \limits _{j}u_{j}bav_{j}$ and $y_{2}=g(b)$. By the hypotheses, we get $x_{1}\in {\mathcal {A}}^{nil}$, $x_{2}=af(a)\in {\mathcal {A}}^{d}$ and $x_{1}x_{2}\prod \nolimits _{i=1}^{k}(x_{1}^{p_{i}}x_{2}^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$. Applying Theorem 3.5, we have $x_{1}+x_{2}\in {\mathcal {A}}^{d}$. Similarly, $y_{1}+y_{2}\in {\mathcal {A}}^{d}$. It is clear that $(x_{1}+x_{2})(y_{1}+y_{2})\prod \nolimits _{i=1}^{k}((x_{1}+x_{2})^{p_{i}}(y_{1}+y_{2})^{q_{i}})=0$, for every $(p_{1},q_{1},p_{2},q_{2},\ldots ,p_{k},q_{k})\in U_{k}$. So, $g(a+b) \in {\mathcal {A}}^{d}$, which implies $f(a+b) \in {\mathcal {A}}^{d}$.

“$\Leftarrow$”. Following the same strategy as in the proof of the necessity and using Theorem 3.5, Corollary 4.2(i), we can conclude that the sufficiency holds. $\square$