# Point spectrum of operator on $\ell^2$?

Considere the bounded linear operator $S:\ell^2\longrightarrow \ell^2$ given by $$S(\xi_j)_j:=\left(\frac{\xi_2}{1}, \frac{\xi_3}{2}, \frac{\xi_4}{3}, \ldots\right).$$

How to show the point spectrum of $S$ is trivial, i.e., $\sigma_p(S):=\{0\}$?

For the inclusion $\{0\}\subset \sigma_p(S)$ it suffices taking the vector $\xi:=(1, 0, 0, \ldots)\in \ell^2-\{0\}$ that one has $T\xi=0\xi$.

On the other hand, I guess I must show that $\lambda\neq 0$ implies $S-\lambda I$ is invertible by either showing its kernel is trivial or by exhibiting the inverse but I wasn't able to do any of that.

Thanks

• You just need to show that $S$ has no other eigenvalues than $0$. So suppose $\lambda \neq 0$ and $Sx = \lambda x$. Deduce $x = 0$. – Daniel Fischer May 5 '15 at 11:09
• Cool, I hadn't tried this way, thanks =) – PtF May 5 '15 at 11:13
• @DanielFischer Indeed the reasoning to show what you suggested me is the same for showing $\textrm{ker}(S-\lambda I)=\{0\}$ (what I was stuck on..) – PtF May 5 '15 at 11:18

## 1 Answer

To see that $S - \lambda I$ has trivial kernel for $\lambda \ne 0$, let $x \in \ell^2$ with $Sx = \lambda x$, we have $$\lambda x_j = (Sx)_j = \frac{1}{j}\cdot x_{j+1}$$ That is, for any $j$, we have $$x_j = (j-1)!\lambda^{j-1} x_1$$ As $x \in \ell^2$, we have $$\|x\|^2_2 = \sum_{j=1}^\infty \bigl((j-1)!\lambda^{j-1}\bigr)^2 x_1^2 < \infty$$ But this only holds iff $x_1 = 0$, so $x_j = (j-1)!\lambda^{j-1}x_1 = 0$ for all $j$, that is $x= 0$.

Therefore $0$ is the only eigenvalue of $S$ (note that this does not mean that every $S -\lambda I$, $\lambda \ne 0$ is invertible, as trivial kernel does not imply invertibility for infinite dimensional endomorphisms).