Timeline for How to prove this determinant is positive?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 3, 2015 at 20:19 | comment | added | Lei Wang | @Suvrit: Yes, why things still work out remain a mystery to me. | |
| May 3, 2015 at 20:15 | comment | added | Suvrit | @user23765: I see the difference now. In my answer, I assume a "sufficient" condition, and frequently it is satisfied. In the cases where it is not (the $\pm a + \log(-1)$ case in your bullet point 3), things still work out. So it remains to guarantee that these are the only possible cases. | |
| May 3, 2015 at 19:50 | history | edited | Suvrit | CC BY-SA 3.0 |
fixed a typo, and changed flow a tiny bit.
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| May 3, 2015 at 16:31 | comment | added | Lei Wang | @Suvrit: The complex eigenvalues are indeed in conjugate pairs, but the real ones are not necessarily doubly degenerate. | |
| May 3, 2015 at 15:14 | comment | added | Lei Wang | Is it possible for $e^M$ to have real eigenvalues smaller than -1 that are not paired with others ? | |
| May 3, 2015 at 14:04 | history | undeleted | Suvrit | ||
| May 3, 2015 at 14:04 | history | edited | Suvrit | CC BY-SA 3.0 |
expanded the answer....
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| May 3, 2015 at 2:36 | history | deleted | Suvrit | via Vote | |
| May 3, 2015 at 1:28 | comment | added | Christian Remling | Well, I just reached the opposite conclusion from the same starting point. But of course it wouldn't surprise anyone if I'm wrong again. | |
| May 2, 2015 at 23:43 | history | answered | Suvrit | CC BY-SA 3.0 |