Timeline for Determinant of non-symmetric sum of matrices
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 25, 2013 at 12:04 | |||||
| Jun 10, 2013 at 14:01 | history | edited | user34406 | CC BY-SA 3.0 |
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| May 28, 2013 at 8:56 | history | edited | user34406 | CC BY-SA 3.0 |
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| May 28, 2013 at 7:27 | vote | accept | user34406 | ||
| May 28, 2013 at 4:24 | answer | added | Suvrit | timeline score: 6 | |
| May 27, 2013 at 22:20 | comment | added | user34406 | I'd much appreciate it. But in any case, what's the direction/idea of your proof? | |
| May 27, 2013 at 21:17 | comment | added | Suvrit | The fixed version with $BA$ does hold. The proof is interesting. If I get time, I'll type it out; otherwise someone else may want to do it. | |
| May 27, 2013 at 19:30 | comment | added | user34406 | Of course, you are right, but... now that's highly embarrassing: there's an error in my question, it should have been $BA$ all along! Sorry for wasting you time. :-( | |
| May 27, 2013 at 19:24 | history | edited | user34406 | CC BY-SA 3.0 |
AB -> BA; edited body
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| May 27, 2013 at 19:05 | vote | accept | user34406 | ||
| May 27, 2013 at 19:21 | |||||
| May 27, 2013 at 18:58 | comment | added | Suvrit | It still will not hold! even though $A$ turned out to be semidefinite in my example, you can see that a trivial epsilon perturbation will make it positive definite, but still yield a counterexample. I've included another explicit counterexample, including for the case where all three matrices are positive definite, so that you feel more convinced ;-) | |
| May 27, 2013 at 18:46 | comment | added | user34406 | Thank you very much for your answers. Your counterexample is for the case where $A$ in $(\star)$ has a zero eigenvalue. My apologies, I hadn't made this clear in my initial question, but $A$ should be positive definite. I've edited my question to be more specific. | |
| May 27, 2013 at 18:36 | history | edited | user34406 | CC BY-SA 3.0 |
More specific problem formulation.
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| May 27, 2013 at 18:14 | comment | added | Suvrit | Even the inequality $(\star)$ is false; see below for another counterexample. | |
| May 27, 2013 at 12:01 | comment | added | user34406 | $det(A^2+AB+AC) = det (A^2+A^{1/2}BA^{1/2} + A^{1/2}CA^{1/2})$, which are then all symmetric positive semi definite matrices. | |
| May 27, 2013 at 11:58 | history | edited | user34406 | CC BY-SA 3.0 |
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| May 27, 2013 at 11:52 | history | edited | user34406 | CC BY-SA 3.0 |
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| May 27, 2013 at 0:37 | history | edited | user34406 | CC BY-SA 3.0 |
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| May 26, 2013 at 22:32 | answer | added | Suvrit | timeline score: 3 | |
| May 26, 2013 at 22:04 | history | asked | user34406 | CC BY-SA 3.0 |