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).



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> **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.
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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.



Semi-simple matrices over fields of finite characteristic
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 $A\in V$. 


