Timeline for Rank of a fat random matrix
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2016 at 21:40 | answer | added | Jean-Luc Bouchot | timeline score: 0 | |
| Jun 30, 2016 at 15:31 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 31, 2016 at 14:47 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 1, 2016 at 14:43 | comment | added | Gro-Tsen | To put Robert Israel's answer differently, non-full-rank matrices are a (singular) algebraic subvariety of $\mathbb{C}^{n\times k}$ that is not the full space, so it has codimension at least $1$ and Lebesgue measure zero: with probability $1$ a random matrix has full rank, you don't need to take a limit. (Over finite fields, of course, things would be different.) | |
| Nov 3, 2015 at 11:11 | answer | added | Igor Rivin | timeline score: 2 | |
| Nov 3, 2015 at 10:22 | comment | added | Jeff | @Robert, I see. So it means that the rank of any $n \times m $ random matrix with i.i.d. entries taken from an absolutely continuous distribution, is $\min (n,m) $, right? | |
| Nov 3, 2015 at 7:26 | comment | added | Robert Israel | For any absolutely continuous distribution of random variables $X_1, \ldots, X_m$,, any nonconstant polynomial in the $X_j$ is a.s. nonzero. Apply that to the determinant of an $n \times n$ submatrix. | |
| Nov 3, 2015 at 7:19 | comment | added | Jeff | It was a mistake. The rank was $n$ in those cases. | |
| Nov 3, 2015 at 7:18 | history | edited | Jeff | CC BY-SA 3.0 |
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| Nov 3, 2015 at 7:05 | comment | added | Carlo Beenakker | I am confused: you say the rank is $k$ when $k\geq n$, but how can the rank be larger than $n$? | |
| Nov 3, 2015 at 6:52 | history | asked | Jeff | CC BY-SA 3.0 |