A question related to the eigenvalues of a shift operator

Question

Suppose $\{e_k\}^{\infty}_{k=1}$ be an orthonormal basis for $\mathcal{H}$ (Hilbert space), and let $T$ be the shift operator defined by $$Te_k=e_{k+1}$$ for $k=1,2,...$

I have shown that $T$ has no eigenvalues, and I'm currently working on computing the eigenvalues of $T^{*}$ (adjoint of $T$) and $\sigma(T)$ (spectrum of $T$) respectively.

So far, I've managed to show that $T^*$ is simply the operator defined by: $T^{*}e_k=e_{k-1}$, for $k=1,2,...$ with $Te_1=0$.

Also, if $\lambda$ is an eigenvalue of $T^*$ and $\phi$ a corresponding eigenvector such that $\phi=\Sigma^{\infty}_{i=1}\beta_ie_i$, then we can see that $\beta_2=\lambda \beta_1, \beta_3=\lambda^2 \beta_1, \beta_4=\lambda^3 \beta_1,...$

Clearly, $\lambda \neq 0$, for otherwise we get $\phi=0$, which is a contradiction. How can I make use of this information to compute the eigenvalues of $T^*$? Also, how do we compute $\sigma(T)$?

Any help or hint will be extremely useful. Thanks.

Answer 1

If $|\lambda| <1$ then $x=\sum \lambda ^{k} e_k \in H$ and $T^{*}(x)=\lambda x$. Hence $\lambda$ is an eigen value of $T^{*}$. I will let you check that ther are no other eigen values.

The spectrum of $T$ is the closed unit disk. It is contained in the closed unit disk because $\|T\|=1$. If $|\lambda| <1$ then there is no $x$ such that $Tx-\lambda x =e_1$: indeed, writing $x=\sum a_k e_k$ we get $a_k-\lambda a_{k+1}=0$ for all $k >1$. But then $a_{k+1}=\frac 1 {\lambda} a_k$. Iterating this we see that $\sum |a_k|^{2}=\infty$, a contradiction. This proves that $T-\lambda I$ is not invertible whenever $|\lambda| <1$, so $\sigma (T)=\{\lambda : |\lambda |\leq 1\}$.