Timeline for Simultaneous triangularisation of an exterior power of a set of matrices
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2017 at 17:23 | vote | accept | Ian Morris | ||
| S Mar 15, 2017 at 10:49 | history | bounty ended | CommunityBot | ||
| S Mar 15, 2017 at 10:49 | history | notice removed | CommunityBot | ||
| Mar 7, 2017 at 17:29 | answer | added | Robert Bryant | timeline score: 8 | |
| Mar 7, 2017 at 12:46 | history | edited | Ian Morris |
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| S Mar 7, 2017 at 9:23 | history | bounty started | Ian Morris | ||
| S Mar 7, 2017 at 9:23 | history | notice added | Ian Morris | Draw attention | |
| Mar 4, 2017 at 9:17 | comment | added | Ian Morris | That approach works fine if the matrix has rank k, but the invertible case is surprisingly different. | |
| Mar 3, 2017 at 18:11 | comment | added | Anthony Quas | Oops... sorry... I'm definitely off the mark here. | |
| Mar 3, 2017 at 15:03 | comment | added | Ian Morris | Anthony, I'm not sure I follow. In that case there'll be a one-dimensional invariant subspace in the exterior power, but not in general an invariant complete flag, I think. Can you give a concrete example with k=2 and d=4? | |
| Mar 3, 2017 at 14:50 | comment | added | Anthony Quas | For Q1, why can't you take $X$ to be the collection of all matrices with an invertible $k\times k$ block in the top left, a $(d-k)\times(d-k)$ block in the bottom right and 0's elsewhere? The $k$th exterior powers are simultaneously diagonal, but $X$ is just $GL(k)$. | |
| Mar 3, 2017 at 14:32 | history | asked | Ian Morris | CC BY-SA 3.0 |