Being far from analysis, I recently learned about the Invariant subspace problem and came up with the following (perhaps simple or well-known) question.

Let $H$ be a separable complex Hilbert space and $T:H\to H$ a bounded operator. Assume that the spectrum of $H$ is $\{0\}$, i.e. $T-\lambda I$ has a bounded inverse for every $\lambda\in\mathbb C\setminus\{0\}$. In finite dimensions, this would imply that $T$ is nilpotent ($T^n=0$ for some $n$). I wonder if there is something similar in the infinite dimensional case. The precise formulation I have in mind is the following.

It is easy to see that there is a maximal subspace $X\subset H$ such that $T(X)=X$. This is a purely set-theoretic fact, a sort of explicit construction is the intersection of the images of iterations of $T$ up to and beyond infinity (via transfinite induction).

**Question:** can it happen that $X\ne\{0\}$?

Here are some observations that I made:

$T$ cannot be onto. (I derived this from some random Wikipedia quotes so there are high chances of error; please correct me if I am wrong.) It follows that a nontrivial $X$ cannot be closed.

If there is an example, then there is one where $X$ is dense. Just take the closure of $X$ for $H$.

It is possible that $T(H)$ is dense. My example is the shift in $\ell^2(\mathbb Z)$ composed with a mutiplication by a positive function (sequence) that goes to zero at both ends. Again, please correct me if I am wrong.

quasinilpotent. I don't know much about them, but I guess that knowing the terminology will help you to find some references. $\endgroup$ – Ian Morris May 15 '10 at 9:23