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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

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  • $\begingroup$ You just need to show that $S$ has no other eigenvalues than $0$. So suppose $\lambda \neq 0$ and $Sx = \lambda x$. Deduce $x = 0$. $\endgroup$ – Daniel Fischer May 5 '15 at 11:09
  • $\begingroup$ Cool, I hadn't tried this way, thanks =) $\endgroup$ – PtF May 5 '15 at 11:13
  • $\begingroup$ @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..) $\endgroup$ – PtF May 5 '15 at 11:18
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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).

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