Let $(a_{n})$ be a bounded sequence of complex numbers and $\mathscr{H}$ be a Hilbert space with orthonormal basis $\{ e_{n} \}$.
Show that for every complex number $\lambda$ with $|\lambda| = 1$ there is a unitary (bounded) operator $U = U_{\lambda}$ such that $UAU^{-1} = \lambda A$, where $A$ is the unilateral weighted shift such that $A e_{n} = a_{n} e_{n + 1}$.
Also deduce that the spectrum of a weighted shift must be the union of concentric circles about $z = 0$.
I already showed that there is a unilateral weighted shift $A$ and that it is unique. Stuck on the unitary operator $U$.
Best Answer
Let $Ue_n:=\lambda^ne_n$, where $\lambda=e^{i\theta}$, $\theta\in\mathbb{R}$.
Then its adjoint is $U^*e_n=\lambda^{-n}e_n$, since $\langle U^*e_n,e_m\rangle=\langle e_n,\lambda^m e_m\rangle=\bar{\lambda}^m\delta_{nm}$.
So $U^*Ue_n=\lambda^n \lambda^{-n}e_n=e_n=UU^*e_n$. Hence $U$ is unitary; moreover,
\begin{align}UAU^{-1}e_n&=UA(\bar{\lambda}^ne_n)\\ &= \bar{\lambda}^na_nUe_{n+1}\\ &=\bar{\lambda}^n\lambda^{n+1}a_ne_{n+1}\\ &=\lambda a_ne_{n+1}=\lambda Ae_n\end{align} Since they agree on all basis vectors, $UAU^{-1}=\lambda A$.