# Upper triangular operator matrices, asymptotic intertwining and Browder, Weyl theorems

## Abstract

Given a Banach space $\mathcal{X}$, let ${M}_{C}\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ denote the upper triangular operator matrix ${M}_{C}=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, and let ${\delta }_{AB}\in B\left(B\left(\mathcal{X}\right)\right)$ denote the generalized derivation ${\delta }_{AB}\left(X\right)=AX-XB$. If ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$, then ${\sigma }_{x}\left({M}_{C}\right)={\sigma }_{x}\left({M}_{0}\right)$, where ${\sigma }_{x}$ stands for the spectrum or a distinguished part thereof (but not the point spectrum); furthermore, if $R={R}_{1}\oplus {R}_{2}\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ is a Riesz operator which commutes with ${M}_{C}$, then ${\sigma }_{x}\left({M}_{C}+R\right)={\sigma }_{x}\left({M}_{C}\right)$, where ${\sigma }_{x}$ stands for the Fredholm essential spectrum or a distinguished part thereof. These results are applied to prove the equivalence of Browder’s (a-Browder’s) theorem for ${M}_{0}$, ${M}_{C}$, ${M}_{0}+R$ and ${M}_{C}+R$. Sufficient conditions for the equivalence of Weyl’s (a-Weyl’s) theorem are also considered.

MSC:47B40, 47A10, 47B47, 47A11.

## 1 Introduction

A Banach space operator $T\in B\left(\mathcal{X}\right)$, the algebra of bounded linear transformations from a Banach space $\mathcal{X}$ into itself, satisfies Browder’s theorem if the Browder spectrum ${\sigma }_{b}\left(T\right)$ of T coincides with the Weyl spectrum ${\sigma }_{w}\left(T\right)$ of T; T satisfies Weyl’s theorem if the complement of ${\sigma }_{w}\left(T\right)$ in $\sigma \left(T\right)$ is the set ${\mathrm{\Pi }}_{0}\left(T\right)$ of finite multiplicity isolated eigenvalues of T. Weyl’s theorem implies Browder’s theorem, but the converse is generally false (see [13]). Let ${M}_{0}$ and ${M}_{C}\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ denote, respectively, the upper triangular operators ${M}_{0}=A\oplus B$ and ${M}_{C}=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$ for some operators $A,C,B\in B\left(\mathcal{X}\right)$. It is well known that ${\sigma }_{x}\left({M}_{0}\right)={\sigma }_{x}\left(A\right)\cup {\sigma }_{x}\left(B\right)={\sigma }_{x}\left({M}_{C}\right)\cup \left\{{\sigma }_{x}\left(A\right)\cap {\sigma }_{x}\left(B\right)\right\}$ for , and ${\sigma }_{w}\left({M}_{0}\right)\subseteq {\sigma }_{w}\left(A\right)\cup {\sigma }_{w}\left(B\right)={\sigma }_{w}\left({M}_{C}\right)\cup \left\{{\sigma }_{w}\left(A\right)\cap {\sigma }_{w}\left(B\right)\right\}$. The problem of finding sufficient conditions ensuring the equality of the spectrum (and certain of its distinguished parts) of ${M}_{0}$ and ${M}_{C}$, along with the problem of finding sufficient conditions for ${M}_{0}$ satisfies Browder’s theorem and/or Weyl’s theorem to imply ${M}_{C}$ satisfies Browder’s theorem and/or Weyl’s theorem (and vice versa), has been considered by a number of authors in the recent past (see [3], and some of the references cited there). For example, if either ${A}^{\ast }$ or B has the single-valued extension property, SVEP for short, then $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)=\sigma \left(A\right)\cup \sigma \left(B\right)$. Again, if ${\sigma }_{w}\left({M}_{C}\right)={\sigma }_{w}\left(A\right)\cup {\sigma }_{w}\left(B\right)$, then $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)=\sigma \left(A\right)\cup \sigma \left(B\right)$ [[3], Proposition 3.2] and ${M}_{0}$ satisfies Browder’s theorem if and only if ${M}_{C}$ satisfies Browder’s theorem [[3], Theorem 4.8]; furthermore, in such a case, ${M}_{0}$ satisfies Weyl’s theorem if and only if ${M}_{C}$ satisfies Weyl’s theorem if and only if ${\mathrm{\Pi }}_{0}\left({M}_{0}\right)={\mathrm{\Pi }}_{0}\left({M}_{C}\right)$ [[3], Theorem 5.1]. The equality ${\sigma }_{w}\left({M}_{C}\right)={\sigma }_{w}\left(A\right)\cup {\sigma }_{w}\left(B\right)$ may be achieved in a number of ways: if either A and ${A}^{\ast }$, or A and B, or ${A}^{\ast }$ and ${B}^{\ast }$, or B and ${B}^{\ast }$ have SVEP, then ${\sigma }_{w}\left({M}_{C}\right)={\sigma }_{w}\left(A\right)\cup {\sigma }_{w}\left(B\right)$ [[3], Proposition 4.5]. In this paper we consider conditions of another kind, conditions which do not assume SVEP.

Given $S,T\in B\left(\mathcal{X}\right)$, S and T are said to be asymptotically intertwined by $X\in B\left(\mathcal{X}\right)$ if ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{ST}^{n}\left(X\right)\parallel }^{\frac{1}{n}}=0$. Here ${\delta }_{ST}\in B\left(B\left(\mathcal{X}\right)\right)$ is the generalized derivation ${\delta }_{ST}\left(X\right)=SX-XT$ and ${\delta }_{ST}^{n}={\delta }_{ST}\left({\delta }_{ST}^{n-1}\right)$. Evidently, S and T asymptotically intertwined by X does not imply T and S asymptotically intertwined by X. Furthermore, S and T asymptotically intertwined by X does not imply $\sigma \left(S\right)=\sigma \left(T\right)$, not even $\sigma \left(S\right)\subseteq \sigma \left(T\right)$; see [[4], Example 3.5.9]. However, as we shall see, if A, B, C are as in the definition of ${M}_{C}$ above, then A and B asymptotically intertwined by C implies the equality of the spectra, and many distinguished parts thereof to spectrum of ${M}_{0}$ and ${M}_{C}$. We prove in the following that if ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$, then ${M}_{C}$ satisfies Browder’s theorem if and only if ${M}_{0}$ satisfies Browder’s theorem. If, additionally, the isolated points of $\sigma \left({M}_{0}\right)$ are poles of the resolvent of ${M}_{0}$, then ${M}_{c}$ satisfies Weyl’s theorem if and only if ${M}_{0}$ satisfies Weyl’s theorem. Extensions to a-Browder’s theorem, a-Weyl’s theorem and perturbations by Riesz operators are considered.

## 2 Notation and complementary results

For a bounded linear Banach space operator $S\in B\left(\mathcal{X}\right)$, let $\sigma \left(S\right)$, ${\sigma }_{p}\left(S\right)$, ${\sigma }_{a}\left(S\right)$, ${\sigma }_{s}\left(S\right)$ and $iso\sigma \left(S\right)$ denote, respectively, the spectrum, the point spectrum, the approximate point spectrum, the surjectivity spectrum and the isolated points of the spectrum of S. Let $\alpha \left(S\right)$ and $\beta \left(S\right)$ denote the nullity and the deficiency of S, defined by

$\alpha \left(S\right)=dim{S}^{-1}\left(0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\beta \left(S\right)=codimS\left(\mathcal{X}\right).$

If the range $S\left(\mathcal{X}\right)$ of S is closed and $\alpha \left(S\right)<\mathrm{\infty }$ (resp. $\beta \left(S\right)<\mathrm{\infty }$), then S is called an upper semi-Fredholm (resp. a lower semi-Fredholm) operator. If $S\in B\left(\mathcal{X}\right)$ is either upper or lower semi-Fredholm, S is called a semi-Fredholm operator, and $ind\left(S\right)$, the index of S, is then defined by $ind\left(S\right)=\alpha \left(S\right)-\beta \left(S\right)$. If both $\alpha \left(S\right)$ and $\beta \left(S\right)$ are finite, then S is a Fredholm operator. The ascent, denoted $asc\left(S\right)$, and the descent, denoted $dsc\left(S\right)$, of S are given by

$asc\left(S\right)=inf\left\{n:{S}^{-n}\left(0\right)={S}^{-\left(n+1\right)}\left(0\right)\right\},\phantom{\rule{2em}{0ex}}dsc\left(S\right)=inf\left\{n:{S}^{n}\left(\mathcal{X}\right)={S}^{n+1}\left(\mathcal{X}\right)\right\}$

(where the infimum is taken over the set of non-negative integers); if no such integer n exists, then $asc\left(S\right)=\mathrm{\infty }$, respectively $dsc\left(S\right)=\mathrm{\infty }$. Let

Here ${\sigma }_{w}\left(S\right)$ is the Weyl spectrum, ${\sigma }_{aw}\left(S\right)$ denotes the Weyl (essential) approximate point spectrum, ${\sigma }_{sw}\left(S\right)$ the Weyl (essential) surjectivity spectrum, ${\sigma }_{b}\left(S\right)$ the Browder spectrum, ${\sigma }_{ab}\left(S\right)$ the Browder (essential) approximate point spectrum, ${\sigma }_{sb}\left(S\right)$ the Browder (essential) surjectivity spectrum, and ${H}_{0}\left(S\right)$ the quasi-nilpotent part of S [1]. Recall, [1], that ${H}_{0}\left(S\right)$ and $K\left(S\right)$, where $K\left(S\right)$ denotes the analytic core

are hyper-invariant (generally non-closed) subspaces of S such that ${S}^{-p}\left(0\right)\subseteq {H}_{0}\left(S\right)$ for every integer $p\ge 0$ and $SK\left(S\right)=K\left(S\right)$. Recall also that if $0\in iso\sigma \left(S\right)$, then $\mathcal{X}={H}_{0}\left(S\right)\oplus K\left(S\right)$.

We say that S has the single valued extension property, or SVEP, at $\lambda \in \mathtt{C}$ if for every open neighborhood U of λ, the only analytic solution f to the equation $\left(S-\mu \right)f\left(\mu \right)=0$ for all $\mu \in U$ is the constant function $f\equiv 0$; we say that S has SVEP if S has a SVEP at every $\lambda \in \mathtt{C}$. It is well known that finite ascent implies SVEP; also, an operator has SVEP at every isolated point of its spectrum (as well as at every isolated point of its approximate point spectrum).

$S\in B\left(\mathcal{X}\right)$ satisfies Browder’s theorem, shortened to S satisfies Bt, if ${\sigma }_{w}\left(S\right)={\sigma }_{b}\left(S\right)$ (if and only if $\sigma \left(S\right)\setminus {\sigma }_{w}\left(S\right)={p}_{0}\left(S\right)$, see [[1], p.156]); S satisfies Weyl’s theorem, shortened to S satisfies Wt, if $\sigma \left(S\right)\setminus {\sigma }_{w}\left(S\right)={\mathrm{\Pi }}_{0}\left(S\right)$ (if and only if S satisfies Bt and ${p}_{0}\left(S\right)={\mathrm{\Pi }}_{0}\left(S\right)$) [[1], p.177]. The implication $\text{Wt}⟹\text{Bt}$ is well known.

An isolated point $\lambda \in iso\sigma \left(S\right)$ is a pole (of the resolvent) of $S\in B\left(\mathcal{X}\right)$ if $asc\left(S-\lambda \right)=dsc\left(S-\lambda \right)<\mathrm{\infty }$. In such a case we say that S is polar at λ; we say that S is polaroid (resp., polaroid on a subset F of the set of isolated points of $\sigma \left(S\right)$) if S is polar at every $\lambda \in iso\sigma \left(S\right)$ (resp., at every $\lambda \in F$). Let $p\left(S\right)$ denote the set of poles of S.

Throughout the following, ${M}_{0}\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ shall denote the diagonal operator ${M}_{0}=A\oplus B$ and ${M}_{C}\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ shall denote the upper triangular operator matrix $\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$, for some operators $A,B,C\in B\left(\mathcal{X}\right)$. Recall, [[5], Exercise 7, p.293], that $asc\left(A\right)\le asc\left({M}_{C}\right)\le asc\left(A\right)+asc\left(B\right)$ and $dsc\left(B\right)\le dsc\left({M}_{C}\right)\le dsc\left(A\right)+dsc\left(B\right)$.

Lemma 2.1 If $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)$, then $p\left({M}_{0}\right)=p\left({M}_{C}\right)$.

Proof Since $\sigma \left({M}_{C}\right)=\sigma \left({M}_{0}\right)=\sigma \left(A\right)\cup \sigma \left(B\right)$, if a complex number $\lambda \in p\left({M}_{C}\right)$ or $p\left({M}_{0}\right)$ then $\lambda \in iso\left(\sigma \left(A\right)\cup \sigma \left(B\right)\right)$. We consider the case in which $\lambda \in iso\sigma \left(A\right)\cap iso\sigma \left(B\right)$: the argument works just as well for the case in which $\lambda \in \rho \left(A\right)$ ($=\mathtt{C}\setminus \sigma \left(A\right)$) or $\lambda \in \rho \left(B\right)$. Let $\lambda \in p\left({M}_{C}\right)$. Then

$asc\left(A-\lambda \right)\le asc\left({M}_{C}-\lambda \right)<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}dsc\left(B-\lambda \right)\le dsc\left({M}_{C}-\lambda \right)<\mathrm{\infty }.$

If $\lambda \in iso\sigma \left(B\right)$ and $dsc\left(B-\lambda \right)<\mathrm{\infty }$, then $asc\left(B-\lambda \right)=dsc\left(B-\lambda \right)<\mathrm{\infty }$ and B is polar at λ [[1], Theorem 3.81]. Now let $\lambda \in iso\sigma \left(A\right)$. Since ${M}_{C}$ is polar at λ, ${H}_{0}\left({M}_{C}-\lambda \right)={\left({M}_{C}-\lambda \right)}^{-p}\left(0\right)$ for some integer $p\ge 1$. Observe that

${H}_{0}\left(A-\lambda \right)={H}_{0}\left({M}_{C}-\lambda \right)\cap \mathcal{X}={\left({M}_{C}-\lambda \right)}^{-p}\left(0\right)\cap \mathcal{X}={\left(A-\lambda \right)}^{-p}\left(0\right).$

Hence, if $\lambda \in iso\sigma \left(A\right)$, then

$\begin{array}{c}\mathcal{X}={H}_{0}\left(A-\lambda \right)\oplus K\left(A-\lambda \right)={\left(A-\lambda \right)}^{-p}\left(0\right)\oplus K\left(A-\lambda \right)\hfill \\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{\left(A-\lambda \right)}^{p}\mathcal{X}=0\oplus {\left(A-\lambda \right)}^{p}K\left(A-\lambda \right)=K\left(A-\lambda \right)\hfill \\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\mathcal{X}={\left(A-\lambda \right)}^{-p}\left(0\right)\oplus {\left(A-\lambda \right)}^{p}\mathcal{X},\hfill \end{array}$

i.e., A is polar at λ. Now, since

$asc\left({M}_{0}-\lambda \right)\le asc\left(A-\lambda \right)+asc\left(B-\lambda \right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}dsc\left({M}_{0}-\lambda \right)\le dsc\left(A-\lambda \right)+dsc\left(B-\lambda \right),$

we have

$asc\left({M}_{0}-\lambda \right)=dsc\left({M}_{0}-\lambda \right)<\mathrm{\infty },$

i.e., ${M}_{0}$ is polar at λ. Conversely, if $\lambda \in p\left({M}_{0}\right)$, then $asc\left({M}_{0}-\lambda \right)=max\left\{asc\left(A-\lambda \right),asc\left(B-\lambda \right)\right\}$ and $dsc\left({M}_{0}-\lambda \right)=max\left\{dsc\left(A-\lambda \right),dsc\left(B-\lambda \right)\right\}$ implies $asc\left({M}_{C}-\lambda \right)\le asc\left(A-\lambda \right)+asc\left(B-\lambda \right)$ and $dsc\left({M}_{C}-\lambda \right)\le dsc\left(A-\lambda \right)+dsc\left(B-\lambda \right)$ are both finite, hence equal. Thus ${M}_{C}$ is polar at λ. □

Remark 2.2 A number of conditions guaranteeing (the spectral equality) $\sigma \left({M}_{C}\right)=\sigma \left({M}_{0}\right)$ are to be found in the literature. Thus, for example, if ${A}^{\ast }$ or B has SVEP, or if ${\sigma }_{w}\left({M}_{C}\right)={\sigma }_{w}\left(A\right)\cup {\sigma }_{w}\left(B\right)$, or ${\sigma }_{aw}\left({M}_{C}\right)={\sigma }_{aw}\left(A\right)\cup {\sigma }_{aw}\left(B\right)$ [[3], (I) p.5 and Proposition 3.2], then $\sigma \left({M}_{C}\right)=\sigma \left({M}_{0}\right)$. Compact operators have SVEP; hence, if either of A or B is compact, then $\sigma \left({M}_{C}\right)=\sigma \left({M}_{0}\right)$.

Lemma 2.1 shows that if B is a compact operator then $p\left({M}_{0}\right)=p\left({M}_{C}\right)$. A proof of the following lemma may be obtained from that of Lemma 2.1: we give here an independent proof, exploiting the additional information contained in the hypothesis.

Lemma 2.3 If $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)$, then ${p}_{0}\left({M}_{0}\right)={p}_{0}\left({M}_{c}\right)$.

Proof Once again we consider points $\lambda \in iso\sigma \left(A\right)\cap iso\sigma \left(B\right)$. Let $\lambda \in {p}_{0}\left({M}_{C}\right)$. Then $\alpha \left({M}_{C}-\lambda \right)=\beta \left({M}_{C}-\lambda \right)<\mathrm{\infty }$ implies ${M}_{C}-\lambda \in \mathrm{\Phi }$, and this in turn implies $A-\lambda \in {\mathrm{\Phi }}_{+}$ and $B-\lambda \in {\mathrm{\Phi }}_{-}$. Since λ is isolated in $\sigma \left(A\right)$ and $\sigma \left(B\right)$, $\lambda \in {p}_{0}\left(A\right)\cap {p}_{0}\left(B\right)$ [[1], Theorem 3.77]. Consequently, $\lambda \in p\left({M}_{0}\right)$; furthermore, since $\alpha \left({M}_{0}-\lambda \right)\le \alpha \left(A-\lambda \right)+\alpha \left(B-\lambda \right)$, $\lambda \in {p}_{0}\left({M}_{0}\right)$. Conversely, if $\lambda \in {p}_{0}\left({M}_{0}\right)$, then $A-\lambda$ and $B-\lambda \in \mathrm{\Phi }$, and hence (since λ is isolated in $\sigma \left(A\right)$ and $\sigma \left(B\right)$) $\lambda \in {p}_{0}\left(A\right)\cap {p}_{0}\left(B\right)$. This, as above, implies $\lambda \in {p}_{0}\left({M}_{C}\right)$. □

The following technical lemma will be required in the sequel.

Lemma 2.4 If A is polaroid on ${\mathrm{\Pi }}_{0}\left({M}_{C}\right)$ and $\sigma \left({M}_{C}\right)=\sigma \left({M}_{0}\right)$, then ${\mathrm{\Pi }}_{0}\left({M}_{C}\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{0}\right)$.

Proof Evidently, ${\left({M}_{C}-\lambda \right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$ implies ${\left({M}_{0}-\lambda \right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$, and $\alpha \left({M}_{C}-\lambda \right)<\mathrm{\infty }$ implies $\alpha \left(A-\lambda \right)<\mathrm{\infty }$. Let $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{C}\right)$; then $\lambda \in iso\sigma \left({M}_{0}\right)$. We prove that $\alpha \left(B-\lambda \right)<\mathrm{\infty }$. Suppose to the contrary that $\alpha \left(B-\lambda \right)=\mathrm{\infty }$. Since

$\left({M}_{C}-\lambda \right)\left(x\oplus y\right)=\left\{\left(A-\lambda \right)x+Cy\right\}\oplus \left(B-\lambda \right)y,$

either $dim\left(C{\left(B-\lambda \right)}^{-1}\left(0\right)\right)<\mathrm{\infty }$ or $dim\left(C{\left(B-\lambda \right)}^{-1}\left(0\right)\right)=\mathrm{\infty }$. If $dim\left(C{\left(B-\lambda \right)}^{-1}\left(0\right)\right)<\mathrm{\infty }$, then (since $\alpha \left(B-\lambda \right)=\mathrm{\infty }$) ${\left(B-\lambda \right)}^{-1}\left(0\right)$ contains an orthonormal sequence $\left\{{y}_{j}\right\}$ such that $\left({M}_{C}-\lambda \right)\left(0\oplus {y}_{j}\right)=0$ for all $j=1,2,\dots$ . But then $\alpha \left({M}_{C}-\lambda \right)=\mathrm{\infty }$, a contradiction. Hence $dim\left(C{\left(B-\lambda \right)}^{-1}\left(0\right)\right)=\mathrm{\infty }$. Since $\lambda \in \rho \left(A\right)\cup iso\sigma \left(A\right)$ and A is (by hypothesis) polar at λ (with, as observed above, $\alpha \left(A-\lambda \right)<\mathrm{\infty }$) $\alpha \left(A-\lambda \right)=\beta \left(A-\lambda \right)<\mathrm{\infty }$. Thus $dim\left\{C{\left(B-\lambda \right)}^{-1}\left(0\right)\cap \left(A-\lambda \right)\mathcal{X}\right\}=\mathrm{\infty }$, and so there exists a sequence $\left\{{x}_{j}\right\}$ such that $\left(A-\lambda \right){x}_{j}=C{y}_{j}$ for all $j=1,2,\dots$ . But then $\left({M}_{C}-\lambda \right)\left({x}_{j}\oplus -{y}_{j}\right)=0$ for all $j=1,2,\dots$ , and hence $\alpha \left({M}_{C}-\lambda \right)=\mathrm{\infty }$. This contradiction implies that we must have $\alpha \left(B-\lambda \right)<\mathrm{\infty }$. Since $\alpha \left({M}_{0}-\lambda \right)\le \alpha \left(A-\lambda \right)+\alpha \left(B-\lambda \right)$, we conclude that $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{0}\right)$. □

Let ${\delta }_{ST}\in B\left(B\left(\mathcal{X}\right)\right)$ denote the generalized derivation ${\delta }_{ST}\left(X\right)=SX-XT$, and define ${\delta }_{ST}^{n}$ by ${\delta }_{ST}^{n-1}\left({\delta }_{ST}\right)$. The operators $S,T\in B\left(\mathcal{X}\right)$ are said to be asymptotically intertwined by the identity operator $I\in B\left(\mathcal{X}\right)$ if ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{ST}^{n}\left(I\right)\parallel }^{\frac{1}{n}}=0$; S, T are said to be quasi-nilpotent equivalent if ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{ST}^{n}\left(I\right)\parallel }^{\frac{1}{n}}={lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{TS}^{n}\left(I\right)\parallel }^{\frac{1}{n}}=0$ [[4], p.253]. Quasi-nilpotent equivalence preserves a number of spectral properties [[4], Proposition 3.4.11]. In particular:

Lemma 2.5 Quasi-nilpotent equivalent operators have the same spectrum, the same approximate point spectrum and the same surjectivity spectrum.

## 3 Results

Let $\mathcal{K}\left(\mathcal{X}\right)$ denote the ideal of compact operators in $B\left(\mathcal{X}\right)$. The following construction, known in the literature as the Sadovskii/Buoni, Harte and Wickstead construction [[6], p.159], leads to a representation of the Calkin algebra $B\left(\mathcal{X}\right)/\mathcal{K}\left(\mathcal{X}\right)$ as an algebra of operators on a suitable Banach space. Let $S\in B\left(\mathcal{X}\right)$. Let ${\ell }^{\mathrm{\infty }}\left(\mathcal{X}\right)$ denote the Banach space of all bounded sequences $x={\left({x}_{n}\right)}_{n=1}^{\mathrm{\infty }}$ of elements of $\mathcal{X}$ endowed with the norm ${\parallel x\parallel }_{\mathrm{\infty }}:={sup}_{n\in \mathtt{N}}\parallel {x}_{n}\parallel$, and write ${S}_{\mathrm{\infty }}$, ${S}_{\mathrm{\infty }}x:={\left(S{x}_{n}\right)}_{n=1}^{\mathrm{\infty }}$ for all $x={\left({x}_{n}\right)}_{n=1}^{\mathrm{\infty }}$, for the operator induced by S on ${\ell }^{\mathrm{\infty }}\left(\mathcal{X}\right)$. The set $m\left(\mathcal{X}\right)$ of all precompact sequences of elements of $\mathcal{X}$ is a closed subspace of ${\ell }^{\mathrm{\infty }}\left(\mathcal{X}\right)$ which is invariant for ${S}_{\mathrm{\infty }}$. Let ${\mathcal{X}}_{q}:={\ell }^{\mathrm{\infty }}\left(\mathcal{X}\right)/m\left(\mathcal{X}\right)$, and denote by ${S}_{q}$ the operator ${S}_{\mathrm{\infty }}$ on ${\mathcal{X}}_{q}$. The mapping $S↦{S}_{q}$ is then a unital homomorphism from $B\left(\mathcal{X}\right)\to B\left({\mathcal{X}}_{q}\right)$ with kernel $\mathcal{K}\left(\mathcal{X}\right)$ which induces a norm decreasing monomorphism from $B\left(\mathcal{X}\right)/\mathcal{K}\left(\mathcal{X}\right)$ to $B\left({\mathcal{X}}_{q}\right)$ with the following properties (see [[6], Section 17] for details):

1. (i)

S is upper semi-Fredholm, $S\in {\mathrm{\Phi }}_{+}$, if and only if ${S}_{q}$ is injective, if and only if ${S}_{q}$ is bounded below;

2. (ii)

S is lower semi-Fredholm, $S\in {\mathrm{\Phi }}_{-}$, if and only if ${S}_{q}$ is surjective;

3. (iii)

S is Fredholm, $S\in \mathrm{\Phi }$, if and only if ${S}_{q}$ is invertible.

Lemma 3.1 For every $S\in B\left(\mathcal{X}\right)$, ${\sigma }_{e}\left(S\right)=\sigma \left({S}_{q}\right)$, ${\sigma }_{S{F}_{+}}\left(S\right)={\sigma }_{a}\left({S}_{q}\right)$ and ${\sigma }_{S{F}_{-}}\left(S\right)={\sigma }_{s}\left({S}_{q}\right)$.

Proof The following implications hold:

□

The following theorem is essentially known [7] we provide here an alternative proof, using quasi-nilpotent equivalence and the construction above. Let ${\mathrm{\Sigma }}_{0}$ denote either of ${\sigma }_{e}$, ${\sigma }_{S{F}_{+}}$, ${\sigma }_{S{F}_{-}}$, ${\sigma }_{w}$, ${\sigma }_{aw}$, ${\sigma }_{sw}$, ${\sigma }_{b}$, ${\sigma }_{ab}$ and ${\sigma }_{sb}$.

Theorem 3.2 Let $S,R\in B\left(\mathcal{X}\right)$. If R is a Riesz operator which commutes with S, then ${\sigma }_{x}\left(S+R\right)={\sigma }_{x}\left(S\right)$, where ${\sigma }_{x}\in {\mathrm{\Sigma }}_{0}$.

Proof It is clear from the definition of a Riesz operator $R\in B\left(\mathcal{X}\right)$ that $R-\mu$ is Browder (i.e., $\mu \notin {\sigma }_{b}\left(R\right)$), and a-Browder and s-Browder, for all non-zero $\mu \in \sigma \left(R\right)$ (see, for example, [[1], Theorem 3.111]). Hence $\sigma \left({R}_{q}\right)=\left\{0\right\}$, i.e., ${R}_{q}\in B\left({\mathcal{X}}_{q}\right)$ is quasi-nilpotent. Let $t\in \left[0,1\right]$; then S commutes with tR and ${\left(S+tR\right)}_{q}={S}_{q}+t{R}_{q}$. It follows that

$\underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{\left(S+tR\right)}_{q}{S}_{q}}^{n}\left({I}_{q}\right)\parallel }^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{S}_{q}{\left(S+tR\right)}_{q}}^{n}\left({I}_{q}\right)\parallel }^{\frac{1}{n}}=0,$

i.e., ${S}_{q}$ and ${S}_{q}+t{R}_{q}$ are quasi-nilpotent equivalent operators for all $t\in \left[0,1\right]$. Thus ${\sigma }_{x}\left({\left(S+R\right)}_{q}\right)={\sigma }_{x}\left({S}_{q}\right)$, where . Hence

The semi-Fredholm index being a continuous function, we also have from the above that

To complete the proof, we prove next that ${\sigma }_{b}\left(S+R\right)={\sigma }_{b}\left(S\right)$; the proof for ${\sigma }_{ab}$ and ${\sigma }_{sb}$ is similar, and left to the reader. It would suffice to prove that $0\in {\sigma }_{b}\left(S\right)⟺0\in {\sigma }_{b}\left(S+R\right)$. Suppose that $0\notin {\sigma }_{b}\left(S\right)$. Then $S\in \mathrm{\Phi }$ (and $asc\left(S\right)=dsc\left(S\right)<\mathrm{\infty }$), hence $S+tR\in \mathrm{\Phi }$ for all $t\in \left[0,1\right]$. For an operator T, let $\overline{{\mathcal{N}}^{\mathrm{\infty }}\left(T\right)}$ and ${T}^{\mathrm{\infty }}\left(\mathcal{X}\right)$ denote, respectively, the closure of the hyper kernel and the hyper range of T. Then $\overline{{\mathcal{N}}^{\mathrm{\infty }}\left(S+tR\right)}\cap {\left(S+tR\right)}^{\mathrm{\infty }}\left(\mathcal{X}\right)$ is constant on $\left[0,1\right]$, and so, since $\overline{{\mathcal{N}}^{\mathrm{\infty }}\left(S\right)}\cap {S}^{\mathrm{\infty }}\left(\mathcal{X}\right)={\mathcal{N}}^{\mathrm{\infty }}\left(S\right)\cap {S}^{\mathrm{\infty }}\left(\mathcal{X}\right)=\left\{0\right\}$, it follows that ${\mathcal{N}}^{\mathrm{\infty }}\left(S+R\right)\cap {\left(S+R\right)}^{\mathrm{\infty }}\left(\mathcal{X}\right)=\left\{0\right\}$. Consequently, $S+R$ has SVEP at 0 [[1], Corollary 2.26]. But then since $S+R\in \mathrm{\Phi }$, $S+R$ is Browder. Considering $S=\left(S+R\right)-R$ proves $0\notin {\sigma }_{b}\left(S+R\right)⟹0\notin {\sigma }_{b}\left(S\right)$. □

The following lemma appears in [[8], Lemma 2.3]. Let ${\mathrm{\Pi }}_{0f}\left(S\right)=\left\{\lambda \in iso\sigma \left(S\right):\alpha \left(S-\lambda \right)<\mathrm{\infty }\right\}$. Clearly, ${\mathrm{\Pi }}_{0}\left(S\right)\subseteq {\mathrm{\Pi }}_{0f}\left(S\right)$.

Lemma 3.3 If $S,R\in B\left(\mathcal{X}\right)$, and R is a Riesz operator which commutes with S, then ${\mathrm{\Pi }}_{0f}\left(S+R\right)\cap \sigma \left(S\right)\subseteq iso\sigma \left(S\right)$.

Let $\mathrm{\Sigma }={\mathrm{\Sigma }}_{0}\cup \sigma \cup {\sigma }_{a}\cup {\sigma }_{s}$.

Theorem 3.4 If ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$, then ${\sigma }_{x}\left({M}_{C}\right)={\sigma }_{x}\left({M}_{0}\right)$, where ${\sigma }_{x}\in \mathrm{\Sigma }$.

Proof A straightforward calculation shows that

${\delta }_{{M}_{C}{M}_{0}}^{n}\left(I\right)=-{\delta }_{{M}_{0}{M}_{C}}^{n}\left(I\right)=\left(\begin{array}{cc}0& {\delta }_{AB}^{n-1}\left(C\right)\\ 0& 0\end{array}\right).$

Hence

$\underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{M}_{C}{M}_{0}}^{n}\left(I\right)\parallel }^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{M}_{0}{M}_{C}}^{n}\left(I\right)\parallel }^{\frac{1}{n}}\le \underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{AB}^{n-1}\left(C\right)\parallel }^{\frac{1}{n}}=0,$

i.e., ${M}_{C}$ and ${M}_{0}$ are quasi-nilpotent equivalent. Similarly, writing ${M}_{C\left(q\right)}$ for ${\left({M}_{C}\right)}_{q}$ and ${M}_{0\left(q\right)}$ for ${\left({M}_{0}\right)}_{q}$,

$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{M}_{C\left(q\right)}{M}_{0\left(q\right)}}^{n}\left({I}_{q}\right)\parallel }^{\frac{1}{n}}& =& \underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{M}_{0\left(q\right)}{M}_{C\left(q\right)}}^{n}\left({I}_{q}\right)\parallel }^{\frac{1}{n}}\\ \le & \underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{{A}_{q}{B}_{q}}^{n-1}\left({C}_{q}\right)\parallel }^{\frac{1}{n}}\\ =& \underset{n\to \mathrm{\infty }}{lim}{\parallel {\delta }_{AB}^{n-1}\left(C\right)\parallel }^{\frac{1}{n}}=0,\end{array}$

i.e., ${M}_{C\left(q\right)}$ and ${M}_{0\left(q\right)}$ are quasi-nilpotent equivalent (in $B\left({\left(\mathcal{X}\oplus \mathcal{X}\right)}_{q}\right)$). Hence ${\sigma }_{x}\left({M}_{C}\right)={\sigma }_{x}\left({M}_{0}\right)$, where . Since

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

and

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

where $\left(\begin{array}{cc}I& C\\ 0& I\end{array}\right)$ is invertible, and since $\lambda \notin {\sigma }_{e}\left({M}_{C}\right)⟺\lambda \notin {\sigma }_{e}\left({M}_{0}\right)⟹A-\lambda ,B-\lambda \in \mathrm{\Phi }$ (similarly, $\lambda \notin {\sigma }_{S{F}_{+}}\left({M}_{C}\right)⟹A-\lambda ,B-\lambda \in {\mathrm{\Phi }}_{+}$ and $\lambda \notin {\sigma }_{S{F}_{-}}\left({M}_{C}\right)⟹A-\lambda ,B-\lambda \in {\mathrm{\Phi }}_{-}$), $ind\left({M}_{C}-\lambda \right)=ind\left(A-\lambda \right)+ind\left(B-\lambda \right)=ind\left({M}_{0}-\lambda \right)$. Hence ${\sigma }_{x}\left({M}_{C}\right)={\sigma }_{x}\left({M}_{0}\right)$, where . Observe that

and

[[1], Corollary 3.23, Theorem 3.23 and Theorem 3.27]. Hence ${\sigma }_{x}\left({M}_{C}\right)={\sigma }_{x}\left({M}_{0}\right)$, where . □

Remark 3.5 If $M\in B\left(\mathcal{X}\oplus \mathcal{X}\right)$ is the operator $M=\left(\begin{array}{cc}A& C\\ D& B\end{array}\right)$ such that the entries A, B, C and D mutually commute, then ${\sigma }_{x}\left(M\right)=\left\{\lambda \in \mathtt{C}:0\in {\sigma }_{x}\left(\left(A-\lambda \right)\left(B-\lambda \right)-CD\right)\right\}$ [[9], Theorem 2.3], where . Dispensing with the mutual commutativity hypothesis and assuming instead that $CD=DC=0$, C commutes with A and B, and ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(D\right)\parallel }^{\frac{1}{n}}=0$, an argument similar to that used to prove Theorem 3.4 shows that ${\sigma }_{x}\left(M\right)={\sigma }_{x}\left({M}_{C}\right)$, where .

Theorem 3.6 Suppose that ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$. Then:

1. (a)

${M}_{C}$ satisfies Bt if and only if ${M}_{0}$ satisfies Bt.

2. (b)

Let ${R}_{i}\in B\left(\mathcal{X}\right)$, $i=1,2$, be Riesz operators such that $R={R}_{1}\oplus {R}_{2}$ commutes with ${M}_{C}$. Then ${M}_{0}$ satisfies $\mathit{\text{Bt}}⟺{M}_{C}+R$ satisfies $\mathit{\text{Bt}}⟺{M}_{0}+R$ satisfies $\mathit{\text{Bt}}⟺{M}_{C}$ satisfies Bt.

Proof The hypothesis R commutes with ${M}_{C}$ implies R commutes with ${M}_{0}$, ${R}_{1}C=C{R}_{2}$ and ${\delta }_{\left({M}_{C}+R\right)\left({M}_{0}+R\right)}^{n}\left(I\right)={\delta }_{{M}_{C}M}^{n}\left(I\right)$.

1. (a)

Recall that an operator S satisfies Bt if and only if ${\sigma }_{w}\left(S\right)={\sigma }_{b}\left(S\right)$. Hence the following implications hold:

2. (b)

The hypothesis ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$ implies that ${M}_{C}+R$ and ${M}_{0}+R$ are quasi-nilpotent equivalent ( by Theorem 3.4 that ${\sigma }_{x}\left({M}_{C}+R\right)={\sigma }_{x}\left({M}_{0}+R\right)$, where ${\sigma }_{x}\in \mathrm{\Sigma }$). The operator R being Riesz, Theorem 3.2 implies ${\sigma }_{x}\left(T+R\right)={\sigma }_{x}\left(T\right)$, where $T={M}_{C}$ or ${M}_{0}$ and ${\sigma }_{x}={\sigma }_{w}$ or ${\sigma }_{b}$. The (two way) implications

now complete the proof. □

Remark 3.7 (i) $S\in B\left(\mathcal{X}\right)$ satisfies a-Browder’s theorem, a-Bt, if and only if ${\sigma }_{aw}\left(S\right)={\sigma }_{ab}\left(S\right)$ (equivalently, if and only if ${\sigma }_{a}\left(S\right)\setminus {\sigma }_{aw}\left(S\right)={p}_{0}^{a}\left(S\right)$ = $\left\{\lambda \in iso{\sigma }_{a}\left(S\right):S-\lambda \in {\mathrm{\Phi }}_{+}\right\}$ = $\left\{\lambda \in {\sigma }_{a}\left(S\right):S-\lambda \in {\mathrm{\Phi }}_{+},asc\left(S-\lambda \right)<\mathrm{\infty }\right\}$ [[2], Theorem 3.3]). Theorem 3.6 holds with Bt replaced by a-Bt. (Thus, if either ${M}_{0}$ or ${M}_{C}$ satisfies a-Bt, then ${M}_{0}$, ${M}_{C}$, ${M}_{0}+R$ and ${M}_{C}+R$ all satisfy a-Bt.) Furthermore, since S satisfies generalized Browder’s theorem, gBt, if and only if it satisfies Bt and S satisfies generalized a-Browder’s theorem, a-gBt, if and only if it satisfies a-Bt [10], Bt may be replaced by gBt or a-gBt in Theorem 3.6. Here, we refer the interested reader to consult [2, 10] for information about gBt and a-gBt.

1. (ii)

The equivalence S satisfies $\text{Bt}⟺{S}^{\ast }$ satisfies Bt is well known. This does not hold for a-Bt: S satisfies a-Bt does not imply ${S}^{\ast }$ satisfies a-Bt (or vice versa). We say that S satisfies s-Bt if ${S}^{\ast }$ satisfies a-Bt (equivalently, if ${\sigma }_{sb}\left(S\right)={\sigma }_{sw}\left(S\right)$). It is easily seen, we leave the verification to the reader, if either ${M}_{0}$ or ${M}_{C}$ satisfies s-Bt, then (in Theorem 3.6) ${M}_{0}$, ${M}_{C}$, ${M}_{0}+R$ and ${M}_{C}+R$ all satisfy s-Bt.

We consider next a sufficient condition for the equivalence of Weyl’s theorem for operators ${M}_{0}$ and ${M}_{C}$ such that ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$. We say in the following that an operator S is finitely polaroid on a subset $F\subseteq iso\sigma \left(S\right)$ if every $\lambda \in F$ is a finite rank pole of S. Evidently, ${M}_{0}$ is finitely polaroid if and only if A and B are finitely polaroid.

Theorem 3.8 Suppose that ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$.

1. (a)

If A is polaroid, then ${M}_{C}$ satisfies Wt if and only if ${M}_{0}$ satisfies Wt.

2. (b)

Let ${R}_{i}\in B\left(\mathcal{X}\right)$, $i=1,2$, be Riesz operators such that $R={R}_{1}\oplus {R}_{2}$ commutes with ${M}_{C}$. A sufficient condition for the equivalence ${M}_{C}+R$ satisfies $\mathit{\text{Wt}}⟺{M}_{0}+R$ satisfies Wt is that ${M}_{0}$ is finitely polaroid.

Proof (a) If ${M}_{C}$ satisfies Wt, then $\sigma \left({M}_{C}\right)\setminus {\sigma }_{w}\left({M}_{C}\right)={p}_{0}\left({M}_{C}\right)={\mathrm{\Pi }}_{0}\left({M}_{C}\right)$. Since $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)$ and ${\sigma }_{w}\left({M}_{C}\right)={\sigma }_{w}\left({M}_{0}\right)$ (Theorem 3.4) and since Wt implies Bt, Theorem 3.6(a) implies $\sigma \left({M}_{0}\right)\setminus {\sigma }_{w}\left({M}_{0}\right)={p}_{0}\left({M}_{0}\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{0}\right)$. Consequently, ${\mathrm{\Pi }}_{0}\left({M}_{C}\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{0}\right)$. Let $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{0}\right)$. Then $\lambda \in iso\sigma \left({M}_{C}\right)$, $\alpha \left(A-\lambda \right)<\mathrm{\infty }$ and $\alpha \left(B-\lambda \right)<\mathrm{\infty }$. Hence, since $\alpha \left(A-\lambda \right)\le \alpha \left({M}_{C}-\lambda \right)\le \alpha \left(A-\lambda \right)+\alpha \left(B-\lambda \right)$, $\alpha \left({M}_{C}-\lambda \right)<\mathrm{\infty }$. Evidently, $\lambda \in iso\sigma \left(A\right)\cup \rho \left(A\right)$. If $\lambda \in iso\sigma \left(A\right)$, then A polaroid implies $0<\alpha \left(A-\lambda \right)$, and hence $0<\alpha \left({M}_{C}-\lambda \right)$. If instead $\lambda \in \rho \left(A\right)$, then $-{\left(A-\lambda \right)}^{-1}Cx\oplus x\in {\left({M}_{C}-\lambda \right)}^{-1}\left(0\right)$ for every $x\in {\left(B-\lambda \right)}^{-1}\left(0\right)$; once again, $0<\alpha \left({M}_{C}-\lambda \right)$. Consequently, $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{C}-\lambda \right)={p}_{0}\left({M}_{C}-\lambda \right)={p}_{0}\left({M}_{0}-\lambda \right)$ and hence ${\mathrm{\Pi }}_{0}\left({M}_{0}\right)={p}_{0}\left({M}_{0}\right)⟹{M}_{0}$ satisfies Wt. Conversely, if ${M}_{0}$ satisfies Wt, then $\sigma \left({M}_{C}\right)\setminus {\sigma }_{w}\left({M}_{C}\right)={p}_{0}\left({M}_{C}\right)={p}_{0}\left({M}_{0}\right)={\mathrm{\Pi }}_{0}\left({M}_{0}\right)=\sigma \left({M}_{0}\right)\setminus {\sigma }_{w}\left({M}_{0}\right)$ and ${\mathrm{\Pi }}_{0}\left({M}_{0}\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{C}\right)$. Since A is polaroid (hence polar on ${\mathrm{\Pi }}_{0}\left({M}_{C}\right)$) and $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)$, Lemma 2.4 implies ${\mathrm{\Pi }}_{0}\left({M}_{0}\right)={\mathrm{\Pi }}_{0}\left({M}_{C}\right)$. Thus ${M}_{C}$ satisfies Wt.

1. (b)

Start by observing that $\sigma \left({M}_{0}\right)=\sigma \left({M}_{C}\right)$, and hence ${M}_{C}$ is finitely polaroid if and only if ${M}_{0}$ is finitely polaroid (Lemma 2.3). Suppose ${M}_{0}+R$ satisfies Wt. Then the implication $\text{Wt}⟹\text{Bt}$ combined with Theorem 3.6(b) implies that both ${M}_{0}+R$ and ${M}_{C}+R$ satisfy Bt. As noted in the proof of Theorem 3.6(b), ${\sigma }_{w}\left(T+R\right)={\sigma }_{w}\left(T\right)$, $T={M}_{0}$ or ${M}_{C}$. Furthermore, since ${M}_{0}+R$ and ${M}_{C}+R$ are quasi-nilpotent equivalent, ${\sigma }_{x}\left({M}_{0}+R\right)={\sigma }_{x}\left({M}_{C}+R\right)$, (Theorem 3.4). Hence

$\begin{array}{rcl}{\mathrm{\Pi }}_{0}\left({M}_{0}+R\right)& =& \sigma \left({M}_{0}+R\right)\setminus {\sigma }_{w}\left({M}_{0}+R\right)=\sigma \left({M}_{C}+R\right)\setminus {\sigma }_{w}\left({M}_{C}+R\right)\\ =& {p}_{0}\left({M}_{C}+R\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{C}+R\right).\end{array}$

If $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{C}+R\right)$ and $\lambda \notin \sigma \left({M}_{C}\right)$, then $\left({M}_{C}-\lambda \right)$ is invertible and so ${M}_{C}-\lambda \in \mathrm{\Phi }⟹{M}_{C}+R-\lambda \in \mathrm{\Phi }$. Hence, since $\lambda \in iso\sigma \left({M}_{C}+R\right)$, $\lambda \in {p}_{0}\left({M}_{C}+R\right)$. If, instead, $\lambda \in \sigma \left({M}_{C}\right)$, then $\lambda \in iso\sigma \left({M}_{C}\right)$ (Lemma 3.3) $⟹\lambda \in iso\sigma \left({M}_{0}\right)⟹\lambda \in {p}_{0}\left({M}_{0}\right)$ (since ${M}_{0}$ is finitely polaroid) $⟹\lambda \in {p}_{0}\left({M}_{C}\right)$ (Lemma 2.3) $⟹{M}_{C}-\lambda \in \mathrm{\Phi }$, and this as above implies $\lambda \in {p}_{0}\left({M}_{c}+R\right)$. Hence ${\mathrm{\Pi }}_{0}\left({M}_{C}+R\right)={p}_{0}\left({M}_{C}+R\right)$, and ${M}_{C}+R$ satisfies Wt. The converse, ${M}_{C}+R$ satisfies $\text{Wt}⟹{M}_{0}+R$ satisfies Wt follows from a similar argument (recall that ${M}_{C}$ is finitely polaroid follows from the hypothesis that ${M}_{0}$ is finitely polaroid). □

Remark 3.9 The equivalence of Theorem 3.8(b) extends to

This is seen as follows. The implication ${M}_{0}+R$ satisfies $\text{Wt}⟹{M}_{0}$ satisfies Bt and ${M}_{C}+R$ satisfies $\text{Wt}⟹{M}_{C}$ satisfies Bt are clear from Theorem 3.6(b). If ${M}_{0}$ satisfies Bt, then the hypothesis ${M}_{0}$ is finitely polaroid implies ${M}_{0}$ satisfies Wt. By Theorem 3.6(b), ${M}_{0}+R$ satisfies Bt, i.e., $\sigma \left({M}_{0}+R\right)\setminus {\sigma }_{w}\left({M}_{0}+R\right)={p}_{0}\left({M}_{0}+R\right)\subseteq {\mathrm{\Pi }}_{0}\left({M}_{0}+R\right)$. Let $\lambda \in {\mathrm{\Pi }}_{0}\left({M}_{0}+R\right)$. If $\lambda \notin \sigma \left({M}_{0}\right)$, then (${M}_{0}-\lambda \in \mathrm{\Phi }⟹$) ${M}_{0}+R-\lambda \in \mathrm{\Phi }⟹\lambda \in {p}_{0}\left({M}_{0}+R\right)$ (since $\lambda \in iso\sigma \left({M}_{0}+R\right)$); if $\lambda \in \sigma \left({M}_{0}\right)$, then $\lambda \in iso\sigma \left({M}_{0}\right)$ (by Lemma 3.3) and so (since ${M}_{0}$ is finitely polaroid) $\lambda \in {p}_{0}\left({M}_{0}\right)⟹{M}_{0}-\lambda \in \mathrm{\Phi }⟹{M}_{0}+R-\lambda \in \mathrm{\Phi }⟹\lambda \in {p}_{0}\left({M}_{0}+R\right)$. Thus, in either case, ${\mathrm{\Pi }}_{0}\left({M}_{0}+R\right)\subseteq {p}_{0}\left({M}_{0}+R\right)$, and hence ${M}_{0}+R$ satisfies Wt. The proof for ${M}_{C}$ satisfies $\text{Bt}⟹{M}_{C}+R$ satisfies Wt is similar: recall from Lemma 2.3 that ${M}_{0}$ finitely polaroid implies ${M}_{C}$ finitely polaroid.

a-Wt. $T\in B\left(\mathcal{X}\right)$ satisfies a-Weyl’s theorem, a-Wt for short, if T satisfies a-Bt and ${p}_{0}^{a}\left(T\right)={\mathrm{\Pi }}_{0}^{a}\left(T\right)$ (equivalently, if ${\sigma }_{a}\left(T\right)\setminus {\sigma }_{aw}\left(T\right)={p}_{0}^{a}\left(T\right)={\mathrm{\Pi }}_{0}^{a}\left(T\right)$), where ${\mathrm{\Pi }}_{0}^{a}\left(T\right)=\left\{\lambda \in iso{\sigma }_{a}\left(T\right):0<\alpha \left(T-\lambda \right)<\mathrm{\infty }\right\}$ [1]. We say in the following that T is a-polaroid if T is polar at every $\lambda \in iso{\sigma }_{a}\left(T\right)$. Trivially, a-polaroid implies polaroid (indeed, ${p}_{0}^{a}\left(T\right)={p}_{0}\left(T\right)$ in such a case), but the converse is not true in general. Theorem 3.8 has an a-Wt analogue, which we prove below. We note, however, that the perturbation of an operator by a commuting Riesz operator preserves neither its spectrum nor its approximate point spectrum: this will, per se, force us into making an assumption on the approximate point spectrum of ${M}_{0}$ and ${M}_{0}+R$ in the analogue of Theorem 3.8(b).

Theorem 3.10 Suppose that ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$.

1. (a)

If ${M}_{0}$ is a-polaroid, then ${M}_{C}$ satisfies a-Wt if and only if ${M}_{0}$ satisfies a-Wt.

2. (b)

Let ${R}_{i}\in B\left(\mathcal{X}\right)$, $i=1,2$, be Riesz operators such that $R={R}_{1}\oplus {R}_{2}$ commutes with ${M}_{C}$. If ${\sigma }_{a}\left({M}_{0}\right)={\sigma }_{a}\left({M}_{0}+R\right)$, then a sufficient condition for the equivalence ${M}_{C}+R$ satisfies $a\text{-}\mathit{\text{Wt}}⟺{M}_{0}+R$ satisfies a-Wt is that ${M}_{0}$ is finitely a-polaroid.

Proof (a) We prove ${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{C}\right)$: the proof of (a) would then follow from the fact that if ${M}_{0}$ satisfies a-Wt ($⟹{M}_{0}$ satisfies $a\text{-Bt}⟺{M}_{C}$ satisfies a-Bt), then

${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}\right)={\sigma }_{a}\left({M}_{0}\right)\setminus {\sigma }_{aw}\left({M}_{0}\right)={\sigma }_{a}\left({M}_{C}\right)\setminus {\sigma }_{aw}\left({M}_{C}\right)={p}_{0}^{a}\left({M}_{C}\right)\subseteq {\mathrm{\Pi }}_{0}^{a}\left({M}_{C}\right)$

and if ${M}_{C}$ satisfies a-Wt, then

${\mathrm{\Pi }}_{0}^{a}\left({M}_{C}\right)={\sigma }_{a}\left({M}_{C}\right)\setminus {\sigma }_{aw}\left({M}_{C}\right)={\sigma }_{a}\left({M}_{0}\right)\setminus {\sigma }_{aw}\left({M}_{0}\right)={p}_{0}^{a}\left({M}_{0}\right)\subseteq {\mathrm{\Pi }}_{0}^{a}\left({M}_{0}\right).$

If $\lambda \in {\mathrm{\Pi }}_{0}^{a}\left({M}_{0}\right)$, then

if instead $\lambda \in {\mathrm{\Pi }}_{0}^{a}\left({M}_{C}\right)$, then

1. (b)

If ${\sigma }_{a}\left({M}_{0}+R\right)={\sigma }_{a}\left({M}_{0}\right)$, then it follows from Lemma 2.4 and Theorem 3.4 that

Recall from Remark 3.7 that if either of ${M}_{0}+R$ or ${M}_{C}+R$ satisfies a-Bt, then ${M}_{0}$, ${M}_{0}+R$, ${M}_{C}$ and ${M}_{C}+R$ all satisfy a-Bt. Hence, in view of the spectral equalities above,

${p}_{0}^{a}\left({M}_{0}\right)={p}_{0}^{a}\left({M}_{C}\right)={p}_{0}^{a}\left({M}_{C}+R\right)={p}_{0}^{a}\left({M}_{0}+R\right),$

whenever either of ${M}_{0}$, ${M}_{0}+R$, ${M}_{C}$ and ${M}_{C}+R$ satisfies a-Bt. Observe that the hypothesis ${M}_{0}$ is finitely a-polaroid implies ${p}_{0}^{a}\left({M}_{0}\right)={p}_{0}\left({M}_{0}\right)={p}_{0}\left({M}_{C}\right)={p}_{0}^{a}\left({M}_{0}+R\right)$; hence (since ${p}_{0}^{a}\left({M}_{0}\right)={p}_{0}^{a}\left({M}_{C}\right)={p}_{0}^{a}\left({M}_{C}+R\right)={p}_{0}^{a}\left({M}_{0}+R\right)$) ${p}_{0}^{a}\left(S\right)={p}_{0}^{a}\left(T\right)$ for every choice of . We prove now that if either of ${M}_{0}+R$ and ${M}_{C}+R$ satisfies a-Wt, then ${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)$: this would then imply that if one satisfies a-Wt, then so does the other.

Suppose ${M}_{0}+R$ satisfies a-Wt. Then ${p}_{0}\left({M}_{0}+R\right)={p}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)$ ($⟹{\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}\left({M}_{0}+R\right)$) and ${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)\subseteq {\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)$. Let $\lambda \in {\mathrm{\Pi }}_{0}^{a}\left({M}_{c}+R\right)$; then $\lambda \in iso{\sigma }_{a}\left({M}_{C}+R\right)=iso{\sigma }_{a}\left({M}_{0}\right)$ implies $\lambda \in {p}_{0}\left({M}_{0}\right)={p}_{0}^{a}\left({M}_{C}+R\right)$. Thus ${\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)\subseteq {p}_{0}^{a}\left({M}_{C}+R\right)={p}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)$. Consequently, ${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)$ in this case. Suppose next that ${M}_{C}+R$ satisfies a-Wt. Then ${p}_{0}\left({M}_{C}+R\right)={p}_{0}^{a}\left({M}_{C}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)$ and ${\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)\subseteq {\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)$. Let $\lambda \in {\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)$; then $\lambda \in iso{\sigma }_{a}\left({M}_{0}\right)$ implies $\lambda \in {p}_{0}^{a}\left({M}_{0}\right)={p}_{0}^{a}\left({M}_{C}+R\right)$. As above, this implies ${\mathrm{\Pi }}_{0}^{a}\left({M}_{0}+R\right)={\mathrm{\Pi }}_{0}^{a}\left({M}_{C}+R\right)$. □

The following corollary is immediate from Theorem 3.10(b).

Corollary 3.11 Suppose that ${lim}_{n\to \mathrm{\infty }}{\parallel {\delta }_{AB}^{n}\left(C\right)\parallel }^{\frac{1}{n}}=0$. If ${R}_{i}\in B\left(\mathcal{X}\right)$, $i=1,2$, are quasi-nilpotent operators such that $R={R}_{1}\oplus {R}_{2}$ commutes with ${M}_{C}$, then a sufficient condition for the equivalence ${M}_{C}+R$ satisfies $a\text{-}\mathit{\text{Wt}}⟺{M}_{0}+R$ satisfies a-Wt is that ${M}_{0}$ is finitely a-polaroid.

## Authors’ information

Work carried out together whilst the first author was visiting Korea.

## References

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## Acknowledgements

This work was supported by the Incheon National University Research Grant in 2012.

## Author information

Authors

### Corresponding author

Correspondence to In Hyoun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally.

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Duggal, B.P., Jeon, I.H. & Kim, I.H. Upper triangular operator matrices, asymptotic intertwining and Browder, Weyl theorems. J Inequal Appl 2013, 268 (2013). https://doi.org/10.1186/1029-242X-2013-268