# Eigenvalues of perturbed matrix. Rouché's theorem.

1. Aug 25, 2013

### SrEstroncio

1. The problem statement, all variables and given/known data

Let $\lambda_0 \in \mathbb{C}$ be an eingenvalue of the $n \times n$ matrix $A$ with algebraic multiplicity $m$, that is, is an $m$-nth zero of $\det{A-\lambda I}$. Consider the perturbed matrix $A+ \epsilon B$, where $|\epsilon | \ll 1$ and $B$ is any $n \times n$ matrix.

Show that given $\delta \gt 0$, $\alpha \gt 0$ exists so that, for $| \epsilon | \lt \alpha$, the matrix $A + \epsilon B$ has exactly $m$ eigenvalues (with algebraic multiplicity) inside $| z - \lambda | \lt \delta$

2. Relevant equations

Rouché's theorem states that if $f$ is holomorphic in a region and $|g(z)| \lt |f(z)|$ on a curve (suitable for integration) inside the open region, then $f$ and $f+ g$ have exactly the same amount of zeros (with multiplicity) inside the curve.

3. The attempt at a solution

I first expanding the $\det$ function in power series and tried applying the integral which counts the number of zeros inside a region but I got stuck with several terms I couldn't get rid of. The TA told me to apply Rouché's theorem, but I can't figure out a way to exact the "sum" out of the determinant.
Any ideas? Any help would be appreciated.

2. Aug 28, 2013

### pasmith

Should that not be $|z - \lambda_0| < \delta$?

I think you want to take
$$f(z) = \det (A - zI)$$
and
$$g(z,\epsilon) = \det (A + \epsilon B - zI) - f(z)$$
and then look at the curve $|z - \lambda_0| = \delta$. If you can show that, for all $\delta > 0$, there exists $\alpha > 0$ such that for all $|\epsilon| < \alpha$, $|g(z,\epsilon)| < |f(z)|$ on that curve, then Rouché's theorem will give you the result.