Let $n\geqslant 5$ and let $E\_4(n)$ be a linear subspace of $(n\times n)$- real skew-symmetric matrices such that 
$$
rank(A)=4,\text{ for all }A\in E\_4(n),A\neq 0.
$$
I'm curious about the following question:


QUESTION:


What can be said about the dimension of $E\_4(n)$? Of course, it is easy to check that $\operatorname\*{dim}E\_4(n)\leqslant \binom{n-1}{2}$. But what is the best possible value? I'm specially interested in the case $n=5,6,7$.



Subspace of Skew-symmetric Matrices of Rank Four
Partial progress: It's easy to achieve $n-3$. Consider matrices of the form


$$\begin{pmatrix} 
0 & 0 & r\_1 & r\_2 & \cdots & r\_{n-3} & 0 \\
0 & 0 & 0 & r\_1 & \cdots & r\_{n-4} & r\_{n-3} \\
r\_1 & 0 & & & & & \\
r\_2 & r\_1 & & & & & \\
\vdots & \vdots & & & {\LARGE 0}& & \\
r\_{n-3} & r\_{n-4} & & & & & \\
0 & r\_{n-3} & & & & & \\
\end{pmatrix}$$
where the bottom right square is entirely $0$. If this has rank $<4$, then the upper-left $4 \times 4$-submatrix implies $r\_1^4=0$, so $r\_1=0$. Then inductively $r\_2^4=0$, and so forth.


For $n=5$, this only gives a $2$ dimensional subspace, and I argued in the comments on my other answer that a generic $3$ dimensional subspace of the $5 \times 5$ matrices should work. Right now, though, I can't see how to do better for $n \geq 5$.


Ah, slight improvement. For $n$ even, and taking advantage of the fact that I'm working over the reals, I can do $n-2$:
$$\begin{pmatrix} 
0 & 0 & a\_1 & b\_1 & \cdots & a\_{(n-2)/2} & b\_{(n-2)/2} \\
0 & 0 & -b\_1 & a\_1 & \cdots & -b\_{(n-2)/2} & a\_{(n-2)/2} \\
a\_1 & -b\_1 & & & & & \\
b\_1 & a\_1 & & & & & \\
\vdots & \vdots & & & {\LARGE 0}& & \\
a\_{(n-2)/2} & -b\_{(n-2)/2} & & & & & \\
b\_{(n-2)/2} & a\_{(n-2)/2} & & & & & \\
\end{pmatrix}$$
If this has rank $<4$, then $(a\_i^2+b\_i^2)^2=0$ for all $i$, so over the reals this can only happen when it is $0$.




---


A potential strategy: What are (up to conjugation) the maximal subspaces of $n \times n$ skew-symmetric matrices on which the rank is always $\leq 4$? Then we can focus our efforts on finding large subspaces on each of these which miss the rank $2$ locus. 


So far, I have only been able to find four maximal subspaces. I'll describe them all as block matrices with the size and nature of the blocks indicated:
$$\begin{pmatrix}
5 \times 5 & 0 \\
0 & 0 
\end{pmatrix}$$
$$\begin{pmatrix}
0 & 2 \times (n-2) \\
(n-2) \times 2 & 0 \end{pmatrix}$$
$$\begin{pmatrix}
0 & 3 \times 3,\ \mbox{skew symmetric} & 0 \\
3 \times 3,\ \mbox{skew symmetric} & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$
$$\begin{pmatrix}
3 \times 3 & 0 & 0 \\
0 & 3 \times 3 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$


If these are the only options, then the only one which grows with $n$ is the second case so that's the one we need to concentrate on. 


In that case, the question is equivalent to "what is the largest linear subspace of the $2 \times (n-2)$ matrices which includes no rank $1$ submatrices?" I can show that the answer to that question is $n-3$, over $\mathbb{C}$, and is $2 \lfloor (n-2)/2 \rfloor$ over $\mathbb{R}$; I'll post the argument if anyone cares. 


