Timeline for Schur complement and depermuting an algorithm for $\mathsf{determinant}\bmod2$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2021 at 2:58 | vote | accept | Turbo | ||
| Oct 18, 2021 at 2:52 | comment | added | Turbo | I am looking at $\mathbb F_2$. So perhaps there is a possibility in $\mathbb F_2$ if not in $\mathbb R$? | |
| Feb 3, 2021 at 7:49 | vote | accept | Turbo | ||
| Oct 18, 2021 at 2:51 | |||||
| Feb 3, 2021 at 7:49 | comment | added | Federico Poloni | Yes; my comment refers to step $i=0$ specifically, but then the same holds at each step; either $M_i$ is the zero matrix, or it has a 1 that you can permute into its (1,1) entry. | |
| Feb 3, 2021 at 7:46 | comment | added | Turbo | I think you meant to say 'unless $M_i$ is zero matrix at an $i\in\{0,1,\dots,n-1\}$'? | |
| Feb 3, 2021 at 7:44 | comment | added | Federico Poloni | If I understand correctly what you are asking, yes: unless the matrix is zero, there is always a way to bring an 1 to the top-left corner by permuting rows and columns with $P_{1,1}$ and $P_{2,1}$. Then the following permutations won't affect the (1,1) entry anymore, since they act on $(2,...,n)$ only. | |
| Feb 3, 2021 at 7:41 | comment | added | Turbo | Interesting.. so always first entry is $1\neq0$? | |
| Feb 3, 2021 at 7:40 | history | answered | Federico Poloni | CC BY-SA 4.0 |