0
$\begingroup$

Let $N$ be a normal matrix.

Now I consider a perturbation of the matrix by another matrix $A.$

The perturbed matrix shall be called $M=N+A.$

Now assume there is a normalized vector $u$ such that $\Vert (N-i\lambda)u \Vert \le \varepsilon$

for some $\lambda \in \mathbb R.$

Since $N$ is normal this implies that $d(i\lambda,\sigma(N))\le \varepsilon.$

Moreover, assume that $\Re(\sigma(M)) \le -\delta$ for some $\delta>0.$

If we assume additionally that $Au=0.$ Does this give us any information about how large $\delta$ can be in terms of $\varepsilon$ or are they independent?

$\endgroup$
2
  • $\begingroup$ Have you tried checking what happens with $2\times 2$ matrices, where things should be simpler? $\endgroup$ – Federico Poloni Feb 22 '19 at 14:27
  • $\begingroup$ In general for a diagonal plus rank one there is a more or less explicit formula for the eigenvalues. Maybe that helps $\endgroup$ – lcv Feb 23 '19 at 9:36
0
$\begingroup$

$\epsilon = 0 \implies \exists v: v $ is eigenvector of N and $v$ is orthogonal to $e_1 \implies Mv = Nv \implies v $ is an eigenvector of $M$ with same eigenvalue. So if $\epsilon = 0$ then $\delta$ can be anything. Now, since $\epsilon \ge 0$ does not rule out that $\epsilon = 0$ therefore $\delta$ can be anything in general as well.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy