I am having matrix $M_0$ with coresponding eigenvectors and 4 eigenvalues {0,0,a,-a}. Eigenvalue $\lambda=0$ is double degenerated. Now I am appliing small perturbation $\epsilon M_1$ and want to get the expansion of zero eigenvalues $\lambda$ like series of small parameter $\epsilon$.

My first thought was to apply eigenvalue deriviate theory and get corrections in a form: $\lambda(\epsilon)=\lambda_0+\epsilon\frac{\vec{w}M_1\vec{u}}{\vec u \vec w}$ (here u,w are left and right eigenvectors). However I have read that this method is not appropriate for eigenvalues having multiplivity higher than 1.

The question is how I can get series expansion for zero eigenvalues in my case?

**UPD**. Have looked Kato's book one more time and still can't get the idea how the eigenprojections are obtained. Resolvent of the the perturbed matrix $M(\epsilon)=M_0+\epsilon M_1$ is given by $R(\epsilon,\zeta)=(M(\epsilon)-\zeta)^{-1}$. Next step is obtaining
the eigenprojection : $P=-\frac{1}{2 \pi i} \int{R(\epsilon,\zeta)d\zeta}$ where integration is carried along small circle around eigenvalue of unperturbed matrix $M_0$, $\lambda_0$. At this point I can't understand how to carry out this integration.

Perturbation theory of linear operators? $\endgroup$ – Liviu Nicolaescu Feb 7 '13 at 20:44