Timeline for A simple but curious determinantal inequality

9 events

when toggle format	what		by	license	comment
Jun 27, 2017 at 21:27	history	edited	Suvrit	CC BY-SA 3.0	rewrote the answer a bit; shortened it; hopefully, easier to digest now.
Jun 24, 2017 at 13:12	comment	added	Suvrit		No, I mean, the proof works in general; just the proof technique looks a bit confusing (because it is a chained implication). When I find some time, I can try to expand / edit my explanation above to make it clearer!
Jun 24, 2017 at 12:55	comment	added	Wolfgang		So you mean your proof only works if CBABC≤I? Well, granted, but then that's a very special case only.
Jun 24, 2017 at 12:20	history	edited	Suvrit	CC BY-SA 3.0	reworded a bit to fix explanation.
Jun 24, 2017 at 12:14	comment	added	Suvrit		Actually, I should be a bit more precise, and just write that what I am using is the observation: $(0 \le Y \le I \implies 0 \le X \le I) \implies \lambda_1(X)\le \lambda_1(Y)$
Jun 24, 2017 at 12:10	comment	added	Suvrit		@jjcale Because it's a conditional statement; if we have $CBABC\le I$, then we have $B\le A^{k/2}$. What we really wanted to prove was the eigenvalue domination; we reduced that to showing that $Y\le I \implies X \le I$ must hold (note that $\lambda(\cdot)$ is positively homogeneous).
Jun 24, 2017 at 7:09	comment	added	Wolfgang		Yes, how can you apply your first lemma if $Y\le I$ doesn't hold?
Jun 24, 2017 at 6:33	comment	added	jjcale		How can you see that $B \le A^{k/2}$ ? $B$ is positive definite but otherwise arbitrary .
Jun 24, 2017 at 2:38	history	answered	Suvrit	CC BY-SA 3.0