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I found a couple of proofs for this theorem but only for the case when A and B are diagonalizable, thus the eigenspace that they share is not the generalized one.

Im looking for the proof (or literature that points to the proof) when A and B are non-diagonalizable matrices.

Thanks

Joao

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  • $\begingroup$ Did you mean that "matrices commute if all their eigenspaces are common"? And, respectively, "all generalised eigenspaces" $\endgroup$ – TZakrevskiy Dec 16 '14 at 18:11
  • $\begingroup$ Your proposition, a little more carefully stated, seems reasonable, though I haven't seen it. If you want a theorem about arbitrary commuting matrices, you may note that any family of commuting matrices can be unitarily simultaneously upper-triangularized. $\endgroup$ – Omnomnomnom Dec 16 '14 at 18:20
  • $\begingroup$ Never mind, your statement is false! $\endgroup$ – Omnomnomnom Dec 16 '14 at 18:22
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The statement, as written (and as I could potentially imagine it being rewritten) is false.

For example, the matrices $$ \pmatrix{0&1\\0&0}, \pmatrix{0&0\\1&0} $$ both have the generalized eigenspace $\Bbb R^2$ associated with $\lambda = 0$. However, they do not commute.

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  • $\begingroup$ See the result I mentioned here. This applies to any pair of commuting matrices. $\endgroup$ – Omnomnomnom Dec 16 '14 at 18:29
  • $\begingroup$ Ok, but do these results on triangularisability mean that, if 2 matrices commute then they share the same generalized eigenspace? $\endgroup$ – Peter Dec 16 '14 at 20:30
  • $\begingroup$ I don't think that there's a connection between one and the other. However, I think that this direction of the implication is true. $\endgroup$ – Omnomnomnom Dec 16 '14 at 21:43
  • $\begingroup$ Ye, thats why Im looking for a proof on that direction of the implication, still thankz for the help $\endgroup$ – Peter Dec 17 '14 at 14:47
  • $\begingroup$ Well, that wasn't what you originally said you were looking for (see your title). I might update this answer if I think of something. In the mean time, you might want to post a new question, making it clear exactly what you're looking for now. $\endgroup$ – Omnomnomnom Dec 17 '14 at 18:14

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