I have a problem with the spectrum of this operator:
$(Tx)_1 = x_2$
$(Tx)_2 = x_1$
$(Tx)_n = \frac{1}{n}x_{n+1}$
with $n\ge3$
Find the $||T||$, the point spectrum $\sigma_P(T)$ and $\sigma_P(T^{\dagger})$ and the residual spectrum $\sigma_{\rho}(T)$ and $\sigma_{\rho}(T^{\dagger})$.
For the $||T||$ I have found:
$||T|| = 1$
Then for the point spectrum I try with:
$\lambda x_1 = x_2$
$\lambda x_2 = x_1$
$\lambda x_n = \frac{1}{n}x_{n+1}$
I found easily that some eigenvalues are $\lambda_n =0,\pm1$, but I have a problem, when I study the case $\lambda \neq\lambda_n$, I found the eigenvector:
$v_{\lambda} = (x_1,\lambda x_1, x_3, 3\lambda x_3, 3\cdot4\lambda^2x_3,....)$
But I don't understand the condition that the $\lambda$ have to satisfies for $v_{\lambda} \in \ell_2 $
For adjoint $T^{\dagger}$ I have found:
$\sigma_P(T^{\dagger}) = \{\lambda = \pm 1 \}$
$\sigma_{\rho}(T^{\dagger}) = \{z = 0\}$
It is correct?
The operator consists of two unrelated parts: reflection in the plane $(x_1,x_2)$, which contributes norm $1$ and eigenvalues $\pm 1$, and scaled backward shift, which contributes the eigenvalue $0$. There are no other eigenvalues because if $\lambda \notin \{0,\pm 1\}$ then $x_3,x_4,\dots $ must be nonzero and satisfy $x_{n+1} = n\lambda x_n$ for all $n$. Since $|n\lambda|>1$ for sufficiently large $n$, such a sequence $x$ tends to infinity, so it can't be in $\ell_2$.
The residual spectrum of $T$ is empty: for any $\lambda$, the range of $T-\lambda I$ contains the space $c_{00}$ of the sequences that are eventually zero; therefore, the range is dense.
Adjoint
The adjoint $T^*$ consists of a planar reflection combined with a scaled forward shift. You correctly found its eigenvalues to be $\pm 1$ and its residual spectrum to be $\{0\}$, since $\operatorname{ran} T^* \subset \{x_3=0\}$ and for $\lambda\ne 0$, the range of $T-\lambda I$ contains $c_{00}$.