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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.

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When you say: "do you know a reference", are you implying that the result is known to be true? –  Igor Rivin Aug 31 '12 at 20:14
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It should be true. For any field $k$, "most" pairs $(A,B)$ of matrices generate $\mathrm{Mat}_n(k)$. Notice that the dimension of the subspace of $\mathrm{Mat}_n(k)$ spanned the monomials in $A$ and $B$ stays the same if we enlarge the field, so we may take $k$ algebraically closed. Then most $A$ will be diagonalizable with distinct eigenvalues. Polynomials in $A$ then give the matrices $e_{ii}$. For most $B$, $e_{ii} B e_{jj}$ is nonzero for all pairs $(i,j)$, and then the $e_{ii} B e_{jj}$ span all matrices. (continued) –  David Speyer Aug 31 '12 at 23:49
    
This argument is easy to make precise over an infinite field. With more work, one can also use it over a finite field. I spent some time try to make it work in the setting of this problem, where we have the Hermitian condition to deal with and have the extra complication of working both with $\mathbb{F}_p$ and $\mathbb{F}_{q^2}$, but I ran out of energy. –  David Speyer Aug 31 '12 at 23:51

1 Answer 1

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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 $\mathbb F_{q^2}$.

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