Timeline for A curious determinantal inequality
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2015 at 11:45 | comment | added | GH from MO | Very nice proof! It would be interesting to see a "coordinate free" proof in the spirit of the Schur-Horn theorem, but probably it would not be any simpler than this one (especially that your proof is self-contained). | |
| Jun 28, 2015 at 11:34 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed a typo: $\lambda_n\geq 0$ instead of $\lambda_n=0$
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| Jun 28, 2015 at 2:51 | comment | added | Terry Tao | I found this argument after playing around with the k=1 and k=2 cases for a while. Roughly speaking, $\lambda_k I + D$ represents the "largest" or "worst" that $A+B$ can be if one only constrains the top $k$ eigenvalues of $A+B$, which is what one is doing when trying to prove majorisation. (When $k=1$, $D$ is not present, and when $k=2$, $D$ is a rank one operator; after seeing these two cases I was able to extrapolate to the general case.) | |
| Jun 28, 2015 at 2:39 | comment | added | M. Lin | My original goal is to prove the majorization relation, then as a byproduct, the determinantal inequality. Could the determinantal inequality be proved without appealing to majorization? I guess other MO readers would like to have such attempts. | |
| Jun 28, 2015 at 2:35 | comment | added | M. Lin | Great, thank you for your detailed and clean proof. I use majorization a lot in my study, but still I do not play it at the same level as you do. After reading your proof, I started asking why I did not find a proof myself. Aha, I failed to observe the "key" step $A+B \leq \lambda_k I + D$. Has a similar construction $D$ been used before? Maybe in this context, such a trick is novel. | |
| Jun 28, 2015 at 2:24 | vote | accept | M. Lin | ||
| Jun 28, 2015 at 1:17 | history | answered | Terry Tao | CC BY-SA 3.0 |