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So far I know every matrix $M$ has left ( $\vec{g}_{k}$ ) and right ( $\vec{e}_{k}$ ) eigenvectors defined via $M \cdot \vec{e}_{k} = \lambda_{k}\vec{e}_{k}$ and $\vec{g}^{t}_{k} \cdot M = \lambda_{k}\vec{g}^{t}_{k}$ where $\lambda_{k}$ is the corresponding eigenvalue. I know also that $\langle \vec{g}^{t}_{k},\vec{e}_{j} \rangle = 0$ if $\lambda_{k} \neq \lambda_{j}$. However, is it possible that in some cases the scalar product between a left and a right eigenvector is zero even is they correspond to the same eigenvalue which is of multiplicity one?

I would be really grateful if someone can give me an example where this is the case or a proof that this cannot happen.

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Of course $\vec{e}_{k}$ are the right eigenvectors and $\vec{g}_{k}$ are the left ones. I interchanged them in the first sentence. Sorry for that. – physicits Feb 16 '12 at 15:19
You can edit your question to fix that. – Alex Becker Feb 16 '12 at 15:25
Sorry that my answer was blst, sorry for wasting your time. – draks ... Feb 16 '12 at 21:06

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