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Oct 9, 2013 at 22:02 comment added SashaKolpakov @sasquires: all right, I see what you mean. I just picked up the above two matrices virtually at random (I mean I did several tries, but without too much consideration). Your example is much more clever in this regard and finally dots the "i".
Oct 9, 2013 at 21:17 comment added sasquires The only thing which struck me about this example is that $A$ has the eigenvalue $0$ with multiplicity 2, and I was hoping that the "converse" discussed above might be true when all eigenvalues have multiplicity 1 (the typical case I'm interested in). But I've just constructed a simple example where it doesn't: $A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$, $B=A^T$. In this case, $\textrm{det}([A,B])=0$ but none of the eigenvectors are equal.
Oct 9, 2013 at 20:49 history answered SashaKolpakov CC BY-SA 3.0