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Earlier, I had asked a similar question but that was not the correct problem where I got stuck. After a few quick answer, I realized that and I apologize for that.

Let $B_m$ be the space of all skew-symmetric matrices of size $m$ over the finite field $\mathbb{F}_q$ of $q$ elements. Let $E$ be a subspace of $B_m$ of dimension $r$ containing atleast one rank $2$ matrix. Write $E$ as $E= E_1 \bigoplus E_2$ with $ \dim E_i= r_i$ for $i=1,2$ and $E_1$ is a maximal subpace of $E$ containing only rank $2$ matrices. Now for a given rank $4$ matrix $Q\in E_2$ how many matrices $P\in E_1$ exist such that $P +Q$ is again of rank $2$. My Guess is $q^2$ and also that $q^2$ is a strict upper bound.

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  • $\begingroup$ Link to the previous question: mathoverflow.net/questions/288156 You should rather have edited it, it hasn't been closed (at this time). You should delete the old question now. $\endgroup$
    – YCor
    Dec 15, 2017 at 11:33

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This is more easily understood as a question about alternating tensors. The space of skew-symmetric $m\times m$ matrices over a field $k$ is isomorphic to $\Lambda^2(k^m)$, via $u\wedge v\mapsto uv^T-vu^T$. Then a matrix is of rank $\leq 2$ if and only if it is a pure alternating tensor $u\wedge v$. Now it is easy to check that a vector space of skew-symmetric matrices consists only of matrices of rank $\leq 2$ if and only if it is of the form $e_1\wedge U$ where $U$ is a subspace of $k^m$. Let us fix once and for all a splitting $k^m=ke_1+W$. We can clearly assume that $U\subset W$.

A skew-symmetric matrix $Q$ has rank $4$ if and only if it is of the form $a\wedge b+c\wedge d$ where $a,b,c,d$ are linearly independent. Now if $Q-e_1\wedge e_2$ has rank 2 then $$a\wedge b+c\wedge d=e_1\wedge e_2+e_3\wedge e_4$$ for some $e_3,e_4$ and $a,b,c,d$ span the same subspace as $e_1,e_2,e_3,e_4$. More precisely, these expressions are equal if and only if the matrix with columns $a,b,c,d$, expressed in terms of $e_1,e_2,e_3,e_4$ is a symplectic $4\times 4$ matrix (where the symplectic form is determined by $e_1\wedge e_2+e_3\wedge e_4$). After applying a suitable symplectic transformation, we may assume that $a=e_1$ and $b,c,d,e_2,e_3,e_4\in W$.

So the problem has been reduced to the following: given $b,c,d\in W$ linearly independent, how many $u\in U$ are there such that $e_1\wedge b+c\wedge d+e_1\wedge u$ has rank 2? Obviously this is equivalent to $e_1,(b+u),c,d$ being linearly dependent. But this is equivalent to $b+u,c,d$ being linearly dependent, so we obtain our answer. The set of all such $u$ is given as the intersection $$(-b+kc+kd)\cap U.$$ In the case of a finite field $k={\mathbb F}_q$, this set consists of at most $q^2$ elements, and that happens precisely when $b,c,d\in U$.

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  • $\begingroup$ Thanks! @Paul Levy , I am not convinced why the span of $ a, b, c, d$ is same as the span of $ e_1, e_2, e_3, e_4$. Can we see it without going into symplectic matrices? $\endgroup$
    – Singh
    Dec 18, 2017 at 13:21

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