Let $H$ be an infinite dimensional, separable, complex Hilbert space.
Let $T \in B(H)$ be a unilateral weighted shift. i.e. $T e_n= t_n e_{n+1}$, where $(t_n)_n \in \ell^{\infty}$.
Suppose $\lambda \in \sigma(T)$. Show that $\{z\in \mathbb{C}: |z|= \lambda \} \subset \sigma(T)$.
I showed that if $\lambda=1$, then we have $UTU^{-1}=\lambda T$, where $U$ is a unitary. Not sure if this is related to what we want to show.
Thank you!
Best Answer
As per this answer, for any $\mu=e^{i\theta}$ there is a unitary operator such that $UTU^{-1}=\mu T$.
Applying the spectrum to both sides, $$\sigma(T)=\sigma(UTU^{-1})=\mu\sigma(T)$$ This means that $\sigma(T)$ is rotationally invariant. Hence if $\lambda\in\sigma(T)$ and $|z|=|\lambda|$ then $z=\frac{z}{\lambda}\lambda\in\frac{z}{\lambda}\sigma(T)=\sigma(T)$ as well.