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?



Perturbing a normal matrix
$\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.


