5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Have you tried checking what happens with $2\times 2$ matrices, where things should be simpler? $\endgroup$ – Federico Poloni Feb 22 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 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.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.