I did notice one unusual situation:
The characteristic polynomial is x^2  (u + 1)*x + (u  1); the discriminant is u^2  2*u + 5. As long as this is nonzero (mod p), there are two eigenvalues, and it is at least not insane to think about diagonalizing the matrix, which might be probative.
If p == 1 (mod 4) however, i^{2} == 1 (mod p) has two solutions. Taking u = 1 + 2*i makes the discriminant of the characteristic polynomial 0, so it has a repeated factor:
x^2  (2 + 2*i)*x + 2*i = (x  (1 + i))^2.
In this case, there is only one eigenvalue, 1 + i, of multiplicity 2. The matrix A  (1+i)I (I = 2x2 identity matrix) is (using PariGP syntax) N = [i,1;1,i] which is nilpotent of index 2. Either column [i;1] or [1;i] of N serves as an eigenvector of A.
Thus A = (1 + i)*I + N. Also, I and N commute, so A^p == (1+i)^p*I + N^p (mod p), and N^p is the 2x2 zero matrix.
Thus, A^p == (1+i)^p*I == (1 + i)*I (mod p) in this case.
Examples:
p = 5, i = 2, u = 1 + 2*2 = 0, 1 + i = 3; or i = 3, u = 1 + 2*3 = 2, 1 + i = 4
p = 13, i = 5, u = 1 = 2*5 = 11, 1 + i = 6; or i = 8, u = 1 + 2*8 = 4, 1 + i = 9
Last fiddled with by Dr Sardonicus on 20200912 at 20:08
Reason: xingif ostpy
