Timeline for For a set of matrices $S$, find $X$ such that the elements of $SX$ commute
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 14, 2013 at 17:05 | vote | accept | Vedran Šego | ||
| Nov 12, 2013 at 19:06 | comment | added | Lev Borisov | Well, my answer gives no help if all $A_i$ are uppertriangular with zeroes on the diagonal :) | |
| Nov 11, 2013 at 14:38 | comment | added | Vedran Šego | You make plenty of sense, and this answer seems to be quite complete. It doesn't cover the case when all $A_i$ are singular, but given that any set of singular matrices is "thin", I don't see it as a problem. I will give it another day (or few) before accepting it, to see if any more answers pop up, possibly with some additional insights. Thank you for your help! | |
| Nov 10, 2013 at 22:19 | comment | added | Lev Borisov | If $B_j$ is invertible, then $B_kB_j=B_jB_k$ implies $B_j^{-1}B_k=B_kB_j^{-1}$. Then you have $B_iB_j^{-1}B_kB_j^{-1} = B_iB_k B_j^{-2}$. So if $B_iB_k=B_kB_i$, you have the same thing for $B_iB_j^{-1}$. Does this make sense? | |
| Nov 10, 2013 at 16:39 | comment | added | Vedran Šego | Thank you, this is an interesting idea. I'm having trouble seeing why the commutativity of $\{B_i\}$ implies commutativity of $\{B_iB_j^{-1}\}$. If all $B_i$ are nonsingular, it is trivial, but how do you show that if some of them are singular? | |
| Nov 10, 2013 at 13:12 | history | answered | Lev Borisov | CC BY-SA 3.0 |