compute the spectrum of the sum of shift operator $S$ and its dual $S^*$ is [-2,2]

Question

How to compute the spectrum of one sided shift operator $S$ on $l^2(N)$ plus its dual $S^*$, based on the definition of spectrum, I have no idea how to do it. Any hints will be appreciated

Answer 1

I'm not aware of a straighforward computation (not saying there isn't, but I can't imagine what it would be).

If you look at the upper left corners of $S+S^*$, you get $S_n+S_n^*$, where $S_n$ is the shift on $M_n(\mathbb C)$. It is well known that the eigenvalues of $S_n+S_n^*$ are $2\cos\frac{k\pi}{n+1}$, $k=1,\ldots,n$. It is possible to use this to show that all those numbers are approximate eigenvalues for $S+S^*$, from where you can deduce that $\sigma(S+S^*)=[-2,2]$.

Another way is to look at the numerical range $W(S+S^*)$. It is well-known that for any selfadjoint operator $T$, we have $W(T)=\overline{\text{conv}}\sigma(T)$. So if $W(T)=[-2,2]$, the necessarily $\sigma(T)=[-2,2]$. To show this, using the basis $\{e_n\}$ in which $S$ is a shift, let $x=ae_1+be_2$, with $|a|^2+|b|^2=1$. Then $$\tag1 \langle (S+S^*)x,x\rangle=2\text{Re}\,a\overline{b}. $$ It is very easy to see that the possible numbers in $(1)$ are $[-2,2]$.