I'm trying to figure out some properties of Hermitian matrices over finite fields. Namely, let $K_0=\mathbb{F}_q$ be a field with $q$ elements, and let $K=\mathbb{F}_{q^2}$. The matrix algebra $\newcommand{\M}{\mathrm{M}} \M_n(K)$ is endowed with the conjugate-transpose map, given by $(x_{i,j})^\circ:=(x_{j,i}^\sigma)$, where $x\mapsto x^\sigma$ is the non-trivial automorphism of $K/K_0$.

We have a map from the algebra $\M_n(K)$ to the $K$-vector space of hermitian matrices (i.e. matrices $X$ such that $X^\circ=X$) over $K$, given by $Y\mapsto Y^\circ Y$.

My question is the following- under what circumstances is this map surjective? That is- when is it true that any hermitian matrix is of the form $Y^\circ Y$ for some $Y\in\M_n(K)$?

I know that in the analogous case of $K=\mathbb{C}$ and $K_0=\mathbb{R}$ for $X$ to be of the form $Y^\circ Y$, one must also require that $X$ is a positive-definite matrix. However, I have seen in article that this fact should hold for finite fields. The reference in the article was to page 16 of Dieudonne's La Geometrie des groupes classiques, but I could not find it there (possibly due to poor french-reading skills). If someone could help me find the proof in this text that would be also great.

Thank you.

  • $\begingroup$ Over finite fields, there is no concept of positive semidefinite. $\endgroup$ Dec 20 '15 at 15:27
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    $\begingroup$ Related: math.stackexchange.com/questions/1492444/… The situation is similar for Hermitian and skew-Hermitian matrices. $\endgroup$ Dec 22 '15 at 17:30
  • $\begingroup$ Excellent, Thank you :) @MorganRodgers $\endgroup$
    – kneidell
    Dec 24 '15 at 9:06

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