Spectrum of bilateral shift

Question

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

Answer 1

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.

Answer 2

Here is a slick way to prove that $\sigma (T)=S^1$.

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

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

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

4. 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.

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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 $UTU^{-1}=\mu T$.