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Nov 29, 2013 at 2:29 vote accept Federico Magallanez
Nov 8, 2013 at 16:45 history edited Federico Magallanez CC BY-SA 3.0
clarify the question
Nov 8, 2013 at 16:20 history edited Federico Magallanez CC BY-SA 3.0
fixed typos
Nov 8, 2013 at 15:40 answer added Suvrit timeline score: 11
Nov 8, 2013 at 15:31 answer added alvarezpaiva timeline score: 4
Nov 8, 2013 at 13:51 comment added Federico Magallanez @FedericoPoloni I don't have much background on this topic, and I don't understand any of closed, smooth, convex, and non-empty interior yet. I'm trying to catch up all the comments / answers so far.
Nov 8, 2013 at 13:48 comment added Joris Bierkens @Misha: do I understand correctly that you mean simultaneous diagonalization by congruence, as described e.g. in Horn, Johnson - Matrix Analysis, Section 4.5 (in particular Theorem 4.5.15)?
Nov 8, 2013 at 13:41 comment added Federico Poloni What are your attempts up to now? What is the part where you are stuck? Closed, smooth, convex or non-empty interior?
Nov 8, 2013 at 13:13 answer added Igor Rivin timeline score: 8
Nov 8, 2013 at 13:12 comment added Misha @FedericoMagallanez: This is the unitary diagonalization (not the conjugation); you should think of the quadratic forms defined via the Hermitian matrices and make a simultaneous change of variables to make both forms diagonal.
Nov 8, 2013 at 13:08 comment added ofer zeitouni write X=aT, Y=(1-a)S where T,S belong to P. Then use the inequality together with a^{1/n}+(1-a)^{1/n}>1
Nov 8, 2013 at 13:05 comment added Federico Magallanez @Misha I don't understand your comment. Tow matrices can be simultaneously diagonalizable if and only if they commute. I don't see any reason why two positive definite matrices must commute. Would you please be more specific?
Nov 8, 2013 at 12:49 comment added Misha Use simultaneous diagonalization to reduce the problem to the case of diagonal matrices.
Nov 8, 2013 at 12:35 history asked Federico Magallanez CC BY-SA 3.0