Timeline for The sum of same powers of all matrices modulo p
Current License: CC BY-SA 3.0
8 events
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| Apr 7, 2013 at 21:37 | comment | added | Will Sawin |
The command sum([m^6 for m in MatrixSpace(GF(2),2,2)]) returns the identity matrix. As does m^7. This completely solves $p=2$.
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| Apr 7, 2013 at 21:35 | history | edited | Ilya Bogdanov | CC BY-SA 3.0 |
edited body
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| Apr 7, 2013 at 21:30 | history | edited | Ilya Bogdanov | CC BY-SA 3.0 |
deleted 25 characters in body
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| Apr 7, 2013 at 21:24 | comment | added | ya-tayr | Replace 26 by 78,80,82,84, or 340 and sage still returns 0. | |
| Apr 7, 2013 at 21:15 | comment | added | Will Sawin | You can go further. You've shown that the number of subcycles is at least $p^p-1$. But $k$ is certainly larger than the number of subcycles. In fact, I think the same logic shows that the number coming out of $2$, the number coming out of $3$, and so on, are all at least $p^p-1$. So you get $p^{p+1} -p$. Or for $d\times d$ matrices, $p^{d+1}-p$. | |
| Apr 7, 2013 at 21:05 | comment | added | Ilya Bogdanov | Does it remain for larger $k$? | |
| Apr 7, 2013 at 20:56 | comment | added | ya-tayr | For p = 3, k = 26, the sage command sum([m^26 for m in MatrixSpace(GF(3),3,3)]) also returns the 0 matrix. | |
| Apr 7, 2013 at 20:32 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |