Square matrices A and B commute if and only if they share the same generalized eigenspace.

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

• Did you mean that "matrices commute if all their eigenspaces are common"? And, respectively, "all generalised eigenspaces" – TZakrevskiy Dec 16 '14 at 18:11
• 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. – Omnomnomnom Dec 16 '14 at 18:20
• Never mind, your statement is false! – Omnomnomnom Dec 16 '14 at 18:22

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.