Timeline for Why does this antisymmetric product factor out a determinant?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2014 at 23:41 | vote | accept | Greg Egan | ||
| Jul 28, 2014 at 14:41 | answer | added | darij grinberg | timeline score: 8 | |
| Jul 28, 2014 at 14:16 | history | edited | Greg Egan | CC BY-SA 3.0 |
Edited to change question from sum to individual components.
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| Jul 28, 2014 at 14:07 | comment | added | Greg Egan | You're absolutely right, there's no need to sum. The summation arose in the problem I was tackling and I assumed it was crucial. I should have checked! The separate results do indeed look like higher-order minors (though it's not yet clear to me why). | |
| Jul 28, 2014 at 13:44 | comment | added | darij grinberg | Am I being stupid or is the thing polynomial even if you don't sum over $k$ ? And (without summing) it should be an $\left(n-2\right)\times\left(n-2\right)$-minor of $M$ (up to sign). (I am not 100% sure of my proof so far.) | |
| Jul 28, 2014 at 12:58 | history | edited | Greg Egan | CC BY-SA 3.0 |
Corrected bound on sum
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| Jul 28, 2014 at 11:57 | history | edited | Greg Egan | CC BY-SA 3.0 |
Added intermediate step in calculation
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| Jul 28, 2014 at 11:40 | history | asked | Greg Egan | CC BY-SA 3.0 |