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The problem is this:

Prove that if $A$ and $B$ are linear operators (on a finite dimensional space) that commute, then there exists a basis $X$ in which the matrices $[A]_X$ and $[B]_X$ are Jordan canonical forms.

What I have noted is that if $K_\lambda$ is the generalized eigenspace of $A$ corresponding to the eigenvalue $\lambda$, then $B$ maps $K_{\lambda}$ into $K_{\lambda}$. Thus, one can find a Jordan basis for the restriction of $B$ to $K_{\lambda}$; vectors of which basis will be generalized eigenvectors of $A$, but not necessarily (or at least I haven't proved that that will be the case) a disjoint union of cycles.

Thank you.

Weltschmerz
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  • That is a really complex sentence!: ...if... , then ... , so... , and ... , but ... . Of course there is no law like this in English, but a good rule for composition is to avoid conjoining more than two sentences together so that the reader doesn't get lost. Three can be done too (I did that in the last sentence!), but four is starting to get a little silly :) I like the question, too +1 – rschwieb Apr 04 '13 at 17:47
  • rschwieb: thank you. I tried to make it better. – Weltschmerz Apr 04 '13 at 17:51
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    The problem is that the problem is wrong. – darij grinberg Apr 04 '13 at 17:55
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    Take a look at this question. Georges Elencwajg provides a counterexample: Matrices which commute but cannot be simultaneously "Jordanized". – EuYu Apr 04 '13 at 17:57

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