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
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).