How can I get eigenvalues of infinite dimensional linear operator?

Question

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.

Answer 1

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.

Answer 2

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space and let $S$ be any operator on $\mathcal{H}$ that has no eigenvalues [for example, take $\mathcal{H} = L^2[0, 1]$ and let $S$ be the operator on $\mathcal{H}$ defined by $(Sf)(x) = xf(x)$.] Now define an operator $T$ on $\mathcal{H} \times \mathcal{H}$ by $$T(f, g) = (0, Sg).$$ Then $0$ is the only eigenvalue of $T$, but $T$ is not nilpotent.