# Perturbing a normal matrix

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?

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

$$\epsilon = 0 \implies \exists v: v$$ 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.