TITLE: Are operators with trivial spectrum nilpotent in a sense?
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: Just some remarks. A natural example of a quasinilpotent operator, which may be of interest to you, is 
Tu(x):=∫0xu(s)ds 
on C[0,1], or on Lp[0,1], for 1≤p≤∞. In all these cases T is compact. However note that a noncompact example can be easily defined e.g. on a Hilbert sum of infinitely many Hilbert copies of a Hilbert space H, as the direct sum of infinitely many copies of a non-zero quasinilpotent T:H→H (that is, just taking {hj} to {Thj}).