Timeline for Set of Positive Definite matrices with determinant > 1 forms a convex set
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2013 at 19:20 | comment | added | alvarezpaiva | Yes Deane. Sorry I was being sloppy. I felt the person who proposed the question had to do some work in answering it. | |
| Nov 8, 2013 at 19:18 | comment | added | Suvrit | @Deane: That's what I meant by "simultaneous congruence" :-), though it suffices to just turn $A$ into $I$, as you note. | |
| Nov 8, 2013 at 19:15 | comment | added | Deane Yang | The $C^{-1}$ should be $C^t$. In fact, you can take $C$ to be the square root of $A$ (positive symmetric matrices can be raised to any real power). Then the rest follows. | |
| Nov 8, 2013 at 18:55 | comment | added | Suvrit | But the idea of reducing the problem to where $A=I$ and $B=D$ (diagonal) does work, if one follows the simultaneous congruence lemma, see e.g., mathoverflow.net/questions/71678/… | |
| Nov 8, 2013 at 16:59 | comment | added | Suvrit | $CAC^{-1}=I \implies CA=C \implies C^{-1}CA=C^{-1}C=I$, i.e., $A=I$. Perhaps I'm being really silly here... | |
| Nov 8, 2013 at 15:48 | comment | added | alvarezpaiva | You don't have to simultaneously diagonalize $A$ and $B$: $\det(A + B) = \det(C(A + B)C^{-1})$ for any invertible $C$ and now choose $C$ so that $CAC^{-1} = I$. This leaves you with having to prove the case where $A = I$. | |
| Nov 8, 2013 at 15:41 | comment | added | Suvrit | I am not fully sure, because if $A$ and $B$ don't commute, you cannot simultaneously diagonalize them. However, Minkowski's inequality essentially reduces to AM-GM after some manipulation, see e.g., mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n for more details. | |
| Nov 8, 2013 at 15:31 | history | answered | alvarezpaiva | CC BY-SA 3.0 |