Timeline for Packing symmetric matrices in spectral norm, and defining measures on symmetric matrices
Current License: CC BY-SA 3.0
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| Nov 8, 2013 at 18:51 | comment | added | martin | @Suvrit oh I see, yes that would be an upper bound, unfortunately in this example I think it's too loose because $a$ and $b$ are fixed, hence for small enough $\epsilon$ I can't guarantee $\|\Lambda_1^{\downarrow}-\Lambda_2^{\uparrow}\|\leq \epsilon$.. | |
| Nov 8, 2013 at 13:59 | comment | added | Suvrit | what I meant was that $\|\Lambda_1^\downarrow-\Lambda_2^\uparrow\|$ is an upper bound (the arrows denote sorting)---this might be too loose, but with suitable choice of matrices, equality can be achieved in this....hence.... | |
| Nov 8, 2013 at 8:30 | comment | added | martin | @Suvrit Thanks. Yes, $a\preceq A$ is shorthand for $aI\preceq A$, and good means that the resulting upper bound should scale correctly in terms of $\epsilon$. So for instance if it were true that $\|U_1\Lambda_1 U_1^T-U_2\Lambda_2 U_2^T\|_2\leq\|U_1-U_2\|_2+\|\Lambda_1-\Lambda_2\|_2$, that would probably be good enough. On the other hand, $\|U_1\Lambda_1 U_1^T-U_2\Lambda_2 U_2^T\|_2\leq d \|U_1-U_2\|_{\infty,\infty}+d \|\Lambda_1-\Lambda_2\|_{\infty,\infty}$ (I'm just making that up, as an example) probably would be too loose. | |
| Nov 7, 2013 at 22:43 | comment | added | Suvrit | Martin, maybe the following similar question is sorta helpful (though it did not see much activity): mathoverflow.net/questions/122934/… --- also, by $a \preceq A$ do you mean $a I \preceq A$? Also, as for "good" upper bound on the last term you wrote, what does "good" mean? E.g., it can be bounded by $\|\Lambda_1 - \Lambda_2\|_2$, but that is not "good" enough I guess? | |
| Nov 7, 2013 at 21:28 | review | First posts | |||
| Nov 7, 2013 at 21:29 | |||||
| Nov 7, 2013 at 21:10 | history | asked | martin | CC BY-SA 3.0 |