Semi-simple matrices over fields of finite characteristic Well-known and useful facts are: 


*

*any symmetric matrix over $\mathbb R$ is semi-simple (i.e. diagonalizable), and

*any hermitean matrix over $\mathbb C$ is semi-simple.


I will loosely speak about the shape of a matrix and mean the existence of some (linear) relations between matrix-entries (or functions of the matrix-entries).

Question: Let $k$ be an algebraically closed field of characteristic $p$. Is there any result whatsoever, which says that a rich class of matrices of a given shape consists only of semi-simple matrices.

Since I am more interested in positive results, the notion of shape is kept flexible. However, if it could be proved that semi-simplicity is not implied by any shape in some reasonable class of shapes, this would be interesting as well.
 A: A basic observation: If $K$ is a field where $0$ is a sum of nonzero squares, say $0=\sum_{i=1}^n x_i^2$, then $\left( x_i x_j \right)_{1 \leq i,j \leq n}$ is a symmetric, nonzero, matrix with square $0$. Such a matrix cannot be semisimple.
So the implication "symmetric implies semisimple" only works over formally real fields.
A: This is only a hint, not an answer.
There is a simple characterization of semisimple matrices over finite fields. Namely, if $A\in M_n(F_q)$, its eigenvalues lie in $F_{q^m}$, $m=lcm(2,\dots,n)$, and there is $P\in GL_n(F_{q^m})$ such that $P^{-1}AP$ is a diagonal of Jordan blocks $\lambda_i I + N_i$, $i=1,\dots,s$. But it is easy to see that $(\lambda_i I+N_i)^{q^m}=\lambda_i I$ (note that $q^m \gt n$), so that $A$ is semisimple if and only if $A^{q^m}=A$.
Now you might want to start to study the possibilities for vector spaces $V\subset  M_n(F_q)$ (or other subvarieties) such that $A^{q^m}=A$ for all $A\in V$.  
A: Well, distinct eigenvalues over the algebraic closure is enough to ensure semisimplicity, so the discriminant of the characteristic polynomial is a polynomial in the matrix entries whose non-vanishing ensures semisimplicity. If you prefer a closed set, you could require it to have a specific non-0 value, say 1. Whether this qualifies as a "shape" is another matter.
A: There is a kind of spectral theorem describing a class of linear operators on Banach spaces over non-Archimedean fields possessing orthogonal (in the non-Archimedean sense) spectral decompositions. See A. N. Kochubei, Non-Archimedean normal operators, J. Math. Phys. 51 (2010), article 023526 (or ArXiv: 0908.4381).
