Spectrum of bilateral shift Let $T:l^2(\mathbb{Z})\longrightarrow l^2(\mathbb{Z})$ and define $T(\{x_n\})=\{x_{n-1}\}.$ Some reference tells me that the spectrum of $T$ is $\mathbb{T}=\{\lambda\in\mathbb{C}:|\lambda|=1 \}.$ What I'm sure is that for $\lambda=1$ then $T-\lambda I$ is not invertible.
But, I'm not quite sure that for $\lambda\in\mathbb{T}$ other than $\lambda=1$, $T-\lambda I$ also not invertible. For example, even for $\lambda=-1$ or $\lambda=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2},$ I can't see why $T-\lambda I$ is not invertible and (for example) if $\lambda=\sqrt{2}+i\sqrt{2}\notin\mathbb{T}$, then $T-\lambda I$ (perhaps) invertible.
Any help would be appreciated.
Thank you
 A: Observe that $TT^* = 1 = T^*T$, i.e. $T$ is a unitary. Then note that the spectrum of a unitary is contained in the unit circle $\mathbb{T}$.
Conversely, let $\lambda \in \mathbb{T}$, i.e. $|\lambda|=1$. We show that $T-\lambda I$ is not invertible. Consider $$e_0 = \{\dots, 0,\underbrace{1}_{position \ 0},0, \dots, \} \in \ell^2(\mathbb{Z})$$
Assume to the contrary that $T-\lambda I$ would be invertible, so in particular it is surjective and we can choose $\{a_n\}_n \in \ell^2(\mathbb{Z})$ with $(T-\lambda I)(\{a_n\}_n) = e_0$, i.e.
we have
$$a_{n-1}-\lambda a_n= 0, \quad n \neq 0, \quad \quad a_{-1}- \lambda a_0 = 1.$$
Thus $\{a_n\}_n = \{\dots, a_0, \lambda^{-1}a_0, \lambda^{-2} a_0, \lambda^{-3} a_0, \dots \}$ and consequently
$$\|\{a_n\}_n\|_2^2 \ge \sum_{n=0}^\infty |\lambda^{-n} a_0|^2= \sum_n |a_0|^2 = \infty $$
which contradicts the fact that $\{a_n\}_n \in \ell^2(\mathbb{Z}).$ Hence, we must have that $T-\lambda I$ is not invertible so $\lambda$ is in the spectrum.
A: Here is a slick way to prove that $\sigma (T)=S^1$.

*

*Recall that,  since $T$ is unitary,  one has that $\sigma (T)\subseteq S^1$.


*Prove (see later) that,  for each $\mu \in S^1$, there is an invertible operator  $U $ such that
$$
  U TU ^{-1}=\mu T.
  $$


*With $U $, as above, notice that
$$
  \sigma (T) =   \sigma (U TU ^{-1}) = \sigma (\mu T) = \mu \sigma (T).
  $$


*Since $\sigma (T)$ is nonempty,  choose any $\lambda _1$ in $\sigma (T)$.  For every $\lambda \in S^1$, set $\mu =\lambda \lambda _1^{-1}$,  and notice that  by (3) we have
$$
  \lambda  = \mu \lambda _1 \in \mu \sigma (T)=\sigma (T),
  $$
concluding the proof.

Here is the proof of (2):  Given $\mu $ in $S^1$,  consider the operator
$$
  U :\ell ^2({\mathbb Z})\to \ell ^2({\mathbb Z})
  $$
given by $U (\{x_n\}) = \{\mu ^nx_n\}$.  Then clearly $U $ is invertible with inverse
$$
  U^{-1} (\{x_n\}) = \{\mu^{-n}x_n\}.
  $$
Moreover, for each
vector $e_n$ in the canonical orthonormal basis of $\ell ^2({\mathbb Z})$, one has that
$$
  U T(e_n)=U (e_{n+1}) = \mu ^{n+1}e_{n+1},
  $$
while
$$
  TU (e_n) = T(\mu ^ne_n) = \mu ^ne_{n+1},
  $$
thus proving that $UT=\mu TU$,  whence $UTU^{-1}=\mu T$.
