Discs of the The Bauer–Fike theorem The Bauer–Fike theorem states that if $A = VDV^{-1}$ for $D = \operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$ is a diagonalizable matrix and $\tilde{A}=A+E$ is a perturbed matrix than the eigenvalues of $\tilde{A}$ lie in the union of the discs $D_{i}$ where every $D_{i}$ is centered at $\lambda_{i}$ and has radius $\kappa(V)\|E\|$. My question is: Is it possible that some disc is empty? That is, can there be an $i$ for which $D_{i}$ contains \emph{no eigenvalues} of $\tilde{A}$? I think not, but could not prove it.

If you expand the radius by a factor of roughly $2n$ than the answer is indeed that a disc cannot be empty.

The Bauer-Fike theorem tells us that the distance measure $d_1 = \max_{i}\min_{j}|\lambda_{i}-\mu_{j}|$ is small. What I wanted to infer is that the distance measure $d_2 = \min_{\sigma \in S_{n}}\max_{i}|\lambda_{i}-\mu_{\sigma(i)}|$ is small (where $S_n$ is the permutations on $n$ elements).

It turns out that under plausible assumptions, $d_{2} \le 2n d_{1}$, as proved here, by Elsner.