Timeline for An "almost" true inequality for Hermitian matrices
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19 at 5:11 | comment | added | WunderNatur | @NarutakaOZAWA Thanks! This is a good point. | |
| Jun 18 at 3:08 | comment | added | Narutaka OZAWA | $e_k$ is for the standard basis and $f_i(k)$ is the value of $f_i$ at $k$. $\{ f_i\otimes f_j : i,j\}$ is an orthonormal basis for $\ell_2^N \otimes \ell_2^N$. | |
| Jun 18 at 2:33 | comment | added | WunderNatur | @NarutakaOZAWA This looks interesting, but it might be helpful if your notations ($e_k$, $f_i(k)$, overline, etc.) could be explained a little more. | |
| Jun 17 at 0:57 | comment | added | Narutaka OZAWA | Let $f_i$ denote the unit eigenvectors for $A$ for $\lambda_i$. Then $e_k\otimes e_k=\sum_{i,j} \overline{f_i(k)f_j(k)}(f_i\otimes f_j)$. Thus the LHS is $N^{-1}\sum_{i,j,k} \lambda_i^p\lambda_j^q |f_i(k)|^2|f_j(k)|^2$ while the RHS is $N^{-2}\sum_{i,j,k,l}\lambda_i^p\lambda_j^q|f_i(k)|^2|f_j(l)|^2$. If $f_i$ (or the unitary element $[f_i(k)]_{i,k}$) are random, then by Levy's lemma, $|f_i(k)|$ are close to $N^{-1/2}$ and LHS $\approx$ RHS. For the $i=j$ summand, LHS $\geq$ RHS by Cauchy--Schwarz. This perhaps hints why LHS is likely larger than RHS. | |
| Jun 13 at 1:30 | comment | added | WunderNatur | @Malkoun That's a good point. In this case, A can be scaled to a projector. Whenever A is a projector, we have $A^p=A$ for any p, and this inequality always holds. | |
| Jun 12 at 20:03 | comment | added | Malkoun | Did you consider the rank 1 case? I would start by looking at that first. Just a comment. | |
| Jun 12 at 10:25 | history | asked | WunderNatur | CC BY-SA 4.0 |