Generalized Drazin spectrum of upper triangular matrices in Banach algebras

Abstract

Let \mathcal{A} be a Banach algebra with unit e and a,b,c\in \mathcal{A},M_c=\left(\begin{array}{cc}a\hfill & c\hfill \\ 0\hfill & b\hfill \end{array}\right)\in M_2\left(\mathcal{A}\right). The concepts of left and right generalized Drazin invertible of elements in a Banach algebra are proposed. A generalized Drazin spectrum of \alpha is defined by {\sigma }_{\mathrm{g}\mathrm{D}}\left(\alpha \right)=\{\lambda \in \mathbb{C}:\alpha -\lambda e\mathrm{i}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{d}\mathrm{D}\mathrm{r}\mathrm{a}\mathrm{z}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{b}\mathrm{l}\mathrm{e}\}. It is shown that

　　　　　　　　　　 {\sigma }_{\mathrm{g}\mathrm{D}}\left(a\right)\cup {\sigma }_{\mathrm{g}\mathrm{D}}\left(b\right)={\sigma }_{\mathrm{g}\mathrm{D}}\left(M_c\right)\cup W_2,

where W_g is a union of certain holes {\sigma }_{\mathrm{g}\mathrm{D}} and W_g\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}\left(a\right)\cap {\sigma }_{\mathrm{g}\mathrm{D}}\left(b\right), or more finely W_g\subseteq {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}\left(a\right)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}\left(b\right). In addition, some properties of generalized Drazin spectrum of elements in a Banach algebra are studied.

Keywords

Banach algebra / generalized Drazin inverse / generalized Drazin spectrum / upper / triangular matrices

1 Introduction

In recent years, the spectral theory of operator matrices, especially the triangular operator matrix on 2×2, has become an international research hot spot. Among them, Han et al. [6] discussed the spectral hole-filling interval of triangular operator matrices on 2×2 in Banach space. Zhang Shifang et al. [11] discussed the Drazin spectrum of triangular operator matrices on 2×2 in Banach space. Zhang Haiyan et al. [10] discussed the correlation problem of Drazin spectrum of triangular operator matrix on 2×2 in Hilbert space. Zhang Shifang et al. [12] discussed the correlation problem of the generalized Drazin spectrum of triangular operator matrices on 2×2 in Banach space. Lin [8] discussed the interproblem of spectral hole filling of triangular matrices on 2×2 in Banach algebra. More interproblems of triangular matrix spectra can be found in [1, 5]. Inspired by the above literature, this paper discusses the correlation between the generalized Drazin spectrum of trigonometric matrices on 2×2 in Banach algebra. The Drzain inverse is one of the main concepts of generalized inverse theory which has important applications in disciplines such as cryptography, singular differential, and multi-body dynamics.

Based on the concept of generalized Drazin inverse of elements, this paper presents the left generalized Drazin inverse and the right generalized Drazin inverse. The properties and hole-filling interval of the upper triangular matrix g-Drazin spectrum in M_2(\mathcal{A}) are discussed, and the other properties of the upper triangular matrix g-Drazin spectrum are also studied.

2 Preparatory knowledge

Let \mathcal{A} be a Banach algebra with unit element e over a complex domain \mathcal{C}, a\in \mathcal{A}. \sigma (a) and r(a) denote the spectrum and the spectral radius of a, respectively. r(a)=\underset{\lambda \in \sigma (a)}{sup}|\lambda |, {\sigma }_{\ell }(a)\mathrm{a}\mathrm{n}\mathrm{d}{\sigma }_{\tau }(a) denote the left and right spectra of a, respectively. If there exists k\in \mathbb{N} such that a^k=0, then a is said to be a nilpotent element. The smallest natural number k satisfying the above condition is called the nilpotency index of a. If \sigma (a)=\{0\}, then a is called the quasi-nilpotent. Let p\in \mathcal{A}. If p^2=p, then p is said to be a nilpotent element. QN(\mathcal{A}) denotes the set of all proposed nilpotent elements in \mathcal{A}. Let M be a nonempty subset of \mathcal{C}. Then \mathrm{a}\mathrm{c}\mathrm{c}(M),\mathrm{i}\mathrm{n}\mathrm{t}(M),\mathrm{i}\mathrm{s}\mathrm{o}(M), and \mathrm{\partial }M represent the set of M consisted by all clusters, interior points, isolated points and boundary points, respectively.

Let the parametrization in the Banach space \mathcal{A}\times \mathcal{A} be \Vert (a_1,a_2)\Vert =max(\Vert a_1\Vert ,\Vert a_2\Vert ), (a_1,a_2)\in \mathcal{A}\times \mathcal{A},B(\mathcal{A}\times \mathcal{A}) is the set consisting of all bounded linear operators on \mathcal{A}\times \mathcal{A}. Let M_2(\mathcal{A})=\{\left(\begin{array}{cc}a& c\\ d& b\end{array}\right):a,b,c,d\in \mathcal{A}\}. M_2(\mathcal{A}) can be considered as a subspace of B(\mathcal{A}\times \mathcal{A}). Then M_2(\mathcal{A}) is a Banach algebra with unit element \left(\begin{array}{ll}e& 0\\ 0& 0\end{array}\right).

Let a\in \mathcal{A}. a is left invertible if there exists b\in \mathcal{A} such that ba=e. a is right invertible if there exists c\in \mathcal{A} such that ca=e. If a is both left and right reversible, then a is reversible. Denote Inv(\mathcal{A}) as the set of all reversible elements in \mathcal{A}, then L-Inv(\mathcal{A}) represents the set of all left-reversible elements and R-lnv(\mathcal{A}) represents the set of all right-reversible elements in \mathcal{A}, respectively.

Drazin [4] introduces the concept of Drazin inverse. Let a\in \mathcal{A}. If there exists b\in \mathcal{A} such that

ab=ba,b=a{b}^2,a^{k+1}b=a^k,

then b is said to be the Drazin inverse of a, where k\in \mathbb{N}. The smallest natural number k satisfying the above condition is called the Drazin indicator of a, denoted as i(a). If a is Drazin invertible, then the Drazin inverse of a is unique, denoted as a^D.

Koliha [7] generalizes the concept of Drazin inverse. Let a\in \mathcal{A}. If there exists b\in \mathcal{A} such that

ab=ba,b=a{b}^2,a-a^{2}b\in \mathrm{Q}\mathrm{N}(\mathcal{A}),

then a is said to be generalized Drazin invertible, or g-Drazin invertible, and b is the generalized Drazin inverse of a. If a is generalized Drazin invertible, then the generalized Drazin inverse of a is unique.

Let a\in \mathcal{A}. Then the set {\sigma }_{\mathrm{g}\mathrm{D}}(a)=\{\lambda \in \mathbb{C}:a-\lambda e\text{is not generalized Drazin}\text{invertible}\} is called the generalized Drazin spectrum of a. The set {\rho }_{\mathrm{g}\mathrm{D}}(a)=\mathbb{C}\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(a) is called the generalized Drazin pre-solutions of a. Thus,

\sigma (a)={\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)).

By [9], if a\in \mathcal{A}, then

r(a)=\underset{n⩾1}{inf}\Vert a^n{\Vert }^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}\Vert a^n{\Vert }^{\frac{1}{n}}.

Let a,b,c\in \mathcal{A}, M_c=\left(\begin{array}{ll}a& c\\ 0& b\end{array}\right). We focus on the correlation interval of the generalized Drazin spectrum of M_c in this paper.

3 Main results and proofs

Lemma 3.1 [7] Let \mathcal{A} be a Banach algebra with unit element e, a\in \mathcal{A}. Then the following conditions are equivalent:

(1) 0\notin \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a));

(2) There exists a curtain equivalent element p\in \mathcal{A} exchangeable with a such that ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}),p+a\in \mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A});

(3) a is g-Drazin invertible.

Lemma 3.2 [8]　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If a and b are invertible, then M_c is invertible and M_c^{-1}=\left(\begin{array}{cc}{a}^{-1}& -a^{-1}c{b}^{-1}\\ 0& b^{-1}\end{array}\right).

Lemma 3.3　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If a and b are g-Drazin invertible, then M_c is g-Drazin invertible.

Proof　From Lemma 3.1 and the fact that a and b are both g-Drazin invertible, it follows that \mathrm{\exists }\epsilon >0, when , a-\lambda e is invertible and b-\lambda e is also invertible. Since then, when , there is \lambda \notin \sigma (M_c), and 0\notin \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (M_c)). By Lemma 3.1, we know that M_c is g-Drazin invertible.

Lemma 3.4 [3]　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If M_c and a are g-Drazin invertible, then b is g-Drazin invertible.

Lemma 3.4 is equivalent to {\sigma }_{\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(a). Similarly, it follows that {\sigma }_{\mathrm{g}\mathrm{D}}(a)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Lemma 3.5 [8]　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If M_c and a are invertible, then b is invertible.

Lemma 3.5 is equivalent to \sigma (b)\subseteq \sigma (M_c)\cup \sigma (a). Similarly, it follows that \sigma (a)\subseteq \sigma (M_c)\cup \sigma (b).

Lemma 3.6　 Let \mathcal{A} be a Banach algebra with unit element e, a\in \mathcal{A}. If a is g-Drazin invertible, then there exists \delta >0 such that for any \lambda \in \mathbb{C} satisfying , there is a-\lambda e invertible.

Proof　Since a is g-Drazin invertible, 0\notin \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a)), and 0\in \rho (a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)). If 0\in \rho (a), then 0\notin \sigma (a). Thus, a is invertible. Let {\delta }_1=\frac{1}{\Vert a^{-1}\Vert }. When , there is , then and e-\lambda a^{-1} is invertible. Since a is invertible, then a(e-\lambda a^{-1}) is invertible, i.e., a-\lambda e is invertible. If 0\in \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)), there exists {\delta }_2>0 such that U(0,{\delta }_2)\cap \sigma (a)=\mathrm{\varnothing }. If \lambda \in \mathbb{C} and satisfying , there is \lambda \in U(0,{\delta }_2), then \lambda \notin \sigma (a). Since then, \lambda \in \rho (a). Thus, a-\lambda ϵ is invertible.

Take \delta =min\{{\delta }_1,{\delta }_2\}. Then for any \lambda \in \mathbb{C} satisfying , a-\lambda e is invertible.

Definition 3.1　Let \mathcal{A} be a Banach algebra with unit element e, a,p\in \mathcal{A} and p is a curtain equal element exchangeable with a. If a+p\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) and ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}), then a is said to be left generalized Drazin invertible. If a+p\in \mathrm{R}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) and ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}), then a is said to be right generalized Drazin invertible.

The left generalized Drazin spectrum and the right generalized Drazin spectrum of a are defined as follows:

\begin{array}{l}{\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)=\{\lambda \in \mathbb{C}:a-\lambda e\text{is not left generalized invertible}\},\\ {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)=\{\lambda \in \mathbb{C}:a-\lambda e\text{is not right generalized invertible}\}.\end{array}

From Definition 3.1, it follows that

(3.1){\sigma }_{\mathrm{g}\mathrm{D}}(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\subseteq \sigma (a).

The following proves the equation part of this conclusion (3.1).

Proof of Eq. (3.1)　If a is generalized Drazin invertible, then there exist curtain equivalences p exchangeable with a such that a+p\in \mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}),ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}). Since a+p\in \mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}), there is a+p\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) and a+p\in \mathrm{R}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}). By Definition 3.1, a is left generalized Drazin invertible and right generalized Drazin invertible, hence {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a).

Conversely, let a be left g-Drazin invertible. Then there exist curtain equivalents p exchangeable with a such that a+p\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) and ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}). Let \lambda \in \mathbb{C}. Then \lambda e-a=(\lambda e-ap)p+(\lambda e-(p+a))(e-p). So there exists r>0 so that \lambda e-(p+a)\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) when . From ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}), \lambda e-ap\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}), we have \lambda e-a\in \mathrm{L}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}). As p\ne 0, 0\in \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)), then 0\notin \mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_l(a)). Similarly, a is right g-Drazin invertible, then 0\notin \mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_r(a)). From \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a))=\mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_l(a)\cup {\sigma }_r(a))=\mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_l(a))\cup \mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_r(a)), we have 0\notin \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a)). By Lemma 3.1, a is generalized Drazin invertible. Therefore, {\sigma }_{\mathrm{g}\mathrm{D}}(a)\subseteq {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a). Then {\sigma }_{\mathrm{g}\mathrm{D}}(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a).

Theorem 3.1　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If M_c is g-Drazin invertible, then a is left g-Drazin invertible and b is right g-Drazin invertible.

Proof　Since M_c is g-Drazin invertible, by Lemma 3.1, it is known that there exist curtain equivalence elements E=\left(\begin{array}{ll}p& r\\ 0& q\end{array}\right), where p,q,r\in \mathcal{A} such that M_{c}E=E{M}_c,M_{c}E\in \mathrm{Q}\mathrm{N}(\mathcal{A}),M_c+E\in \mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}). As \left(\begin{array}{cc}{p}^2& pr+rq\\ 0& q^2\end{array}\right)=E^2=E=\left(\begin{array}{ll}p& r\\ 0& q\end{array}\right), p^2=p,q^2=q, so p and q are curtain equivalents in \mathcal{A}. As \left(\begin{array}{cc}ap& ar+cq\\ 0& bq\end{array}\right)=M_{c}E=E{M}_c=\left(\begin{array}{cc}pa& pc+rb\\ 0& qb\end{array}\right), ap=pa,bq=qb. By \left(\begin{array}{cc}a-p& c+r\\ 0& b+q\end{array}\right)=M_c+E\in \mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}) and M_c+E=\left(\begin{array}{cc}a+p& c+r\\ 0& b+q\end{array}\right)=\left(\begin{array}{cc}e& 0\\ 0& b+q\end{array}\right)\left(\begin{array}{cc}c& c+r\\ 0& c\end{array}\right)\left(\begin{array}{cc}a+p& 0\\ 0& e\end{array}\right), we have a+p\in \mathrm{L}-\mathrm{l}\mathrm{n}\mathrm{v}(\mathcal{A}),b+q\in \mathrm{R}-\mathrm{I}\mathrm{n}\mathrm{v}(\mathcal{A}).

Since M_{c}E\in \mathrm{Q}\mathrm{N}(\mathcal{A}), \sigma (M_{c}E)=\{0\}. By \left(\begin{array}{cc}ap& ar+cq\\ 0& bq\end{array}\right)=M_{c}E and Lemma 3.5, there is \sigma (ap)\subseteq \sigma (M_{c}E)\cup \sigma (bq)\mathrm{a}\mathrm{n}\mathrm{d}\sigma (bq)\subseteq \sigma (M_{c}E)\cup \sigma (ap). Then \sigma (ap)\subseteq \{0\}\cup \sigma (bq)\mathrm{a}\mathrm{n}\mathrm{d}\sigma (bq)\subseteq \{0\}\cup \sigma (ap). Finally, we have \sigma (ap)=\{0\}=\sigma (bq), or \sigma (ap)=\sigma (bq).

The following proof is for when \sigma (ap)=\sigma (bq), \sigma (ap)=\{0\}=\sigma (bq).

By \sigma (M_{c}E)\subseteq \sigma (ap)\cup \sigma (bq), \{0\}\subseteq \sigma (ap)\cup \sigma (bq). Since \sigma (M_{\mathrm{c}}E)=\{0\}, there is r(M_{c}E)=\underset{\lambda \in \sigma (M_{c}E)}{sup}|\lambda |=0. As \Vert (a_1,a_2)\Vert =max\{\Vert a_1\Vert ,\Vert a_2\Vert \},(a_1,a_2)\in \mathcal{A}\times \mathcal{A}, we have

\begin{array}{rl}\Vert M_{c}E\Vert & =sup\left\{\Vert \left(\begin{array}{cc}ap& ar+cq\\ 0& bq\end{array}\right)\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\Vert :\Vert x\Vert =1,x=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\right\}\\ & =sup\left\{\Vert \left(\begin{array}{c}ap{x}_1+(ar+cq)y_1\\ bq{y}_1\end{array}\right)\Vert :\Vert x\Vert =1,x=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\right\}.\end{array}

Since then,

\begin{array}{rl}{M}_{\mathrm{c}}E\Vert & ⩾sup\left\{\Vert \left(\begin{array}{c}ap{x}_1+(ar+cq)y_1\\ bq{y}_1\end{array}\right)\Vert :x=\left(\begin{array}{c}0\\ f\end{array}\right),\Vert f\Vert =1\right\}\\ & =sup\left\{\Vert \left(\begin{array}{c}(ar+cq)f\\ bqf\end{array}\right)\Vert :\Vert f\Vert =1\right\}\\ & ⩾max\{\Vert ar+cq\Vert ,\Vert bq\Vert \},\end{array}

\begin{array}{rl}\Vert M_{c}E\Vert & ⩾sup\left\{\Vert \left(\begin{array}{c}ap{x}_1+(ar+cq)y_1\\ bq{y}_1\end{array}\right)\Vert :x=\left(\begin{array}{c}h\\ 0\end{array}\right),\Vert h\Vert =1\right\}\\ & =sup\left\{\Vert \left(\begin{array}{c}aph\\ 0\end{array}\right)\Vert :\Vert h\Vert =1\right\}\\ & ⩾\Vert ap\Vert .\end{array}

Then \Vert M_{c}E\Vert ⩾max\{\Vert ap\Vert ,\Vert bq\Vert ,\Vert ar+cq\Vert \}.

(M_{c}E{)}^n={\left(\begin{array}{cc}ap& (ar+cq)\\ 0& bq\end{array}\right)}^n=\left(\begin{array}{cc}(ap{)}^n& \sum _{k=1}^n(ap{)}^{n-k}(ar+cq)(bq{)}^{k-1}\\ 0& (bq{)}^n\end{array}\right).

Let \begin{array}{l}{S}_n=\sum _{k=1}^n(ap{)}^{n-k}(ar+cq)(bq{)}^{k-1}\end{array}. Then

\begin{array}{rl}\Vert (M_{c}E{)}^n\Vert & =sup\left\{\Vert \left(\begin{array}{cc}(ap{)}^n& S_n\\ 0& (bq{)}^n\end{array}\right)\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\Vert :\Vert x\Vert =1,x=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\right\}\\ & =sup\left\{\Vert \left(\begin{array}{c}(ap{)}^n{x}_1+S_n{y}_1\\ (bq{)}^n{y}_1\end{array}\right)\Vert :\Vert x\Vert =1,x=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)\right\}\\ & ⩾max\{\Vert (ap{)}^n\Vert ,\Vert (bq{)}^n\Vert ,\Vert S_n\Vert \}.\end{array}

So \Vert (M_{c}E{)}^n{\Vert }^{\frac{1}{n}}⩾max\{\Vert (ap{)}^n{\Vert }^{\frac{1}{n}},\Vert (bq{)}^n{\Vert }^{\frac{1}{n}},\Vert S_n{\Vert }^{\frac{1}{n}}\}.

As \Vert (ap{)}^n{\Vert }^{\frac{1}{n}}⩾0\mathrm{a}\mathrm{n}\mathrm{d}\underset{n\to \mathrm{\infty }}{lim}\Vert (M_{c}E{)}^n{\Vert }^{\frac{1}{n}}=r(M_{c}E)=0, there is r(ap)=\underset{n\to \mathrm{\infty }}{lim}\Vert (ap{)}^n{\Vert }^{\frac{1}{n}}=0, then \sigma (ap)=0. Since \sigma (ap)=\sigma (bq), then \sigma (ap)=\sigma (bq)=\{0\}. Therefore, ap\in \mathrm{Q}\mathrm{N}(\mathcal{A}),bq\in \mathrm{Q}\mathrm{N}(\mathcal{A}).

In short, a is left g-Drazin invertible and b is right g-Drazin invertible.

From the above theorem, we have {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Definition 3.2　Let \mathcal{A} be a Banach algebra with unit element e, a\in \mathcal{A}.

(1) If there exists \{b_n\}\subseteq \mathcal{A}\mathrm{a}\mathrm{n}\mathrm{d}\Vert b_n\Vert =1,\mathrm{\forall }n\in \mathbb{N}\text{such that}\underset{n\to \mathrm{\infty }}{lim}(a{b}_n)=0, then a is called a generalized left zero divisor.

(2) If there exists \{b_n\}\subseteq \mathcal{A}\mathrm{a}\mathrm{n}\mathrm{d}\Vert b_n\Vert =1,\mathrm{\forall }n\in \mathbb{N}\text{such that}\underset{n\to \mathrm{\infty }}{lim}(b_{n}a)=0, then a is called a generalized left zero divisor.

(3) a is called a generalized zero divisor if a is both a generalized left zero divisor and a generalized right zero divisor. If a is both a generalized left zero divisor and a generalized right zero divisor, then a is neither left invertible nor right invertible.

Lemma 3.7 [2]　 Let \mathcal{A} be a Banach algebra with unit element e,a\in \mathcal{A}. If \lambda \in \mathrm{\partial }\sigma (a), then a-\lambda c is the generalized zero divisor and hence \lambda \in {\sigma }_l(a)\cap {\sigma }_r(a).

Lemma 3.8　 Let \mathcal{A} be a Banach algebra with unit element e,a\in \mathcal{A}, then \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)⟩=\mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)).

Proof　As \sigma (a)={\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)), {\sigma }_l(a)\cup {\sigma }_r(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)) = {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)\cup {\sigma }_r(a))\subseteq {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)). a is left g-Drazin invertible, 0\notin \mathrm{a}\mathrm{c}\mathrm{c}({\sigma }_l(a)), then 0\in {\rho }_l(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)). Hence, {\rho }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\subseteq {\rho }_l(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)). Similarly, {\sigma }_r(a)\subseteq {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)). Conversely, let \lambda \notin {\sigma }_l(a), as {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\subseteq {\sigma }_l(a), then \lambda \notin {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a). As \lambda \notin {\sigma }_l(a), \lambda \notin \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_{\ell }(a)), then {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\subseteq {\sigma }_l(a). Similarly, we can get {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a))\subseteq {\sigma }_r(a). Therefore, {\sigma }_l(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)),{\sigma }_r(a)={\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)).

Since {\sigma }_{\mathrm{g}\mathrm{D}}(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a) and \sigma (a)={\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)), \sigma (a)={\sigma }_l(a)\cup {\sigma }_r(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a))={\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup (\mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a))). Since {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))=\mathrm{\varnothing }, \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))=\mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_i(a))\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_{\tau }(a)).

Lemma 3.9　 Let \mathcal{A} be a Banach algebra with unit element e,a\in \mathcal{A}. If \lambda \in \mathrm{\partial }{\sigma }_{\mathrm{g}\mathrm{D}}(a), then \lambda \in {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a).

Proof　From \lambda \in \mathrm{\partial }{\sigma }_{\mathrm{g}\mathrm{D}}(a)/\text{and}\mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))=\mathrm{\partial }(\mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a)))\subseteq \mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a))\setminus \mathrm{i}\mathrm{n}\mathrm{t}(\sigma (a))\subseteq \mathrm{\partial }\sigma (a), we have \lambda \in \mathrm{\partial }\sigma (a) and \lambda \notin \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)). By Lemma 3.7, \lambda \in {\sigma }_l(a)\cap {\sigma }_r(a). Then \lambda \in {\sigma }_l(a)\mathrm{a}\mathrm{n}\mathrm{d}\lambda \in {\sigma }_{\mathrm{r}}(\mathrm{a}). The proof of Lemma 3.8 shows that {\sigma }_l(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)),{\sigma }_r(a)={\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)), then \lambda \in {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a))\mathrm{a}\mathrm{n}\mathrm{d}\lambda \in {\sigma }_{\mathrm{t}\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)). Due to the fact that \lambda \notin \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)) and by Lemma 3.8, there is \lambda \notin \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_l(a)) and \lambda \notin \mathrm{i}\mathrm{s}\mathrm{o}({\sigma }_r(a)), then \lambda \in {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\text{and}\lambda \in {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a). Therefore, \lambda \in {\sigma }_{\mathrm{I}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a).

Theorem 3.2　 Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}, M_c=\left(\begin{array}{cc}a& c\\ 0& b\end{array}\right)\in M_2(\mathcal{A}), then

(3.2){\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\cup W_g,

where W_g is a union of certain holes of {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), and W_g\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Proof　From Lemma 3.3, {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b).

\mathrm{\forall }\lambda \in ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b)), then \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(a)\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(b) or \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(b)\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(a). If \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(a)\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(b), then \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(a) and \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(b). Therefore, b-\lambda e is generalized Drazin invertible while a-\lambda e is not. From Lemma 3.5, M_c-\lambda e is not generalized Drazin invertible, then \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). Similarly, if \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(b)\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(a), then \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Therefore, \mathrm{\forall }\lambda \in ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b)) implies \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), i.e.,

({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

From Lemma 3.2, we have

({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Next, we give the proof of

\eta ({\sigma }_{\mathrm{g}\mathrm{D}}(M_c))=\eta ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)),

where \eta (S) is a polynomial convex package of tight subsets S in \mathbb{C}.

From {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b) and the maximum mode principle, we only need to prove \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(M_c)). Since

\mathrm{i}\mathrm{n}\mathrm{t}({\sigma }_{\mathrm{g}\mathrm{D}}(M_c))\subseteq \mathrm{i}\mathrm{n}\mathrm{t}({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)),

it is sufficient to prove \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). Now the following equation can be proved:

\mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))\cup \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

The first conclusion obviously holds and the third conclusion follows from Theorem 3.1. Now prove the second conclusion.

\mathrm{\forall }\lambda \in \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))\cup \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(b)), two cases are discussed in the following.

(1) When \lambda \in \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)), from Lemma 3.9 and Theorem 3.1, we can get

\mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))\subseteq {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Therefore, \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

(2) When \lambda \in \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(b)), similarly, \lambda \in {\sigma }_{\mathrm{g}\mathrm{D}}(M_c) can be obtained.

Therefore, \lambda \in \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))\cup \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). Then \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a))\cup \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). Thus, \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_{\mathrm{c}}).

From \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq \mathrm{\partial }({\sigma }_{\mathrm{g}\mathrm{D}}(M_c)) and {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b), it is clear that from {\sigma }_{\mathrm{g}\mathrm{D}}(M_c) to {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b), we need to fill a union of certain holes for {\sigma }_{\mathrm{g}\mathrm{D}}(M_{\mathrm{c}}), i.e., {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\cup W_g. It can be known from ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_{\mathrm{c}}) that W_g\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Corollary 3.1　 If \mathrm{i}\mathrm{n}\mathrm{t}({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))=\mathrm{\varnothing }, then {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_{\mathrm{c}}).

Proof　From \mathrm{i}\mathrm{n}\mathrm{t}({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))=\mathrm{\varnothing } and Theorem 3.2, we can get W_g=\mathrm{\varnothing }. Hence, {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Property 3.1　Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}, then {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b)\subseteq \sigma (a)\cap \sigma (b).

Proof　As \sigma (a)\cup \sigma (b)={\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b)), then

\begin{array}{rl}\sigma (a)& \cap \sigma (b)\\ & =({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a)))\cap ({\sigma }_{\mathrm{g}\mathrm{D}}(b)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b)))\\ & =({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\cup ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b)))\cup (\mathrm{i}\mathrm{f}(a)\cap \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b))).\end{array}

Thus, {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b)\subseteq \sigma (a)\cap \sigma (b).

Property 3.2　Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. If \sigma (a)\cup \sigma (b)=\sigma (M_c)\cup W and W\subseteq \sigma (a)\cap \sigma (b), then W\subseteq W_g, where W_g is defined by Eq. (3.2).

Proof　 \mathrm{\forall }\lambda \in (\sigma (a)\cup \sigma (b))\setminus \sigma (M_c), then \lambda \notin \sigma (M_c). Thus, \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). Now prove the following conclusion holds true.

\mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b))\subseteq \mathrm{\partial }(\sigma (a))\cup \mathrm{\partial }(\sigma (b))\subseteq {\sigma }_l(a)\cup {\sigma }_r(b)\subseteq \sigma (M_c).

From [12], it follows that \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))\cup \mathrm{\partial }(\mathrm{a}\mathrm{c}\mathrm{c}(\sigma (a)))=\mathrm{\partial }(\sigma (a)), then \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))\subseteq \mathrm{\partial }(\sigma (a)), the first inclusion relation of the above equations holds. By Lemma 3.7 and [8], the second and third inclusion relations of the above equations also hold. Therefore,

\begin{array}{rl}& (\sigma (a)\cup \sigma (b))\setminus \sigma (M_c)\\ & =[(\sigma (a)\cup \sigma (b))\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)]\setminus \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (M_c))\\ & =\{[{\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (a))\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)\cup \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (b))]\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\}\setminus \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (M_c))\\ & =[({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)]\setminus \mathrm{i}\mathrm{s}\mathrm{o}(\sigma (M_c)).\end{array}

Thus, (\sigma (a)\cup \sigma (b))\setminus \sigma (M_c)\subseteq ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), then \lambda \in ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), i.e., W\subseteq W_g.

Property 3.3　Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}, M_c=\left(\begin{array}{ll}a& c\\ 0& b\end{array}\right). Then

{\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\cup W_g,

where W_g is a union of certain holes of {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), and W_g\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Proof　First, we prove that the following conclusion holds.

({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b).

The second inclusion relation is obvious can be obtained from Lemma 3.3. Now prove the first inclusion relation. Assuming that \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(M_c), then \lambda \notin {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b). If \lambda \notin {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b), then \lambda \notin {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a) or \lambda \notin {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b). Thus, \lambda \notin {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a), \lambda \notin {\sigma }_{\mathrm{i}\mathrm{g}\mathrm{D}}(b)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b). By {\sigma }_{\mathrm{g}\mathrm{D}}(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a),{\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b), there is \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(a), \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(b). From \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(M_c) and Lemma 3.4, we can get \lambda \notin {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b).

Therefore, \lambda \notin ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b)). From Theorem 3.2, it is known that \eta ({\sigma }_{\mathrm{g}\mathrm{D}}(M_c))=\eta ({\sigma }_{\mathrm{g}\mathrm{D}}(a))\cup \eta ({\sigma }_{\mathrm{g}\mathrm{D}}(b)). Thus, W_g\subseteq {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b).

In particular, when \mathrm{i}\mathrm{n}\mathrm{t}({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))=\mathrm{\varnothing }, {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b)={\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Property 3.4　Let \mathcal{A} be a Banach algebra with unit element e,a,b,c\in \mathcal{A}. Then

({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Proof　Since {\sigma }_{\mathrm{g}\mathrm{D}}(a)={\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a), then

\begin{array}{rl}& {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b)\\ & =({\sigma }_{\mathrm{k}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{r}\mathrm{K}\mathrm{D}}(a))\cap ({\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b)\cup {\sigma }_{\mathrm{r}\mathrm{k}\mathrm{D}}(b))\\ & =({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\cup ({\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b))\cup ({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\cup ({\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(b)).\end{array}

Therefore, {\sigma }_{\mathrm{r}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b)\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b).

By Theorem 3.2, we have ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\mathrm{\setminus }({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c). By Theorem 3.3, we have ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cup {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{z}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

Therefore, ({\sigma }_{\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{g}\mathrm{D}}(b))\setminus ({\sigma }_{\mathrm{z}\mathrm{g}\mathrm{D}}(a)\cap {\sigma }_{\mathrm{l}\mathrm{g}\mathrm{D}}(b))\subseteq {\sigma }_{\mathrm{g}\mathrm{D}}(M_c).

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