Point spectrum (eigenvalues) of two operators in $l^2$.

Question

Let the following operators of $l^2$ to $l^2$

$S(x_0,x_1, x_2,...)=(0,x_0, x_1/2, x_2/3,...)$

$T(x_0,x_1, x_2,...)=(x_1/2, x_2/3, x_3/4,...)$

I need to find the point spectrum(eigenvalues set) $\sigma_p(S)$ and $\sigma_p(T)$.

I showed that $\sigma_p(S)=\varnothing$.

For $\sigma_p(T)$, I have this:

\begin{align} T(x)=\lambda x\,\,\,&\Leftrightarrow\,\,\,(x_1/2, x_2/3, x_3/4,...)=\lambda(x_0,x_1, x_2,...)\\ &\Leftrightarrow\,\,\, x_n=(n+1)!\lambda x_0,\,\,\,\,\,\, n\geq 1 \end{align} Then $(x_0,x_1, x_2,x_4...)=x_0(1, 2\lambda, 6\lambda^2, 24\lambda^3,...)$.

But from here I do not know how to conclude which are the eigenvalues ​​of T.

I appreciate the help you can give me.

Answer 1

In order that the sequence $(x_n)$ you obtained is in $l^{2}$ we must have $\sum ((n+1)!)^{2}|\lambda|^{2} <\infty$. This is possible only for $\lambda =0$. But $\lambda =0$ gives $x_n=0$ for all $n$. Hence $T$ has no eigen values.