Consider the Hilbert space $H=\ell^2(\mathbb{Z}\times\mathbb{Z})$ with an orthonormal basis $\{e_{i,j}\}_{i,j\in\mathbb{Z}}$ and fix a complex number $\mu$ such that $|\mu|\in(0,1).$ Define the operator $T$ on $H$ by
$$T(e_{i,j})=\mu^i e_{i,j+1}$$
Then $T$ is a densely defined unbounded normal operator on $H$ with spectrum $\mathbb{C}^{\mu}\colon =\{\lambda\in\mathbb{C}\mid |\lambda|\in\mu^{\mathbb{Z}}\}$ where $\mu^{\mathbb{Z}}\colon=\{\mu^i\mid i\in\mathbb{Z}\}$. It is clear that $T$ cannot be bounded as the indices $i,j$ runs over the set of all integers. My questions are the following:
$1$. I suspect the spectrum of $T$ is $\textit{contained}$ in $\mathbb{C}^{\mu}$ and not equal to $\mathbb{C}^{\mu}$. Clearly, $\mu^i$ is in the spectrum for all $i\in\mathbb{Z}$ and real $\mu\in(0,1)$. I can't see why $\lambda$ is in the spectrum with $|\lambda|=|\mu|^i$ when $\mu$ is complex number such that $0<|\mu|<1.$
$2$.If $U$ is a unitary operator on $H$ which commutes with $T$, then how are the spectra of $T$ and $UT$ related? are they equal?
Any suggestion will be highly appreciated.
Best Answer
Let $$ H_k=\overline{\operatorname{span}}\{e_{k,j}:\ j\in\mathbb Z\}. $$ Then $H=\bigoplus_kH_k$. More importantly, each $H_k$ is invariant for $T$, so $T=\bigoplus_k T_k$, where $T_k=T|_{H_k}$. Then $\sigma(T)=\overline{\bigcup_k\sigma(T_k)}$.
On $H_k$, we have $$ T\sum_j x_je_{k,j}=\sum_j \mu^kx_je_{k,j+1}. $$ That is, $T_k=\mu^k\,B$, where $B$ is the bilateral shift. Thus $$\sigma(T_k)=\mu^k\,\sigma(B)=\mu^k\,\mathbb T=|\mu|^k\,\mathbb T. $$ Thus $$ \sigma(T)=\overline{\{\lambda\in\mathbb C:\ \exists k\in\mathbb Z, |\lambda|=|\mu|^k\}}. $$ The closure is definitely needed, as $0\in\sigma(T)$.
As for your second question, the answer is "not related". For instance if $T=I$, you have $\sigma(T)=\{1\}$ and you are free to choose $\sigma(UT)=\sigma(U)$.