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The following question came up as a though while I was reading. I cannot see how to proceed on it.

Let us have $M_1,\ldots,M_n$ be commuting matrices. I know that that the generalized eigenspaces are the same across the $M_i$. However, is it true that for each generalized eigenspace, $V$, where $V$ is the generalized eigenspace of $\lambda_i$ for $M_i$, there exists $v\in V$ such that $v$ is an eigenvector for all the $M_i$.

A related question is if it is even true that the $M_i$ need have a common eigenvector (I see how an answer to this question implies an answer to my first question, so these seem equivalent)

I have tried to come up with examples for which what I said is not true, but this has been to no avail. From the generalized eigenspaces I do not see how to proceed.

Thank you for any help.

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What do you mean that the generalized eigenspaces are the "same"? The most you can conclude is that they are invariant subspaces of the other matrices. – EuYu Feb 26 '13 at 4:45
up vote 5 down vote accepted

Hopefully, this might help you to successfully resolve your problem. :)

Let $ M_{1},\ldots,M_{n} $ be commuting $ (k \times k) $-matrices over an algebraically closed field $ \mathbb{F} $ of characteristic $ 0 $. Then $ {\frak{g}} := \text{Span}(\{ M_{1},\ldots,M_{n} \}) $ can be viewed as an abelian, hence solvable, Lie-subalgebra of $ \text{End}(V) \cong {\mathbf{M}_{k \times k}}(\mathbb{F}) $. Applying Lie’s Theorem to $ {\frak{g}} $, we see that there exists a vector $ v \in \mathbb{F}^{k} $ that is a simultaneous eigenvector of $ M_{1},\ldots,M_{n} $.

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Functioning link for Lie's theorem:… . – darij grinberg Dec 19 '14 at 5:10

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