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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.

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    $\begingroup$ My intuition is that this should be false, but true if you restrict to compact operators. I have no idea how I would want to formalize this or prove it, though. $\endgroup$
    – user98602
    Apr 6, 2015 at 3:32
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    $\begingroup$ Maybe this will be of some interest: by Weyl-von Neumann-Berg theorem, every normal operator $N$ on a (separable) Hilbert space is of the form $N = D + K$, where $D$ is diagonalisable (in some orthonormal basis) and $K$ is compact with arbitrarily small norm. $\endgroup$ Apr 6, 2015 at 13:22
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    $\begingroup$ I'm voting to close this question because it has been asked an answered on MathOverflow. $\endgroup$
    – user642796
    Apr 16, 2015 at 4:58
  • $\begingroup$ I would hesitate to label operators like the ones you define as 'diagonalizable.' $\endgroup$ Jul 5, 2017 at 18:31

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