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I tried asking this question on Math StackExchange and didn't get any replies. I was hoping that maybe someone here could help, sorry for duplicating.

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.

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    $\begingroup$ The relevant fact seems to be on p. 17 rather than p. 16, where it is written that over finite fields, any two hermitian forms of the same rank are always equivalent (and hence of maximum possible index, namely $\lfloor n/2 \rfloor$). $\endgroup$
    – Tom De Medts
    Dec 21, 2015 at 16:37

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