Timeline for Condition for two matrices to share at least one eigenvector?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 9, 2013 at 19:19 | comment | added | sasquires | Thanks, I think that this is both intuitive and computationally straightforward, although the question of the converse is still open. I'm kicking myself for not thinking of it, because it's the natural generalization of the standard theorem that $A$ and $B$ can be simultaneously diagonalized if $[A,B]=0$ (although I suppose that this corresponds to the converse). | |
| Oct 9, 2013 at 16:55 | history | edited | Mustafa Said | CC BY-SA 3.0 |
Added Remark
|
| Oct 9, 2013 at 16:05 | comment | added | Mustafa Said | @SashaKolpkov: I just put the restriction (that the shared eigenvector has a shared eigenvalue of 1) that sasquires suggests in his problem. You are right, however, that the converse should be written up differently. | |
| Oct 9, 2013 at 15:52 | comment | added | SashaKolpakov | If Ax=ax and Bx=bx, in'nit ABx= A(bx) = b Ax = ba x = ab x = a Bx = B(ax) = BAx, for every matrices A, B, and eigenvector x with eigenvalues a, b with respect to these matrices? Then, a, b, should not be necessarily 1. | |
| Oct 9, 2013 at 15:42 | history | answered | Mustafa Said | CC BY-SA 3.0 |