Timeline for Well known matrix inequality?

Current License: CC BY-SA 4.0

12 events

when toggle format	what		by	license	comment
May 21, 2018 at 3:54	history	edited	Hammerhead	CC BY-SA 4.0	took out parts from the question that later proved to be irrelevant.
May 14, 2018 at 5:18	vote	accept	Hammerhead
S May 13, 2018 at 18:29	history	suggested	Mahdi - Free Palestine		adding a new tag
May 13, 2018 at 18:27	review	Suggested edits
S May 13, 2018 at 18:29
May 13, 2018 at 16:20	answer	added	jjcale		timeline score: 7
May 13, 2018 at 5:51	comment	added	Hammerhead		@Alan, if you think you have a solution, please write it out. I will gladly accept it. Saying that it is trivial and leaving a few superficial comments does not seem to serve any purpose.
May 13, 2018 at 5:33	comment	added	Alan		Seems to be trivial, denote by a=log A and by b=logB and use the first inequality and the additivity of the trace.
May 13, 2018 at 3:21	history	edited	Hammerhead	CC BY-SA 4.0	added 89 characters in body
May 12, 2018 at 22:39	history	edited	Hammerhead	CC BY-SA 4.0	I adjusted the formulation to only deal with symmetric matrices.
May 12, 2018 at 22:38	comment	added	Hammerhead		What you write is worth considering. Unless I am mistaken, $BA^{-1}$ and $A^{-1/2}BA^{-1/2}$ seem to have the same eigenvalues. This is the main reason behind my formulation. The other was the fact that I wanted to have a statement similar to the one with real numbers. But I do understand the logic behind working with "only" symmetric matrices. Now that I write this, your point of view seems more appealing. Edited.
May 12, 2018 at 22:24	comment	added	amakelov		just a little note: $BA^{-1}$ is not necessarily symmetric - was that the intention? If not, maybe you can replace it by something like $A^{-1/2}BA^{-1/2}$ (of course this still works fine but you need to define the log without appealing to the diagonalization)
May 12, 2018 at 22:17	history	asked	Hammerhead	CC BY-SA 4.0