Bauer–Fike theorem

Remarks

If $\mu$ is an eigenvalue of $A$, there is nothing to prove. Otherwise $\Lambda - \mu I$ is invertible, and $A + E - \mu I = V(\Lambda - \mu I)\big(I + (\Lambda - \mu I)^{-1} V^{-1} E V\big) V^{-1}$ is singular. So $1 \le \norm{(\Lambda - \mu I)^{-1} V^{-1} E V}_p \le \kappa_p(V) \norm{E}_p / \min_\lambda \abs{\lambda - \mu}$, which is the claim.

If $A$ is normal, $V$ can be chosen unitary, so $\kappa_2(V) = 1$ and each eigenvalue of $A + E$ is within $\norm{E}_2$ of an eigenvalue of $A$.

The diagonalisability hypothesis is needed for a bound linear in $\norm{E}$. For $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $E = \begin{pmatrix} 0 & 0 \\ \varepsilon & 0 \end{pmatrix}$ with $\varepsilon > 0$, the eigenvalues of $A + E$ are $\pm \sqrt{\varepsilon}$, much larger than $\varepsilon$ when $\varepsilon$ is small.

Going deeper

For a student

The theorem is a basic tool of eigenvalue perturbation theory and of the error analysis of eigenvalue algorithms. It shows that the eigenvalue problem of a normal matrix is well conditioned. When the eigenvectors are nearly parallel, $\kappa_p(V)$ is large and the eigenvalues can be very sensitive.