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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.

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Have you checked Kato's book Perturbation theory of linear operators? –  Liviu Nicolaescu Feb 7 '13 at 20:44
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en.wikipedia.org/wiki/… –  Carlo Beenakker Feb 7 '13 at 22:59
    
I have cheked Kato's book but find it to difficult for me to understand. I have tried to apply quantum mechanics perturbation theory in degenerate case but the first step is failed: corrections to the degenerate eigenvalues in my case are complex, this have no sence. –  Denys Feb 8 '13 at 6:15
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1 Answer

Part (L) of the theorem in the following paper answers your question in the case that of normal matrices.

Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. (pdf)

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