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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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 |
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