# Weyl-type theorems and *k*-quasi-*M*-hyponormal operators

- Fei Zuo†
^{1}and - Hongliang Zuo†
^{1}Email author

**2013**:446

https://doi.org/10.1186/1029-242X-2013-446

© Zuo and Zuo; licensee Springer. 2013

**Received: **24 April 2013

**Accepted: **29 August 2013

**Published: **1 November 2013

## Abstract

In this paper, we show that if *E* is the Riesz idempotent for a non-zeroisolated point *λ* of the spectrum of a*k*-quasi-*M*-hyponormal operator *T*, then *E* isself-adjoint, and $R(E)=N(T-\lambda )=N{(T-\lambda )}^{\ast}$. Also, we obtain that Weyl-type theorems hold foralgebraically *k*-quasi-*M*-hyponormal operators.

**MSC:** 47B20, 47A10.

### Keywords

*k*-quasi-

*M*-hyponormal operators Riesz idempotent operators Weyl-type theorems

## 1 Introduction

Let *T* be a bounded linear operator on a complex Hilbert space *H*,write it for $T\in B(H)$, take a complex number *λ* in ℂ, and,henceforth, shorten $T-\lambda I$ to $T-\lambda $. One of recent trends in operator theory is studying naturalextensions of normal operators. We introduce some of these operators as follows.

*T* is said to be a hyponormal operator if ${T}^{\ast}T\ge T{T}^{\ast}$;

*T*is

*M*-hyponormal [1] if there exists a real positive number

*M*such that

*T*is quasi-

*M*-hyponormal [2] if there exits a real positive number

*M*such that

*T*is

*k*-quasi-

*M*-hyponormal [3] if there exists a real positive number

*M*such that

where *k* is a natural number.

It is clear that $\text{hyponormal}\Rightarrow M\text{-hyponormal}\Rightarrow k\text{-quasi-}M\text{-hyponormal}$.

We give the following example to indicate that there exists an *M*-hyponormaloperator, which is not hyponormal.

**Example 1.1** Consider the unilateral weighted shift operator as aninfinite-dimensional Hilbert space operator. Recall that given a bounded sequence ofpositive numbers *α*: ${\alpha}_{1},{\alpha}_{2},{\alpha}_{3},\dots $ (called weights), the unilateral weighted shift ${W}_{\alpha}$ associated with *α* is the operator on $H={l}_{2}$ defined by ${W}_{\alpha}{e}_{n}:={\alpha}_{n}{e}_{n+1}$ for all $n\ge 1$, where ${\{{e}_{n}\}}_{n=1}^{\mathrm{\infty}}$ is the canonical orthogonal basis for ${l}_{2}$. It is well known that ${W}_{\alpha}$ is hyponormal if and only if *α* is monotonicallyincreasing. Also, ${W}_{\alpha}$ is *M*-hyponormal if and only if *α* iseventually increasing. Hence, if we take the weights *α* such that ${\alpha}_{1}=2$, ${\alpha}_{2}=1$, ${\alpha}_{3}=2$, ${\alpha}_{3}={\alpha}_{4}=\cdots $ , then ${W}_{\alpha}$ is an *M*-hyponormal operator, but it is nothyponormal.

Next, we give a 2-quasi-*M*-hyponormal operator, which is not*M*-hyponormal.

**Example 1.2** Let $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ defined on ${\mathbb{C}}^{2}$. Then by simple calculations, we see that *T* is a2-quasi-*M*-hyponormal operator, but is not *M*-hyponormal.

If $T\in B(H)$, we shall write $N(T)$ and $R(T)$ for the null space and the range space of *T*. Also, let $\alpha (T):=dimN(T)$, $\beta (T):=dimN({T}^{\ast})$, $\sigma (T)$ and $iso\sigma (T)$ for the spectrum and the isolated points of the spectrum of*T*, respectively.

Let $\lambda \in iso\sigma (T)$. The Riesz idempotent *E* of *T* with respect to*λ* is defined by $E=\frac{1}{2\pi i}{\int}_{\partial D}{(\mu -T)}^{-1}\phantom{\rule{0.2em}{0ex}}d\mu $, where *D* is a closed disk, centered at *λ*,which contains no other points of $\sigma (T)$. It is well known that the Riesz idempotent satisfies ${E}^{2}=E$, $ET=TE$, $\sigma (T{|}_{R(E)})=\{\lambda \}$, and $N(T-\lambda )\subseteq R(E)$. Stampfli [4] showed that if *T* satisfies the growth condition ${G}_{1}$, then *E* is self-adjoint and $R(E)=N(T-\lambda )$. Recently, Chō and Tanahashi [5] obtained an improvement of Stampfli’s result to*p*-hyponormal operators or log-hyponormal operators. Furthermore, Chōand Han extended it to *M*-hyponormal operators as follows.

**Proposition 1.3** [[6], Theorem 4]

*Let* *T* *be an* *M*-*hyponormal operator*, *and let* *λ* *be an isolated point of*$\sigma (T)$. *If* *E* *is the Riesz idempotent for* *λ*, *then* *E* *is self*-*adjoint*, *and*$R(E)=N(T-\lambda )=N{(T-\lambda )}^{\ast}$.

## 2 Isolated point of spectrum of *k*-quasi-*M*-hyponormaloperators

**Lemma 2.1** *Let* *T* *be a* *k*-*quasi*-*M*-*hyponormal operator*. *If*$0\ne \lambda \in \mathbb{C}$, *and assume that*$\sigma (T)=\{\lambda \}$, *then*$T=\lambda I$.

*Proof* If $\lambda \ne 0$ and $\sigma (T)=\{\lambda \}$, then *T* is invertible, so *T* is an*M*-hyponormal operator, and hence, $T=\lambda I$ by [6]. □

**Lemma 2.2** *Let* *T* *be a* *k*-*quasi*-*M*-*hyponormal operator and*$0\ne \lambda \in \mathbb{C}$. *Then*$Tx=\lambda x$*implies that*${T}^{\ast}x=\overline{\lambda}x$.

*Proof* Suppose that $Tx=\lambda x$. Since *T* is a *k*-quasi-*M*-hyponormaloperator, $M\parallel (T-\alpha ){T}^{k}y\parallel \ge \parallel {(T-\alpha )}^{\ast}{T}^{k}y\parallel $ for all vectors $y\in H$ and $\alpha \in \mathbb{C}$. In particular, $M\parallel (T-\lambda ){T}^{k}x\parallel \ge \parallel {(T-\lambda )}^{\ast}{T}^{k}x\parallel $. Since $Tx=\lambda x$, $0=M{|\lambda |}^{k}\parallel (T-\lambda )x\parallel =M\parallel (T-\lambda ){T}^{k}x\parallel \ge \parallel {(T-\lambda )}^{\ast}{T}^{k}x\parallel ={|\lambda |}^{k}\parallel {(T-\lambda )}^{\ast}x\parallel $. $|\lambda |\ne 0$, therefore $\parallel {(T-\lambda )}^{\ast}x\parallel =0$. □

**Theorem 2.3**

*Let*

*T*

*be a*

*k*-

*quasi*-

*M*-

*hyponormal operator*,

*and let*

*λ*

*be a non*-

*zero isolated point of*$\sigma (T)$.

*Then the Riesz idempotent*

*E*

*for*

*λ*

*is self*-

*adjoint*,

*and*

*Proof* We can derive the result from Lemma 2.2, [[3], Theorem 2.5] and [[7], Lemma 5.2]. □

## 3 Weyl-type theorems of algebraically *k*-quasi-*M*-hyponormaloperators

We say that *T* is an algebraically *k*-quasi-*M*-hyponormaloperator if there exists a nonconstant complex polynomial *p* such that $p(T)$ is a *k*-quasi-*M*-hyponormal operator. From thedefinition above, *T* is an algebraically*k*-quasi-*M*-hyponormal operator, then so is $T-\lambda $ for each $\lambda \in \mathbb{C}$.

An operator *T* is called Fredholm if $R(T)$ is closed, and both $N(T)$ and $N({T}^{\ast})$ are finite-dimensional. The index of a Fredholm operator*T* is given by $i(T)=\alpha (T)-\beta (T)$. An operator *T* is called Weyl if it is Fredholm of indexzero. The Weyl spectrum of *T*[8] is defined by $w(T):=\{\lambda \in \mathbb{C}:T-\lambda \text{is not Weyl}\}$. Following [9], we say that Weyl’s theorem holds for *T* if $\sigma (T)\mathrm{\setminus}w(T)={\pi}_{00}(T)$, where ${\pi}_{00}(T):=\{\lambda \in iso\sigma (T):0<\alpha (T-\lambda )<\mathrm{\infty}\}$.

More generally, Berkani investigated the *B*-Fredholm theory (see [10–12]). We define $T\in SB{F}_{+}^{-}(H)$ if there exists a positive integer *n* such that $R({T}^{n})$ is closed, ${T}_{[n]}:R({T}^{n})\ni x\to Tx\in R({T}^{n})$ is upper semi-Fredholm (*i.e.*, $R({T}_{[n]})=R({T}^{n+1})$ is closed, $dimN({T}_{[n]})=dimN(T)\cap R({T}^{n})<\mathrm{\infty}$) and $i({T}_{[n]})\le 0$[12]. We define ${\sigma}_{SB{F}_{+}^{-}}(T)=\{\lambda \in \mathbb{C}:T-\lambda \notin SB{F}_{+}^{-}(H)\}$. Let ${E}^{a}(T)$ denote the set of all isolated points *λ* of ${\sigma}_{a}(T)$ with $0<\alpha (T-\lambda )$. We say that generalized *a*-Weyl’s theorem holds for*T* if ${\sigma}_{a}(T)\setminus {\sigma}_{SB{F}_{+}^{-}}(T)={E}^{a}(T)$.

We know that Weyl’s theorem holds for hermitian operators [13], which have been extended to hyponormal operators [14], algebraically hyponormal operators by [15], algebraically *M*-hyponormal operators [6] and algebraically quasi-*M*-hyponormal operators [2], respectively. In this section, we obtain that generalized*a*-Weyl’s theorems hold for algebraically*k*-quasi-*M*-hyponormal operators.

**Lemma 3.1**[3]

*Let*$T\in B(H)$

*be a*

*k*-

*quasi*-

*M*-

*hyponormal operator*,

*let the rangeof*${T}^{k}$

*be not dense and*

*Then*${T}_{1}$*is* *M*-*hyponormal*, ${T}_{3}^{k}=0$*and*$\sigma (T)=\sigma ({T}_{1})\cup \{0\}$.

**Theorem 3.2** *Let* *T* *be a quasinilpotent algebraically* *k*-*quasi*-*M*-*hyponormal operator*. *Then* *T* *is nilpotent*.

*Proof*We first assume that

*T*is a

*k*-quasi-

*M*-hyponormal operator. Consider two cases, Case I: If therange of ${T}^{k}$ has dense range, then it is an

*M*-hyponormal operator.Hence, by [[6], Lemma 8],

*T*is nilpotent. Case II: If

*T*does nothave dense range, then by Lemma 3.1, we can represent

*T*as the uppertriangular matrix

*M*-hyponormal operator. Since

*T*isquasinilpotent, $\sigma (T)=\{0\}$. But $\sigma (T)=\sigma ({T}_{1})\cup \{0\}$, hence, $\sigma ({T}_{1})=\{0\}$. Since ${T}_{1}$ is an

*M*-hyponormal operator, ${T}_{1}=0$. Since ${T}_{3}^{k}=0$, simple computation shows that

*T*is an algebraically

*k*-quasi-

*M*-hyponormal operator. Then there exists a nonconstantpolynomial

*p*such that $p(T)$ is a

*k*-quasi-

*M*-hyponormal operator. If $p(T)$ has dense range, then $p(T)$ is an

*M*-hyponormal operator. Thus

*T*is analgebraically

*M*-hyponormal operator. It follows from [[6], Lemma 8] that it is nilpotent. If ${(p(T))}^{k}$ does not have a dense range, then by Lemma 3.1, we canrepresent $p(T)$ as the upper triangular matrix

where $A:=p(T)|\overline{R({(p(T))}^{k})}$ is an *M*-hyponormal operator. Since $\sigma (T)=\{0\}$ and $\sigma (p(T))=p(\sigma (T))=\{p(0)\}$, the operator $p(T)-p(0)$ is quasinilpotent. But $\sigma (p(T))=\sigma (A)\cup \{0\}$, thus $\sigma (A)\cup \{0\}=\{p(0)\}$. So $p(0)=0$, and hence, $p(T)$ is quasinilpotent. Since $p(T)$ is a *k*-quasi-*M*-hyponormal operator, by theprevious argument $p(T)$ is nilpotent. On the other hand, since $p(0)=0$, $p(z)=c{z}^{m}(z-{\lambda}_{1})(z-{\lambda}_{2})\cdots (z-{\lambda}_{n})$ for some natural number *m*. $p(T)=c{T}^{m}(T-{\lambda}_{1})(T-{\lambda}_{2})\cdots (T-{\lambda}_{n})$. $p(T)$ is nilpotent, therefore, *T* isnilpotent. □

Recall that an operator *T* is said to be isoloid if every isolated point of $\sigma (T)$ is an eigenvalue of *T* and polaroid if every isolatedpoint of $\sigma (T)$ is a pole of the resolvent of *T*. In general, if*T* is polaroid, then it is isoloid. However, the converse is not true.In [6], it is showed that every algebraically *M*-hyponormal operator isisoloid, we can prove more.

**Theorem 3.3** *Let* *T* *be an algebraically* *k*-*quasi*-*M*-*hyponormal operator*. *Then* *T* *is polaroid*.

*Proof*Suppose that

*T*is an algebraically

*k*-quasi-

*M*-hyponormal operator. Then $p(T)$ is a

*k*-quasi-

*M*-hyponormal operator for somenonconstant polynomial

*p*. Let $\lambda \in iso\sigma (T)$ and ${E}_{\lambda}$ be the Riesz idempotent associated to

*λ*defined by ${E}_{\lambda}:=\frac{1}{2\pi i}{\int}_{\partial D}{(\mu -T)}^{-1}\phantom{\rule{0.2em}{0ex}}d\mu $, where

*D*is a closed disk of center

*λ*,which contains no other point of $\sigma (T)$. We can represent

*T*as the direct sum in the followingform:

where $\sigma ({T}_{1})=\{\lambda \}$ and $\sigma ({T}_{2})=\sigma (T)\mathrm{\setminus}\{\lambda \}$. Since ${T}_{1}$ is an algebraically *k*-quasi-*M*-hyponormaloperator, so is ${T}_{1}-\lambda $. But $\sigma ({T}_{1}-\lambda )=\{0\}$, it follows from Theorem 3.2 that ${T}_{1}-\lambda $ is nilpotent, thus ${T}_{1}-\lambda $ has finite ascent and descent. On the other hand, since ${T}_{2}-\lambda $ is invertible, clearly, it has finite ascent and descent. $T-\lambda $ has finite ascent and descent, and hence, *λ* is apole of the resolvent of *T*, therefore, *T* ispolaroid. □

**Corollary 3.4** *Let* *T* *be an algebraically* *k*-*quasi*-*M*-*hyponormal operator*. *Then* *T* *is isoloid*.

*T*has the single valued extension property (abbreviated SVEP)if, for every open set

*U*of ℂ, the only analytic solution

*f*: $U\to H$ of the equation

is a zero function on *U*.

**Theorem 3.5** *Let* *T* *be an algebraically* *k*-*quasi*-*M*-*hyponormal operator*. *Then* *T* *has SVEP*.

*Proof* Suppose that *T* is an algebraically*k*-quasi-*M*-hyponormal operator. Then $p(T)$ is a *k*-quasi-*M*-hyponormal operator for somenonconstant complex polynomial *p*, and hence, $p(T)$ has SVEP by [[3], Theorem 2.1]. Therefore, *T* has SVEP by [[16], Theorem 3.3.9]. □

In the following theorem, $H(\sigma (T))$ denotes the space of functions analytic in an open neighborhood of $\sigma (T)$.

**Theorem 3.6** *Let* *T* *or*${T}^{\ast}$*be an algebraically* *k*-*quasi*-*M*-*hyponormal operator*. *ThenWeyl’s theorem holds for*$f(T)$*for every*$f\in H(\sigma (T))$.

*Proof* Firstly, suppose that *T* is an algebraically*k*-quasi-*M*-hyponormal operator. We first show that Weyl’stheorem holds for *T*. Using the fact [[17], Theorem 2.2] that if *T* is polaroid, then Weyl’stheorem holds for *T* if and only if *T* has SVEP at points of $\lambda \in \sigma (T)\setminus w(T)$. We have that *T* is polaroid by Theorem 3.3, and*T* has SVEP by Theorem 3.5. Hence, *T* satisfiesWeyl’s theorem.

Next, suppose that ${T}^{\ast}$ is an algebraically *k*-quasi-*M*-hyponormaloperator. Now we show that Weyl’s theorem holds for *T*. We use thefact [[18], Theorem 3.1] that if *T* or ${T}^{\ast}$ has SVEP, then Weyl’s theorem holds for *T* if andonly if ${\pi}_{00}(T)={p}_{00}(T)$. Since ${T}^{\ast}$ has SVEP, it is sufficient to show that ${\pi}_{00}(T)={p}_{00}(T)$. ${p}_{00}(T)\subseteq {\pi}_{00}(T)$ is clear, so we only need to prove ${\pi}_{00}(T)\subseteq {p}_{00}(T)$. Let $\lambda \in {\pi}_{00}(T)$. Then *λ* is an isolated point of $\sigma (T)$. Hence, *λ* is a pole of the resolvent of *T*,since *T* is polaroid by Theorem 3.3, that is, $p(\lambda -T)=q(\lambda -T)<\mathrm{\infty}$. By assumption, we have $\alpha (\lambda -T)<\mathrm{\infty}$, so $\beta (\lambda -T)<\mathrm{\infty}$. Hence, we conclude that $\lambda \in {p}_{00}(T)$. Therefore, Weyl’s theorem holds for *T*.

Finally, we can derive the result by Theorem 3.5 and [[17], Theorem 2.4]. □

Following [[19], Theorem 3.12], we obtain the following result.

**Theorem 3.7**

*Let*

*f*

*be an analytic function on*$\sigma (T)$,

*and*

*f*

*is not constant on each connected component of the open set*

*U*

*containing*$\sigma (T)$.

- (i)
*If*${T}^{\ast}$*is an algebraically**k*-*quasi*-*M*-*hyponormal operator*,*then*$f(T)$*satisfies a generalized a*-*Weyl’s theorem*. - (ii)
*If**T**is an algebraically**k*-*quasi*-*M*-*hyponormal operator*,*then*$f({T}^{\ast})$*satisfies a generalized a*-*Weyl’s theorem*.

## Notes

## Declarations

### Acknowledgements

We wish to thank the referee for careful reading and valuable comments for theorigin draft. This work is supported by the Basic Science and TechnologicalFrontier Project of Henan Province (No. 132300410261). This work is partiallysupported by the National Natural Science Foundation of China (No. 11271112,11201126).

## Authors’ Affiliations

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