TITLE: Eigenvalues and eigenvectors computation (in infinite dimension)
QUESTION [5 upvotes]: Let $T$ be the backward shift operator: $Tv = T(v_1,v_2,....) = (v_2,v_3,....)$. I would like to determine all the eigenvectors and eigenvalues. So far I have the following:

It is evident that $(\alpha, 0,0,....)$ is an eigenvector for the eigenvalue $0$.
It is also easy to see that $(\alpha, \alpha, \alpha, ...)$ is an eigenvector for the eigenvalue $1$.
Finally I observed that if $\lambda$ is any scalar in the underlying field then $(1,\lambda, \lambda^2, \lambda^3,...) , (\lambda, \lambda^2, \lambda^3,...)$ etc. are all eigenvectors for the eigenvalue $\lambda$.

Now I need to either argue why these are all eigenvectors or find more. But although I think these are all I don't know how to prove it. How to proceed from here?

REPLY [9 votes]: If $v$ satisfies
$$0 = Tv - \lambda v = (v_2 - \lambda v_1, v_3 - \lambda v_2, \dotsc, v_{n+1} - \lambda v_n, \dotsc),$$
then $v_{n+1} = \lambda v_n$ for all $n$, and hence $v_n = \lambda^{n-1}\cdot v_1$. Conversely, if we have $v_n = \lambda^{n-1}\cdot v_1$ for $n \geqslant 2$, then $Tv = \lambda v$, and either $v = 0$ (if and only if $v_1 = 0$) or $v$ is an eigenvector to the eigenvalue $\lambda$.
Therefore all eigenspaces are one-dimensional, a basis for $E_\lambda$ is $\{g^{(\lambda)}\}$ with $g^{(\lambda)}_n = \lambda^{n-1}$.