Timeline for Solve permutation matrix equations of the form: $X^T A X = B_1$ and $X A X^T = B_2$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2023 at 16:43 | answer | added | loup blanc | timeline score: 1 | |
| Oct 19, 2023 at 5:38 | comment | added | Sungjin Kim | $A$, $B_1$, and $B_2$ are permutation matrices for product of disjoint transpositions (or possibly empty product, which gives $I$). Of course, similar, so the number of transpositions must be the same. | |
| Oct 18, 2023 at 21:01 | comment | added | Christian Remling | I think this is better phrased as a question about (conjugacy of) permutations. | |
| Oct 18, 2023 at 20:57 | comment | added | Christian Remling | Moreover, $A,B_1,B_2$ must all have the same cycle structure (as permutations). | |
| Oct 18, 2023 at 14:17 | comment | added | Sungjin Kim | If $B_1=I\neq B_2$, then there is no such $X$ because $B_1$ and $B_2$ are not similar. Note that $X^T=X^{-1}$. So, if the above equations hold, then $B_1$ and $B_2$ must be similar. | |
| S Oct 18, 2023 at 13:49 | review | First questions | |||
| Oct 18, 2023 at 14:43 | |||||
| S Oct 18, 2023 at 13:49 | history | asked | Danish | CC BY-SA 4.0 |