The Perturbation of the Sub-Noncommutative Pseudo-Browder Essential Spectrum of Bounded Upper Triangular Operator Matrices

Abstract

Let $\epsilon >0$ and $\mathcal{TB}(\mathcal{X}\times \mathcal{X})$ be the Banach algebra of all $2\times 2$ bounded upper triangular operator matrices on a separable Hilbert space $\mathcal{X}\times \mathcal{X}$. In this paper, we first establish the spectrum equalities for special cases of upper triangular operator matrices—diagonal block operator matrix $M_0=\left({\displaystyle \genfrac{}{}{0pt}{}{A0}{0B}}\right)$. We obtain that ${\widehat{\Sigma }}_{bi,\epsilon }\left(M_0\right)={\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right)$, $i\in \{1,2,4\}$, where ${\Sigma }_{bi,\epsilon }(·)$ and ${\widehat{\Sigma }}_{bi,\epsilon }(·)$ denote the noncommutative pseudo-upper (resp. lower) semi-Browder essential spectrum, noncommutative pseudo-Browder essential spectrum, sub-noncommutative pseudo-upper (resp. lower) semi-Browder essential spectrum, and sub-noncommutative pseudo-Browder essential spectrum. Secondly, based on Cao and Bai’s works, we study the perturbation of the sub-noncommutative pseudo-Browder essential spectrum ${\widehat{\Sigma }}_{b4,\epsilon }(·)$ of a 2 × 2 bounded upper triangular operator matrix $M_C=\left({\displaystyle \genfrac{}{}{0pt}{}{AC}{0B}}\right)$ on a separable Hilbert space. We obtain that ${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta }$, where $\Delta =\{\lambda \in \mathbb{C}:$ there exist $P_i\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel P_i\parallel <\epsilon ,i\in \{1,2\},$ such that $\alpha (A+P_1-\lambda I)+\alpha (B+P_2-\lambda I)\ne \beta (A+P_1-\lambda I)+\beta (B+P_2-\lambda I)\}$. Finally, we obtain ${\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right)={\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)\cup W,i\in \{1,2,4\}$, where W is the union of certain holes in $({\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right))\backslash {\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)$.

1. Introduction

Let $\mathcal{X}$ be a separable Hilbert space, and $\mathcal{B}\left(\mathcal{X}\right)$ and $G\left(\mathcal{X}\right)$ denote the set of all bounded linear operators and invertible operators on $\mathcal{X}$. If $T\in \mathcal{B}\left(\mathcal{X}\right)$, then denote $T^{\ast }$, $D\left(T\right)$, $N\left(T\right)$ and $R\left(T\right)$ as the adjoint operator, domain, kernel and range of T. The nullity of T is defined as the dimension of $N\left(T\right)$, denoted by $\alpha \left(T\right)$. The deficiency of T is defined as the codimension of $R\left(T\right)$, denoted by $\beta \left(T\right)$. The ascent and descent of T, denoted by $asc\left(T\right)$ and $dsc\left(T\right)$, are defined as follows

$$asc\left(T\right)=inf\{n\ge 0:N\left(T^n\right)=N\left(T^{n+1}\right)\};dsc\left(T\right)=inf\{n\ge 0:R\left(T^n\right)=R\left(T^{n+1}\right)\},$$

where n is a nonnegative integer. In particular, the infimum over the empty set is taken to be ∞. If $R\left(T\right)$ is closed and $\alpha \left(T\right)<\infty$ (resp. $\beta \left(T\right)<\infty$), then T is called an upper (resp. a lower) semi-Fredholm operator. T is called a semi-Fredholm operator if it is an upper or lower semi-Fredholm operator, and it is called a Fredholm operator if it is both an upper and lower semi-Fredholm operator. Let ${\Phi }_{+}\left(\mathcal{X}\right)$, ${\Phi }_{-}\left(\mathcal{X}\right)$, ${\Phi }_{\pm }\left(\mathcal{X}\right)$ and $\Phi \left(\mathcal{X}\right)$ denote the set of all upper (resp. lower) semi-Fredholm, semi-Fredholm and Fredholm operators on $\mathcal{X}$, respectively. For $T\in {\Phi }_{\pm }\left(\mathcal{X}\right)$, $ind\left(T\right)=\alpha \left(T\right)-\beta \left(T\right)$ is called the index of T. If $T\in {\Phi }_{+}\left(\mathcal{X}\right)$ (resp. $T\in {\Phi }_{-}\left(\mathcal{X}\right)$) and $asc\left(T\right)<\infty$ (resp. $dsc(T<\infty )$), then T is called an upper (resp. a lower) semi-Browder operator. If $T\in \Phi \left(\mathcal{X}\right)$ and $asc\left(T\right)=dsc\left(T\right)<\infty$, then T is called a Browder operator. Let $B_{+}\left(\mathcal{X}\right)$, $B_{-}\left(\mathcal{X}\right)$, $B\left(\mathcal{X}\right)$ be the set of all upper (resp. lower) semi-Browder and Browder essential operators on $\mathcal{X}$. The upper (resp. lower) semi-Browder essential spectrum and Browder essential spectrum of T are defined as

$$\begin{array}{cc}\hfill & {\sigma }_{b1}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda \begin{array}{c}\hfill I\end{array}\notin B_{+}\left(\mathcal{X}\right)\};\hfill \\ \hfill & {\sigma }_{b2}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda \begin{array}{c}\hfill I\end{array}\notin B_{-}\left(\mathcal{X}\right)\};\hfill \\ \hfill & {\sigma }_{b4}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda \begin{array}{c}\hfill I\end{array}\notin B\left(\mathcal{X}\right)\}.\hfill \end{array}$$

As an important branch of functional analysis, the spectral theory of block operator matrices is devoted to the characterization of the algebraic structure and topological properties of operators via their spectral features. It has wide application prospects in applied disciplines such as differential equations and quantum mechanics [1,2]. For example, the stability and energy decay of coupled systems are essentially governed by the spectral distribution of the corresponding block operator matrices [2]. If $T\in \mathcal{B}\left(\mathcal{X}\right)$ has an invariant subspace ${\mathcal{X}}_1$, then T admits an upper triangular form with respect to the decomposition $\mathcal{X}={\mathcal{X}}_1\times {\mathcal{X}}_1^{\perp }$, given by

$$T=\left(\begin{array}{cc}{T}_{11}& T_{12}\\ 0& T_{22}\end{array}\right):{\mathcal{X}}_1\times {\mathcal{X}}_1^{\perp }\to {\mathcal{X}}_1\times {\mathcal{X}}_1^{\perp }.$$

Consequently, the research problem of the operator T is transformed into that of a 2 × 2 upper triangular block operator matrix. Therefore, how to reduce the spectral problem of block operator matrices to spectral equalities among diagonal entry operators has long been one of the research goals in this field [3].

In recent decades, researchers worldwide have carried out systematic investigations on the full spectrum, various essential spectra, Weyl spectrum and local spectrum of upper triangular block operator matrices. A wealth of theorems concerning spectral inclusions and spectral equalities has been obtained, together with the corresponding sufficient or necessary and sufficient conditions.

As an illustration, for the Browder essential spectrum proposed by Browder in [4,5], Cao [6], Zhang [7,8] and Bai et al. [9,10] successfully derived spectral equality relations between the Browder essential spectrum of upper triangular block operator matrices

$$M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right),A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$$

and their diagonal operators A and B by constructing special operators. The results are as follows:

$$\begin{array}{cc}\hfill \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\sigma }_{b4}\left(M_C\right)& ={\sigma }_{b1}\left(A\right)\cup {\sigma }_{b2}\left(B\right)\hfill \\ \hfill & \cup \left\{\lambda \in \mathbb{C}:\alpha (A-\lambda I)+\alpha (B-\lambda I)\ne \beta (A-\lambda I)+\beta (B-\lambda I)\right\};\hfill \\ \hfill \bigcap _{C\in \Phi \left(\mathcal{X}\right)}{\sigma }_{b4}\left(M_C\right)& ={\sigma }_{b1}\left(A\right)\cup {\sigma }_{b2}\left(B\right)\hfill \\ \hfill & \cup \left\{\lambda \in \mathbb{C}:\alpha (A-\lambda I)+\alpha (B-\lambda I)\ne \beta (A-\lambda I)+\beta (B-\lambda I)\right\};\hfill \\ \hfill \bigcap _{C\in G\left(\mathcal{X}\right)}{\sigma }_{b4}\left(M_C\right)& ={\sigma }_{b1}\left(A\right)\cup {\sigma }_{b2}\left(B\right)\hfill \\ \hfill & \cup \left\{\lambda \in \mathbb{C}:\alpha (A-\lambda I)+\alpha (B-\lambda I)\ne \beta (A-\lambda I)+\beta (B-\lambda I)\right\};\hfill \\ \hfill {\sigma }_{b4}\left(M_C\right)& ={\sigma }_{b4}\left(A\right)\cup {\sigma }_{b4}\left(B\right).\hfill \end{array}$$

These results indicate that, under suitable conditions, the Browder essential spectrum of the full operator can be characterized in terms of the semi-Browder essential spectrum, ascent, descent, nullity and deficiency of its diagonal entries.

Inspired by the concept of pseudospectra [11,12,13], Abdmouleh et al. [14,15] introduced the pseudo-Browder essential spectrum. For any $\epsilon >0$ and $T\in \mathcal{B}\left(\mathcal{X}\right)$, the pseudo-upper (resp. lower) semi-Browder essential spectrum and pseudo-Browder essential spectrum are defined by

$$\begin{array}{cc}\hfill & {\sigma }_{b1,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon ,TE=ET}{\sigma }_{b1}(T+E);\hfill \\ \hfill & {\sigma }_{b2,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon ,TE=ET}{\sigma }_{b2}(T+E);\hfill \\ \hfill & {\sigma }_{b4,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon ,TE=ET}{\sigma }_{b4}(T+E),\hfill \end{array}$$

which extended the classical Browder essential spectrum theory from the perspective of perturbation analysis. Based on the above definitions, Abdmouleh et al. [14,15] obtained the following conclusions

$$\begin{array}{cc}\hfill & {\sigma }_{bi,\epsilon }\left(M_C\right)\subseteq {\sigma }_{bi,\epsilon }\left(A\right)\cup {\sigma }_{bi,\epsilon }\left(B\right);\hfill \\ \hfill & {\sigma }_{bi,\epsilon }(A+B)\backslash \left\{0\right\}=\{{\sigma }_{bi,\epsilon }\left(A\right)\cup {\sigma }_{bi,\epsilon }\left(B\right)\}\backslash \left\{0\right\},i\in \{1,2,4\}.\hfill \end{array}$$

Similarly, for the pseudo-Browder essential spectrum, do spectral equalities still hold, and can they be characterized by its diagonal entries? Accordingly, this study investigates the conditions under which the pseudo-Browder essential spectrum equalities hold for 2 × 2 upper triangular block operator matrices.

According to the given perturbation definition of the pseudo-Browder essential spectrum in [15,16], we first present the relevant definitions of the noncommutative pseudo-Browder essential spectrum. For any $\epsilon >0$ and $T\in \mathcal{B}\left(\mathcal{X}\right)$, the noncommutative pseudo-upper semi-Browder essential spectrum, noncommutative pseudo-lower semi-Browder essential spectrum, and noncommutative pseudo-Browder essential spectrum are defined by

$$\begin{array}{cc}\hfill & {\Sigma }_{b1,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon }{\sigma }_{b1}(T+E);\hfill \\ \hfill & {\Sigma }_{b2,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon }{\sigma }_{b2}(T+E);\hfill \\ \hfill & {\Sigma }_{b4,\epsilon }\left(T\right)=\bigcup _{E\in \mathcal{B}\left(\mathcal{X}\right),\parallel E\parallel <\epsilon }{\sigma }_{b4}(T+E).\hfill \end{array}$$

In particular, for the upper triangular block operator matrices $M_C$, the corresponding noncommutative pseudo-upper (resp. lower) semi-Browder essential spectrum and noncommutative pseudo-Browder essential spectrum are defined as

$$\begin{array}{cc}\hfill & {\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)=\bigcup _{E\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),\parallel E\parallel <\epsilon }{\sigma }_{b1}(M_C+E);\hfill \\ \hfill & {\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)=\bigcup _{E\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),\parallel E\parallel <\epsilon }{\sigma }_{b2}(M_C+E);\hfill \\ \hfill & {\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)=\bigcup _{E\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),\parallel E\parallel <\epsilon }{\sigma }_{b4}(M_C+E),\hfill \end{array}$$

which are called the sub-noncommutative pseudo-upper (resp. lower) semi-Browder essential spectrum and sub-noncommutative pseudo-Browder essential spectrum, where $\mathcal{TB}(\mathcal{X}\times \mathcal{X})$ denotes the Banach algebra of all 2 × 2 bounded upper triangular operator matrices on a separable Hilbert space $\mathcal{X}\times \mathcal{X}$. In particular, the above definition applies when $C=0$.

In this study, firstly, we establish the spectrum equalities for special cases of upper triangular operator matrices—the diagonal block operator matrix

$$M_0=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right),A,B\in \mathcal{B}\left(\mathcal{X}\right),$$

we obtain that

$${\widehat{\Sigma }}_{bi,\epsilon }\left(M_0\right)={\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right),i\in \{1,2,4\}.$$

Secondly, for the upper triangular block operator matrix

$$M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right),A,B,C\in \mathcal{B}\left(\mathcal{X}\right),$$

we derive that

$${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta ,}$$

where $\Delta =\{\lambda \in \mathbb{C}:$ exist $P_i\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel P_i\parallel <\epsilon ,i\in \{1,2\},$ such that $\alpha (A+P_1-\lambda I)+\alpha (B+P_2-\lambda I)\ne \beta (A+P_1-\lambda I)+\beta (B+P_2-\lambda I)\}$. Finally, we obtain

$${\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right)={\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)\cup W,i\in \{1,2,4\},$$

where W is the union of certain holes in $({\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right))\backslash {\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)$.

2. Preliminaries

In this section, we introduce the preliminary results that will be used throughout this paper.

Definition 1

(see [17], Chapter V). Let T be a bounded linear operator on the Hilbert space $\mathcal{X}$, and there exists a constant $m>0$ such that

$$\parallel Tx\parallel \ge m\parallel x\parallel ,\forall x\in \mathcal{D}\left(T\right).$$

Then T is called bounded below.

Definition 2

(see [18]). Let $T\in \mathcal{B}\left(\mathcal{X}\right)$. The reduced minimum modulus of T is defined by the equation

$$\gamma \left(T\right)=\left\{\begin{array}{cc}inf\left\{\parallel Tx\parallel :x\in N{\left(T\right)}^{\perp },\parallel x\parallel =1\right\}\hfill & \mathrm{if}T\ne 0,\hfill \\ 0\hfill & \mathrm{if}T=0.\hfill \end{array}\right.$$

Lemma 1

(see [18,19], Theorem 5.13 of [20]). Let $T,M,N\in \mathcal{B}\left(\mathcal{X}\right)$. Then

(i)

$\gamma \left(T\right)>0$ if and only if $R\left(T\right)$ is closed;

(ii)

$R\left(T\right)$ is closed if and only if $R\left(T^{\ast }\right)$ is closed;

(iii)

If N and M are both invertible operators, then $R\left(MTN\right)$ is closed if and only if $R\left(T\right)$ is closed.

Lemma 2.

Let $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If $R\left(M_C\right)$ is closed, then $R\left(A\right)$ and $R\left(B\right)$ are closed.

Proof. 

Let $R\left(M_C\right)$ be closed. Then

$$\gamma \left(M_C\right)=inf\{\parallel \left(\begin{array}{c}Ax+Cy\\ By\end{array}\right)\parallel :\left(\begin{array}{c}x\\ y\end{array}\right)\in N{\left(M_C\right)}^{\perp },{\parallel x\parallel }^2+{\parallel y\parallel }^2=1\}>0$$

by Lemma 1. On the one hand, if $y=0$, then

$$inf\{\parallel \left(\begin{array}{c}Ax\\ 0\end{array}\right)\parallel :\left(\begin{array}{c}x\\ 0\end{array}\right)\in N{\left(M_C\right)}^{\perp },\parallel x{\parallel }^2=1\}>0,$$

i.e., $\gamma \left(A\right)\ge \gamma \left(M_C\right)>0$; thus $R\left(A\right)$ is closed.

On the other hand, it follows from Lemma 1 that $R\left(M_C^{\ast }\right)$ and the range of

$$M=\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)\left(\begin{array}{cc}{A}^{\ast }& 0\\ C^{\ast }& B^{\ast }\end{array}\right)\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)=\left(\begin{array}{cc}{B}^{\ast }& C^{\ast }\\ 0& A^{\ast }\end{array}\right)$$

are closed. By the same argument as above, we obtain that $\gamma \left(B^{\ast }\right)\ge \gamma \left(M\right)>0$; thus $R\left(B^{\ast }\right)$ is closed, i.e., $R\left(B\right)$ is closed.    □

Lemma 3

(see Chapter VI of [21] and Proposition 2.1 of [10]). Let $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. Then

(i)

$\alpha \left(A\right)\le \alpha \left(M_C\right)\le \alpha \left(A\right)+\alpha \left(B\right)$;

(ii)

$\beta \left(B\right)\le \beta \left(M_C\right)\le \beta \left(A\right)+\beta \left(B\right)$;

(iii)

$asc\left(A\right)\le asc\left(M_C\right)\le asc\left(A\right)+asc\left(B\right)$;

(iv)

$dsc\left(B\right)\le dsc\left(M_C\right)\le dsc\left(A\right)+dsc\left(B\right)$.

In particular, for diagonal block operator matrices, we present the following lemma.

Lemma 4.

Let $M_0=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$, where $A,B\in \mathcal{B}\left(\mathcal{X}\right)$. Then

(i)

$\alpha \left(M_0\right)=\alpha \left(A\right)+\alpha \left(B\right)$;

(ii)

$\beta \left(M_0\right)=\beta \left(A\right)+\beta \left(B\right)$;

(iii)

$asc\left(M_0\right)=max\{asc\left(A\right),asc\left(B\right)\};$

(iv)

$dsc\left(M_0\right)=max\{dsc\left(A\right),dsc\left(B\right)\}$.

Proof. 

$\left(\mathrm{i}\right)$ Let $\alpha \left(A\right)=m$ and $\beta \left(B\right)=n$. Then there exist two bases ${\left\{x_i\right\}}_{i=1}^m$ and ${\left\{y_j\right\}}_{j=1}^n$ such that ${\left\{x_i\right\}}_{i=1}^m\subseteq N\left(A\right)$, ${\left\{y_j\right\}}_{j=1}^n\subseteq N\left(B\right)$. It is known that each element in $N\left(M_0\right)$ can be linearly expressed by

$$\left(\begin{array}{c}{x}_1\\ 0\end{array}\right),\left(\begin{array}{c}{x}_2\\ 0\end{array}\right)\dots \left(\begin{array}{c}{x}_m\\ 0\end{array}\right),\left(\begin{array}{c}0\\ y_1\end{array}\right),\left(\begin{array}{c}0\\ y_2\end{array}\right)\dots \left(\begin{array}{c}0\\ y_n\end{array}\right).$$

(1)

It remains to show that (1) is linearly independent. Let

$$l_1\left(\begin{array}{c}{x}_1\\ 0\end{array}\right)+l_2\left(\begin{array}{c}{x}_2\\ 0\end{array}\right)+\dots +l_m\left(\begin{array}{c}{x}_m\\ 0\end{array}\right)+k_1\left(\begin{array}{c}0\\ y_1\end{array}\right)+k_2\left(\begin{array}{c}0\\ y_2\end{array}\right)+\dots +k_n\left(\begin{array}{c}0\\ y_n\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

Then

$$\left\{\begin{array}{cc}\hfill l_1{x}_1+\dots +l_m{x}_m& =0;\hfill \\ \hfill k_1{y}_1+\dots +k_n{y}_n& =0.\hfill \end{array}\right.$$

i.e.,

$$\left\{\begin{array}{cc}\hfill l_1=l_2=\dots =l_m& =0;\hfill \\ \hfill k_1=k_2=\dots =k_n& =0.\hfill \end{array}\right.$$

Thus (1) is linearly independent, and $\alpha \left(M_0\right)=\alpha \left(A\right)+\alpha \left(B\right)$.

(ii)

Similar to the proof of (i).

(iii)

Similar to the proof of (i).

Then

$$N\left(M_0^p\right)=N\left(A^p\right)\times N\left(B^p\right)=N\left(A^{p+1}\right)\times N\left(B^{p+1}\right)=N\left(M_0^{p+1}\right).$$

Therefore

$$\left\{\begin{array}{cc}\hfill N\left(A^p\right) =& N\left(A^{p+1}\right);\hfill \\ \hfill N\left(B^p\right) =& N\left(B^{p+1}\right);\hfill \\ \hfill asc\left(A\right) \le & p;\hfill \\ \hfill asc\left(B\right) \le & p,\hfill \end{array}\right.$$

i.e., $max\left\{asc\right(A),asc(B\left)\right\}\le p$.

Conversely, let $asc\left(A\right)=p_1$ and $asc\left(B\right)=p_2$. Then $N\left(A^{p_1}\right)=N\left(A^{p_1+1}\right)$, and $N\left(B^{p_2}\right)=N\left(B^{p_2+1}\right)$. Let $p=max\{p_1,p_2\}$. Then

$$N\left(M_0^p\right)=N\left(M_0^{p+1}\right).$$

Hence $asc\left(M_0\right)\le p=max\{p_1,p_2\}=max\{asc\left(A\right),asc\left(B\right)\}$. In summary, we obtain that $asc\left(M_0\right)=max\{asc\left(A\right),asc\left(B\right)\}$.

(iv)

Similar to the proof of (iii).    □

Lemma 5

(see [22], Theorem 3.4). Let $T\in \mathcal{B}\left(\mathcal{X}\right)$.

(i)

If $asc\left(T\right)<\infty$, then $\alpha \left(T\right)\le \beta \left(T\right)$;

(ii)

If $dsc\left(T\right)<\infty$, then $\beta \left(T\right)\le \alpha \left(T\right)$;

(iii)

If $asc\left(T\right)=dsc\left(T\right)<\infty$, then $\alpha \left(T\right)=\beta \left(T\right)$;

(iv)

If $\alpha \left(T\right)=\beta \left(T\right)<\infty$, and if either $asc\left(T\right)$ or $dsc\left(T\right)$ is finite, then $asc\left(T\right)=dsc\left(T\right)$.

Lemma 6.

Let $T\in \mathcal{B}\left(\mathcal{X}\right)$. Then $asc\left(T\right)=dsc\left(T^{\ast }\right)$.

Proof. 

Let $p_1=asc\left(T\right)$. Then $N\left(T^{p_1}\right)=\begin{array}{c}\hfill R{\left({\left(T^{\ast }\right)}^{p_1}\right)}^{\perp }=R{\left({\left(T^{\ast }\right)}^{p_1+1}\right)}^{\perp }\end{array}=N\left(T^{p_1+1}\right)$. Consequently, $dsc\left(T^{\ast }\right)\le asc\left(T\right)$.

Conversely, let $p_2=dsc\left(T^{\ast }\right)$. Then $\begin{array}{c}\hfill R{\left({\left(T^{\ast }\right)}^{p_2}\right)}^{\perp }\end{array}=N\left(T^{p_2}\right)=N\left(T^{p_2+1}\right)=\begin{array}{c}\hfill R{\left({\left(T^{\ast }\right)}^{p_2+1}\right)}^{\perp }\end{array}$. Consequently, $dsc\left(T^{\ast }\right)\ge asc\left(T\right)$. Thus, we conclude that $asc\left(T\right)=dsc\left(T^{\ast }\right)$.    □

Lemma 7

(see [19], Theorems 5, 6 and 12 in Chapter III). Let $\mathcal{X}$ be the Hilbert space, $T\in \mathcal{B}\left(\mathcal{X}\right)$ and $S\in \mathcal{B}\left(\mathcal{X}\right)$. Then

(i)

If T and S are lower semi-Fredholm, then $ST$ is lower semi-Fredholm;

(ii)

If T and S are upper semi-Fredholm, then $ST$ is upper semi-Fredholm;

(iii)

If T and S are Fredholm, then $ST$ is Fredholm;

(iv)

If both T and S are upper semi-Fredholm (or both are lower semi-Fredholm), then $ind\left(TS\right)=ind\left(T\right)+ind\left(S\right)$;

(v)

If $ST$ is lower semi-Fredholm, then S is lower semi-Fredholm;

(vi)

If $ST$ is upper semi-Fredholm, then T is upper semi-Fredholm;

(vii)

If $ST$ is Fredholm, then S is lower semi-Fredholm and T is upper semi-Fredholm.

Similarly to the proof of ([9], Theorem 2.1), we obtain the following lemma.

Lemma 8.

Let $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If $\beta \left(A\right)=\infty$, then $M_C$ is an upper semi-Browder operator for some $C\in \mathcal{B}\left(\mathcal{X}\right)$ if and only if A is an upper semi-Browder operator.

Similarly to the proof of ([9], Theorem 2.2), we obtain the following lemma.

Lemma 9.

Let $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If $\alpha \left(B\right)=\infty$, then $M_C$ is a lower semi-Browder operator for some $C\in \mathcal{B}\left(\mathcal{X}\right)$ if and only if B is a lower semi-Browder operator.

Proposition 1

(see [22], Theorem 3.3). Let $T\in \mathcal{B}\left(\mathcal{X}\right)$. If both $asc\left(T\right)$ and $dsc\left(T\right)$ are finite, then $asc\left(T\right)=dsc\left(T\right)$.

Proposition 2.

Let $T\in \mathcal{B}\left(\mathcal{X}\right)$. Then

(i)

$asc\left(T\right)=0$ if and only if $N\left(T\right)=\left\{0\right\}$;

(ii)

$dsc\left(T\right)=0$ if and only if $R\left(T\right)=\mathcal{X}$.

Proof. 

(i) If $asc\left(T\right)=0$, then $N\left(T^0\right)=N\left(I\right)=N\left(T\right)$, and thus $N\left(T\right)=\left\{0\right\}$. Conversely, if $N\left(T\right)=\left\{0\right\}$, then $N\left(T^0\right)=N\left(T\right)$, and therefore $asc\left(T\right)=0$.

(ii) If $dsc\left(T\right)=0$, then $R\left(T^0\right)=R\left(I\right)=R\left(T\right)$, and thus $R\left(T\right)=\mathcal{X}$. Conversely, if $R\left(T\right)=\mathcal{X}$, then $R\left(T^0\right)=R\left(T\right)=\mathcal{X}$, and therefore $dsc\left(T\right)=0$.    □

3. Noncommutative Pseudo-Browder Essential Spectrum of Bounded Diagonal Block Operator Matrix

In this section, we mainly use the direct sum properties of the kernel and the orthogonal complement of the range to characterize the relationship between the union of the noncommutative pseudo-Browder essential spectrum of the overall operator and its diagonal entries.

Theorem 1.

Let $\epsilon >0$, $M_0=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$, where $A,B\in \mathcal{B}\left(\mathcal{X}\right)$. Then

$$\begin{array}{c}\hfill {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b1,\epsilon }\left(B\right).\end{array}$$

Proof. 

Let $\lambda \notin {\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b1,\epsilon }\left(B\right)$. Then for any

$$E=\left(\begin{array}{cc}{E}_1& E_2\\ 0& E_3\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),$$

with $\parallel E\parallel <\epsilon$, we have $A+E_1-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$, $asc(A+E_1-\lambda I)<\infty$ and $B+E_3-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$, $asc(B+E_3-\lambda I)<\infty$. By Lemmas 4 and 7, we know that $M_0+E-\lambda I\in {\Phi }_{+}(\mathcal{X}\times \mathcal{X})$ and $asc(M_0+E-\lambda I)<\infty$. Hence $\lambda \notin {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$.

Conversely, let $\lambda \in {\Sigma }_{b1,\epsilon }\left(A\right)$. Then there exists $E_1^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1^{\prime }\parallel <\epsilon$, such that $A+E_1^{\prime }-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$ or $asc(A+E_1^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime }=\left(\begin{array}{cc}{E}_1^{\prime }& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime }\parallel =\parallel E_1^{\prime }\parallel <\epsilon$. If $A+E_1^{\prime }-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$, then

$$M_0+E^{\prime }-\lambda I=\left(\begin{array}{cc}A+E_1^{\prime }-\lambda I& 0\\ 0& B-\lambda I\end{array}\right)\notin {\Phi }_{+}(\mathcal{X}\times \mathcal{X}).$$

Hence $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$. If $asc(A+E_1^{\prime }-\lambda I)=\infty$, then $asc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$. Let $\lambda \in {\Sigma }_{b1,\epsilon }\left(B\right)$. Then there exists $E_3^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_3^{\prime }\parallel <\epsilon$, such that $B+E_3^{\prime }-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$ or $asc(B+E_3^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime \prime }=\left(\begin{array}{cc}0& 0\\ 0& E_3^{\prime }\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime \prime }\parallel =\parallel E_3^{\prime }\parallel <\epsilon$. If $B+E_3^{\prime }-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$, then $R(B+E_3^{\prime }-\lambda I)$ is not closed or $\alpha (B+E_3^{\prime }-\lambda I)=\infty$. Hence

$$M_0+E^{\prime \prime }-\lambda I=\left(\begin{array}{cc}A-\lambda I& 0\\ 0& B+E_3^{\prime }-\lambda I\end{array}\right)\notin {\Phi }_{+}(\mathcal{X}\times \mathcal{X}).$$

Therefore $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$. If $asc(B+E_3^{\prime }-\lambda I)=\infty$, then $asc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$.    □

Theorem 2.

Let $\epsilon >0$, $M_0=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$, where $A,B\in \mathcal{B}\left(\mathcal{X}\right)$. Then

$$\begin{array}{c}\hfill {\widehat{\Sigma }}_{b2,\epsilon }\left(M_0\right)={\Sigma }_{b2,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right).\end{array}$$

Proof. 

Let $\lambda \notin {\Sigma }_{b2,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)$. Then for any

$$E=\left(\begin{array}{cc}{E}_1& E_2\\ 0& E_3\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),$$

with $\parallel E\parallel <\epsilon$, we have $A+E_1-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(A+E_1-\lambda I)<\infty$ and $B+E_3-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(B+E_3-\lambda I)<\infty$. By Lemmas 4 and 7, we know that $M_0+E-\lambda I\in {\Phi }_{-}(\mathcal{X}\times \mathcal{X})$, $dsc(M_0+E-\lambda I)<\infty$. Therefore $\lambda \notin {\widehat{\Sigma }}_{b2,\epsilon }\left(M_0\right)$.

Conversely, let $\lambda \in {\Sigma }_{b2,\epsilon }\left(A\right)$. Then there exists $E_1^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1^{\prime }\parallel <\epsilon$, such that $A+E_1^{\prime }-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$ or $dsc(A+E_1^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime }=\left(\begin{array}{cc}{E}_1^{\prime }& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime }\parallel =\parallel E_1^{\prime }\parallel <\epsilon$. If $A+E_1^{\prime }-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$, then $R(A+E_1^{\prime }-\lambda I)$ is not closed or $\beta (A+E_1^{\prime }-\lambda I)=\infty$. Hence

$$M_0+E^{\prime }-\lambda I=\left(\begin{array}{cc}A+E_1^{\prime }-\lambda I& 0\\ 0& B-\lambda I\end{array}\right)\notin {\Phi }_{-}(\mathcal{X}\times \mathcal{X}).$$

Therefore $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_0\right)$. If $dsc(A+E_1^{\prime }-\lambda I)=\infty$, then $dsc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_0\right)$. Let $\lambda \in {\Sigma }_{b2,\epsilon }\left(B\right)$. Then there exists $E_3^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_3^{\prime }\parallel <\epsilon$, such that $B+E_3^{\prime }-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$ or $dsc(B+E_3^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime \prime }=\left(\begin{array}{cc}0& 0\\ 0& E_3^{\prime }\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime \prime }\parallel =\parallel E_3^{\prime }\parallel <\epsilon$. If $B+E_3^{\prime }-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$, then

$$M_0+E^{\prime \prime }-\lambda I=\left(\begin{array}{cc}A-\lambda I& 0\\ 0& B+E_3^{\prime }-\lambda I\end{array}\right)\notin {\Phi }_{-}(\mathcal{X}\times \mathcal{X}).$$

Hence $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_0\right)$. If $dsc(B+E_3^{\prime }-\lambda I)=\infty$, then $dsc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_0\right)$.    □

Theorem 3.

Let $\epsilon >0$, $M_0=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$, where $A,B\in \mathcal{B}\left(\mathcal{X}\right)$. Then

$$\begin{array}{c}\hfill {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)={\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right).\end{array}$$

Proof. 

Let $\lambda \notin {\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right)$. Then for any

$$E=\left(\begin{array}{cc}{E}_1& E_2\\ 0& E_3\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X}),$$

with $\parallel E\parallel <\epsilon$, we have $A+E_1-\lambda I\in \Phi \left(\mathcal{X}\right)$, $asc(A+E_1-\lambda I)<\infty$, $dsc(A+E_1-\lambda I)<\infty$ and $B+E_3-\lambda I\in \Phi \left(\mathcal{X}\right)$, $asc(B+E_3-\lambda I)<\infty$, $dsc(B+E_3-\lambda I)<\infty$. By Lemmas 4 and 7, we know that $M_0+E-\lambda I\in \Phi (\mathcal{X}\times \mathcal{X})$, $asc(M_0+E-\lambda I)<\infty$, $dsc(M_0+E-\lambda I)<\infty$. Thus $\lambda \notin {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)$.

Conversely, let $\lambda \in {\Sigma }_{b4,\epsilon }\left(A\right)$. Then there exists $E_1^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1^{\prime }\parallel <\epsilon$, such that $A+E_1^{\prime }-\lambda I\notin \Phi \left(\mathcal{X}\right)$ or $asc(A+E_1^{\prime }-\lambda I)=\infty$ or $dsc(A+E_1^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime }=\left(\begin{array}{cc}{E}_1^{\prime }& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime }\parallel =\parallel E_1^{\prime }\parallel <\epsilon$. If $A+E_1^{\prime }-\lambda I\notin \Phi \left(\mathcal{X}\right)$, then $R(A+E_1^{\prime }-\lambda I)$ is not closed or $\alpha (A+E_1^{\prime }-\lambda I)=\infty$ or $\beta (A+E_1^{\prime }-\lambda I)=\infty$. Hence

$$M_0+E^{\prime }-\lambda I=\left(\begin{array}{cc}A+E_1^{\prime }-\lambda I& 0\\ 0& B-\lambda I\end{array}\right)\notin \Phi (\mathcal{X}\times \mathcal{X}).$$

Therefore $\lambda \in {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)$. If $asc(A+E_1^{\prime }-\lambda I)=\infty$ or $dsc(A+E_1^{\prime }-\lambda I)=\infty$, then $asc(M_0+E^{\prime }-\lambda I)=\infty$ or $dsc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)$. Let $\lambda \in {\Sigma }_{b1,\epsilon }\left(B\right)$. Then there exists $E_3^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_3^{\prime }\parallel <\epsilon$, such that $B+E_3^{\prime }-\lambda I\notin \Phi \left(\mathcal{X}\right)$ or $asc(B+E_3^{\prime }-\lambda I)=\infty$ or $dsc(B+E_3^{\prime }-\lambda I)=\infty$. Let

$$E^{\prime \prime }=\left(\begin{array}{cc}0& 0\\ 0& E_3^{\prime }\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).$$

Then $\parallel E^{\prime \prime }\parallel =\parallel E_3^{\prime }\parallel <\epsilon$. If $B+E_3^{\prime }-\lambda I\notin \Phi \left(\mathcal{X}\right)$, then $R(B+E_3^{\prime }-\lambda I)$ is not closed or $\alpha (B+E_3^{\prime }-\lambda I)=\infty$ or $\beta (B+E_3^{\prime }-\lambda I)=\infty$. Hence

$$M_0+E^{\prime \prime }-\lambda I=\left(\begin{array}{cc}A-\lambda I& 0\\ 0& B+E_3^{\prime }-\lambda I\end{array}\right)\notin \Phi (\mathcal{X}\times \mathcal{X}).$$

Therefore $\lambda \in {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)$. If $asc(B+E_3^{\prime }-\lambda I)=\infty$ or $dsc(B+E_3^{\prime }-\lambda I)=\infty$, then $asc(M_0+E^{\prime }-\lambda I)=\infty$ or $dsc(M_0+E^{\prime }-\lambda I)=\infty$. Thus $\lambda \in {\widehat{\Sigma }}_{b4,\epsilon }\left(M_0\right)$.    □

4. Noncommutative Pseudo-Browder Essential Spectrum of Bounded Upper Triangular Block Operator Matrix

In this section, by using the properties of the kernel, orthogonal complements of ranges, and ascent and descent of diagonal entries, we characterize the spectral relation between the noncommutative Browder essential spectrum of the whole operator and those of the diagonal entry operators.

Theorem 4.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, $\begin{array}{c}\hfill whereA,B,C\in \mathcal{B}\left(\mathcal{X}\right)\end{array}$. Then

$${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta ,}$$

where $\Delta =\{\lambda \in \mathbb{C}:$ there exist $P_i\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel P_i\parallel <\epsilon ,\begin{array}{c}\hfill i\in \{1,2\}\end{array},$ such that $\alpha (A+P_1-\lambda I)+\alpha (B+P_2-\lambda I)\ne \beta (A+P_1-\lambda I)+\beta (B+P_2-\lambda I)\}.$

Proof. 

Claim 1. ${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\subseteq {\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta }$. Let $\lambda \notin {\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta$. Without loss of generality, we may assume that $\lambda =0$. Then for any

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{E}_1& E_2\\ 0& E_3\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X})\end{array}$$

with $\parallel E\parallel <\epsilon$, we have $A+E_1\in {\Phi }_{+}\left(\mathcal{X}\right),asc(A+E_1)<\infty ,B+E_3\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(B+E_3)<\infty$ and $\alpha (A+E_1)+\alpha (B+E_3)=\beta (A+E_1)+\beta (B+E_3)$.

Case 1. If $\beta (A+E_1)<\infty$, then $\alpha (B+E_3)<\infty$. Consequently, $A+E_1\in \Phi \left(\mathcal{X}\right)$, $B+E_3\in \Phi \left(\mathcal{X}\right)$. Therefore

$$M_C+E=\left(\begin{array}{cc}A+E_1& C+E_2\\ 0& B+E_3\end{array}\right)\in \Phi (\mathcal{X}\times \mathcal{X}).$$

To conclude that $M_C+E$ is a Browder operator, it suffices to show that either $asc(M_C+E)<\infty$ or $asc(M_C^{\ast }+E^{\ast })<\infty$. We now distinguish the following two cases:

(1)

$\alpha (B+E_3)\le \beta (A+E_1);$

(2)

$\alpha (B+E_3)>\beta (A+E_1).$

If (1) holds, we can define a bounded below operator $T_1:N(B+E_3)\to R{(A+E_1)}^{\perp }$. Let

$$E_2=\left(\begin{array}{cc}{E}_{21}& E_{22}\\ E_{23}& E_{24}\end{array}\right):\left(\begin{array}{c}N(B+E_3)\\ N{(B+E_3)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R(A+E_1)\\ R{(A+E_1)}^{\perp }\end{array}\right),$$

and define operator

$$C=\left(\begin{array}{cc}{E}_{21}& -E_{22}\\ T_1-E_{23}& -E_{24}\end{array}\right):\left(\begin{array}{c}N(B+E_3)\\ N{(B+E_3)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R(A+E_1)\\ R{(A+E_1)}^{\perp }\end{array}\right).$$

Then $M_C+E$ can be represented as

$$M_C+E=\left(\begin{array}{ccc}{A}_1& 0& 0\\ 0& T_1& 0\\ 0& 0& B_1\end{array}\right):\left(\begin{array}{c}\mathcal{X}\\ N(B+E_3)\\ N{(B+E_3)}^{\perp }\end{array}\right)\to \left(\begin{array}{c}R(A+E_1)\\ R{(A+E_1)}^{\perp }\\ \mathcal{X}\end{array}\right),$$

where $A_1:\mathcal{X}\to R(A+E_1)$, and $B_1:N{(B+E_3)}^{\perp }\to \mathcal{X}$. It is evident that $B_1$ and $T_1$ are injective; consequently $asc\left(B_1\right)=asc\left(T_1\right)=0$. It follows from Lemma 4 that $asc(M_C+E)=asc(A+E_1)<\infty$. And since $\lambda \notin \Delta$, it follows that $\alpha (M_C+E)=\beta (M_C+E)<\infty$. Thus, $M_C+E$ is a Browder operator by Lemma 5.

If (2) holds, we can define a bounded below operator $T_2:N(A^{\ast }+E_1^{\ast })\to R{(B^{\ast }+E_3^{\ast })}^{\perp }$. Let

$$E_2^{\ast }=\left(\begin{array}{cc}{E}_{21}^{\ast }& E_{23}^{\ast }\\ E_{22}^{\ast }& E_{24}^{\ast }\end{array}\right):\left(\begin{array}{c}N{(A^{\ast }+E_1^{\ast })}^{\perp }\\ N(A^{\ast }+E_1^{\ast })\end{array}\right)\to \left(\begin{array}{c}R{(B^{\ast }+E_3^{\ast })}^{\perp }\\ R(B^{\ast }+E_3^{\ast })\end{array}\right),$$

and define operator

$$C^{\ast }=\left(\begin{array}{cc}-E_{21}^{\ast }& T_2-E_{23}^{\ast }\\ -E_{22}^{\ast }& -E_{24}^{\ast }\end{array}\right):\left(\begin{array}{c}N{(A^{\ast }+E_1^{\ast })}^{\perp }\\ N(A^{\ast }+E_1^{\ast })\end{array}\right)\to \left(\begin{array}{c}R{(B^{\ast }+E_3^{\ast })}^{\perp }\\ R(B^{\ast }+E_3^{\ast })\end{array}\right).$$

Then $M_C^{\ast }+E^{\ast }$ can be represented as

$$M_C^{\ast }+E^{\ast }=\left(\begin{array}{ccc}{A}_1^{\prime \ast }& 0& 0\\ 0& T_2& 0\\ 0& 0& B_1^{\prime \ast }\end{array}\right):\left(\begin{array}{c}N{(A^{\ast }+E_1^{\ast })}^{\perp }\\ N(A^{\ast }+E_1^{\ast })\\ \mathcal{X}\end{array}\right)\to \left(\begin{array}{c}\mathcal{X}\\ R{(B^{\ast }+E_3^{\ast })}^{\perp }\\ R(B^{\ast }+E_3^{\ast })\end{array}\right),$$

where $A_1^{\prime \ast }:N{(A^{\ast }+E_1^{\ast })}^{\perp }\to \mathcal{X}$, and $B_1^{\prime \ast }:\mathcal{X}\to R(B^{\ast }+E_3^{\ast })$. It is evident that $A_1^{\prime \ast }$ and $T_2$ are injective; consequently $asc\left(A_1^{\prime \ast }\right)=asc\left(T_2\right)=0$. It follows from Lemma 4 that $asc(M_C^{\ast }+E^{\ast })=asc(B^{\ast }+E_3^{\ast })<\infty$. And since $\lambda \notin \Delta$, it follows that $\alpha (M_C+E)=\beta (M_C+E)<\infty$. Therefore, $M_C+E$ is a Browder operator by Lemma 5.

Case 2. If $\beta (A+E_1)=\infty$, then $\alpha (B+E_3)=\infty$. By Lemmas 8 and 9, we find that $M_C+E$ is a Browder operator.

Claim 2. ${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\supseteq {\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta }$. Let $\lambda \in {\Sigma }_{b1,\epsilon }\left(A\right)$. Without loss of generality, we may assume that $\lambda =0$. Then there exists an $E_1\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1\parallel <\epsilon$ such that either $A+E_1\notin {\Phi }_{+}\left(\mathcal{X}\right)$ or $asc(A+E_1)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{E}_1& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_1\parallel <\epsilon$. If $A+E_1\notin {\Phi }_{+}\left(\mathcal{X}\right)$, then $M_C+E\notin {\Phi }_{+}\left(\mathcal{X}\right)$; hence $0\in \bigcap _{C\in \mathcal{B}\left(X\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)$. If $asc(A+E_1)=\infty$, then by Lemma 3, we know that $asc(M_C+E)=\infty$; thus $0\in \bigcap _{C\in \mathcal{B}\left(X\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)$.

Let $0\in {\Sigma }_{b2,\epsilon }\left(B\right)$. Then there exists an $E_2\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_2\parallel <\epsilon$ such that either $B+E_2\notin {\Phi }_{-}\left(\mathcal{X}\right)$ or $dsc(B+E_2)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}0& 0\\ 0& E_2\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_2\parallel <\epsilon$. If $B+E_2\notin {\Phi }_{-}\left(\mathcal{X}\right)$, then $M_C+E\notin {\Phi }_{-}(\mathcal{X}\times \mathcal{X})$; hence $0\in \bigcap _{C\in \mathcal{B}\left(X\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)$. If $dsc(B+E_2)=\infty$, then by Lemma 3, we know that $dsc(M_C+E)=\infty$; therefore $0\in \bigcap _{C\in \mathcal{B}\left(X\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)$.

Let $0\in \Delta$. Then there exist $P_1,P_2\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel P_1\parallel <\epsilon$, $\parallel P_2\parallel <\epsilon$ such that

$$\alpha (A+P_1)+\alpha (B+P_2)\ne \beta (A+P_1)+\beta (B+P_2),$$

i.e., $ind(A+P_1)+ind(B+P_2)\ne 0$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{P}_1& 0\\ 0& P_2\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =max\{\parallel P_1\parallel ,\parallel P_2\parallel \}<\epsilon$. Since $ind(M_C+E)=ind(A+P_1)+ind(B+P_2)\ne 0$, we conclude that $0\in \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)$.    □

Remark 1.

This theorem extends Corollary 2.5 in [6], with the key improvement that the equality remains valid under small perturbations of the noncommutative pseudo-Browder essential spectrum.

Corollary 1.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, $\begin{array}{c}\hfill whereA,B,C\in \mathcal{B}\left(\mathcal{X}\right)\end{array}$. If $\Delta =\varnothing$, then

$${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right).}$$

Theorem 5.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, $\begin{array}{c}\hfill whereA,B,C\in \mathcal{B}\left(\mathcal{X}\right)\end{array}$. Then

$${\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b1,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)\cup N_{\infty }^{\epsilon }\left(B\right)\cup {\sigma }_{asc}^{\epsilon }\left(B\right),$$

where

$$\begin{array}{c}{N}_{\infty }^{\epsilon }(·)=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right)\mathit{with}\parallel P\parallel <\epsilon \mathit{such}\mathit{that}\alpha (·+P-\lambda I)=\infty \right\}.\\ {\sigma }_{\mathrm{asc}}^{\epsilon }(·)=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right)\mathit{with}\parallel P\parallel <\epsilon \mathit{such}\mathit{that}asc(·+P-\lambda I)=\infty \right\}.\end{array}$$

Proof. 

Let $\lambda \in {\Sigma }_{b1,\epsilon }\left(A\right)$. Then there exists an $E_1\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1\parallel <\epsilon$ such that either $A+E_1-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$ or $asc(A+E_1-\lambda I)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{E}_1& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_1\parallel <\epsilon$. If $A+E_1-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$, then $M_C+E-\lambda I\notin {\Phi }_{+}(\mathcal{X}\times \mathcal{X})$; therefore $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)$. If $asc(A+E_1-\lambda I)=\infty$, then by Lemma 3, we have $asc(M_C+E-\lambda I)=\infty$; hence $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)$. Let $\lambda \in {\Sigma }_{b1,\epsilon }\left(B\right)$. Then there exists an $E_3\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_3\parallel <\epsilon$ such that either $B+E_3-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$ or $asc(B+E_3-\lambda I)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}0& 0\\ 0& E_3\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_3\parallel <\epsilon$. If $B+E_3-\lambda I\notin {\Phi }_{+}\left(\mathcal{X}\right)$, then $R(B+E_3-\lambda I)$ is not closed or $\alpha (B+E_3-\lambda I)=\infty$. If $R(B+E_3-\lambda I)$ is not closed, then $R(M_C+E-\lambda I)$ is not closed by Lemma 2; therefore $\lambda \in {\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)$. If $\alpha (B+E_3-\lambda I)=\infty$ or $asc(B+E_3-\lambda I)=\infty$, then $\lambda \in N_{\infty }^{\epsilon }\left(B\right)\cup {\sigma }_{asc}^{\epsilon }\left(B\right)$.

Conversely, let $\lambda \notin {\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b1,\epsilon }\left(B\right)$. Then for any $E_3^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_3^{\prime }\parallel <\epsilon$, we have $B+E_3^{\prime }-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$ and $asc(B+E_3^{\prime }-\lambda I)<\infty$; thus $\lambda \notin N_{\infty }^{\epsilon }\left(B\right)\cup {\sigma }_{asc}^{\epsilon }\left(B\right)$. And for any

$$\begin{array}{c}\hfill E^{\prime \prime }=\left(\begin{array}{cc}{E}_1^{\prime \prime }& E_2^{\prime \prime }\\ 0& E_3^{\prime \prime }\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X})\end{array}$$

with $\parallel E^{\prime \prime }\parallel <\epsilon$, since $\parallel E_1^{\prime \prime }\parallel <\epsilon ,\parallel E_3^{\prime \prime }\parallel <\epsilon$, we have $A+E_1^{\prime \prime }-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$, $asc(A+E_1^{\prime \prime }-\lambda I)<\infty$ and $B+E_3^{\prime \prime }-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$, $asc(B+E_1^{\prime \prime }-\lambda I)<\infty$. By Lemmas 3 and 7, we know that $M_C+E^{\prime \prime }-\lambda I$ is an upper semi-Browder operator, i.e., $\lambda \notin {\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right)$.    □

Corollary 2.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If $N_{\infty }^{\epsilon }\left(B\right)\cup {\sigma }_{asc}^{\epsilon }\left(B\right)=\varnothing$, then

$${\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b1,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b1,\epsilon }\left(M_C\right).$$

Theorem 6.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. Then

$${\Sigma }_{b2,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)\cup R_{\infty }^{\epsilon }\left(A\right)\cup {\sigma }_{dsc}^{\epsilon }\left(A\right),$$

where

$$\begin{array}{c}{R}_{\infty }^{\epsilon }(·):=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right),\parallel P\parallel <\epsilon \mathit{such}\mathit{that}\beta (·+P-\lambda I)=\infty \right\}.\\ {\sigma }_{\mathrm{dsc}}^{\epsilon }(·):=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right),\parallel P\parallel <\epsilon \mathit{such}\mathit{that}dsc(·+P-\lambda I)=\infty \right\}.\end{array}$$

Proof. 

Let $\lambda \in {\Sigma }_{b2,\epsilon }\left(A\right)$. Then there exists an $E_1\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1\parallel <\epsilon$ such that either $A+E_1-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$ or $dsc(A+E_1-\lambda I)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{E}_1& 0\\ 0& 0\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_1\parallel <\epsilon$. When $A+E_1-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$, $R(A+E_1-\lambda I)$ is not closed or $\beta (A+E_1-\lambda I)=\infty$. If $R(A+E_1-\lambda I)$ is not closed, then $R(M_C+E-\lambda I)$ is not closed by Lemma 2; hence $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)$. If $\beta (A+E_1-\lambda I)=\infty$ or $dsc(A+E_1-\lambda I)=\infty$, then $\lambda \in R_{\infty }^{\epsilon }\left(A\right)\cup {\sigma }_{dsc}^{\epsilon }\left(A\right)$. Let $\lambda \in {\Sigma }_{b2,\epsilon }\left(B\right)$. Then there exists an $E_2\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_2\parallel <\epsilon$ such that $B+E_2-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$ or $dsc(B+E_2-\lambda I)=\infty$. Let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}0& 0\\ 0& E_2\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then $\parallel E\parallel =\parallel E_2\parallel <\epsilon$. If $B+E_2-\lambda I\notin {\Phi }_{-}\left(\mathcal{X}\right)$, then $M_C+E-\lambda I\notin {\Phi }_{-}(\mathcal{X}\times \mathcal{X})$; therefore $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)$. If $dsc(B+E_2-\lambda I)=\infty$, then by Lemma 3, we know that $dsc(M_C+E-\lambda I)=\infty$; thus $\lambda \in {\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)$.

Conversely, let $\lambda \notin {\Sigma }_{b2,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)$. Then for any $E_1^{\prime }\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1^{\prime }\parallel <\epsilon$, we have $A+E_1^{\prime }-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$ and $dsc(A+E_1^{\prime }-\lambda I)<\infty$; thus $\lambda \notin R_{\infty }^{\epsilon }\left(A\right)\cup {\sigma }_{dsc}^{\epsilon }\left(A\right)$. And for any

$$\begin{array}{c}\hfill E^{\prime \prime }=\left(\begin{array}{cc}{E}_1^{\prime \prime }& E_2^{\prime \prime }\\ 0& E_3^{\prime \prime }\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X})\end{array}$$

with $\parallel E^{\prime \prime }\parallel <\epsilon$, we have $A+E_1^{\prime \prime }-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(A+E_1^{\prime \prime }-\lambda I)<\infty$ and $B+E_3^{\prime \prime }-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(B+E_1^{\prime \prime }-\lambda I)<\infty$. By Lemmas 3 and 7, we know that $M_C+E^{\prime \prime }-\lambda I$ is a lower semi-Browder operator, i.e., $\lambda \notin {\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right)$.    □

Corollary 3.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If $R_{\infty }^{\epsilon }\left(A\right)\cup {\sigma }_{dsc}^{\epsilon }\left(A\right)=\varnothing$, then

$${\Sigma }_{b2,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b2,\epsilon }\left(M_C\right).$$

Theorem 7.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. Then

$${\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\cup ({\sigma }_{dsc}^{\epsilon }\left(A\right)\cap {\sigma }_{asc}^{\epsilon }\left(B\right)),$$

where

$$\begin{array}{c}{\sigma }_{asc}^{\epsilon }(·)=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right)\mathrm{with}\parallel P\parallel <\epsilon \mathit{such}\mathit{that}asc(·+P-\lambda I)=\infty \right\}.\\ {\sigma }_{dsc}^{\epsilon }(·)=\left\{\lambda \in \mathbb{C}:\mathit{there}\mathit{exists}\mathrm{a}P\in \mathcal{B}\left(\mathcal{X}\right)\mathrm{with}\parallel P\parallel <\epsilon \mathit{such}\mathit{that}dsc(·+P-\lambda I)=\infty \right\}.\end{array}$$

Proof. 

Let $\lambda \notin {\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\cup ({\sigma }_{dsc}^{\epsilon }\left(A\right)\cap {\sigma }_{asc}^{\epsilon }\left(B\right))$. Then at least one of the following holds:

(1)

$\lambda \notin {\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\cup {\sigma }_{dsc}^{\epsilon }\left(A\right)$;

(2)

$\lambda \notin {\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\cup {\sigma }_{asc}^{\epsilon }\left(B\right)$.

If (1) holds, then for any $E_1,E_2\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel E_1\parallel <\epsilon$, $\parallel E_2\parallel <\epsilon$, let

$$\begin{array}{c}\hfill E=\left(\begin{array}{cc}{E}_1& 0\\ 0& E_2\end{array}\right)\in \mathcal{B}(\mathcal{X}\times \mathcal{X}).\end{array}$$

Then

$$M_C+E-\lambda I=\left(\begin{array}{cc}A+E_1-\lambda I& C\\ 0& B+E_2-\lambda I\end{array}\right)\in \Phi (\mathcal{X}\times \mathcal{X}),$$

and $asc(M_C+E-\lambda I)<\infty$, $dsc(M_C+E-\lambda I)<\infty$. Consequently, $A+E_1-\lambda I\in {\Phi }_{+}\left(\mathcal{X}\right)$, $asc(A+E_1-\lambda I)<\infty$ and $B+E_2-\lambda I\in {\Phi }_{-}\left(\mathcal{X}\right)$, $dsc(B+E_2-\lambda I)<\infty$. Considering that $\lambda \notin {\sigma }_{dsc}^{\epsilon }\left(A\right)$, we have $dsc(A+E_1-\lambda I)<\infty$ for all $\parallel E_1\parallel <\epsilon$. Therefore, $\beta (A+E_1-\lambda I)<\infty$ by Lemma 5, which means that $\lambda \notin {\Sigma }_{b4,\epsilon }\left(A\right)$. And since

$$0=ind(M_C+E-\lambda I)=ind(A+E_1-\lambda I)+ind(B+E_2-\lambda I)=ind(B+E_2-\lambda I),$$

we have $\alpha (B+E_2-\lambda I)=\beta (B+E_2-\lambda I)<\infty$. By Lemma 5, we get $asc(B+E_2-\lambda I)<\infty$; thus $\lambda \notin {\Sigma }_{b4,\epsilon }\left(B\right)$.

If (2) holds, similarly we obtain that $\lambda \notin {\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right)$.

Conversely, let $\lambda \notin {\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right)$, then for any

$$\begin{array}{c}\hfill E^{\prime }=\left(\begin{array}{cc}{E}_1^{\prime }& E_2^{\prime }\\ 0& E_3^{\prime }\end{array}\right)\in \mathcal{TB}(\mathcal{X}\times \mathcal{X})\end{array}$$

with $\parallel E^{\prime }\parallel <\epsilon$, we have $A+E_1^{\prime }-\lambda I\in \Phi \left(\mathcal{X}\right)$, $asc(A+E_1^{\prime }-\lambda I)=dsc(A+E_1^{\prime }-\lambda I)<\infty$ and $B+E_3^{\prime }-\lambda I\in \Phi \left(X\right)$, $asc(B+E_3^{\prime }-\lambda I)=dsc(B+E_3^{\prime }-\lambda I)<\infty$. By Lemmas 3 and 7, we have $asc(M_C+E^{\prime }-\lambda I)=dsc(M_C+E^{\prime }-\lambda I)<\infty$, $M_C+E^{\prime }-\lambda I\in \Phi (\mathcal{X}\times \mathcal{X})$. Thus $\lambda \notin {\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)\cup ({\sigma }_{dsc}^{\epsilon }\left(A\right)\cap {\sigma }_{asc}^{\epsilon }\left(B\right))$.    □

Corollary 4.

Let $\epsilon >0$, $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, where $A,B,C\in \mathcal{B}\left(\mathcal{X}\right)$. If ${\sigma }_{asc}^{\epsilon }\left(A\right)\cap {\sigma }_{dsc}^{\epsilon }\left(B\right)=\varnothing$, then

$${\Sigma }_{b4,\epsilon }\left(A\right)\cup {\Sigma }_{b4,\epsilon }\left(B\right)={\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right).$$

5. Discussion

This paper investigates the spectral equality problems of the noncommutative pseudo-Browder essential spectrum for 2 × 2 upper triangular block operator matrices on separable Hilbert spaces. Firstly, the relevant concepts of noncommutative pseudo-Browder operators are introduced. By virtue of the properties of the null spaces and orthogonal complements of the ranges of the diagonal entries, the conditions for the validity of the spectral equality

$${\widehat{\Sigma }}_{bi,\epsilon }\left(M_0\right)={\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right),i\in \{1,2,4\}$$

for special cases of upper triangular operator matrices—the diagonal block operator matrix $M_0$—are obtained. Secondly, the spectral equality for the upper triangular block operator matrix $M_C$ is established as follows:

$${\displaystyle \bigcap _{C\in \mathcal{B}\left(\mathcal{X}\right)}{\widehat{\Sigma }}_{b4,\epsilon }\left(M_C\right)={\Sigma }_{b1,\epsilon }\left(A\right)\cup {\Sigma }_{b2,\epsilon }\left(B\right)\cup \Delta ,}$$

where $\Delta =\{\lambda \in \mathbb{C}:\mathrm{exist}{P}_i\in \mathcal{B}\left(\mathcal{X}\right)$ with $\parallel P_i\parallel <\epsilon ,i\in \{1,2\},$ such that $\alpha (A+P_1-\lambda I)+\alpha (B+P_2-\lambda I)\ne \beta (A+P_1-\lambda I)+\beta (B+P_2-\lambda I)\}$. This result refines earlier works by Cao [6], Zhang [8], and Bai [9,10], who studied similar equalities for the Browder essential spectrum (i.e., the case $\epsilon =0$). Our analysis reveals that when $\epsilon >0$, the interaction between diagonal entries A and B becomes more intricate due to the independent perturbations allowed within the noncommutative framework. Finally, the conditions under which the equality

$${\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right)={\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)\cup W,i\in \{1,2,4\}$$

holds are presented, where W is the union of certain holes in $({\Sigma }_{bi,\epsilon }\left(A\right)\cup {\Sigma }_{bi,\epsilon }\left(B\right))\backslash {\widehat{\Sigma }}_{bi,\epsilon }\left(M_C\right)$. This result provides detailed characterizations of the relationships between the pseudo-semi-Browder spectra of $M_C$ and its diagonal components. Our results indicate that for the pseudo-Browder spectrum, these relationships involve additional sets related to infinite ascent, descent, $N_{\infty }^{\epsilon }\left(B\right)$, $R_{\infty }^{\epsilon }\left(A\right)$, ${\sigma }_{dsc}^{\epsilon }\left(A\right)$ and ${\sigma }_{asc}^{\epsilon }\left(B\right)$. These terms quantify the extent to which small perturbations can destroy the semi-Browder property of individual operators, thereby affecting the global stability of the matrix operator.

Due to the limitations of the authors’ time and research capacity, numerous issues remain to be further investigated, such as:

(i)

The above-mentioned spectral equality problems for the unbounded operator matrix $M_C$.

(ii)

The pseudo-Browder essential spectral equality problems of bounded and unbounded upper triangular block operator matrices under commutative perturbations.

(iii)

The pseudo-Browder essential spectrum problems of the general block operator matrix $T=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$.

In summary, this paper conducts a preliminary study on the noncommutative Browder essential spectral equalities of block operator matrices. Although some achievements were made, there are still certain limitations. On this basis, further in-depth exploration will be carried out in the future to improve the relevant theories and methods, striving to obtain more systematic and complete research results.