# 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 $\mathrm{rank}(g)=k$?

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*}

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If k is odd, I think that will give very strict conditions on A. You might consider asking the question for k=1 or k=3, and perhaps doing a computer enumeration leveraged by whatever thery you can derive. Gerhard "Ask Me About System Design" Paseman, 2012.10.24 – Gerhard Paseman Oct 24 '12 at 13:37
I need the above as a lemma to deduce some deaper results about distribution of special matrices over finite rings. – micha Nov 6 '12 at 15:05
Try the following: look at the map which takes a matrix of your form to D=[B,A,,A^t,-C]. Ideally the map is bijective and rank preserving. Now if I have not messed up, your problem reduces to counting order 2n matrices which are of rank k and symmetric, for which you may find assistance in the literature. Gerhard "Check All These Suggestions Carefully" Paseman, 2012.11.06 – Gerhard Paseman Nov 6 '12 at 16:56
Thank you for the very helpful hint. It works. – micha Nov 7 '12 at 16:07