Timeline for On a matrix problem in the field $\mathbb F_2$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2021 at 18:00 | comment | added | Antoine Labelle | The condition $PMP'=(M+J+I)$ basically says that $M_{i+1,j+1}=M_{i,j}+1$ for $i\ne j$, (indices taken mod $n$). If $n$ is odd you can deduce $M_{i,j}=M_{i,j}+1$ by repeating this, which is impossible. Similarly if $n=2k$ with $k$ odd we find $M_{i,i+k}=M_{i+k,i}+1$ which is imposibble as $M$ is symmetric. | |
| Mar 19, 2021 at 14:04 | comment | added | Turbo | Why is it easy to see? | |
| Mar 19, 2021 at 13:10 | comment | added | Antoine Labelle | Yes unless specified otherwise it's over $\mathbb{F}_2$. Also for cyclic permutations it's easy to see that the matrix $M$ can only exist if $4|n$, so $a=1$ is impossible. | |
| Mar 19, 2021 at 12:20 | comment | added | Turbo | In the comment on $8|n$ are you talking det mod $2$ or det in reals (in answer you say det is $16$ so I think you imply det mod $2$)? Can you check for $n=4k+a$ where $a\in\{0,1\}$? My guess is det in reals is mostly $0$ (hence $0\bmod2$) if $a=1$. | |
| Mar 17, 2021 at 14:51 | history | edited | Antoine Labelle | CC BY-SA 4.0 |
added 576 characters in body
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| Mar 16, 2021 at 20:30 | history | answered | Antoine Labelle | CC BY-SA 4.0 |