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The number of symmetric matrices of order $n$ and rank $r$ over finite fields has been counted e.g.

http://www.math.clemson.edu/~kevja/REU/2004/SymmetricRankRMatrices.pdf

Is the number of skew-symmetric matrices over finite fields of order $n$ and rank $r$ is known? If yes please provide some reference.

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    $\begingroup$ This problem is much easier than the symmetric case. As long as the characteristic of the field $\mathbb{F}$ is not $2$, the space $A_{n,r}$ of anti-symmetric $n$-by-$n$ matrices of rank $r$ (necessarily even) is a homogeneous space $\mathrm{GL}(n,\mathbb{F})/P_{n,r}$ where $P_{n,r}$ is the subgroup that stabilizes some particular element of $A_{n,r}$. Thus, you are essentially asking for the order of this subgroup, which is an easy exercise, so I expect one would find it as a remark in some paper about something else, rather than as a main result in something. Characteristic 2 may be harder. $\endgroup$
    – Robert Bryant
    Commented May 14, 2016 at 12:17
  • $\begingroup$ Thank you for your comments. But I still have some doubts. I am not sure how to prove that the $GL(n,\mathbb{F})$ action is transitive. Could you please suggest me some reference. $\endgroup$
    – Singh
    Commented Jun 6, 2016 at 18:11
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    $\begingroup$ It is known that the following three numbers are equal, but no combinatorial proof is known that any two are equal: (1) number of symmetric matrices in $\mathrm{GL}(2n,q)$ with zero diagonal, (2) number of symmetric matrices in $\mathrm{GL}(2n-1,q)$, (3) number of skew-symmetric matrices in $\mathrm{GL}(2n,q)$. See Enumerative Combinatorics, vol. 1, 2nd ed., Exercise 1.199. $\endgroup$
    – Richard Stanley
    Commented Jul 10, 2016 at 22:14

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Let me make Robert's answer a bit more concrete. The theory of skewsymmteric forms is much nicer than the symmetric one since they are classified by their rank for any field in any characteristic (including $p=2$). Now the stabilizer of the standard form $$ e_1\wedge e_2+\ldots+e_{2r-1}\wedge e_{2r} $$ is $Sp(2r)\times GL(n-2r)\ltimes U$. Thus the number of rank-$2r$-skewsymmetric forms in $n$-space is $$ a(n,r)=\frac{|GL(n)|}{|Sp(2r)|\cdot|GL(n-2r)|\cdot|U|} $$ where $$ |GL(n)|=q^{\frac12 n(n-1)}\prod_{i=1}^n(q^i-1),\quad |Sp(2r)|=q^{r^2}\prod_{i=1}^r(q^{2i}-1),\quad |U|=q^{2r(n-2r)}. $$ Thus $$ a(n,r)=q^{r(r-1)}\ \prod_{i=1}^r(q^{2i-1}-1)\ \left[\matrix{n\\2r}\right]_q $$

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Offhand, I don't know a reference, but I'm sure that the transitivity result can be found in some book that treats Lie algebras over finite fields. However, it's easy enough that you don't need to go looking:

Basically, there is a 1-to-1 correspondence between elements of $A_{n,r}$ and anti-symmetric bilinear pairings $a:F^n\times F^n\to F$ that have a kernel $K_a\subset F^n$ of dimension $n-r$. Such an $a$ induces a nondegenerate, anti-symmetric pairing $\bar a: Q\times Q\to F$ where $Q = F^n/K_a\simeq F^r$, and the data $(K_a,\bar a)$ is equivalent to specifying $a$. (Of course, $r$ must be even if the characteristic of the field is not $2$.)

Now, all nondegenerate anti-symmetric pairings on $F^r$ are equivalent under $\mathrm{GL}(r,F)$, which is an easy exercise, and the stabilizer of any one of them is a group that is sometimes denoted $\mathrm{Sp}(r,F)$. You can look up its order in any book that treats Lie groups over finite fields. Probably it's in Humphries' book, but I don't have it here with me, so I can't say for sure.

Using the above, you should have no trouble working out the order of the set $A_{n,r}$ since the above argument shows that $$ |A_{n,r}| = |\mathrm{GL}(r,F)|-|\mathrm{Sp}(r,F)| + \left|\mathrm{Gr}_{n-r}(F^n)\right|, $$ where $\left|\mathrm{Gr}_{n-r}(F^n)\right|$ is the number of subspaces of $F^n$ that have dimension $n{-}r$.

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