Timeline for The number of invertible 4×4 circulant matrices over the ring Z
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2022 at 6:50 | answer | added | Maciej Ulas | timeline score: 1 | |
| Aug 17, 2022 at 12:40 | comment | added | Katy | Thanks for the explanation | |
| Aug 17, 2022 at 12:35 | comment | added | Olivier Bégassat | My bad, I forgot some minus signs... The eigenvalues are $(a+c) + (b+d)$, $(a+c) - (b+d)$, $(a-c) -i(b-d)$ and $(a-c) + i(b-d)$ so you'll need ($c = a \pm 1$ and $b=d$) or ($c = a$ and $d=b\pm 1$). In the first case you also need $\pm 1 + 2a + 2b = \pm 1$ and $\pm 1 + 2a - 2b = \pm 1$ so that it seems only a few possiibilities for a and b are allowed. Something analoguous holds in the second case. There are thus only finitely many possibilities. | |
| Aug 17, 2022 at 12:16 | comment | added | Olivier Bégassat | As far as I can tell this should work because the complex eigenvalues of a 4x4 circulant matrix with first row $(a,b,c,d)$ are $(a+c) + (b+d)$, $(a+c) - (b+d)$, $(a+c) +i(b+d)$ and $(a+c) - i(b+d)$ which will all be in $\{\pm1, \pm i\}$ unless I'm mistaken. | |
| Aug 17, 2022 at 12:05 | comment | added | Katy | I tried it for k=2 and n=3 it does not work..the determinant is not 1 or _1 so it is not invertible in Z?Am I wrong? | |
| Aug 17, 2022 at 11:58 | comment | added | Olivier Bégassat | It seems that there are four infinite families: those with first row $(n\pm1,k,-n,-k)$ and those with first row $(n,k\pm1,-n,-k)$, with $n,k\in\Bbb{Z}$. | |
| S Aug 17, 2022 at 11:48 | review | First questions | |||
| Aug 17, 2022 at 14:39 | |||||
| S Aug 17, 2022 at 11:48 | history | asked | Katy | CC BY-SA 4.0 |