Timeline for When can I find a continuously-varying basis of eigenvectors for a non-simple eigenvalue?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Sep 9, 2016 at 6:53 | comment | added | Gordon Royle | @KeithMcClary Familiar is probably too strong a word to describe my relationship with Kato, but I do know of its existence. | |
| Sep 9, 2016 at 2:50 | history | edited | Gordon Royle | CC BY-SA 3.0 |
added image
|
| Sep 8, 2016 at 22:49 | comment | added | Gordon Royle | @WillieWong Changing dimension is definitely one case where the answer is "can't be done". Fortunately I can now show that my matrix avoids this problem. | |
| Sep 8, 2016 at 19:27 | comment | added | Keith McClary | Are you familiar with Kato's Perturbation Theory for Linear Operators (Ch 2), [downloadable from U. of Edinburgh](www.maths.ed.ac.uk/~aar/papers/kato1.pdf) . | |
| Sep 8, 2016 at 17:40 | answer | added | Peter Michor | timeline score: 1 | |
| Sep 8, 2016 at 15:49 | comment | added | Willie Wong | This may be relevant. I think it is more worrisome that the eigenvectors may collide: consider the 2x2 matrix $[x, 1; 0,0]$. Away from $x = 0$ you have two distinct eigenvectors. At $x = 0$ you only have one eigenvector $[1;0]$. Do you count this as having "continuous eigenvectors"? Also, since in your previous question you make no assumption that your matrix is symmetric: you are not even guaranteed the existence of an eigenbasis. | |
| Sep 8, 2016 at 14:31 | history | edited | Gordon Royle | CC BY-SA 3.0 |
added 368 characters in body
|
| Sep 8, 2016 at 14:24 | history | asked | Gordon Royle | CC BY-SA 3.0 |