Timeline for Linearly independent functions evaluated at random points create full rank matrices
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2019 at 22:59 | comment | added | mohi | $x_1,...,x_n\in\mathbb{R}^d$ are n points in $\mathbb{R}^d$ none of which can be written as a positive multiple of the other (i.e. $x_i=\alpha x_j$ for some positive $\alpha$. See mathoverflow.net/questions/336543/… for a related question of mine about the independent part. | |
| Jul 20, 2019 at 21:31 | comment | added | Kostya_I | and what are $x_i$? | |
| Jul 20, 2019 at 21:05 | comment | added | mohi | yes no linear combination that vanishes identically. To avoid pathological cases like you state above. Let's assume f is sufficiently nice that this doesn't happen e.g. f is smooth and differentiable. Specifically I'm interested in the case where $f_i(w_j)=\phi'(w_j^Tx_i)x_i$ with $\phi(z)=\log(1+e^z)$ | |
| Jul 20, 2019 at 20:46 | comment | added | Kostya_I | What do you mean by "linearly independent functions"? If it's just that there's no linear combination that vanishes identically, then it can still be that $f_n$ vanishes in a ball of radius $e^n$ around the origin. Which means that with overwhelming probability your matrix will have a zero column. | |
| Jul 19, 2019 at 18:41 | history | asked | mohi | CC BY-SA 4.0 |