Cowen-Douglas operators and the third of Halmos' ten problems

TL;DR

This paper investigates Halmos' third problem on intransitive operators with inverses, providing affirmative answers under spectral conditions using Cowen-Douglas operators and spectral analysis, advancing understanding in operator theory.

Contribution

The paper offers new theoretical results affirming that intransitivity is preserved under inversion for certain classes of operators, addressing a long-standing open problem in operator theory.

Findings

Affirmative answer under spectral conditions for intransitivity preservation.

Inversion of hyponormal operators retains intransitivity if spectral conditions are met.

Intransitivity is preserved for operators with specific spectral and invariant subspace properties.

Abstract

Let $T$ be a bounded linear operator on a complex separable infinite dimensional Hilbert space $\mathcal{H}$. $T$ is called intransitive if it leaves invariant spaces other than 0 or the whole space $\mathcal{H}$; otherwise it is transitive. In 1970, P. R. Halmos raised ten open problems on operator theory. In the past more than 50 years, nine of Halmos' ten problems were answered, but only the third one has made little progress. The third problem of Halmos is the following: if an intransitive operator has an inverse, is its inverse also intransitive? In this paper, we establish a set of theoretical systems with the help of Cowen-Douglas operators and spectral analysis. We give an affirmative answer to this problem under certain spectral conditions, which make essential progress in the research of Halmos' third problem. As the first application, we show that for an invertible hyponormal operator $T$, if $T^{-1}$ is intransitive and int$\sigma(T^{-1})^{\land}$ is not connected, then $T$ is also intransitive. As the second application, we show that if $T^{-1}$ has a proper strictly cyclic invariant subspace and there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap \mathcal{U}_0=\emptyset$, where $\mathcal{U}_0$ is the connected component of $int(\sigma(T^{-1})^\land)$ containing zero point, then $T$ is intransitive.

Equations

$$K^{\wedge }\triangleq \left\{z\in \mathbb{C}:|f(z)|\le \underset{w\in K}{\sup }|f(w)|\text{for all polynomials f}\right\}.$$

$$\ker (T^{\ast }-w{)}^{\perp }\in Lat(T^{-1})\cap Lat(T).$$

$$Lat(T)=Lat(T^{-1})\text{mbox}and\mathcal{C}(T)=\mathcal{C}(T^{-1}).$$

$$T=\left(\begin{array}{cc}{T}_1& X\\ 0& T_2\end{array}\right)\begin{array}{c}\mathcal{K}\\ {\mathcal{K}}^{\perp }\end{array}.$$

$$\begin{array}[]{lll}B_{n}(\Omega):=\{T\in\mathcal{B}(\mathcal{H}):&(1)\,\,\Omega\subset\sigma(T):=\{w\in\mathbb{C}:T-w~{}~{}\mbox{is not invertible}\},\\ &(2)\,\,\mbox{span}\{\mbox{ker}(T-w),\,w\in\Omega\}=\mathcal{H},\\ &(3)\,\,\mbox{Ran}(T-w)=\mathcal{H},\\ &(4)\,\,\mbox{dim ker}(T-w)=n,\forall~{}w\in\Omega.\}\end{array}$$

$$span\{\text{mbox}ker(T-w_0{)}^k,k\ge 1\}=span\{\text{mbox}ker(T-w),w\in \Omega \}=\mathcal{H}.$$

$$span\{(T-w_1)e(w),\omega \in \Omega \}=span\{(w-w_1)e(w),\omega \in \Omega \}=span\{e(w),\omega \in \Omega \}=\mathcal{H}.$$

$$A_1=\left(\begin{array}{ccccc}0& {\alpha }_1& {\alpha }_2& {\alpha }_3& \cdots \\ 0& & & & \\ 0& & A& & \\ ⋮& & & & \end{array}\right)\begin{array}{cc}{e}_0& \\ & \\ \mathcal{H}& \\ & \end{array}.$$

$$\begin{array}[]{lll}A_{1}X&=&A_{1}\cdot A_{1}|_{(\ker A_{1})^{\bot}}\\ &=&A_{1}(P_{\ker A_{1}}A_{1}|_{(\ker A_{1})^{\bot}}+P_{(\ker A_{1})^{\bot}}A_{1}|_{(\ker A_{1})^{\bot}})\\ &=&A_{1}P_{(\ker A_{1})^{\bot}}A_{1}|_{(\ker A_{1})^{\bot}}\end{array}$$

$$\begin{array}[]{lll}XA&=&A_{1}|_{(\ker A_{1})^{\bot}}\cdot A\\ &=&A_{1}|_{(\ker A_{1})^{\bot}}P_{(\ker A_{1})^{\bot}}A_{1}|_{(\ker A_{1})^{\bot}},\end{array}$$

$$\ker A^k=\{e_0,B{e}_0,\cdots ,B^{(k-1)}{e}_0\}.$$

$$span\{\ker A^k,k\ge 0\}=\mathcal{H}.$$

$$span\{B^k{e}_0,k\ge 0\}=\mathcal{H}.$$

$$(\ker (T-\lambda {)}^{\ast }{)}^{\perp }=\overline{Ran(T-\lambda )}\in Lat(T).$$

$$T_1=T{|}_{\overline{Ran(T-\lambda )}},T_2=P_{\ker (T-\lambda {)}^{\ast }}T{|}_{\ker (T-\lambda {)}^{\ast }},T_{12}=P_{\overline{Ran(T-\lambda )}}T{|}_{\ker (T-\lambda {)}^{\ast }}.$$

$$T_2=\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right),T=\left(\begin{array}{cc}{T}_1& T_{12}\\ 0& T_2\end{array}\right)\begin{array}{cc}\overline{Ran(T-\lambda )}& \\ & \\ \ker (T-\lambda {)}^{\ast }& \end{array},T^k=\left(\begin{array}{cc}{T}_1^k& \ast \\ 0& T_2^k\end{array}\right)\begin{array}{cc}\overline{Ran(T-\lambda )}& \\ & \\ \ker (T-\lambda {)}^{\ast }& \end{array}.$$

$$P_{\ker (T-\lambda {)}^{\ast }}y={\alpha }_1{f}_1+{\alpha }_2{f}_2\ne 0.$$

$$\begin{array}[]{lll}span\{T_{2}^{k}P_{\ker(T-\lambda)^{*}}y,\,k\geq 0\}&=&span\{\lambda^{k}(\alpha_{1}f_{1}+\alpha_{2}f_{2}),\,k\geq 0\}\\ &\subseteq&\mathbb{C}\{\alpha_{1}f_{1}+\alpha_{2}f_{2}\}\neq\ker(T-\lambda)^{*}.\end{array}$$

$$\text{mbox}dim\text{mbox}ker(T^{\ast }-\lambda )<1,\lambda \in {\Omega }^{\prime }.$$

$$T_x=\left(\begin{array}{cc}T& x\otimes e_0\\ 0& S_1^{\ast }\end{array}\right).$$

$$\left(\begin{array}{cc}T-w& x\otimes e_0\\ 0& S_1^{\ast }-w\end{array}\right)\left(\begin{array}{c}\xi \\ \eta \end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

$$\left(\begin{array}{cc}T-w& x\otimes e_0\\ 0& S_1^{\ast }-w\end{array}\right)\left(\begin{array}{c}\xi \\ \eta \end{array}\right)=\left(\begin{array}{c}{\xi }^{\prime }\\ {\eta }^{\prime }\end{array}\right).$$

$$span\{y(w)\oplus e(w),w\in \Sigma \}=\mathcal{H}\oplus \mathcal{H}.$$

$$⟨y(w),x_1⟩=-⟨e(w),x_2⟩,r(T)<|w|<1.$$

$$\underset{|w|\to +\infty }{\lim }⟨y(w),x_1⟩=0,$$

$$⟨y(w),x_1⟩=⟨e(w),x_2⟩=0.$$

$$⟨\sum _{n=0}^{\infty }\left(\frac{T^{n}x}{w^{n+1}}\right),x_1⟩=\sum _{n=0}^{\infty }⟨T^{n}x,x_1⟩\frac{1}{w^{n+1}}=0,|w|>r(T).$$

$$S=\left(\begin{array}{cc}{T}^{-1}& 0\\ 0& S_1\end{array}\right).$$

$$T_x{S}_x=\left(\begin{array}{cc}T& x\otimes e_0\\ 0& S_1^{\ast }\end{array}\right)\left(\begin{array}{cc}{T}^{-1}& 0\\ 0& S_1\end{array}\right)=\left(\begin{array}{cc}I& (x\otimes e_0)S_1\\ 0& I\end{array}\right)=\left(\begin{array}{cc}I& 0\\ 0& I\end{array}\right).$$

Full text

1. Introduction

Let $\mathcal{H}$ be a complex separable infinite dimensional Hilbert space and $\mathcal{B}({\mathcal{H}})$ denote the set of bounded linear operators on $\mathcal{H}$. An operator $T\in\mathcal{B}({\mathcal{H}})$ is called intransitive if it leaves invariant spaces other than 0 or the whole space $\mathcal{H}$; otherwise it is transitive.

**Invariant subspace problem: **Let $T\in\mathcal{B}({\mathcal{H}})$. Does $T$ have a nontrivial invariant subspace?

It has been more than a century since the invariant subspace problem was raised, and it is still an open question.

A problem closely related to the famous invariant subspace problem is the third problem of Halmos. In 1970, P. R. Halmos raised ten open problems on operator theory. The third problem of Halmos is the following.

Problem 3   If an intransitive operator has an inverse, is its inverse also intransitive?

The third problem of Halmos can be expressed in the following way: If an invertible operator T has no nontrivial invariant subspaces, does $T^{-1}$ also have no nontrivial invariant subspaces? Or equivalently, if every nonzero vector $x$ is a cyclic vector of $T$, is $x$ also a cyclic vector of $T^{-1}$.

A nonzero vector $x$ in $\mathcal{H}$ is called a cyclic vector of $T$ if $span\{T^{k}x\}_{k=0}^{\infty}=\mathcal{H}$, where $span\{T^{k}x,k\geq 0\}$ is the norm closure of the linear span of elements in $\{T^{k}x\}_{k=0}^{\infty}$. In 2007, R. G. Douglas, C. Foias and C. Pearcy showed in [9] that Problem 3 cannot be answered by proving that $T$ and $T^{-1}$ have the same set of cyclic vectors: three kinds of operators are given for which these two sets of cyclic vectors differ. In 2004, the first author discussed Halmos’ third problem with R.G.Douglas and came up Lemma 5.1 of this paper, which plays a key role in this paper. The first author thanks R.G.Douglas for many stimulating discussions on Halmos’ third problem.

R. G. Douglas, C. Foias and C. Pearcy also pointed out in [9] that the third problem of Halmos is the last unresolved open problem of the ten problems of Halmos. Unfortunately, we only find part of the literature. In 1972, Problem 1 was solved by P. A. Fillmore, J. G. Stampfli and J. P. Williams in [10]. In 1974, Problem 7 was solved by C. Apostol and D. Voiculescu in [2]. In 1976, Problem 8 was solved by D. Voiculescu in [34]. In 1983, Problem 5 was resolved by C.C.Cowen, J.Long and S. Sun and the readers refer to references [7], [32], [33]. In 1997, Problem 6 was solved by G. Pisier in [28].

For $T\in\mathcal{B}(\mathcal{H})$, let $\sigma(T)$ denote the spectrum of $T$ and $\rho(T)=\mathbb{C}\setminus\sigma(T)$. Let $LatT$ denote the invariant subspaces lattice of $T$. For a Hilbert space $\mathcal{K}\subseteq\mathcal{H}$, $P_{\mathcal{K}}$ denote the orthogonal projection from $\mathcal{H}$ onto $\mathcal{K}$. Let $\sigma(T)^{\land}$ denote the polynomial convex hull of $\sigma(T)$. For a compact subset $K$ of $\mathbb{C}$, the polynomial convex hull $K^{\land}$ of $K$ is defined by ( [31], p23)

[TABLE]

By ([31], p24), $\sigma(T)^{\land}$ is the union of $\sigma(T)$ and all of the bounded connected components of $\mathbb{C}\setminus\sigma(T)$.

In the past more than 50 years, nine of Halmos’ ten problems were answered, but only the third one has made little progress. In this paper, we have established a set of theoretical systems with the help of Cowen-Douglas operators and spectral analysis, which has made essential progress in the research of Halmos’ third problem. The specific results are as follows:

Theorem A. (Theorem 6.1) Suppose $T\in\mathcal{B}(\mathcal{H})$ is an invertible operator, $x$ is a nonzero noncyclic vector of $T^{-1}$, and $\mathcal{N}(x)=span\{T^{-k}x:\,k\geq 1\}.$ Let $\overline{T}_{1}(x)=P_{\mathcal{N}(x)}T^{-1}P_{\mathcal{N}(x)}$. If $\sigma(\overline{T}_{1}(x))^{\land}\cap\rho_{F}(\overline{T}_{1}(x))$ has a connected component which does not contain zero point, then $T$ is intransitive.

Theorem B. (Theorem 6.4) Suppose $T\in\mathcal{B}(\mathcal{H})$ is an invertible operator, $x$ is a nonzero noncyclic vector of $T^{-1}$, and $\mathcal{N}(x)=span\{T^{-k}x:\,k\geq 1\}.$ Let $\overline{T}_{2}(x)=P_{\mathcal{N}(x)^{\perp}}T^{-1}P_{\mathcal{N}(x)^{\perp}}$. If $\sigma(T^{-1})\subseteq\sigma(\overline{T}_{2}(x))$ and there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap\mathcal{U}_{0}=\emptyset$, where $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land})$ containing zero point, then $T$ is intransitive.

As applications of Theorem A and Theorem B, we have the following results.

Theorem C. (Theorem 7.7) For an invertible hyponormal operator $T\in\mathcal{B}(\mathcal{H})$, if $T^{-1}$ is intransitive and int$\sigma(T^{-1})^{\land}$ is not connected, then $T$ is also intransitive.

Theorem D. (Theorem 7.11) For an invertible operator $T\in\mathcal{B}(\mathcal{H})$, if $T^{-1}$ has a proper strictly cyclic invariant subspace and there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap\mathcal{U}_{0}=\emptyset$, where $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land})$ containing zero point, then $T$ is intransitive.

The difficulty in solving the third problem of Halmos is that we only have the information that $T$ is invertible. Therefore, if we are able to answer this problem affirmatively, we need to give the operator $T$ more structure; on the other hand, if the answer to this problem is negative, it means that the answer to the invariant subspace problem is also negative. In view of this reason, we introduce the Cowen-Douglas operators as a bridge to solve this problem. In this paper, we use rigid structures of Cowen-Douglas operators to give partial answers to this problem.

This paper is divided into seven sections, organized as follows. In Section 2, we introduce some basic knowledge about operator spectrum theory. In Section 3, we introduce some results about Cowen-Douglas operators. In Section 4, we discuss cyclic vectors. In Section 5, we will study the operator structures of operators $T$ and $T^{-1}$ and build up connections between $T$, $T^{-1}$ and Cowen-Douglas operators. In Section 6, the proofs of the main theorems are given. In the last section, we give some applications of the main theorems including the descriptions of invariant subspace of hyponormal operators whose spectrums are not “thick” (In [5], Brown proved that each hypernormal operator with “thick” spectrum has a nontrivial invariant subspace) and invertible operators which have proper strictly cyclic invariant subspaces.

We refer to [3], [5], [11], [12], [14], [15], [17], [24], [25], [30] for more information on Halmos’ third problem and the invariant subspace problem. We refer to [18], [21], [27] for operator theory on Hilbert spaces.

Acknowledgements: C.Jiang, J.Fang and K.Ji contributed equally to this work. The authors thank Professor Cheng Lixin, Ji Youqing, Wang Kai, Wu jinsong, and Zhang yuanhang for many valuable discussions on the paper.

2. Preliminaries

In this paper, $\mathcal{H}$ is a complex separable infinite dimensional Hilbert space and $\mathcal{B}({\mathcal{H}})$ is the set of bounded linear operators on $\mathcal{H}$. An operator $T\in\mathcal{B}({\mathcal{H}})$ is called compact if the image of the unit ball of $\mathcal{H}$ under $T$ is a compact subset of $\mathcal{H}$. Let ${\mathcal{K}}(\mathcal{H})$ be the ideal of compact operators in $\mathcal{B}({\mathcal{H}})$, and $\pi:\mathcal{B}(\mathcal{H})\rightarrow\mathcal{B}(\mathcal{H})/{\mathcal{K}}(\mathcal{H})$ (Calkin algebra) be the natural homomorphism projection. The essential spectrum of $T$, $\sigma_{e}(T)$, is the spectrum of $\pi(T)$ in $\mathcal{B}(\mathcal{H})/{\mathcal{K}}(\mathcal{H})$ and $\mathbb{C}\backslash\sigma_{e}(T)$ is called the Fredholm domain of $T$ and is denoted by $\rho_{F}(T)$. Operator $T$ is called semi-Fredholm, if $\mbox{ran}(T)$ (the range of $T$) is closed, and either dim$\,\ker T$ or dim$\,\ker T^{*}$ is finite. In this case, $\mbox{ind}\,T=\mbox{dim}\,\ker T-\mbox{dim}\,\ker T^{*}$. A semi-Fredholm operator $T$ is called Fredholm if $\mbox{ind}\,T$ is finite. By the famous Atkinson theorem, we know $T$ is Fredholm if and only if $\pi(T)$ is invertible in $\mathcal{B}(\mathcal{H})/{\mathcal{K}}(\mathcal{H})$.

For $T\in\mathcal{B}(\mathcal{H})$, let $\sigma(T)$ denote the spectrum of $T$ and $\rho(T)=\mathbb{C}\setminus\sigma(T)$. The point spectrum of $T$, $\sigma_{p}(T)$, is defined by $\sigma_{p}(T)=\{\lambda:\mbox{dim}\,\ker(T-\lambda)\geq 1\}$. Let $LatT$ denote the invariant subspaces lattice of $T$.

The definition of strongly irreducible operators was introduced by Gilfeather [13] and Jiang Zejian [22] in 1970’s, respectively.

Definition 2.1**.**

An operator $T\in\mathcal{B}({\mathcal{H}})$ is called strongly irreducible, if the commutant of $T$, $\mathcal{A}^{\prime}(T)=\{X\in\mathcal{B}({\mathcal{H}})|XT=TX\}$, doesn’t have any nontrivial idempotent operator. Otherwise, it is called strongly reducible.

Recall that an operator $P\in\mathcal{B}({\mathcal{H}})$ is called an idempotent if $P^{2}=P$ and a nontrivial idempotent if $P\neq 0$ or $I$.

Definition 2.2**.**

A nonzero vector $x\in\mathcal{H}$ is called cyclic of $T$ if $span\{T^{k}x,\,k\geq 0\}=\mathcal{H}$, where “span” denote the norm closure of linear span of the set.

Let $\mathcal{C}(T)\triangleq\{x|x$ is a cyclic vector of $T\}$. An operator $T\in\mathcal{B}({\mathcal{H}})$ is transitive if and only if $\mathcal{C}(T)=\mathcal{H}\backslash\{0\}$. It is clear that an operator $T\in\mathcal{B}({\mathcal{H}})$ is strongly reducible, then $T$ is intransitive. When do $T$ and $T^{-1}$ have a common nontrivial invariant subspace? We have the following results.

Proposition 2.3**.**

Let $T\in\mathcal{B}({\mathcal{H}})$ be an invertible operator. If $T$ is strongly reducible, then so is $T^{-1}$. Furthermore, $T$ and $T^{-1}$ have a common nontrivial invariant subspace.

Proof.

Suppose that $P\in\mathcal{A}^{\prime}(T)$ is a nontrivial idempotent operator. Then $P\in\mathcal{A}^{\prime}(T^{-1})$. Thus $Ran\,P$ and $Ran\,(I-P)$ are both nontrivial invariant subspaces of $T$ and $T^{-1}$. ∎

Corollary 2.4**.**

Let $T\in\mathcal{B}({\mathcal{H}})$ be an invertible operator. If $\sigma(T)$ is not connected, then $T$ and $T^{-1}$ have a common nontrivial invariant subspace.

Proof.

Suppose $\sigma(T)$ is not connected. By the Riesz functional calculus, there is a nontrivial idempotent $P$ such that $P\in\mathcal{A}^{\prime}(T)\cap\mathcal{A}^{\prime}(T^{-1})$. Therefore, $RanP\in Lat(T)\cap Lat(T^{-1})$. ∎

Proposition 2.5**.**

Let $T\in\mathcal{B}({\mathcal{H}})$ be invertible. If $\sigma_{p}(T)\cup\sigma_{p}(T^{*})\neq\emptyset$, then $T$ and $T^{-1}$ have a common nontrivial invariant subspace.

Proof.

Without loss of generality, we assume that $T$ is not a scalar operator. Suppose that $w\in\sigma_{p}(T)\cup\sigma_{p}(T^{*})$. We have two cases.

(1) If $w\in\sigma_{p}(T)$, then we can find some $0\neq x\in\mathcal{H}$ such that $(T-w)x=0$. This means $T^{-1}((T-w)x)=(I-wT^{-1})x=0$. Then $x\in\ker(T^{-1}-w^{-1})$. Thus, $\ker(T-w)\in Lat(T^{-1})\cap Lat(T)$.

(2) If $w\in\sigma_{p}(T^{*})$, then for $x\in\ker(T^{*}-w)$, we have $x\in\ker((T^{*})^{-1}-w^{-1})$. Thus,

[TABLE]

∎

Let $\sigma(T)^{\land}$ denote the polynomial convex hull of $\sigma(T)$. Note that $\sigma(T)^{\land}$ is the union of $\sigma(T)$ and all of the bounded connected components of $\mathbb{C}\setminus\sigma(T)$.

Proposition 2.6**.**

[9]** Let $T\in\mathcal{B}({\mathcal{H}})$ be an invertible operator. If $0\notin\sigma(T)^{\land}$, then

[TABLE]

At the end of this section, we give the following well-known result.

Lemma 2.7**.**

Let $T\in\mathcal{B}(\mathcal{H})$ and $\mathcal{K}$ be an invariant subspace of $T$. Write

[TABLE]

Then $\sigma(T_{1})\subseteq\sigma(T)^{\land}$ and $\sigma(T_{2})\subseteq\sigma(T)^{\land}$.

3. Cowen-Douglas operators

In this section, we introduce basic properties and techniques of Cowen-Douglas operators, which play important role in the proof of our main theorem. Let $\Omega$ be a bounded open connected subset of the complex plane $\mathbb{C}$. In [8], M. J. Cowen and R. G. Douglas introduced a class of operators denoted by $B_{n}(\Omega)$ which contains $\Omega$ as eigenvalues of constant multiplicity $n$. The class of Cowen-Douglas operators with rank $n$, $B_{n}(\Omega)$ is defined as follows [8]:

[TABLE]

We recall the following results in [8].

Proposition 3.1** (1.7.1, [8]).**

Let $T\in B_{n}(\Omega)$ and $w_{0}\in\Omega$. Then

[TABLE]

Lemma 3.2**.**

[8]** If $\Omega_{1}\subset\Omega_{2}$, then $B_{1}(\Omega_{2})\subseteq B_{1}(\Omega_{1})$, where $\Omega_{1},\ \Omega_{2}$ are bounded connected open subsets of the complex plane $\mathbb{C}$.

For Cowen-Douglas operators with index one, we have the following observation.

Lemma 3.3**.**

Let $T\in B_{1}(\Omega)$. Then $\sigma_{p}(T^{*})=\emptyset$.

Proof.

Suppose that $e(w)$ is a non-zero vector such that $T(e(w))=we(w)$. By Proposition 3.1, we know $span\{e(w),\,w\in\Omega\}=\mathcal{H}$. Note that $(T-w_{1})e(w)=(w-w_{1})e(w),\,w_{1}\in\mathbb{C}$. When $w_{1}\in\Omega$, we have $Ran(T-w_{1})=\mathcal{H}$ by the definition of $B_{1}(\Omega)$. If $w_{1}\notin\Omega$, it follows that $w-w_{1}$ is non-zero. Thus,

[TABLE]

This shows that $Ran(T-w_{1})$ is dense in $\mathcal{H}$ and $\ker(T-w_{1})^{*}=\{0\}$. Therefore, $\sigma_{p}(T^{*})=\emptyset$.

∎

Remark 3.4**.**

For any $T\in B_{n}(\Omega)$, $\sigma_{p}(T^{*})=\emptyset,\,n\geq 1$.

Lemma 3.5**.**

For $T\in B_{1}(\Omega)$, there exists a connected open subset $\Phi$ of $\mathbb{C}$ such that $\Omega\subseteq\Phi$ and

1. (1)

$B_{1}(\Phi)\subseteq B_{1}(\Omega)$; 2. (2)

$\partial\Phi\subset\sigma_{e}(T)$.

In this case, we call $\Phi$ is the maximal domain of $T$.

Proof.

By Zorn’s Lemma, we can find a maximal bounded connected open subset $\Phi$ of $\mathbb{C}$ such that $T\in B_{1}(\Phi)$. Now, we need to prove $\partial\Phi\subset\sigma_{e}(T)$. Otherwise, there exists $\lambda\in\partial\Phi$ such that $\lambda\in\rho_{F}(T)$. That means ind$\,(T-\lambda)=1$ and there exists a neighborhood $O_{\lambda}$ of $\lambda$ such that ind$\,(T-\lambda)=1,\,\,\lambda\in O_{\lambda}$. By Lemma 3.3, $\sigma_{p}((T-\lambda)^{*})=\emptyset$. We have dim$\,\ker(T-\lambda)=1$. Let $\Phi^{\prime\prime}=\Phi\cup O_{\lambda}$. Then $\Phi^{\prime\prime}$ is connected. Thus, we have $T\in B_{1}(\Phi^{\prime\prime})$. This contradicts to the maximality of $\Phi$. ∎

Remark 3.6**.**

If $\Omega$ is the maximal domain of Cowen-Douglas operator $T$, then $\partial\Omega\subseteq\sigma_{e}(T).$

For two operators $A,B\in\mathcal{B}(\mathcal{H})$, we call $A\sim_{s}B$ if there exists an invertible operator $X\in\mathcal{B}(\mathcal{H})$ such that $X^{-1}AX=B$.

Lemma 3.7**.**

Suppose that $A\in\mathcal{B}({\mathcal{H}})$, $A_{1}\in{\mathcal{B}}({\mathcal{H}}\oplus\mathbb{C}e_{0})$, and

[TABLE]

If $A_{1}$ is surjective and $\mbox{dimker}A_{1}=1$, then $A\sim_{s}A_{1}$. In particular, if $A_{1}\in B_{1}(\Omega)$, $0\in\Omega$ and $A_{1}e_{0}=0$, then $A\sim_{s}A_{1}$.

Proof.

Since $\mbox{dimker}A_{1}=1$, we have $\ker A_{1}=\mathbb{C}e_{0}$. Thus, $(\ker A_{1})^{\bot}=\mathcal{H}$. Let $X=A_{1}|_{(\ker A_{1})^{\bot}}$. Then $X:\,\mathcal{H}=(\ker A_{1})^{\bot}\rightarrow\mathcal{H}\oplus\mathbb{C}e_{0}$ is an invertible bounded linear operator.

By the next two equations,

[TABLE]

and

[TABLE]

we obtain $A_{1}X=XA$, that is, $A\sim_{s}A_{1}$. ∎

For more on Cowen-Douglas operators, we refer to [6], [19], [20], [21].

4. Cyclic vectors

In this section $\Omega$ is a connected open subset of $\mathbb{C}$.

Lemma 4.1** ([23]).**

Suppose that $T\in B_{n}(\Omega)$. Then $\mathcal{C}(T)\neq\emptyset$, where $\mathcal{C}(T)$ is the set of cyclic vectors of $T$.

Lemma 4.2**.**

Suppose that $A\in B_{1}(\Omega)$, $0\in\Omega$, and $AB=I$. If $0\neq e_{0}\in\ker A$, then $e_{0}\in\mathcal{C}(B)$. In particular, $\mathcal{C}(B)\neq\emptyset$.

Proof.

Since $AB=I$,

[TABLE]

Since $A\in B_{1}(\Omega)$,

[TABLE]

It follows that

[TABLE]

Thus, we have $e_{0}\in\mathcal{C}(B)$. ∎

The following result is due to P. A. Fillmore, J. G. Stampfli, J. P. Williams [10]. For the convenience of readers, we give a simple proof.

Proposition 4.3**.**

[10]** Let $T\in\mathcal{B}(\mathcal{H})$. If there is a $\lambda\in\mathbb{C}$ such that $dim\,\ker(T-\lambda)^{*}\geq 2$, then $\mathcal{C}(T)=\emptyset$.

Proof.

Without loss of generality, we assume $dim\,\ker(T-\lambda)^{*}=2$. Suppose that $\{f_{1},f_{2}\}$ is an ONB of $\ker(T-\lambda)^{*}$. Note that

[TABLE]

Let

[TABLE]

We have

[TABLE]

If $\mathcal{C}(T)\neq\emptyset$, then we can find some non-zero vector $y\in\mathcal{C}(T)$. There exist $\alpha_{1},\,\alpha_{2}\in\mathbb{C}$ such that

[TABLE]

Notice that $span\{T^{k}y,\,k\geq 0\}=\mathcal{H}$. On the other hand,

[TABLE]

This contradicts to the fact that $y\in\mathcal{C}(T)$. ∎

The following proposition is due to D. A. Herrero.

Proposition 4.4**.**

[17]** Let $T\in\mathcal{B}(\mathcal{H})$. If $\mathcal{C}(T)\neq\emptyset$, then $\rho_{s-F}^{-1}(T)$ is simply connected.

Proposition 4.5**.**

Suppose $T\in\mathcal{B}(\mathcal{H})$, $T^{*}\in B_{1}(\Omega)$, and $\Omega$ is the maximal domain of $T^{*}$. If $\mathcal{C}(T)\neq\emptyset$, then $\Omega=\rho_{s-F}^{-1}(T)$ and $\rho_{s-F}(T)=\rho_{F}(T)$. Furthermore, assume $\Omega^{\prime}$ is a bounded connected component of $\rho_{F}(T^{*})=\rho_{s-F}(T^{*})$ such that $\Omega^{\prime}\cap\Omega=\emptyset$. Then $\Omega^{\prime}\subset\rho(T^{*})$.

Proof.

Clearly, we have $\Omega\subseteq\rho_{s-F}^{-1}(T)$. Since $\mathcal{C}(T)\neq\emptyset$, $\rho_{s-F}^{-1}(T)$ is simply connected by Proposition 4.4. Therefore, $\Omega=\rho_{s-F}^{-1}(T)$. Since $\mathcal{C}(T)\neq\emptyset$, $\rho_{s-F}(T)=\rho_{F}(T)$ is a corollary of Lemma 3.3 and Proposition 4.3.

By Lemma 3.3, $\sigma_{p}(T)=\emptyset$. Since $\Omega^{\prime}$ is a bounded connected component of $\rho_{F}(T)=\rho_{s-F}(T)$ and $\Omega^{\prime}\cap\Omega=\emptyset$, By Proposition 4.3 and Proposition 4.4, we have that

[TABLE]

This means $\mbox{dimker}(T^{*}-\lambda)=0,\lambda\in\Omega^{\prime}.$ Hence, $\Omega^{\prime}\subset\rho(T^{*}).$ ∎

5. Spectral structures of $T$ and $T^{-1}$

In this section, we first introduce a $2\times 2$ matrix technique to obtain the spectral structures of $T$ and $T^{-1}$.

Let $\{e_{k}\}_{k=0}^{\infty}$ be an orthogonal normal basis (denoted by “ONB”) of a Hilbert space $\mathcal{H}$ and let $S_{1}^{*}$ be the backward shift operator defined as $S_{1}^{*}(e_{0})=0$, $S_{1}^{*}(e_{k+1})=e_{k}$, $k=0,1,2,\cdots$. For $T\in\mathcal{B}(\mathcal{H})$ and $x\in\mathcal{H}$, $x\neq 0$, define an operator $T_{x}\in\mathcal{B}(\mathcal{H}\oplus\mathcal{H})$ as the following

[TABLE]

The following lemma plays a key role throughout this section.

Lemma 5.1**.**

Let $T\in\mathcal{B}(\mathcal{H})$ with spectral radius $r(T)<1$ and let

[TABLE]

Let $\Sigma$ be the connected component of $\mathbb{D}\setminus\sigma(T)$ which contains $\{w\in\mathbb{D}:r(T)<|w|<1\}$. Then we have the following:

1. (1)

For every $w\in\mathbb{D}\setminus\sigma(T)$, $dimker(T_{x}-w)=1$ and $Ran(T_{x}-w)=\mathcal{H}\oplus\mathcal{H}$; 2. (2)

$T_{x}\in B_{1}(\Sigma)$* if and only if $x$ is a cyclic vector of $T$, i.e., $span\{T^{n}x:\,n\geq 0\}=\mathcal{H}$.*

Proof.

(1). Let $w\in\mathbb{D}\setminus\sigma(T)$ and $e(w)\triangleq\sum\limits_{k=0}^{\infty}w^{k}e_{k}$. Then $e(w)\in ker(S_{1}^{*}-w)$, $w\in\mathbb{D}$. We calculate

[TABLE]

This is equivalent to $(S_{1}^{*}-w)\eta=0$ and $(T-w)\xi=-\langle\eta,e_{0}\rangle x$. Since $dimker(S_{1}^{*}-w)=1$, there is a $w_{1}\in\mathbb{C}$ such that $\eta=w_{1}e(w)$. Without loss of generality, we assume that $w_{1}=1$. Note that $\langle e(w),e_{0}\rangle=1$ and $T-w$ is invertible for $w\in\mathbb{D}\setminus\sigma(T)$. We have $\xi=-(T-w)^{-1}x$. This proves that $dimker(T_{x}-w)=1$.

To show that $T_{x}-w$ is surjective for $w\in\mathbb{D}\setminus\sigma(T)$, we need to find for every $\xi^{\prime}\oplus\eta^{\prime}\in\mathcal{H}\oplus\mathcal{H}$ a vector $\xi\oplus\eta\in\mathcal{H}\oplus\mathcal{H}$ satisfying the following equation:

[TABLE]

This is equivalent to $(S_{1}^{*}-w)\eta=\eta^{\prime}$ and $(T-w)\xi=\xi^{\prime}-\langle\eta,e_{0}\rangle x$. Note that both $S_{1}^{*}-w$ and $T-w$ are surjective for $w\in\mathbb{D}\setminus\sigma(T)$. The existence of $\xi$ and $\eta$ is clear.

(2). For $w\in\Sigma$, define $y(w)=-(T-w)^{-1}x$. Then by (1), $y(w)\oplus e(w)$ is in $ker(T_{x}-w)$. To show $T_{x}\in B_{1}(\Sigma)$, we need to prove that

[TABLE]

Suppose that there exists an $x_{1}\oplus x_{2}\in\mathcal{H}\oplus\mathcal{H}$ such that $\langle y(w)\oplus e(w),x_{1}\oplus x_{2}\rangle=0$. Then $\langle y(w),x_{1}\rangle+\langle e(w),x_{2}\rangle=0$. Since $\langle e(w),x_{2}\rangle$ is analytic on $\mathbb{D}$ and $(w-T)^{-1}=\frac{1}{w}\sum_{n=0}^{\infty}\left(\frac{T}{w}\right)^{n}$ for $|w|>r(T)$, $\langle-(T-w)^{-1}x,x_{1}\rangle$ is analytic when $|w|>r(T)$. Furthermore,

[TABLE]

Thus, by the analytic continuation theorem, we know that $\langle y(w),x_{1}\rangle$ is analytic on $\mathbb{C}$. Since

[TABLE]

we see that $\langle y(w),x_{1}\rangle$ is a bounded entire function on $\mathbb{C}$. Thus we have

[TABLE]

Note that $\langle(w-T)^{-1}x,x_{1}\rangle=0$ for all $|w|>r(T)$, we have

[TABLE]

It follows that $\langle T^{n}x,x_{1}\rangle=0$ for $n=0,1,\cdots$. Suppose $x$ is a cyclic vector of $T$. Then $x_{1}=0$. Since $S_{1}^{*}\in B_{1}(\mathbb{D})\subset B_{1}(\Sigma)$, we have $span\{e(w):w\in\Sigma\}=\mathcal{H}$. This means $x_{2}=0$. Thus $span\{y(w)\oplus e(w):\,w\in\Sigma\}=\mathcal{H}\oplus\mathcal{H}$. Suppose $x$ is not a cyclic vector of $T$. Let $0\neq x_{1}\perp\{T^{n}x:\,n\geq 0\}$. Then $(x_{1}\oplus 0)\perp span\{y(w)\oplus e(w):\,w\in\Sigma\}$ and therefore $span\{y(w)\oplus e(w):\,w\in\Sigma\}\neq\mathcal{H}\oplus\mathcal{H}$. This implies that $T_{x}\notin B_{1}(\Sigma)$.

∎

Lemma 5.2**.**

For every $x\in\mathcal{H}$, $x\neq 0$, define $S\in\mathcal{B}(\mathcal{H}\oplus\mathcal{H})$ as follows

[TABLE]

Then $S_{x}\triangleq S$ is a right inverse of $T_{x}$.

Proof.

Notice that $S_{1}^{*}(e_{0})=0$. Then we have $(x\otimes e_{0})S_{1}=x\otimes S^{*}_{1}(e_{0})=0,$ and

[TABLE]

∎

Define

[TABLE]

Then

[TABLE]

for $n\geq 1$ and

[TABLE]

for all $n\geq 0$. Thus $\mathcal{M}(x)\in LatT_{x}\cap LatS_{x}$.

We set

[TABLE]

and

[TABLE]

Lemma 5.3**.**

Let $a_{n}=P_{\mathcal{M}(x)}(0\oplus e_{n})$ and $b_{n}=P_{\mathcal{M}(x)^{\perp}}(0\oplus e_{n})$ for $n\geq 0$. Then we have

[TABLE]

and

[TABLE]

Proof.

Since

[TABLE]

we have

[TABLE]

Thus $T_{2}b_{n}=b_{n-1}$, $\forall n\geq 1$. A similar calculation shows that $T_{2}b_{0}=P_{\mathcal{M}(x)^{\perp}}(x\oplus 0)$.

Since

[TABLE]

we have

[TABLE]

Thus $\hat{T}^{*}_{x}a_{n}=a_{n+1}$, $\forall n\geq 0$.

Since

[TABLE]

we have

[TABLE]

Thus $S_{2}b_{n}=b_{n+1}$, $\forall n\geq 0$.

Since

[TABLE]

where we define $e_{-1}=0$, we have

[TABLE]

Thus $\hat{S}_{x}^{*}a_{n}=a_{n-1}$, $\forall n\geq 1$ and $\hat{S}_{x}^{*}a_{0}=0$. ∎

Note that

[TABLE]

In particular, $a_{n}\neq 0$ for all $n\geq 0$.

Lemma 5.4**.**

Let $T\in\mathcal{B}(\mathcal{H})$, $w_{0}\in\mathbb{C}$ and $\delta>0$. Suppose $\forall w$, $|w-w_{0}|<\delta$, and

1. (1)

$Ran(T-w)=\mathcal{H}$; 2. (2)

$dimker(T-w)=1$.

Then $span\{ker(T-w_{0})^{n},\,n\geq 1\}=span\{ker(T-w),\,|w-w_{0}|<\delta\}$.

Proof.

We may assume that $w_{0}=0$. Then $RanT=\mathcal{H}$ and $dimkerT=1$. Assume that $T\xi=0$ for some $\xi\neq 0$. Let $\mathcal{K}=span\{ker(T-w),\,|w|<\delta\}$. Then $\mathcal{K}\in LatT$ and $\xi\in\mathcal{K}$. Write

[TABLE]

Claim $T_{2}$ is injective. Otherwise $T_{2}\eta=0$ for some $\eta\neq 0$. Let $e(w)$ be a nonzero vector such that $(T-w)e(w)=0$. Then $RanT_{1}$ contains $span\{we(w):\,{|w|<\delta}\}$ and $RanT_{1}$ is dense in $\mathcal{K}$. Therefore, there exists a sequence of vectors $\xi_{n}\in\mathcal{K}$ such that

[TABLE]

Write

[TABLE]

as an orthogonal decomposition. Note that $T$ is an invertible map from $\left[\xi\oplus 0\right]^{\perp}$ onto $RanT$. Therefore, there exists a $K>0$ such that

[TABLE]

So

[TABLE]

implies that

[TABLE]

Therefore, $\eta=0$. This is a contradiction and thus $T_{2}$ is injective.

Since $RanT=\mathcal{H}$, there is an operator

[TABLE]

satisfying

[TABLE]

This implies that $T_{2}S_{21}=0$ and $T_{2}S_{22}=1$. Since $T_{2}$ is injective, $S_{21}=0$ and $T_{2}$ is invertible. Then $T_{1}S_{11}=1$. So $RanT_{1}=\mathcal{K}$. Similarly, $Ran(T_{1}-w)=\mathcal{K}$ for all $|w|<\delta$. Therefore, $T_{1}\in B_{1}(\Omega)$.

Note that for $n\geq 1$,

[TABLE]

So

[TABLE]

implies that $T_{2}^{n}\eta=0$ and therefore $\eta=0$. Furthermore, $T_{1}^{n}\xi=0$ and $kerT^{n}=kerT_{1}^{n}$. By Proposition 3.1,

[TABLE]

∎

Lemma 5.5**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be invertible, $r(T)<1$, and

[TABLE]

Let $\Phi_{0}$ be the connected component of $\mathbb{D}\setminus\sigma(T)$ with $0\in\Phi_{0}$. Let $\Sigma$ be the connected component of $\mathbb{D}\setminus\sigma(T)$ which contains $\{w\in\mathbb{D}:r(T)<|w|<1\}$. If $x$ is a cyclic vector of $T$ and $\mathcal{M}(x)\neq\mathcal{H}\oplus\mathcal{H}$, then

1. (1)

$\hat{T}_{x}\in B_{1}(\Phi_{0})$* and $\sigma(\hat{T}_{x})\subset\sigma(T)^{\land}$ ;* 2. (2)

$T_{2}$* is invertible and $T_{2}\in B_{1}(\Sigma)$.*

Proof.

(1). By (1) of Lemma 5.1, $\forall w\in\Phi$, $Ran(T_{x}-w)=\mathcal{H}\oplus\mathcal{H}=\mathcal{M}(x)\oplus\mathcal{M}(x)^{\perp}$ and $dimker(T_{x}-w)=1$. By Lemma 5.4,

[TABLE]

Therefore, $ker(T_{x}-w)\subset\mathcal{M}(x)$ for $w\in\Phi_{0}$. So $ker(\hat{T}_{x}-w)=ker(T_{x}-w)=\mathbb{C}f(w)\subset\mathcal{M}(x)$, where $f(w)=y(w)\oplus e(w)$ as in the proof of Lemma 5.1. Since

[TABLE]

$Ran(\hat{T}_{x}-w_{1})$ is dense in $\mathcal{M}(x)$, $w_{1}\in\Phi_{0}$.

Suppose $Ran(\hat{T}_{x}-w)\neq\mathcal{M}(x)$ for some $w\in\Phi_{0}$. Then there exists a $y\in\mathcal{M}(x)$ such that $y\notin Ran(\hat{T}_{x}-w)$ and $y\oplus 0\in Ran(T_{x}-w)$. Therefore, there exists a vector $\xi\oplus\eta\in\mathcal{M}(x)\oplus\mathcal{M}(x)^{\perp}$ such that $\eta\neq 0$ and

[TABLE]

i.e.,

[TABLE]

Since $Ran(\hat{T}_{x}-w)$ is dense in $\mathcal{M}(x)$, there exists a sequence of vectors $\xi_{n}$ such that

[TABLE]

i.e.,

[TABLE]

Write

[TABLE]

as an orthogonal decomposition. Note that $(T_{x}-w)$ is an invertible map from $\left[f(w)\oplus 0\right]^{\perp}$ onto $Ran(T_{x}-w)=\mathcal{H}\oplus\mathcal{H}$. Therefore, there exists a $K>0$ such that

[TABLE]

So $\lim\limits_{n\rightarrow+\infty}(T_{x}-w)\left(\begin{matrix}\xi_{n}-\alpha_{n}f(w)\\ \eta\end{matrix}\right)=\left(\begin{matrix}0\\ 0\end{matrix}\right)$ implies that

[TABLE]

Thus $\eta=0$. This is a contradiction. Hence, $\hat{T}_{x}\in B_{1}(\Phi_{0})$.

Since $\hat{T}_{x}=T_{x}|_{\mathcal{M}(x)}$, $\sigma(\hat{T}_{x})\subseteq\bar{\mathbb{D}}$. In the following we show $\forall w\in\mathbb{D}\setminus\sigma(T)^{\land}$, $w\in\rho(\hat{T}_{x})$.

By (1) of Lemma 5.1, $\forall w\in\mathbb{D}\setminus\sigma(T)^{\land}$,

[TABLE]

Let $f(w)$ be a nonzero eigenvector of $T_{x}-w$. Then

[TABLE]

with respect to the decomposition $\mathcal{H}\oplus\mathcal{H}=\mathcal{M}(x)\oplus\mathcal{M}(x)^{\perp}$.

Claim:  $f_{2}(w)\neq 0$.

Suppose $f_{2}(w)=0$. Then $f(w)\in\mathcal{M}(x)$. Note that

[TABLE]

and $Ran(\hat{T}_{x}-w)$ is dense in $\mathcal{M}(x)$. So same argument as above shows that $Ran(\hat{T}_{x}-w)=\mathcal{M}(x)$. Since $T_{x}-w$ is surjective, $T_{2}-w$ is also surjective. Assume that $T_{2}-w$ is not injective. Then there exists an $\eta\in\mathcal{M}(x)^{\perp}$, $\eta\neq 0$, such that $(T_{2}-w)\eta=0$. Let $\xi\in\mathcal{M}(x)$ be such that $(\hat{T}_{x}-w)\xi=-T_{12}\eta$. Then

[TABLE]

So $\xi\oplus\eta,f(w)\in ker(T_{x}-w)$ are linearly independent. This contradicts to $dimker(T_{x}-w)=1$. Thus $T_{2}-w$ is injective and therefore $T_{2}-w$ is invertible. So there is a $\delta>0$ such that $T_{2}-w^{\prime}$ is invertible for all $w^{\prime}$ satisfying $|w^{\prime}-w|<\delta$. Suppose

[TABLE]

Then $(T_{2}-w^{\prime})\eta=0$ implies that $\eta=0$. So $ker(T_{x}-w^{\prime})\subseteq\mathcal{M}(x)$. Since $w\in\mathbb{D}\setminus\sigma(T)^{\land}$, $w\in\Sigma$. By (2) of Lemma 5.1,

[TABLE]

This contradicts to the assumption of the lemma.

Thus

[TABLE]

and $f_{2}(w)\neq 0$. Since $dimker(T_{x}-w)=1$, $\hat{T}_{x}-w$ is injective for $w\in\mathbb{D}\setminus\sigma(T)^{\land}$. Since $\hat{T}_{x}\in B_{1}(\Omega)$, $Ran(\hat{T}_{x}-w)$ is dense in $\mathcal{H}$ by Lemma 3.3. Claim $\hat{T}_{x}-w$ is surjective. Otherwise, there exists a sequence of unit vectors $\xi_{n}\in\mathcal{M}$ such that $\lim\limits_{n\rightarrow+\infty}(\hat{T}_{x}-w)\xi_{n}=0$. So

[TABLE]

Let

[TABLE]

be an orthogonal decomposition. Then $T_{x}-w$ is an invertible map from $\left[f_{1}(w)\oplus f_{2}(w)\right]^{\perp}$ onto $Ran(T_{x}-w)=\mathcal{H}\oplus\mathcal{H}$. Therefore,

[TABLE]

implies that

[TABLE]

Notice that

[TABLE]

So $\lim\limits_{n\rightarrow+\infty}\alpha_{n}=0$. Thus $\lim\limits_{n\rightarrow+\infty}\left\|\xi_{n}\oplus 0\right\|=0$, which contradicts to the assumption $\|\xi_{n}\|=1$ for all $n$. This proves that $\forall w\in\mathbb{D}\setminus\sigma(T)^{\land}$, $w\in\rho(\hat{T}_{x})$. Therefore, $\sigma(\hat{T}_{x})\subset\sigma(T)^{\land}$.

(2). By Lemma 5.1, $T_{x}\in B_{1}(\Sigma)$. Therefore, $Ran(T_{x}-w)=\mathcal{H}\oplus\mathcal{H}=\mathcal{M}\oplus\mathcal{M}^{\perp}$, $\forall w\in\Sigma$. It is clear that $Ran(T_{2}-w)=\mathcal{M}(x)^{\perp}$, $\forall w\in\Sigma$. Let $f(w)$ be a nonzero eigenvector of $T_{x}-w$, $w\in\Sigma$. Then

[TABLE]

and $f_{2}(w)\neq 0$. It is clear that $(T_{2}-w)f_{2}(w)=0$. Assume that $(T_{2}-w)\eta=0$. Then

[TABLE]

Since $dimker(T_{x}-w)=1$, $f_{1}(w)\oplus f_{2}(w)$ and $(-(\hat{T}_{x}-w)^{-1}T_{12}(\eta))\oplus\eta$ are linearly dependent. Therefore $f_{2}(w)$ and $\eta$ are linearly dependent. This implies that $dimker(T_{2}-w)=1$. Since $span\{f(w):\,w\in\Sigma\}=\mathcal{H}\oplus\mathcal{H}$, $span\{f_{2}(w):\,w\in\Sigma\}=\mathcal{M}(x)^{\perp}$. Hence, $T_{2}\in B_{1}(\Sigma)$.

By Lemma 5.1, $RanT_{x}=\mathcal{H}\oplus\mathcal{H}=\mathcal{M}(x)\oplus\mathcal{M}(x)^{\perp}$, so $RanT_{2}=\mathcal{M}(x)^{\perp}$. Also we have $dimkerT_{x}=1$. This implies that $kerT_{x}=ker\hat{T}_{x}$. Since $Ran\hat{T}_{x}=\mathcal{M}(x)$, $T_{2}$ is injective. This proves that $T_{2}$ is invertible. ∎

Lemma 5.6**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be an invertible bounded linear operator. For $x\in\mathcal{H}$, let $\mathcal{N}_{-k}(x)=span\{T^{-n}x:\,n\geq k\}$. Then $\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)\in Lat(T)$.

Proof.

Choose $y\in\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)$, we prove $Ty\in\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)$. Since $y\in\mathcal{N}_{-2}(x)$, $\forall\epsilon>0$, there exists $\sum\limits_{i=2}^{m_{2}}\alpha_{i}T^{-i}x$, $\alpha_{i}\in\mathbb{C}$, such that

[TABLE]

Thus

[TABLE]

It follows that $Ty\in\mathcal{N}_{-1}(x)$. Similarly, we can show that $Ty\in\mathcal{N}_{-k}(x)$ for all $k\geq 1$ and therefore, $Ty\in\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)$. ∎

Lemma 5.7**.**

Suppose $T\in\mathcal{B}(\mathcal{H})$ is an invertible transitive operator. Let $\mathcal{N}_{-k}(x)=span\{T^{n}x:n\leq-k\}$ for $k\geq 1$. If $\mathcal{N}(x)\triangleq\mathcal{N}_{-1}(x)\neq\mathcal{H}$, then $\{\mathcal{N}_{-k}(x)\}$ is a strictly decreasing sequence and $\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)=\{0\}$.

Proof.

By Lemma 5.6, $\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)\in LatT$. Since $\mathcal{N}(x)=\mathcal{N}_{-1}(x)\neq\mathcal{H}$, $\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)=\{0\}$. Suppose $\mathcal{N}_{-k}(x)=\mathcal{N}_{-(k+1)}(x)$ for some $k\geq 1$. Then $T^{-k}x\in\mathcal{N}_{-(k+1)}(x)$. Hence, $T^{-(k+1)}x\in T^{-1}\mathcal{N}_{-(k+1)}(x)=\mathcal{N}_{-(k+2)}(x)$. This implies that $\mathcal{N}_{-(k+2)}(x)=\mathcal{N}_{-(k+1)}(x)=\mathcal{N}_{-k}(x)$. By induction, we have $\mathcal{N}_{-n}(x)=\mathcal{N}_{-k}(x)$ for all $n\geq k$. This contradicts to $\cap_{k=1}^{\infty}\mathcal{N}_{-k}(x)=\{0\}$. ∎

Lemma 5.8**.**

Suppose an invertible operator $T\in\mathcal{B}(\mathcal{H})$ is transitive, $r(T)<1$, and

[TABLE]

Then the map $B_{x}:(-T^{-(n+1)}x)\oplus e_{n}\rightarrow-T^{-(n+1)}x$, $n\geq 0$, extends to a bounded linear isomorphism from $\mathcal{M}(x)=span\{(-T^{-(n+1)}x)\oplus e_{n}:\,n\geq 0\}$ onto $\mathcal{N}(x)$.

Proof.

Recall that $a_{n}=P_{\mathcal{M}(x)}(0\oplus e_{n})$ (see Lemma 5.3). For $n\geq 0$, define $f_{n}(x)=\langle x,a_{n}\rangle$ for $x\in\mathcal{H}\oplus\mathcal{H}$. By Lemma 5.5, $\sigma(\hat{T}_{x}^{*})\subset\sigma(T)^{\land}$, so the spectral radius of $\hat{T}_{x}^{*}$ is strictly less than a positive number $r<1$. By Lemma 5.3, $\hat{T}_{x}^{*}a_{n}=a_{n+1}$ and $a_{n}=\left(\hat{T}_{x}^{*}\right)^{n}a_{0}$ for $n\geq 0$. Thus $\|f_{n}\|=\|a_{n}\|\leq\left(r+\epsilon\right)^{n}\|a_{0}\|$ for sufficient large $N$ such that $\sum\limits_{n=N+1}^{\infty}\|a_{n}\|=\theta<1$. For each $y\in\mathcal{H}\oplus\mathcal{H}$, put

[TABLE]

Note that

[TABLE]

[TABLE]

Then for $k\geq N+1$, we have

[TABLE]

[TABLE]

[TABLE]

Note that

[TABLE]

So $A$ is an invertible bounded linear operator on $\mathcal{H}\oplus\mathcal{H}$ and $A$ maps

[TABLE]

onto $span\{-T^{-(n+1)}x:n\geq N+1\}$.

Claim   $span\{T^{-1}x,\cdots,T^{-(N+1)}x\}\cap span\{T^{-(N+2)}x,T^{-(N+3)}x,\cdots\}=\{0\}$. Suppose

[TABLE]

and

[TABLE]

satisfy $w=v$. If $w\neq 0$, then $w=\sum\limits_{i=1}^{N+1}\alpha_{i}T^{-i}x=z$. We may assume that $\alpha_{j}=-1$ and $\alpha_{i}=0$ for $1\leq i<j$. Then

[TABLE]

This implies that $\mathcal{N}_{-j}(x)=\mathcal{N}_{-(j+1)}(x)$, which contradicts to Lemma 5.7. Let $x^{\prime}\in\mathcal{N}(x)$. Then there exist a sequence of vectors $\{x_{n}\}\in\mathcal{N}(x)$ such that each $x_{n}$ can be written as a finite linear combinations of $\{T^{-n}x\}_{n=1}^{\infty}$ and $\lim\limits_{n\rightarrow\infty}\|x_{n}-x^{\prime}\|=0$. In particular, there is $\{y_{n}\}\in span\{T^{-1}x,T^{-2}x,\cdots,T^{-(N+1)}x\}$ and $z_{n}\in span\{T^{-(N+2)}x,T^{-(N+3)}x,\cdots\}$ such that $x_{n}=y_{n}+z_{n}$. Let $\pi$ be the quotient map from $\mathcal{N}$ onto $\mathcal{L}/span\{T^{-(N+2)}x,T^{-(N+3)}x,\cdots\}$. Then $\pi(y_{n})=\pi(x_{n})$ is a Cauchy sequence in $\mathcal{L}/span\{T^{-(N+2)}x,T^{-(N+3)}x,\cdots\}$. Clearly, $\pi$ is a surjective map from $span\{T^{-1}x,T^{-2}x,\cdots,T^{-(N+1)}x\}$ onto $span\{\pi(T^{-1}x),\pi(T^{-2}x),\cdots,\pi(T^{-(N+1)}x)\}$. Suppose $\pi(w_{1}T^{-1}x+w_{2}T^{-2}x+\cdots+w_{N+1}T^{-(N+1)}x)=0$. Then there exists a

[TABLE]

such that $w_{1}T^{-1}x+w_{2}T^{-2}x+\cdots+w_{N+1}T^{-(N+1)}x=z$. By the above argument, $w_{1}T^{-1}x+w_{2}T^{-2}x+\cdots+w_{N+1}T^{-(N+1)}x=z=0$. So $\pi$ is an injective map from

[TABLE]

onto

[TABLE]

Since $span\{T^{-1}x,T^{-2}x,\cdots,T^{-(N+1)}x\}$ is finite dimensional, $\{y_{n}\}$ is a Cauchy sequence. Thus $\{z_{n}\}$ is also a Cauchy sequence. Let $y=\lim\limits_{n\rightarrow\infty}y_{n}$ and $z=\lim\limits_{n\rightarrow\infty}z_{n}$. Then

[TABLE]

and

[TABLE]

such that $x^{\prime}=y+z$. This implies that

[TABLE]

Similarly,

[TABLE]

Hence $B_{x}:(-T^{-(n+1)}x)\oplus e_{n}\rightarrow-T^{-(n+1)}x$ extends to a bounded linear isomorphism from $\mathcal{M}(x)$ onto $\mathcal{N}(x)$. ∎

Lemma 5.9**.**

Suppose an invertible operator $T\in\mathcal{B}(\mathcal{H})$, $r(T)<1$, $x$ is a cyclic vector of $T$ and $\mathcal{M}(x)\neq\mathcal{H}\oplus\mathcal{H}$. If $b_{0}\neq 0$, then $P_{\mathcal{M}(x)^{\perp}}$ is an invertible bounded linear operator from $0\oplus\mathcal{H}$ onto $\mathcal{L}(x)=span\{b_{0},b_{1},\cdots\}$, where $b_{n}=P_{\mathcal{M}(x)^{\perp}}(0\oplus e_{n})$ (see in Lemma 5.3).

Proof.

By Lemma 5.3, $\hat{T}_{x}^{*}a_{n}=a_{n+1}$ for $n\geq 0$. Then we have $a_{n}=\left(\hat{T}_{x}^{*}\right)^{n}a_{0}$ for $n\geq 0$ and

[TABLE]

By Lemma 5.5, $\sigma(\hat{T}_{x}^{*})\subset\sigma(T)^{\land}$, so the spectral radius of $\hat{T}_{x}^{*}$ is a positive number $r<1$. For arbitrary $\epsilon>0$, there exists an $N_{1}$ such that $\|a_{n}\|\leq\left|\left(r+\epsilon\right)^{n}\right|\|a_{0}\|$ for $n\geq N_{1}$. This implies that there exists an $N$ sufficiently large such that

[TABLE]

By Theorem 1.3.9 of [1], there exists an invertible bounded linear operator $A$ from

[TABLE]

onto

[TABLE]

such that $A(e_{k})=b_{k}$ for $k=N+1,N+2,\cdots$. So $A$ is the restriction of $P_{\mathcal{M}(x)^{\perp}}$ onto $span\{e_{N+1},e_{N+2},\cdots\}$ and $\{b_{N+1},b_{N+2},\cdots\}$ is a basic sequence.

Suppose $P_{\mathcal{M}(x)^{\perp}}(0\oplus z)=0$. Write $z=\sum\limits_{n=0}^{\infty}w_{n}(0\oplus e_{n})$, where $\sum\limits_{n=0}^{\infty}|w_{n}|^{2}<\infty$. Then

[TABLE]

If $z\neq 0$, then $w_{k}\neq 0$ for some $k$ and $w_{j}=0$ for all $j<k$. We may assume that $w_{k}=-1$. Then $b_{k}=\sum\limits_{n=k+1}^{\infty}w_{n}b_{n}$. Recall that $S_{2}b_{n}=b_{n+1}$ for all $n\geq 0$ (see Lemma 5.3). Thus

[TABLE]

This contradicts to the fact that $\{b_{N+1},b_{N+2},\cdots\}$ is a basic sequence. Hence, $z=0$. So we have $P_{\mathcal{M}(x)^{\perp}}$ is an injective bounded linear operator from $0\oplus\mathcal{H}$ into $\mathcal{L}=span\{b_{0},b_{1},\cdots\}$. We need only to show the map is surjective.

Suppose $w\in span\{b_{0},b_{1},\cdots,b_{N}\}$ and $v\in span\{b_{N+1},b_{N+2},\cdots\}$ satisfy $w=v$. Write $w=\sum\limits_{i=0}^{N}\alpha_{i}b_{i}$ and $v=\sum\limits_{i=N+1}^{\infty}\alpha_{i}b_{i}$, where $\sum\limits_{i=N+1}^{\infty}|\alpha_{i}|^{2}<\infty$. Let

[TABLE]

Then $P_{\mathcal{M}^{\perp}}(0\oplus z)=w-v=0$. This implies that $z=0$ and therefore, $w=v=0$. So

[TABLE]

Let $x^{\prime}\in\mathcal{L}$. Then there exists a sequence of vectors $\{x_{n}\}\in\mathcal{L}$ such that each $x_{n}$ can be written as a finite linear combinations of $\{b_{n}\}_{n=1}^{\infty}$ and $\lim\limits_{n\rightarrow\infty}\|x_{n}-x^{\prime}\|=0$. In particular, there is $y_{n}\in span\{b_{0},b_{1},\cdots,b_{N}\}$ and $z_{n}\in span\{b_{N+1},b_{N+2},\cdots\}$ such that $x_{n}=y_{n}+z_{n}.$

Let $\pi$ be the quotient map from $\mathcal{L}(x)$ onto $\mathcal{L}(x)/span\{b_{N+1},b_{N+2},\cdots\}$. Then $\pi(y_{n})=\pi(x_{n})$ is a Cauchy sequence in $\mathcal{L}(x)/span\{b_{N+1},b_{N+2},\cdots\}$. Clearly, $\pi$ is a surjective map from $span\{b_{0},b_{1},\cdots,b_{N}\}$ onto $span\{\pi(b_{0}),\pi(b_{1}),\cdots,\pi(b_{N})\}$.

Suppose $\pi(w_{0}b_{0}+w_{1}b_{1}+\cdots+w_{N}b_{N})=0$. Then there exists a $v\in span\{b_{N+1},b_{N+2},\cdots\}$ such that

[TABLE]

Therefore,

[TABLE]

So $\pi$ is an injective map from $span\{b_{0},b_{1},\cdots,b_{N}\}$ onto $span\{\pi(b_{0}),\pi(b_{1}),\cdots,\pi(b_{N})\}$.

Since $span\{b_{0},b_{1},\cdots,b_{N}\}$ is finite dimensional, $\{y_{n}\}$ is a Cauchy sequence. Thus $\{z_{n}\}$ is also a Cauchy sequence. Let $y=\lim\limits_{n\rightarrow\infty}y_{n}$ and $z=\lim\limits_{n\rightarrow\infty}z_{n}$. Then $y\in span\{b_{0},b_{1},\cdots,b_{N}\}$ and $z\in span\{b_{N+1},b_{N+2},\cdots\}$ such that $x^{\prime}=y+z$. This implies that

[TABLE]

and $P_{\mathcal{M}(x)^{\perp}}$ is onto. ∎

Lemma 5.10**.**

Let $\mathcal{L}(x)=span\{b_{n},\,n\geq 0\}$, $\mathcal{N}(x)=span\{T^{-(n+1)}x,\,n\geq 0\}$. Then

[TABLE]

where $b_{n}=P_{\mathcal{M}(x)^{\perp}}(0\oplus e_{n})$ (see in Lemma 5.3.)

Proof.

We need only to prove $\mathcal{M}(x)\oplus\mathcal{L}(x)=\mathcal{N}(x)\oplus\mathcal{H}$. First, we show that

[TABLE]

Clearly, $(-T^{-(n+1)}x)\oplus e_{n}\in\mathcal{N}(x)\oplus\mathcal{H}$ for all $n\geq 0$. Thus $\mathcal{M}(x)\subset\mathcal{N}(x)\oplus\mathcal{H}$. Note that $b_{n}=(0\oplus e_{n})-a_{n}\in\mathcal{N}(x)\oplus\mathcal{H}$. Thus $\mathcal{L}(x)\subset\mathcal{N}(x)\oplus\mathcal{H}$ and $\mathcal{M}(x)\oplus\mathcal{L}(x)\subseteq\mathcal{N}(x)\oplus\mathcal{H}$. Second, we show that $\mathcal{N}(x)\oplus\mathcal{H}\subseteq\mathcal{M}(x)\oplus\mathcal{L}(x)$. Note that for $n\geq 0$,

[TABLE]

Also for $n\geq 0$, $(-T^{-(n+1)}x)\oplus 0=(-T^{-(n+1)}x)\oplus e_{n}-(0\oplus e_{n})\in\mathcal{M}(x)\oplus\mathcal{L}(x)$. Thus

[TABLE]

∎

Corollary 5.11**.**

$\mathcal{L}(x)\subset\mathcal{M}(x)^{\perp}$* and $\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)=\mathcal{N}(x)^{\perp}\oplus 0$.*

By equation (5.2), Lemma 5.10 and Corollary 5.11, we have the following

[TABLE]

[TABLE]

In particular,

[TABLE]

where $\bar{T}_{2}(x)=P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}$.

Recall that

[TABLE]

and

[TABLE]

Lemma 5.12**.**

$\hat{S}_{x}^{*}\in B_{1}(s\mathbb{D})$, where $s=r(T)^{-1}$.

Proof.

Since $T_{x}S_{x}=I$, $S^{*}_{x}T^{*}_{x}=I$. We have

[TABLE]

This implies that $\hat{S}^{*}_{x}\hat{T}^{*}_{x}=I$. By Lemma 5.5 and Theorem 16.12 of [27],

[TABLE]

For $\lambda\in\mathbb{C}$, we have

[TABLE]

By Lemma 5.5, $\sigma\left(\hat{T}^{*}_{x}\right)\subseteq\sigma(T)^{\land}\subseteq r(T)\bar{\mathbb{D}}$. Hence, for $|\lambda|<s$, $|\lambda^{-1}|>r(T)$ and $I-\lambda\hat{T}^{*}_{x}$ is invertible. This implies that $\hat{S}^{*}_{x}-\lambda$ is surjective for $|\lambda|<s$. Therefore, $dimker(\hat{S}_{x}-\lambda)=0$ for $|\lambda|<s$. By the continuity of index, $ind(\hat{S}^{*}_{x}-\lambda)=1$ for all $|\lambda|<s$. So $dimker(\hat{S}^{*}_{x}-\lambda)=1$ for all $|\lambda|<s$.

By Lemma 5.3, $\hat{S}^{*}_{x}a_{0}=0$. Let

[TABLE]

Since $r(\hat{T}^{*}_{x})\leq r(T)$, $e(w)$ is well-defined for $|w|<s$. We have

[TABLE]

Suppose

[TABLE]

for some $|w_{0}|<s$. Then

[TABLE]

Hence

[TABLE]

Therefore,

[TABLE]

This is a contradiction. By Lemma 5.4,

[TABLE]

we conclude that $\hat{S}^{*}_{x}\in B_{1}(s\mathbb{D})$. ∎

Lemma 5.13**.**

Let $\Sigma$ be the connected component of $\mathbb{D}\setminus\sigma(T)$ which contains $\{w\in\mathbb{D}:r(T)<|w|<1\}$ and let $\Sigma^{-1}=\{w^{-1}:\,w\in\Sigma\}$. Then $S_{2}\in B_{1}(\Sigma^{-1})$.

Proof.

Since $T_{x}S_{x}=I$, we have

[TABLE]

This implies that $T_{2}S_{2}=I$. By Lemma 5.5, for $w\in\Sigma^{-1}$, we have $w^{-1}\in\Sigma$ and

[TABLE]

is surjective. By Lemma 5.5, $T_{2}$ is invertible, so $S_{2}-w$ is surjective for $w\in\Sigma^{-1}$. Also note that $\forall w\in\Sigma^{-1}$, $w^{-1}\in\Sigma$,

[TABLE]

So $Ran(S_{2}^{*}-\bar{w})=Ran(w^{-1}-T_{2})^{*}$. This implies that

[TABLE]

is dimensional one and by Lemma 5.5,

[TABLE]

So $S_{2}\in B_{1}(\Sigma^{-1})$. ∎

Lemma 5.14**.**

Let $S_{2}=\left(\begin{matrix}S_{2}|_{\mathcal{L}(x)}&G_{12}\\ 0&P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}\end{matrix}\right)\begin{matrix}\mathcal{L}(x)\\ \mathcal{N}(x)^{\perp}\oplus 0\end{matrix}.$ Then $S_{2}|_{\mathcal{L}(x)}$ is similar to the unilateral shift operator and $P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}\in B_{1}(\Sigma^{-1})$.

Proof.

By Lemma 5.9,

[TABLE]

for all $n\geq 0$. So $S_{2}|_{\mathcal{L}(x)}$ is similar to the unilateral shift operator.

By Lemma 5.13, $S_{2}-w$ is surjective for $w\in\Sigma^{-1}$. Therefore, $\left(P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}-w\right)$ is also surjective. For $w\in\Sigma^{-1}$, there exists a nonzero vector $f(w)=f_{1}(w)\oplus f_{2}(w)$ such that

[TABLE]

Suppose $f_{2}(w)=0$. Then $f_{1}(w)\neq 0$ and $(S_{2}|_{\mathcal{L}(x)}-w)f_{1}(w)=0$. So $S_{2}|_{\mathcal{L}(x)}-w$ is not invertible. This is a contradiction. Thus $f_{2}(w)\neq 0$ and $P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}(S_{2}-w)P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}(f_{2}(w))=0$. Suppose $\eta_{1},\eta_{2}\in ker\left(P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}-w\right)$. Then there exist $\xi_{1},\xi_{2}\in\mathcal{L}(x)$ such that

[TABLE]

So $\xi_{1}\oplus\eta_{1},\xi_{2}\oplus\eta_{2}\in ker(S_{2}-w)$. Since $dimker(S_{2}-w)=1$, $\xi_{1}\oplus\eta_{1},\xi_{2}\oplus\eta_{2}$ are linearly dependent. Therefore, $\eta_{1},\eta_{2}$ are linearly dependent. So

[TABLE]

Since $span\left\{f_{1}(w)\oplus f_{2}(w):\,{w\in\Omega_{1}^{-1}}\right\}=\mathcal{M}(x)^{\perp}$, $span\{f_{2}(w):\,{w\in\Omega_{1}^{-1}}\}=\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)$. This implies that

[TABLE]

∎

Lemma 5.15**.**

Write

[TABLE]

Then $\bar{T}_{2}(x)=P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}S_{2}P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}\in B_{1}(\Sigma^{-1})$ and $\bar{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$, where $\Omega_{0}$ is the connected component of $\mathbb{C}\setminus\sigma(T^{-1})$ containing 0.

Proof.

By Lemma 5.8,

[TABLE]

This implies that $\bar{T}_{1}(x)=B_{x}\hat{S}_{x}B_{x}^{-1}$. By Lemma 5.12, we have $\bar{T}_{1}^{*}(x)\in B_{1}(s\mathbb{D})$, where $s=r(T)^{-1}$. For $\lambda\in\Omega_{0}$, $\lambda\in\rho(T^{-1})$. Hence, $\lambda\in\rho_{s-F}(\bar{T}_{1}(x))$. Since $ind\bar{T}_{1}(x)=0$. The continuity of index implies that $ind\left(\bar{T}_{1}(x)-\lambda\right)=-1$ for all $\lambda\in\Omega_{0}$. Since $\bar{T}_{1}^{*}\in B_{1}(s\mathbb{D})$, Lemma 3.3 implies that $ker\left(\bar{T}_{1}(x)-\lambda\right)=\{0\}$ and $dimker\left(\bar{T}_{1}^{*}(x)-\bar{\lambda}\right)=1$ for all $\lambda\in\Omega_{0}$. Thus $\bar{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$. By Lemma 5.14, $\bar{T}_{2}(x)=P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}S_{2}P_{\mathcal{M}(x)^{\perp}\ominus\mathcal{L}(x)}=P_{\mathcal{N}(x)^{\perp}\oplus 0}S_{2}P_{\mathcal{N}(x)^{\perp}\oplus 0}\in B_{1}(\Sigma^{-1})$. This proves the lemma. ∎

Recall that $\Phi_{0}$ is the connected component of $\mathbb{C}\setminus\sigma(T)$ containing 0 and $\Omega_{0}$ is the connected component of $\mathbb{C}\setminus\sigma(T^{-1})$ containing 0. Note that

[TABLE]

By Lemma 5.8, the map $B_{x}:(-T^{-(n+1)}x)\oplus e_{n}\rightarrow-T^{-(n+1)}x$, $n\geq 0$, extends to a bounded linear isomorphism from $\mathcal{M}(x)=span\{(-T^{-(n+1)}x)\oplus e_{n}:\,n\geq 0\}$ onto $\mathcal{N}(x)=span\{T^{-(n+1)}x:\,n\geq 0\}$.

Proposition 5.16**.**

Suppose $T$ is transitive, $r(T)<1$, $x$ is a unit noncyclic vector of $T^{-1}$. Then

[TABLE]

satisfy

1. (1)

$\bar{T}_{1}(x)=B_{x}\hat{S}_{x}B_{x}^{-1}$, $\bar{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$, and $T^{-1}x\in\mathcal{C}(\bar{T}_{1}(x))$; 2. (2)

$\bar{T}_{2}(x)\in B_{1}(\Omega_{0})$, $\bar{T}_{2}(x)H_{2}(x)=I_{\mathcal{N}(x)^{\perp}}$, and $\mathcal{C}(H_{2}(x))\neq\emptyset$; 3. (3)

$\bar{H}_{1}(x)\triangleq B_{x}\hat{T}_{x}B_{x}^{-1}\in B_{1}(\Phi_{0})$, $\bar{H}_{1}(x)\bar{T}_{1}(x)=I_{\mathcal{N}(x)}$, and $\mathcal{C}(\bar{H}_{1}^{*}(x))\neq\emptyset$; 4. (4)

$H_{21}(x)$* is a rank one operator.*

Proof.

(1) $\bar{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$ and $\bar{T}_{1}(x)=B_{x}\hat{S}_{x}B_{x}^{-1}$ follow from Lemma 5.15 and its proof. Note that

[TABLE]

Therefore, $T^{-1}x\in\mathcal{C}(\bar{T}_{1}(x))$.

(2)By Lemma 5.14 and Lemma 5.15, $\overline{T}_{2}(x)\in B_{1}(\Phi)$, where

[TABLE]

Note that $\Phi\subseteq\Omega_{0}$. Since $\Omega_{0}\subset\rho(T^{-1})$ and $\overline{T}_{1}^{*}(x)\in B_{1}(\Omega_{0})$, $\Omega_{0}\subseteq\rho_{F}(\overline{T}_{2}(x))$. By the continuity of index, we deduce that $\overline{T}_{2}(x)\in B_{1}(\Omega_{0})$. Since $\bar{T}_{2}(x)H_{2}(x)=I_{\mathcal{N}(x)^{\perp}}$, by lemma 4.2, $\mathcal{C}(H_{2}(x))\neq\emptyset$.

(3) $\bar{H}_{1}(x)\triangleq B_{x}\hat{T}_{x}B_{x}^{-1}\in B_{1}(\Phi_{0})$ follows from Lemma 5.5.

[TABLE]

Also

[TABLE]

Then $\bar{T}^{*}_{1}(x)\bar{H}^{*}_{1}(x)=I_{\mathcal{N}(x)}$. By Lemma 4.2, $\mathcal{C}(\bar{H}_{1}^{*}(x))\neq\emptyset$.

(4) Since $H_{21}(x)\bar{T}_{1}(x)=0$, $\bar{T}_{1}^{*}(x)H_{21}^{*}(x)=0$. So $H_{21}(x)$ is a rank one operator. ∎

6. The main theorems

In this section, it is helpful to keep the following special case in mind: $\sigma(T^{-1})$ is the union of two circles, one is with center $(0,0)$ and radius $2$ and the other is with center $\left(4,0\right)$ and radius $2$. In this special case

[TABLE]

and

[TABLE]

Theorem 6.1**.**

Suppose $T\in\mathcal{B}(\mathcal{H})$ is an invertible operator, $x$ is a nonzero noncyclic vector of $T^{-1}$, and $\mathcal{N}(x)=span\{T^{-k}x:\,k\geq 1\}.$ Let $\overline{T}_{1}(x)=P_{\mathcal{N}(x)}T^{-1}P_{\mathcal{N}(x)}$. If $\sigma(\overline{T}_{1}(x))^{\land}\cap\rho_{F}(\overline{T}_{1}(x))$ has a connected component which does not contain zero point, then $T$ is intransitive.

Proof.

Suppose $T$ is transitive. By (1) of Proposition 5.16, $\overline{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$ and $0\in\Omega_{0}$. By Lemma 3.5, let $\overline{T}_{1}^{*}(x)\in B_{1}(L_{0}^{*})$ such that $L_{0}^{*}\supseteq\Omega_{0}^{*}$ is maximal. Suppose $\Omega$ is a connected component of $\sigma(\overline{T}_{1}(x))^{\land}\cap\rho_{F}(\overline{T}_{1}(x))$ such that $0\notin\Omega$. Then $\Omega\cap L_{0}=\emptyset$. By Proposition 5.16, $T^{-1}x\in\mathcal{C}(\overline{T}_{1}(x))$. By Proposition 4.5, $\Omega\subseteq\rho(\overline{T}_{1}(x))$. Note that $\partial\Omega\subseteq\sigma_{e}(\overline{T}_{1}(x))$. Since $\overline{H}_{1}(x)\overline{T}_{1}(x)=I$ and $\overline{T}_{1}(x)^{*}\in B_{1}(\Omega_{0}^{*})$, $\pi(\overline{H}_{1}(x))=\pi(\overline{T}_{1}(x))^{-1}$. Then $(\partial\Omega)^{-1}\subseteq\sigma_{e}(\overline{H}_{1}(x))$ and $\Omega^{-1}\subseteq\rho_{F}(\overline{H}_{1}(x))$. By (3) of Proposition 5.16, we have $\overline{H}_{1}(x)\in B_{1}(\Phi_{0})$ and $0\in\Phi_{0}$. By Lemma 3.5, let $\overline{H}_{1}(x)\in B_{1}(L_{1})$ such that $L_{1}\supseteq\Phi_{0}$ is maximal. Then $\Omega^{-1}\cap L_{1}=\emptyset$. By (3) of Proposition 5.16, $\mathcal{C}(\overline{H}_{1}^{*}(x))\neq\emptyset$. By Proposition 4.5, $\Omega^{-1}\subseteq\rho(\overline{H}_{1}(x))$. Note that $\overline{H}_{1}(x)\overline{T}_{1}(x)=I_{\mathcal{N}(x)}$. For $\lambda\in\Omega$, $\frac{1}{\lambda}\in\Omega^{-1}$ and

[TABLE]

Note that both $\overline{H}_{1}(x)-\frac{1}{\lambda}$ and $\lambda-\overline{T}_{1}(x)$ are invertible. Thus $\overline{H}_{1}(x)$ is invertible. On the other hand $\overline{H}_{1}(x)\in B_{1}(\Phi_{0})$ and $0\in\Phi_{0}$. So $\overline{H}_{1}(x)$ is not invertible. This is a contradiction. ∎

Lemma 6.2**.**

Suppose $T$ is transitive, $x$ is a nonzero noncyclic vector of $T^{-1}$, and $\mathcal{N}(x)=span\{T^{-k}x:\,k\geq 1\}$. Let $\mathcal{U}_{0}$ denote the connected component of int$(\sigma(T^{-1})^{\land})$ containing [math] and let $\Omega$ be a connected open subset of $\rho(T^{-1})$. If $\Omega\cap\mathcal{U}_{0}=\emptyset$, then $\Omega\subseteq\rho(\overline{T}_{1}(x))\cap\rho(\overline{T}_{2}(x))$.

Proof.

Recall that

[TABLE]

By Proposition 5.16, $\bar{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$, $\bar{T}_{2}(x)\in B_{1}(\Omega_{0})$ and $0\in\Omega_{0}$. By Lemma 3.3,

[TABLE]

Suppose $\lambda\in\rho(T^{-1})$. Write

[TABLE]

Then we have

[TABLE]

This implies that $(\overline{T}_{2}(x)-\lambda)D=I_{\mathcal{N}(x)^{\perp}}$ and $A(\overline{T}_{1}(x)-\lambda)=I_{\mathcal{N}(x)}$ or equivalently

[TABLE]

Therefore, $Ran(\overline{T}_{2}(x)-\lambda)$ and $Ran(\overline{T}_{1}(x)-\lambda)^{*}$ are closed. Thus $Ran(\overline{T}_{1}(x)-\lambda)$ is also closed. Since

[TABLE]

$\lambda\in\rho_{s-F}(\overline{T}_{1}(x))\cap\rho_{s-F}(\overline{T}_{2}(x))$. We have

[TABLE]

Since $\Omega\subseteq\rho(T^{-1})$, $\Omega\subseteq\rho_{s-F}(\overline{T}_{1}(x))\cap\rho_{s-F}(\overline{T}_{2}(x))$. Note that $\overline{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$ and $0\in\Omega_{0}$. By Lemma 3.5, $\overline{T}_{1}^{*}(x)\in B_{1}(L_{0}^{*})$ such that $L_{0}^{*}\supseteq\Omega_{0}^{*}$ is maximal. Note that $\mathcal{N}(x)\in Lat(T^{-1}).$ By Lemma 2.7, we have $\sigma(\overline{T}_{1}(x))\subseteq\sigma(T^{-1})^{\land}$ and thus $\sigma(\overline{T}_{1}(x))^{\land}\subseteq\sigma(T^{-1})^{\land}$. Then the connected component of int$\sigma(\overline{T}_{1}(x))^{\land}$ containing zero point is a subset of $\mathcal{U}_{0}$. Since $L_{0}\subseteq\sigma(\overline{T}_{1}(x))$ is a connected open set containing zero point, $L_{0}\subseteq\mathcal{U}_{0}$. By the assumption of Lemma 6.2, $\Omega\cap\mathcal{U}_{0}=\emptyset$. Thus $\Omega\cap L_{0}=\emptyset$. By Proposition 5.16, $T^{-1}x\in\mathcal{C}(\overline{T}_{1}(x))$. By Proposition 4.5, $\Omega\subset\rho(\overline{T}_{1}(x))$.

Notice that $\Omega\subseteq\rho(T^{-1})\cap\rho(\overline{T}_{1}(x)).$ Therefore,

[TABLE]

Thus, we have $\mbox{ind}(\overline{T}_{2}(x)-\lambda)=0,\forall\lambda\in\Omega.$ By (2) of Proposition 5.16, $\overline{T}_{2}(x)\in B_{1}(\Omega_{0})$. By Lemma 3.3, $\sigma_{p}(\overline{T}_{2}^{*}(x))=\emptyset$. Thus $dimker(\overline{T}_{2}(x)-\lambda)=0$, $\forall\lambda\in\Omega$, and $\Omega\subseteq\rho(\overline{T}_{2}(x))$. ∎

Lemma 6.3**.**

Let $\sigma_{e}(T^{-1})=\sigma(T^{-1})\subseteq\sigma(\overline{T}_{2}(x))$ and let

[TABLE]

[TABLE]

Then the following statements hold:

1. (1)

$\Phi_{0}=\left\{\frac{1}{\lambda}:\,\lambda\in\mathbb{C}\setminus\sigma(\overline{T}_{2}(x))^{\land}\cup\{\infty\}\right\};$** 2. (2)

$\Phi_{0}\subseteq R_{0}$; 3. (3)

$\overline{H}_{1}(x)\in B_{1}(R_{0})$, $R_{0}$ is maximal and $R_{0}^{*}$ is the connected component of $\rho_{F}^{-1}(\overline{H}_{1}^{*}(x_{1}))$ containing zero point.

Proof.

(1). By the assumption of Lemma 6.3 and Lemma 2.7,

[TABLE]

Then $\sigma(\overline{T}_{2}(x))^{\land}=\sigma(T^{-1})^{\land}$ and (1) holds.

(2). Since $\mathcal{N}(x)\in Lat(T^{-1})$, $\sigma(\overline{T}_{1}(x))\subseteq\sigma(T^{-1})^{\land}$. Therefore, $\sigma(\overline{T}_{1}(x))^{\land}\subseteq\sigma(T^{-1})^{\land}$. Then $\Phi_{0}\subseteq R_{0}$.

(3). By (3) of Proposition 5.16, $\overline{H}_{1}(x)\in B_{1}(\Phi_{0})$. Since $\overline{H}_{1}(x)\overline{T}_{1}(x)=I_{\mathcal{N}(x)}$,

[TABLE]

Since

[TABLE]

$R_{0}\subseteq\rho(\pi(\overline{H}_{1}(x)))$. By (2), Lemma 3.3, and continuity of index, $\overline{H}_{1}(x)\in B_{1}(R_{0})$. Since

[TABLE]

we have

[TABLE]

Thus $R_{0}^{*}$ is the connected component of $\rho_{F}^{-1}(\overline{H}_{1}^{*}(x))$ containing zero point.

∎

Theorem 6.4**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be an invertible operator. Suppose $x$ is a nonzero noncyclic vector of $T^{-1}$ and $\mathcal{N}(x)=span\{T^{-k}x:\,k\geq 1\}$. Let $\overline{T}_{2}(x)=P_{\mathcal{N}^{\perp}(x)}T^{-1}P_{\mathcal{N}^{\perp}(x)}$. If $\sigma(T^{-1})\subseteq\sigma(\overline{T}_{2}(x))$, and there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap\mathcal{U}_{0}=\emptyset$, where $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land})$ containing zero point, then $T$ is intransitive.

Proof.

Suppose $T$ is transitive. Then $\sigma(T^{-1})=\sigma_{e}(T^{-1})$ and $\rho(T^{-1})=\rho_{F}(T^{-1})$. By Lemma 6.2, since $\Omega\cap\mathcal{U}_{0}=\emptyset$,

(1.1) $\Omega\subseteq\rho(\overline{T}_{1}(x))\cap\rho(\overline{T}_{2}(x))$.

By Proposition 5.16, $\overline{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$ and $0\in\Omega_{0}$. By Lemma 3.5, let $\overline{T}_{1}^{*}(x)\in B_{1}(L_{0}^{*})$, and $L_{0}^{*}\supseteq\Omega_{0}^{*}$ is maximal. By Proposition 5.16, $T^{-1}x\in\mathcal{C}(\overline{T}_{1}(x))$. By Proposition 4.5, $L_{0}$ is the connected component of $\rho_{F}^{-1}(\overline{T}_{1}(x))$ containing zero point.

Claim 1: $L_{0}\subseteq\mathcal{U}_{0}$. By Lemma 2.7, $\sigma(\overline{T}_{1}(x))\subseteq\sigma(T^{-1})^{\land}$. Thus

[TABLE]

Since $0\in L_{0}$, and $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land}$ containing zero point, it follows that $L_{0}\subseteq\mathcal{U}_{0}$ and Claim 1 holds.

By the assumption of Theorem 6.4, $\Omega$ is a bounded connected component of $\rho(T^{-1})$. Since $\sigma(T^{-1})\subseteq\sigma(\overline{T}_{2}(x))$, $\rho(\overline{T}_{2}(x))\subseteq\rho(T^{-1})$. By (1.1), $\Omega\subseteq\rho(\overline{T}_{2}(x))\subseteq\rho(T^{-1})$ is a bounded connected component of $\rho(\overline{T}_{2}(x))$ and $\partial\Omega\subseteq\sigma_{e}(\overline{T}_{2}(x))$. By Proposition 5.16, $\overline{T}_{2}(x)\in B_{1}(\Omega_{0})$ and $0\in\Omega_{0}$. Since $\overline{T}_{2}(x)H_{2}(x)=I_{\mathcal{N}(x)^{\perp}}$,

[TABLE]

Thus

[TABLE]

and $\partial\Omega^{-1}\subseteq\sigma_{e}(H_{2}(x))$. Since $\overline{T}_{2}(x)\in B_{1}(\Omega_{0})$ and $0\in\Omega_{0}$, $ind(\overline{T}_{2}(x))=1$. Since $\overline{T}_{2}(x)H_{2}(x)=I$, $indH_{2}(x)=-1$. Note that $0\notin\Omega^{-1}$ and $\mathcal{C}(H_{2}(x))\neq\emptyset$ by Proposition 5.16. We have $\Omega^{-1}\cap\rho_{F}^{-1}(H_{2}(x))=\emptyset$. By Proposition 4.3,

[TABLE]

Thus $ind(H_{2}(x)-\lambda)\geq-1$ for all $\lambda\in\mathbb{C}$. Since $\Omega^{-1}\cap\rho_{F}^{-1}(H_{2}(x))=\emptyset$,

(1.2) $ind(H_{2}(x)-\lambda)\geq 0,\quad\forall\lambda\in\Omega^{-1}$.

Claim 2.  $\Omega^{-1}\subseteq R_{0}=\left\{\frac{1}{\lambda}:\,\lambda\in\mathbb{C}\setminus\sigma(\overline{T}_{1}(x))^{\land}\cup\{\infty\}\right\}$.

Note that either $\Omega^{-1}\cap R_{0}=\emptyset$ or $\Omega^{-1}\subseteq R_{0}$. In fact, since $\overline{H}_{1}(x)\overline{T}_{1}(x)=I_{\mathcal{N}(x)}$,

[TABLE]

By (1.1), $\Omega\subseteq\rho(\overline{T}_{1}(x))$. Thus $\Omega^{-1}\subseteq\rho_{F}(\overline{H}_{1}(x))$. By (3) of Lemma 6.3, $\overline{H}_{1}(x)\in B_{1}(R_{0})$ and $R_{0}$ is maximal. By Proposition 5.16, $\mathcal{C}(\overline{H}_{1}^{*}(x))\neq\emptyset$. By Proposition 4.5, $R_{0}=\rho_{F}^{1}(\overline{H}_{1}(x))$. If $\lambda_{0},\lambda_{1}\in\Omega^{-1}$ and $\lambda_{0}\in R_{0}$, $\lambda_{1}\notin R_{0}$, then

[TABLE]

and

[TABLE]

This contradicts to the continuity of index.

Now suppose $\Omega^{-1}\cap R_{0}=\emptyset$. By Proposition 5.16, $\mathcal{C}(\overline{H}_{1}^{*}(x))\neq\emptyset$. By Proposition 4.5, $\Omega^{-1}\subseteq\rho(\overline{H}_{1}(x))$. For $\lambda\in\Omega$,

[TABLE]

By (1.1), both $\lambda-\overline{T}_{1}(x)$ and $\overline{H}_{1}(x)-\frac{1}{\lambda}I$ are invertible. So $\overline{H}_{1}(x)$ is invertible. By Proposition 5.16, $\overline{H}_{1}(x)\in B_{1}(\Phi_{0})$ and $0\in\Phi_{0}$. So $\overline{H}_{1}(x)$ is not invertible. This is a contradiction. Thus Claim 2 holds.

By Claim 2 and (3) of Lemma 6.3, we have $ind(\overline{H}_{1}(x)-\lambda)=1$ for $\lambda\in\Omega^{-1}$. Since $\Omega\subseteq\rho(T^{-1})$, $\Omega^{-1}\subseteq\rho(T)$. By (1.2),

[TABLE]

This is a contradiction. ∎

7. Applications

7.1. Hyponormal operators

Definition 7.1**.**

Let $T\in\mathcal{B}(\mathcal{H})$. $T$ is called hyponormal if $T^{*}T-TT^{*}\geq 0$.

The following proposition is well-known.

Proposition 7.2**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be an invertible hyponormal operator and $\mathcal{N}\in Lat(T)$. Then $T|_{\mathcal{N}}$ and $T^{-1}$ are both hypernormal.

Suppose $A\in B_{1}(\Omega)$ and $0\in\Omega$. For $n\geq 1$, choose a unit vector $e_{n-1}\in kerA^{n}\ominus kerA^{n-1}$. Then $\{e_{n}\}_{n=0}^{\infty}$ is an ONB of $\mathcal{H}$ and

[TABLE]

Operator $B$ is called the standard right inverse of $A$ if $AB=I$ and

[TABLE]

In order to prove the main result in this subsection, we need the following lemmas.

Lemma 7.3**.**

Suppose that $A\in B_{1}(\Omega)$, $0\in\Omega$ and $B$ is the standard right inverse of $A$. Write

[TABLE]

Then $|a_{k,k+1}|\geq\frac{1}{\|B\|}.$

Proof.

Since $B$ can be regarded as an invertible operator from $\mathcal{H}$ onto $\ker A^{\perp}$. So for any $y\in\ker A^{\perp}$, we have $y=ABy=BAy.$ Then it follows that $\|y\|\leq\|B\|\cdot\|Ay\|$, and $\|Ay\|\geq\frac{1}{\|B\|}\|y\|.$ Note that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $y_{3}$ is in span$\{e_{1},e_{2}\}$. By induction, we have

[TABLE]

where $y_{k+1}$ is in span$\{e_{1},\cdots,e_{k}\}$. Thus, we have that

[TABLE]

∎

Lemma 7.4**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be invertible and transitive. Suppose that $x$ is a nonzero noncyclic vector of $T^{-1}$ and $\mathcal{N}(x)=\mbox{span}\{T^{-k}x,k\geq 1\}$. Then

[TABLE]

Furthermore, there exists an ONB $\{u_{k}\}^{\infty}_{k=-\infty}$ of $\mathcal{H}$, $u_{k}\in ker\overline{T}^{*(k+1)}_{1}(x)\ominus ker\overline{T}^{*k}_{1}(x)$ for $k\geq 0$, $u_{-k}\in ker\overline{T}_{2}^{k}(x)\ominus ker\overline{T}_{2}^{k-1}(x)$ for $k\geq 1$, which satisfies

[TABLE]

and

[TABLE]

where $|t_{-k,-k+1}|\geq\frac{1}{2\|T\|},|t_{k,k+1}|\geq\frac{1}{2\|T\|},k=0,1,2\cdots.$

Proof.

Note that $\overline{T}^{*}_{1}(x)H^{*}_{1}(x)=I_{\mathcal{N}(x)}$. Write

[TABLE]

and

[TABLE]

Then $\overline{T}^{*}_{1}(x)B=I_{\mathcal{N}(x)}$ and $B$ is the standard right inverse of $\overline{T}^{*}_{1}(x)$. By Lemma 7.3, we have that

[TABLE]

Notice that $T_{2}(x)H_{2}(x)=I_{\mathcal{N}(x)^{\perp}}$, we also have that

[TABLE]

∎

Lemma 7.5**.**

Suppose $T\in\mathcal{B}(\mathcal{H})$ is invertible and transitive, and there exists an ONB $\{u_{k}\}^{\infty}_{k=-\infty}$ of $\mathcal{H}$ given as in Lemma 7.4 such that

[TABLE]

[TABLE]

Then $a_{01}\neq 0.$

Proof.

Notice that

[TABLE]

and set

[TABLE]

By Lemma 3.7, there exists an invertible operator $G$ such that

[TABLE]

Since $\mathcal{C}(\overline{T}_{1}(x))\neq\emptyset$, we have $\mathcal{C}(A^{*})\neq\emptyset$. Thus, there exists an $x_{-1}$ such that

[TABLE]

Since $T$ is transitive, by Proposition 5.16, we have that $\overline{T}_{2}(x_{-1})\in B_{1}(\Omega)$. If $\alpha_{01}=0$, we have $\mbox{dim}\ker\overline{T}_{2}(x_{-1})=2,$ which is a contradiction.

∎

Lemma 7.6**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be invertible and transitive. Then there exists an $x_{n}\in\mathcal{H}$ such that $\mathcal{N}(x_{n})=\mbox{span}\{u_{k},k\leq n\},$ and satisfies

1. (1)

$\overline{T}_{1}(x_{n})\sim_{s}\overline{T}_{1}(x_{n-1})$; 2. (2)

$P_{\mathcal{N}(x_{n})}\stackrel{{\scriptstyle SOT}}{{\longrightarrow}}I$; 3. (3)

$P_{\mathcal{N}(x_{n})}T^{-1}|_{\mathcal{N}(x_{n})}\stackrel{{\scriptstyle SOT}}{{\longrightarrow}}T^{-1}$.

Proof.

Firstly, we will find $x_{1}$ such that $\mathcal{N}(x_{1})=\mbox{span}\{u_{k},k\leq 1\}.$ Set

[TABLE]

By Lemma 7.5, we have $a_{01}\neq 0$. By Lemma 3.7, $A_{1}\sim_{s}\overline{T}^{*}_{1}(x_{0})$, where $x_{0}$ is the vector $x$ in Lemma 7.5. Since $T^{-1}x_{0}\in\mathcal{C}(\overline{T}_{1}(x_{0}))$, there exists some $x_{1}\neq 0$ such that

[TABLE]

Meanwhile,

[TABLE]

and $T^{-1}x_{1}\in\mathcal{C}(\overline{T}_{1}(x_{1}))$. Now set

[TABLE]

By Lemma 7.4, we have that $|t_{1,2}|\geq\frac{1}{2\|T\|}$. By Lemma 3.7, we have that $A_{2}\sim_{s}\overline{T}^{*}_{1}(x_{1})$. Since $T^{-1}x_{1}\in\mathcal{C}(\overline{T}_{1}(x_{1}))$, there exists an $x_{2}$ such that

[TABLE]

In this case, set $\overline{T}_{1}(x_{2})=A^{*}_{2}.$ Repeating the steps above, we can find a sequence $\{x_{n}\}^{\infty}_{n=1}.$ Notice that $\mathcal{N}(x_{n-1})\subseteq\mathcal{N}(x_{n})$. Then $\{x_{n}\}^{\infty}_{n=1}$ satisfies the condition of this lemma. ∎

Theorem 7.7**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be an invertible hyponormal operator. If $T^{-1}$ is intransitive and there exist at least two bounded components of $\mbox{int}\sigma(T^{-1})^{\land}$, then $T$ is also intransitive.

Proof.

If $\mbox{int}\sigma(T^{-1})\neq\emptyset,$ by the result of S. Brown [5], we see that $T$ and $T^{-1}$ both have nontrivial invariants subspaces. So in the following we assume $\mbox{int}\sigma(T^{-1})=\emptyset.$

Let $x$ be a nonzero noncyclic vector of $T^{-1}$ and let $\mathcal{N}(x)=span\{T^{-n}x:\,n\geq 1\}$. Denote by $\overline{T}_{1}(x)=P_{\mathcal{N}(x)}T^{-1}P_{\mathcal{N}(x)}$. Suppose $T$ is transitive. By Proposition 5.16, $\overline{T}_{1}^{*}(x)\in B_{1}(\Omega_{0}^{*})$ and $0\in\Omega_{0}$. By Lemma 3.5, let $\overline{T}_{1}^{*}(x)\in B_{1}(L_{0}^{*})$ such that $L_{0}^{*}\supseteq\Omega_{0}^{*}$ is maximal. By Proposition 5.16, $\mathcal{C}(\overline{T}_{1}(x))\neq\emptyset$. By Proposition 4.5, $L_{0}=\rho_{F}^{-1}(\overline{T}_{1}(x))$.

Let $\mathcal{U}_{0}$ and $\mathcal{U}_{1}$ be two bounded components of $\mbox{int}\sigma(T^{-1})^{\land}$ such that $0\in\mathcal{U}_{0}$. By the assumption $\mbox{int}\sigma(T^{-1})=\emptyset,$ we have $\mathcal{U}_{0},\mathcal{U}_{1}\subseteq\rho(T^{-1})$. Since $\mathcal{N}(x)\in Lat(T^{-1})$, $\sigma(\overline{T}_{1}(x))\subseteq\sigma(T^{-1})^{\land}$ and $\sigma(\overline{T}_{1}(x))^{\land}\subseteq\sigma(T^{-1})^{\land}$. Therefore, $L_{0}\subseteq\mathcal{U}_{0}$ and $L_{0}\cap\mathcal{U}_{1}=\emptyset$. Suppose $\partial\mathcal{U}_{1}\subseteq\sigma_{e}(\overline{T}_{1}(x))$. Since $\mathcal{U}_{1}\subseteq\rho(T^{-1})$, $\mathcal{U}_{1}\subseteq\rho_{s-F}(\overline{T}_{1}(x))$. Note that $L_{0}\cap\mathcal{U}_{1}=\emptyset$ and $\mathcal{C}(\overline{T}_{1}(x))\neq\emptyset$. By Proposition 4.5, $\mathcal{U}_{1}\subseteq\rho(\overline{T}_{1}(x))$ and $\sigma(\overline{T}_{1}(x))^{\land}\cap\rho_{F}(\overline{T}_{1}(x))$ has a connected component $\mathcal{U}_{1}$ which does not contain zero point. By Theorem 6.1, $T$ is intransitive. This contradicts to the assumption that $T$ is transitive. Thus there exists a $\lambda\in\partial\mathcal{U}_{1}$ such that $\lambda\in\rho_{F}(\overline{T}_{1}(x))$. Since $\mathcal{C}(\overline{T}_{1}(x))\neq\emptyset$, by Lemma 3.3 and Proposition 4.3, $\lambda\in\rho_{F}^{-1}(\overline{T}_{1}(x))$ or $\lambda\in\rho(\overline{T}_{1}(x))$. Note that $L_{0}=\rho_{F}^{-1}(\overline{T}_{1}(x))$ and $L_{0}\cap\mathcal{U}_{1}=\emptyset$. We have $\lambda\notin\rho_{F}^{-1}(\overline{T}_{1}(x))$. Thus $\lambda\in\rho(\overline{T}_{1}(x))$.

By Lemma 7.6, there exists a sequence of vectors $\{x_{n}\}^{\infty}_{n=0}\in\mathcal{H}$ such that $\mathcal{N}(x_{n-1})\subseteq\mathcal{N}(x_{n})$. Then $P_{\mathcal{N}(x_{n})}\stackrel{{\scriptstyle SOT}}{{\longrightarrow}}I$, $\overline{T}_{1}(x_{n})\stackrel{{\scriptstyle SOT}}{{\longrightarrow}}T^{-1}$ and $\overline{T}_{1}(x_{n})\sim_{s}\overline{T}_{1}(x_{n-1})$ and thus $\sigma(\overline{T}_{1}(x_{n}))=\sigma(\overline{T}_{1}(x_{0})),n\geq 0$. Furthermore, for any $\xi\in\mathcal{H}$, by [Proposition 2.1, page 72] of [26], we have

[TABLE]

Notice that

[TABLE]

It follows that

[TABLE]

Therefore, $(T^{-1}-\lambda)$ is bounded below. On the other hand, $\lambda\in\partial\mathcal{U}_{1}\subseteq\partial\sigma(T^{-1})$. This is a contradiction. ∎

7.2. Strictly cyclic invariant subspaces

Definition 7.8**.**

Let $T\in\mathcal{B}(\mathcal{H})$ and let ${\mathcal{A}}(T)$ be the closed subalgebra of $\mathcal{B}(\mathcal{H})$ generated by $T$ and $I$. A vector $\xi\in\mathcal{H}$ is called a strictly cyclic vector of $T$ if $\{S\xi:\,S\in{\mathcal{A}}(T)\}=\mathcal{H}$. An invariant subspace $\mathcal{K}\subseteq\mathcal{H}$ of $T$ is called a strictly cyclic invariant subspace if $T|_{\mathcal{K}}\in\mathcal{B}(\mathcal{K})$ has a strictly cyclic vector.

The following lemma is due to B.Barnes (Corollary 3 of [4]).

Lemma 7.9**.**

Suppose $T\in\mathcal{B}(\mathcal{H})$ is strictly cyclic. Then $\sigma_{p}(T^{*})=\sigma(T^{*})$.

Lemma 7.10**.**

Let $T\in\mathcal{B}({\mathcal{H}})$ be an invertible operator. If $T^{-1}$ is intransitive and satisfies the following properties:

1. (1)

there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap\mathcal{U}_{0}=\emptyset$, where $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land})$ containing zero point; 2. (2)

$\sigma_{p}((P_{\mathcal{N}(x)}T^{-1}|_{\mathcal{N}(x)})^{*})=\sigma((P_{\mathcal{N}(x)}T^{-1}|_{\mathcal{N}(x)})^{*}),$**

then $T$ is also intransitive.

Proof.

Suppose $T$ is transitive.

Claim:  $\sigma(T^{-1})\subseteq\sigma(\overline{T}_{2}(x))$. Otherwise, there exists a $\lambda\in\sigma(T^{-1})$ such that $\lambda\in\rho(\overline{T}_{2}(x))$. If $\lambda\in\rho(\overline{T}_{1}(x))$, then direct computation shows that $\lambda\in\rho(T^{-1})$. It is a contradiction. Suppose $\lambda\in\sigma(\overline{T}_{1}(x))$. Then $\bar{\lambda}\in\sigma((\overline{T}_{1}(x))^{*})=\sigma_{p}((\overline{T}_{1}(x))^{*})$. Thus there exists a nonzero vector $\xi\in\mathcal{N}(x)$ such that $(\overline{T}_{1}(x)-\lambda)^{*}\xi=0$. Note that $(\overline{T}_{2}(x)-\lambda)^{*}$ is invertible. There exists a vector $\eta\in\mathcal{N}(x)^{\perp}$ such that

[TABLE]

This implies that

[TABLE]

Thus $\sigma_{p}(T^{-1*})\neq\emptyset$ and $T$ is intransitive. It is a contradiction and the Claim is true. By Theorem 6.4, $T$ is intransitive. ∎

Theorem 7.11**.**

Let $T\in\mathcal{B}(\mathcal{H})$ be invertible. If $T^{-1}$ has a proper strictly cyclic invariant subspace and there exists a bounded open set $\Omega$ which is a connected component of $\rho(T^{-1})$ such that $\Omega\cap\mathcal{U}_{0}=\emptyset$, where $\mathcal{U}_{0}$ is the connected component of $int(\sigma(T^{-1})^{\land})$ containing zero point, then $T$ is intransitive.

Proof.

Suppose $\mathcal{N}(x)$ is a strictly cyclic invariant subspace of $T^{-1}$. By Lemma 7.9, $\sigma_{p}(\bar{T}_{1}^{*}(x))=\sigma(\bar{T}_{1}^{*}(x))$. By Lemma 7.10, $T$ has a nontrivial invariant subspace. ∎