Timeline for Is this Hankel matrix in trace class
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Apr 11, 2014 at 5:28 | vote | accept | tao mei | ||
Apr 10, 2014 at 7:52 | comment | added | Mikael de la Salle | Tao, let's continue this discussion by mail. | |
Apr 10, 2014 at 5:17 | comment | added | tao mei | Thanks for the comments, Yemon and Suvrit. Hi, MiKael, Thanks for the answer. I like your clever use of "the elementary inequality". Your answer implies a very interesting result (to me), see the email I am sending to you. | |
Apr 10, 2014 at 4:54 | comment | added | tao mei | Thanks for the comments, Yemon and Suvrit.Hi, MiKael, Thanks for the answer. | |
Apr 9, 2014 at 10:07 | history | edited | Mikael de la Salle | CC BY-SA 3.0 |
added 499 characters in body
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Apr 8, 2014 at 21:17 | comment | added | Suvrit | Thanks for catching the bug; my error lay in calling $e^{4t(i+j+1)}-1$ a rank-1 matrix --- how silly can one be! | |
Apr 8, 2014 at 18:30 | comment | added | Mikael de la Salle | You are right. It also occured to me that my answer (and the question...) was absurd it there was an easy way to prove that $A$ is positive. And the question whether $A$ is positive is even more natural now that we known (from my complicated answer) that the trace of $A$ and its trace norm are comparable. | |
Apr 8, 2014 at 16:13 | comment | added | Suvrit | It the original Hankel matrix is PSD, then we'll immediately get $\|H\| \le e^{-4t} \le 1$, but I haven't checked if it is psd (but clearly my reasoning seems naive to me).. | |
Apr 7, 2014 at 20:42 | history | answered | Mikael de la Salle | CC BY-SA 3.0 |