Find the eigenvalues of the operator $T(x)=\sum_{i=1}^{\infty} \langle x,e_i \rangle e_{i+1}$

Question

Let $X$ be a separable Hilbert space and let $\{e_1,e_2, \ldots \}$ be an orthonormal basis of $X$. Define $$T(x)=\sum_{i=1}^{\infty} \langle x,e_i \rangle e_{i+1}$$ Is $T$ compact or symmetric? Compute the eigenvalues and the norm of $T$

I think for $||T||=1$, I proved that $||T|| \leq 1$, for the eigenvalues I tried to use the spectral decomposition theorem but first I want to see that $T$ was symmetric, any suggestion or help to find the other points of the question I will be very grateful. Edit: I proved that $T$ is not a compact operator

Answer 1

Suppose $$T(x)=\sum_{i=1}^{\infty} \langle x,e_i \rangle e_{i+1}=\lambda x=\lambda \sum_{i=1}^{\infty} \langle x,e_i \rangle e_{i}$$ If $\lambda \neq 0$ this implies that $\langle x,e_1 \rangle=0$ and $\langle x,e_i \rangle=\lambda \langle x,e_{i+1} \rangle$ for $i \geq 1$ which makes $\langle x,e_i \rangle =0$ for all $i$ and $x =0$. Thus, $\lambda =0$ is the only possible eigen value. I wil let you check if $0$ is indeed an eigen value.

Answer 2

$T$ is not compact: $T(e_n)=e_{n+1}$, hence $(T(e_n))$ contains no convergent subsequence.

$T$ is not symmetric: $T^*(e_1)=0$ and $T^*(e_n)=e_{n-1}$ for $n \ge 2.$ Thus $T^* \ne T.$