Timeline for For a set of matrices $S$, find $X$ such that the elements of $SX$ commute
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Nov 14, 2013 at 17:05 | vote | accept | Vedran Šego | ||
| Nov 11, 2013 at 14:35 | comment | added | Vedran Šego | @AtsushiKanazawa I wanted to look closer into ways to solve matrix polynomials problems (at least of low orders, i.e., $2$ or $3$) without doing linearizations, and using only equivalences via constant matrices. One way would be to compute a simultaneous Schur decomposition (my original idea is somewhat more complex than that, but close enough for this comment), so exploring the above properties was a reasonable first step. Given Lev's answer, I now think it is safe to say that this path will not bring anything new to the field. | |
| Nov 10, 2013 at 17:46 | comment | added | Atsushi Kanazawa | I am interested in how you came up with this question. It is quite interesting. | |
| Nov 10, 2013 at 13:12 | answer | added | Lev Borisov | timeline score: 2 | |
| Nov 9, 2013 at 22:12 | comment | added | Vedran Šego | @LevBorisov I'd like to do this with as little assumptions on $A_k$ (or $A_kX$) as possible. If you can say something about this under the assumption that either $A_k$ or $A_kX$ are diagonalizable, I'd still be glad to hear it. | |
| Nov 9, 2013 at 21:19 | comment | added | Lev Borisov | Do you expect $A_iX$ to be (simultaneously) diagonalizable, or are they expected to have nontrivial Jordan blocks? | |
| Nov 9, 2013 at 21:07 | comment | added | Vedran Šego | @LevBorisov Yes, you are right. However, due to the specific form of these equations, I expect that the existence of the solution can be drawn directly from $A_k$, $k=0,\dots,d$, at least for some $d$. | |
| Nov 9, 2013 at 19:42 | comment | added | Lev Borisov | I think that (*) is of order $n^2$ by $n^2d(d+1)/2$ or something like that, because there are $n^2$ unknowns in $X$. Am I right? | |
| Nov 9, 2013 at 15:05 | history | asked | Vedran Šego | CC BY-SA 3.0 |