TITLE: How can I get eigenvalues of infinite dimensional linear operator?
QUESTION [10 upvotes]: What I want to prove is that for infinite dimensional vector space, $0$ is the only eigenvalue doesn't imply $T$ is nilpotent.
But I am not sure how to find eigenvalues of infinite dimensional linear operator $T$. Since we normally find eigenvalues by find zeros of characteristic polynomials, we even cannot find the characteristic polynomial for this situation.
I am specifically interested in the differential operator on the vector space of all formal power series.

REPLY [6 votes]: There's no general way to find eigenvalues of an operator in an infinite dimensional space. Actually, some operators don't have any eigenvalue.
You may consider the vector space $k[X]$ of polynomials over the field $k$, and the operator $D : P \mapsto P'$. This operator has only $0$ as an eigenvalue, but is not nilpotent (although it's not far from being nilpotent since for all $P$, there exists $n_P$ such that $D^{n_P}(P) = 0$ if you take $n_P = \deg P + 1$).
The operator $\sigma : P(X) \mapsto P(X+1)$ is an example of operator with no eigenvalues.