most recent 30 from http://mathoverflow.net 2013-05-22T03:13:37Z http://mathoverflow.net/feeds/question/106035 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106035/generating-of-the-matrix-ring-by-two-hermitian-matices Andrei Jaikin 2012-08-31T14:03:47Z 2013-05-08T22:22:00Z

<p>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 <em>hermitian</em> if $A^\sigma$ coincides with the traspose of $A$.</p> <p>Do you know a reference to the following result?</p> <p><strong>Theorem</strong>. The ring of $m$ by $m$ matrices over $\mathbb F_{q^2}$ can be generated over $\mathbb F_p$ by two hermitian matrices.</p>

http://mathoverflow.net/questions/106035/generating-of-the-matrix-ring-by-two-hermitian-matices/128664#128664 Andrei Jaikin 2013-04-24T21:46:44Z 2013-04-24T21:46:44Z

<p>The statement is false if $m=2$.</p> <p>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 $\mathbb F_{q^2}$. </p>