3
$\begingroup$

Let $\mathcal{H}$ be a Hilbert space and let {$e_j$}$_{j\in \mathbb{Z}}$ be an orthonormal basis for $\mathcal{H}$. Define a linear operator $T$ on $\mathcal{H}$ by $T(e_0) = 0$ and $T(e_j) = e_{j+1}$ for $j \neq$ 0. Define another linear operator on $\mathcal{H}$ by $A(e_0) = e_1$ and $A(e_j) = 0$ for $j \neq 0$. For $z \in \mathbb{C}$, define $T_z = T +zA$.

I'm being asked to find $\sigma(T_z)$ and to say what happens when $z \to 0$ (which I imagine is obvious if I can find $\sigma(T_z)$).

I've tried finding $\sigma(T_z)$ for certain values of $z$ with little success. I know that $\sigma(A)$ = {$0$} since the spectral radius of $A$ is $0$ ($A^n$ for $n>1$ is just going to be the trivial operator) and so $\sigma(zA)$ = {$0$}; I'm not sure if this will be useful. Correct me if I'm wrong, but $||T_z||$ = $1$ $\lor$ $||z||$ so $\sigma(T_z) \subset \overline{B}_{1 \lor ||z||}(0)$. I believe that the point spectrum of $T_z$ will be empty for $z \neq 0$ which doesn't help me either.

For context, the previous part of this problem asked for $\sigma(T)$.

Any pointers in the right direction would be great.

2 Answers 2

1

Hint: if $z \ne 0$, $T_z$ is closely related to the ordinary right shift operator.

  • 0
    Not just superficially. They are similar.2011-10-03
0

First, note that $T_z$ is an invertible weighted bilateral as $T_ze_0=ze_1$ and $T_ze_j=e_{j+1}$ for all $j\not=0$. Moreover, $\|T_z^n\|=\max(1,|z|)$ and $\|T_z^{-n}\|=\max(1,|z|^{-1})$ for all $n\geq1$ and thus $r(T_z)=r(T_z^{-1})=1$. This tells you that $\sigma(T_z)\subset\Bbb{T}$ but, since since $\sigma(T_z)$ is nonempty and is invariant by rotation about the origin, $\sigma(T_z)=\Bbb{T}$. Observe that $U\alpha T_z=(\alpha T_z) U_\alpha$ where $U_\alpha$ is the unitary operator defined by $U_\alpha e_n=\alpha^ne_n$, and $T_z$ and $\alpha T_z$ are unitairement equivalent and thus the invariance by rotation about the origin of the spectra of $T_z$