Timeline for Maximum singular value of a random $\pm 1$ matrix
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 12, 2019 at 17:25 | comment | added | Emilio Pisanty | @Glorfindel Thanks for pointing out this dead link. Boy, this thread is old - it's pretty scary how dust accumulates on what used to be pretty reasonable dates. | |
| May 12, 2019 at 17:20 | history | edited | Emilio Pisanty | CC BY-SA 4.0 |
Pulled in code from now-dead link.
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| S Mar 30, 2019 at 21:21 | history | suggested | Glorfindel | CC BY-SA 4.0 |
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference; image retrieved via Wayback Machine); for more info, see https://meta.mathoverflow.net/a/4058/70594
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| Mar 30, 2019 at 20:52 | review | Suggested edits | |||
| S Mar 30, 2019 at 21:21 | |||||
| May 3, 2012 at 18:49 | comment | added | Kevin P. Costello | (continued) Even if you let both $m$ and $n$ go to infinity, but at different rates (say $m/n \rightarrow c<1$), the smallest singular value will with high probability grow be within a constant of the largest (Bai and Yin, "Limit of the smallest eigenvalue of a large dimensional covariance matrix", see also Rudelson and Vershynin's "The smallest singular value of a random rectangular matrix" ) | |
| May 3, 2012 at 18:40 | comment | added | Kevin P. Costello | In a sense, "local" configurations (those involving only a couple rows) can only get you so close to $0$: If you look at, say, $|| \{1,-1,0,0,0,0,0,0,0,0 \} A||$, it's either going to be $0$ or at least $1$, meaning that this can't be the cause of a positive singular value smaller than $2^{-1/2}$. Intuitively, a $0$ singular value (which requires only $1$ row to go wrong) is much easier than a very small positive singular value (which requires $2$ or more rows to go wrong), though this is far from a rigorous proof. | |
| May 2, 2012 at 20:59 | comment | added | Emilio Pisanty | @KevinPCostello: yes, that sounds simple enough. Why is it separate from the rest of the minimum-singular-value histogram, though? if you fix $m$ and take $n\rightarrow\infty$, does it "leak out" to positive singular values? I guess I'm asking why it is so much more unlikely that a random matrix be "almost singular" than it be just plain singular, which kind of makes sense since it's hard to build a very slim parallelogram with only integer coordinates but you might build a plane now and then. Is my intuition right? | |
| May 2, 2012 at 16:59 | comment | added | Kevin P. Costello | Any matrix with two rows which are multiples of each other will automatically have a $0$ singular value. This accounts for most of the peak you see at $0$ (each pair of rows matches for about $3900$ matrices and there's $10$ pairs of rows), but disappears as $\min\{m,n\}$ tends to infinity. | |
| May 2, 2012 at 16:20 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
added 452 characters in body
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| May 2, 2012 at 0:12 | comment | added | Emilio Pisanty | You're welcome! Mathematica code is normally deceptively simple but this was quite straightforward. Plus, it helped me crystallize the Random[]&/@Range[] trick which had been annoyingly floating around my mind with a "there has to be a simple way" kind of taunt. | |
| May 1, 2012 at 20:53 | comment | added | Kostas | Thank you very much for your time to code this! It was very helpful! | |
| May 1, 2012 at 12:56 | history | answered | Emilio Pisanty | CC BY-SA 3.0 |