This is true for finite-dimensional spaces, of course.
To be precise, let $T$ be an operator on a complex Banach space $X$ which is not finite-dimensional. For each $\lambda \in \mathbb{C}$, let $V_\lambda \subseteq X$ be the subspace $\mathrm{ker} (\lambda I - T)$ on which $T$ acts by the scalar $\lambda$. Say that $T$ is diagonalizable if $\sum_\lambda V_\lambda$ is dense in $X$. Or provide a better definition if this one is deficient!
Are "most" operators diagonalizable? For instance, is the set of diagonalizable operators comeagre? Of course, there are lots of operators which are not diagonalizable, but perhaps, as in the finite-dimensional case, they form a "small" set.
I suppose it's natural to consider just bounded operators, although I'd be interested in results about unbounded operators, too. Of course, if the answer depends on the Banach space $X$, I'd be very interested to learn about that. In a Hilbert space, we can consider the additional condition that the $V_\lambda$ be orthogonal and probably a lot more can be said; I'm interested in this, but I think I'm primarily interested in the more general notion of diagonalizability I gave. Also, the question makes perfect sense for any topological vector space; I'm interested in non-Banach spaces, too.
EDIT I've asked this question on mathoverflow; answers may fit better over there.
