Let $T: \ell^2 \rightarrow \ell^2$ be the unilateral shift, and let $\mathcal{T}= \{ I, T, T^* \}$.
Show that $\mathcal{T}' = \{\lambda I_{\ell^2} : \lambda \in \mathbb{C} \}$ and $\mathcal{T}'' = B(\ell^2)$.
Attempt:
Since $T$ is the unilateral shift, we have that $Te_n = e_{n+1}$.
Also, by definition, we have $\mathcal{T}' = \{ S \in B(\ell^2) : ST =TS, \forall T \in \mathcal{T}\}$.
Let's look at the each element in $T$:
We know that the identity $I$ always has the property $IS=SI$ for $S \in B(\ell^2) $.
Also, let $S \in B(\ell^2)$. Then we have $STe_n= Se_{n+1} =e_{n+1 }S= Te_n S = TSe_{n}$.
For $T^*$, I know that $T^*e_n= e_{n-1}$, for $n>1$. So don't we get a similar result as above, i.e $ST^*e_n =T^* S e_n$?
So I'm a bit confused on how to get the results we want. I feel like I got the definitions of $\mathcal{T}'$ and $\mathcal{T}''$ mixed up somewhere…
Thank you for your help!
Best Answer
Observing that, for every $n\geq 1$, one has that $Te_n=e_{n+1}$, and $$ T^*e_n = \left\{\matrix{0, & \text {if } n=1, \cr e_{n-1}, & \text {if } n>1, }\right. $$ one quicky sees that $$ (I-TT^*)e_n = \left\{\matrix{e_1, & \text {if } n=1, \cr 0, & \text {otherwise.} }\right. $$ So we deduce that the operator $P$ defined by $$ P:=I-TT^* $$ is the orthogonal projection onto the space spanned by $e_1$.
If $S$ is any operator commuting with $T$ and $T^*$, then obviously $S$ also commutes with $P$, so $$ Se_1 = SPe_1 = PSe_1 = \lambda e_1, $$ for some scalar $\lambda $. We then claim that $S$ necessarily coincides with $\lambda I$, the reason being that for all $n>1$, $$ Se_n = ST^{n-1}e_1 = T^{n-1}Se_1 = \lambda T^{n-1}e_1 = \lambda e_n. $$