How can I get eigenvalues of infinite dimensional linear operator? 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. 

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 $\sigma : P(X) \mapsto X P(X)$ is an example of operator with no eigenvalues.
