micromass said:
So I guess that the best way to proceed is to diagonalize this matrix over the reals, do the calculations there. And only then reduce everything to \mathbb{F}_{101}.
Ah! Silly me -- it hadn't crossed my mind to compute the 1870-th power of an integer lift of M by diagonalizing over the reals (or complexes as appropriate), then reducing modulo 101.
Alas, that matrix is going to have very, very large components.
Dollydaggerxo said:
yes i began by just squaring my way up to 1870 using the primes however i dont' really see how this uses Euler's theorem?
Alas, I can't see how Euler's theorem could possibly apply to this problem, except in the special case that the matrix diagonalizes over F
101.
If you know your finite fields, you could show that if M is a 2x2 matrix over
F101 whose characteristic polynomial is irreducible, then M
10200 is the identity matrix, but that's not particularly useful for this problem.
(key proof idea -- the matrix must be diagonalizable over
F1012)
Dollydaggerxo said:
M17 = M x (M16)
or
M17 = (M16) x M
Matrix multiplication is associative, so it doesn't matter where you put the parentheses in the expression
MMMMMMMMMMMMMMMMM (this is a product of 17 M's)
so both of your methods of computing M
17 give the same value.