Take the finite field GF(3), and look at all its 3x3 matrices. How many of these 3

^{9}= 19683 matrices' characteristic functions have all three roots? Or no roots? Brute force search reveals:

roots count () 3456 (0) 2106 (1) 2106 (2) 2106 (0,0,0) 729 (0,0,1) 1054 (0,0,2) 1053 (0,1,1) 1052 (0,1,2) 1404 (0,2,2) 1053 (1,1,1) 729 (1,1,2) 1053 (1,2,2) 1053 (2,2,2) 729

No cubic in GF(3) has just two roots for the same reason as cubics in the reals don't have just two real roots, I think. The element of a field extension in which those two extra roots exist must cancel out. But in GF(3) you've got degree-three extensions too, so zero-root polynomials exist as well.

What's potentially interesting here is the symmetry breaking for (0, 0, 1) and (0, 0, 2). Those cases have +1 and -1 count, respectively, than you would "expect" by comparison with (0, 0, 1), (1, 1, 2), (0, 1, 1), and (1, 2, 2). The single-root cases are perfect symmetric, by contrast.

Anyway, since these roots are the eigenvalues it turns out that just slightly over 50% of the 3x3 matrices in GF(3) have the full set of eigenvalues (and thus might be diagonalizable, although probably a lot are defective.)

Expanding to 4x4 matrices (43 million of them) or GF(5) (about 2 million) would require a lot more processing time, so I'd like to make the computation smarter in some way.

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