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 $n=5,6,7$.
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.
Let $J_4(n)$ be the $n \times n$ matrix $$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ -1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & -1 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$$ where all hidden entries are $0$. Then $E_4(n)$ is the orbit of $J_4(n)$ under $GL_n$ acting by $g : X \mapsto g X g^T$. (See any linear algebra text that works out the classification of skew symmetric forms.)
The stabilizer of $J_4(n)$ is $$\begin{pmatrix} S & T \\ 0 & U \end{pmatrix}$$ where $S$ is in $Sp(4)$, $T$ is an arbitrary $4 \times (n-4)$ matrix and $U$ is an invertible $(n-4) \times (n-4)$ matrix.
The dimension of $GL_n$ is $n^2$. The dimension of $Sp(4)$ is $10$. So the dimension of the orbit is $$n^2 - \left( 10 + 4(n-4) + (n-4)^2 \right) = 4n-10 .$$
Sanity check: A generic $4 \times 4$ skew symmetric matrix is invertible, so $E_4(4)$ is $\binom{4}{2} = 6$ dimensional. A $5 \times 5$ skew symmetric matrix is not invertible, since odd size skew symmetric matrices are always singular; generically there is no reason for it not to have rank $4$. So $E_4(4)$ is $\binom{5}{2} = 10$ dimensional. A $6 \times 6$ skew symmetric matrix has rank $6$ if and only if its Pfaffian does not vanish; if the Pfaffian does vanish than it has rank $\leq 4$ and generically has rank $4$. So $E_4(6)$ is an open dense locus inside a hypersurface in $\mathbb{R}^{15}$, and we conclude that $\dim E_4(6) = 14$.
Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. $\DeclareMathOperator{\rank}{rank}$ Note that for $A\in \uso(n)$
$$ \rank A=n-\dim\ker A, $$
$$\dim \ker A \equiv n\bmod 2. $$
Denote by $ \uso(n)_k $ the space consisting of matrices $X\in\uso(n)$ such that $\dim \ker X= k$, where $k\equiv n\bmod 2$.
You are interested in the space $\uso(n)_{n-4}$.
We have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel. We deduce
$$\dim \uso(n)_k=\dim \uso(n-k)_0=\dim \uso(n-k)=\binom{n-k}{2}. $$
Thus, in your case
$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$
