It is not difficult to get a formula for the number of $n\times n$-matrices 
$g\in\mathrm{Mat}\_{n}(\mathbb{F}\_{q})$ with $\mathrm{rank}(g)=k$. Namely we have got
\begin{align\*}
\mid\{g\in\mathrm{Mat}\_{n}(\mathbb{F}\\_{q})\mid \mathrm{rank}(g)=k\}\mid = \binom{n}{k}\\_{q}\cdot(q^{n}-1)\cdots(q^{n}-q^{k-1}) 
\end{align\*}
with $\binom{n}{k}\_{q}=\frac{(q^{n}-1)\cdots(q^{n}-q^{k-1})}{(q^{k}-1)\cdots(q^{k}-q^{k-1})}$. 
My question is now, if it is possible to get a formula in the same fashion for the following number
\begin{align\*}
f(n,q,k):=\mid\{g\in\mathrm{Mat}\_{2n}(\mathbb{F}\\_{q})\mid g=\begin{pmatrix} A & B \\\ C & -A^{t} \end{pmatrix}, B=B^{t}, C=C^{t} \\text{ and }\mathrm{rank}(g)=k\}\mid
\end{align\*}.
My conjecture (based on computer data for small $n$ and $k$) is the following, but I have no idea for the proof.
\begin{align\*}
f(2n,q,2k)=\binom{2n}{2k}\\_{q}\cdot q^{k(k+1)}\displaystyle\prod\_{i=1}^{k}(q^{2i-1}-1),
\end{align\*}
\begin{align\*}
f(2n,q,2k+1)=\binom{2n}{2k+1}\\_{q}\cdot q^{k(k-1)}\displaystyle\prod\_{i=1}^{k}(q^{2i-1}-1)
\end{align\*}



What is the number of $2n\times 2n$-matrices $g=\begin{pmatrix} A & B \\\ C & -A^{t} \end{pmatrix}$, $B$ and $C$ symmetric, over the finite field $\mathbb{F}\_{q}$ with $\mathrm{rank}(g)=k$?
Just for the (virtual) glory of possible reputation points, here is the comment as an answer.


The columns can be switched to put B and C on the main diagonal, and multiplication
of the bottom rows by -1 should produce a symmetric matrix from one of the desired form.
This should be a bijective rank preserving operation which takes the current problem to one
of counting symmetric matrices of a given rank over the chosen field. I am guessing that this
new version is handled in the literature.


Gerhard "Ask Me About System Design" Paseman, 2012.11.07


