Right. The eigenvalues of a b -b a are a +/- b sqrt(-1) . So if -1 has a square root in F_p (i.e. if -1 is a quadratic residue, which happens if p mod 4 = 1) then we can diagonalize this matrix. More generally, I guess one defines "elliptic" matrices as follows. Let g be a fixed non-quadratic residue, i.e. an element of F_p that doesn't have a square root in F_p. Then consider matrices a , b gb , a where a^2 - gb^2 = 1. (Exercise: show these matrices form a subgroup.) Then the eigenvectors are (1, +/- sqrt(g)) and the eigenvalues are a +/- b sqrt(g). Cris On Mar 22, 2013, at 6:27 AM, Victor Miller wrote:
Cris's canonical matrix ain't elliptic when p mod 4 = 1 --- discriminant -4 b^2 is a quadratic residue, the roots lie in |F_p , and the projectivity is hyperbolic for b /= 0 .