Timeline for Finding the closest matrix to $\text{SO}_n$ with a given determinant
Current License: CC BY-SA 3.0
20 events
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| Nov 5, 2017 at 16:59 | vote | accept | Asaf Shachar | ||
| Aug 29, 2017 at 22:17 | comment | added | Tim Carson | Let us continue this discussion in chat. | |
| Aug 29, 2017 at 21:52 | comment | added | Tim Carson | Here's why if $s \geq 1$ the diagonal matrices win. Recall that lagrange multipliers told us that for any $\alpha$, the only possibilities are $a = b$ or $a = 1-b$. Since $a$ and $b$ are positive, if the second solution exists then $a$ and $b$ are less than 1. Then we find that $H(a,b,\alpha)$ is negative, so $\log(s^{1/n})$ is negative, so $s < 1$. | |
| Aug 29, 2017 at 21:19 | comment | added | Asaf Shachar | Thanks. Your answer is truly amazing (and quite long... it takes me some time to digest it all). Last question(s) for now: What led you to the conclusion that for $s\ge 1$ diagonal matrices win? (I might have lost it inside somewhere). Also, do you have any estimate on when the regime change? (Is it true that for any fixed $s$, for $n$ large enough the optimum is not diagonal?) For a fixed $n$, is there exactly one point of transition (i.e for $s$ below threshold -non-diagonal wins,- above threshold diagonal wins- the phase doesn't change more than once). | |
| Aug 29, 2017 at 20:41 | history | bounty ended | Asaf Shachar | ||
| Aug 29, 2017 at 19:30 | comment | added | Tim Carson | Good point, that is not to trivial to check. See the edit. At this point I've been loose with what I've done by hand and done by CAS and done by numerical computation, but I hope everything can be checked by hand. | |
| Aug 29, 2017 at 19:28 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 29, 2017 at 17:24 | comment | added | Asaf Shachar | Thanks, it's clear now. Something else bothers me now: How do you conclude that taking the smallest possible $a$, amounts to choosing one value $(a)$ once, and the other value $(1-a)$ $n-1$ times? This does not seem to be true, see here. Thanks for your patience. | |
| Aug 29, 2017 at 14:51 | history | edited | Tim Carson | CC BY-SA 3.0 |
Fixed typos in the first section.
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| Aug 29, 2017 at 14:17 | comment | added | Tim Carson | (1) If $f(a)$ is strictly increasing it's clear we want to take the smallest $a$ possible. Now, suppose that $f'(a)$ changes sign at some point $a_*$. Then, on $[a_*, min(e^L, e^{1-L})]$ we will have $f(a)$ decreasing, and so $f(a) > (e^L-1)^2$ by fact (3). So, if you take anything after $a_*$ you're not going to do better than the easy competitor of a diagonal matrix. If you take anything before $a_*$ you want to take the smallest possible $a_*$. (2) You were right about the sign, I fixed it. | |
| Aug 29, 2017 at 14:02 | history | edited | Tim Carson | CC BY-SA 3.0 |
Fixed a minus sign.
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| Aug 29, 2017 at 13:55 | comment | added | Asaf Shachar | Many thanks for your effort. I still have some questions: (1) Why facts $(1)-(3)$ imply the smallest $a$ possible (which makes $n\alpha$ an integer) is the only relevant candidate? If I understood correctly $f(a)$ might increase up to its critical point, then decrease after it. The minimum point can be after the critical point, right? (2) In the expression for the derivative of the constraint, I think it should be $\frac{d\alpha}{da} \left(\log a - \log(1-a) \right) = - \left(\frac{\alpha}{a} -\frac{1-\alpha}{1-a} \right)$ right? (There is a minus sign missing). | |
| Aug 29, 2017 at 2:15 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 29, 2017 at 2:08 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 15:43 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 8:13 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 7:40 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 5:59 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 5:36 | history | edited | Tim Carson | CC BY-SA 3.0 |
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| Aug 26, 2017 at 4:03 | history | answered | Tim Carson | CC BY-SA 3.0 |