Timeline for Bounding the non-multiplicativity of isometric projection
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Sep 22, 2016 at 18:48 | vote | accept | Asaf Shachar | ||
| S Sep 17, 2016 at 17:32 | history | bounty ended | Asaf Shachar | ||
| S Sep 17, 2016 at 17:32 | history | notice removed | Asaf Shachar | ||
| Sep 16, 2016 at 16:15 | answer | added | Asaf Shachar | timeline score: 1 | |
| Sep 16, 2016 at 1:03 | comment | added | fedja | Alas, if we talk about the operator norm, then the constant does depend on the dimension (so, in your notation, $c\approx (\log n)^{-1}$). You seem to care more about the Frobenius norm though. I surmise that it doesn't really make any difference, but am too lazy to try to modify my argument to cover that case now. Let me know if anything is unclear or if you have any other questions :-). | |
| Sep 15, 2016 at 22:03 | answer | added | fedja | timeline score: 5 | |
| Sep 15, 2016 at 15:22 | comment | added | fedja | Indeed :-) Sometimes I'm too hasty in my conclusions. Then, if you go through the routine you suggested, the question becomes if for any two self-adjoint positive definite matrices $P$ and $Q$ the identity matrix $I$ is the closest one to the product $PQ$ among all orthogonal matrices up to a constant factor. I'll try to think of it :-) | |
| Sep 15, 2016 at 14:34 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
added 319 characters in body
|
| Sep 15, 2016 at 14:28 | comment | added | Asaf Shachar | @fedja It turns out the investiagtion of these pairs is non trivial and interesting by itself: see here: mathoverflow.net/questions/248842/… | |
| Sep 15, 2016 at 14:28 | comment | added | Asaf Shachar | Your point seems interesting, though your particular example won't work. It turns out that $O_{A^{-1}}=(O_A)^{-1}$ for every invertible matrix $A$ (not just normal ones). (You can prove this using SVD, or by using the uniqueness of the positive square root). So, we will get that in this case both sides are equal to zero. The problem was that for any $A,B=A^{-1} $ , $O_{AB} =O_AO_B$: In order to try to refute the existence of such a constant we need to examine "bad" pairs where the equality does not hold. | |
| Sep 15, 2016 at 14:16 | comment | added | fedja | $AB$ can be accidentally orthogonal without any of $A$ and $B$ being orthogonal, in which case the left hand side is $0$ but the other one may be not (say,when $B=A^{-1}$ and $A$ fails to commute with $A^T$, in which case you get $A(\sqrt{A^TA})^{-1}A^{-1}\sqrt{AA^T}$ for $O_AO_B$ and I do not see why on Earth that monster should be anywhere near the identity matrix though, to be honest, I didn't try to check formally or even computationally that it isn't.) | |
| S Sep 15, 2016 at 11:53 | history | bounty started | Asaf Shachar | ||
| S Sep 15, 2016 at 11:53 | history | notice added | Asaf Shachar | Draw attention | |
| Sep 12, 2016 at 9:08 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
added 69 characters in body
|
| Sep 12, 2016 at 8:54 | history | asked | Asaf Shachar | CC BY-SA 3.0 |