Timeline for Prove / disprove: If $1 \le n < N$ and $A$ is an $N \times n$ matrix with iid from $\mathcal N(0,1)$, then $s_\min(A) \ge c\sqrt{N}$ w.p $1-2e^{-N}$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 19, 2020 at 23:35 | comment | added | dohmatob | @TerryTao Thanks for the crucial hint (namely, "small-ball probabilities"). Below, I've posted an answer putting every thing together. | |
| Sep 19, 2020 at 23:22 | answer | added | dohmatob | timeline score: 1 | |
| Sep 19, 2020 at 15:04 | comment | added | Terry Tao | It's true (for any $C$, if $c$ is sufficiently small depending on $C$) for each individual column (i.e., for $n$=1) by direct calculation of the chi-squared distribution (or evaluating the measure that an $N$-dimensional gaussian gives to a small ball around the origin), and the epsilon net argument (see e.g., terrytao.wordpress.com/2010/03/05/… ) then extends this bound to tall matrices (again for any $C$, if $c,\lambda_0$ are small enough depending on $C$.) | |
| Sep 19, 2020 at 15:00 | comment | added | dohmatob | Thanks for the input. Could you kindly provide a hight-level justification why this ought to be true for all $n/N < \lambda_0$, for some constant $\lambda_0 \in (0,1)$ ? | |
| Sep 19, 2020 at 2:15 | comment | added | Terry Tao | I believe Rudelson and Vershynin are only claiming this bound in the regime of tall matrices (in which the ratio $n/N$ is assumed to be sufficiently small). | |
| Sep 18, 2020 at 14:19 | history | edited | dohmatob | CC BY-SA 4.0 |
edited title
|
| S Sep 18, 2020 at 14:13 | history | edited | dohmatob | CC BY-SA 4.0 |
edited title
|
| Sep 18, 2020 at 13:55 | review | Suggested edits | |||
| S Sep 18, 2020 at 14:13 | |||||
| Sep 18, 2020 at 13:52 | history | edited | dohmatob | CC BY-SA 4.0 |
added 74 characters in body
|
| Sep 18, 2020 at 13:46 | history | asked | dohmatob | CC BY-SA 4.0 |