Let $p$ be a prime and $q=p^n$. Let $\mathbb F\_{q^2}$ be a field with $q^2$ elements and $\sigma$ its authomorhism of order two. A $m$ by $m$ matrix $A$ over $\mathbb F\_{q^2}$ is *hermitian* if $A^\sigma$ coincides with the traspose of $A$.


Do you know a reference to the following result?


**Theorem**. The ring of $m$ by $m$ matrices over $\mathbb F\_{q^2}$ can be generated over $\mathbb F\_p$ by two hermitian matrices.



Generating of the matrix ring by two hermitian matices
The statement is false if $m=2$.


For $m>2$, let $\alpha$ be a generator of the multiplicative group of $\mathbb F\_{q^2}$. Then the matrices $\alpha E\_{1,2}+\sigma(\alpha) E\_{2,1}$ and $\sum\_{i=1}^{m-1} (E\_{i,i+1}+E\_{i+1,i})$ generate the ring of m by m matrices over $\mathbb F\_{q^2}$. 


