Let $T$ be a compact operator on $l^2$. Let $T_n$ be finite rank operators and $T_n \to T$ in the operator norm. Is it true that the eigenvalues and eigenvectors of $T_n$ converge to eigenvalues and eigenvectors of $T$?

For any compact set K of complex numbers disjoint from the spectrum of T, there is $\epsilon > 0$ such that for every operator S with $\ST\ < \epsilon$, K is disjoint from the spectrum of $S$. Namely, you can take $\epsilon = \inf_{\lambda \in K} \(T\lambda)^{1}\^{1}$. So the eigenvalues of $T_n$ do converge in that sense to the spectrum of $T$ (not necessarily eigenvalues, because $T$ may not have any). 


For eigenvectors there is no chance. One may approximate the identity map $T$ on $\Bbb R^2$ with a symmetric matrix $T_n$ whose eigenvalues are $1$ and $11/n$. The eigenvectors are perpendicular to each other, but otherwise their direction is entirely optional. So by choosing directions erratically one can avoid convergence. Of course some subsequence will converge. 

