Let $ A\_{n}(F) $ be the collection of all skew-symmetric matrices over the field $ F $ ($\operatorname{char} F \neq 2 $). Let M be a subspace of $ A\_{n}(F) $ such that all non zero elements have rank $ 2 $ . Here we consider $ F = \mathbb{R} $ .
Then what will be the maximum dimension of $ M $ when $ n = 6 $ ? I have already got $ \dim M \geq 5 $ .As if consider this subsapce
$$\left[\begin{array}{cc}0 & a & b & c & d & e \\ -a & 0 & 0 & 0 & 0 & 0 \\-b & 0 & 0 & 0 & 0 & 0 \\-c & 0 & 0 & 0 & 0 & 0 \\-d & 0 & 0 & 0 & 0 & 0 \\-e & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right]$$
where $ a,b,c,d,e \in \mathbb{R} $ or . So in this space any element of is of rank $ 2 $.
Does it possible that $ \dim M = 6 $ or $ \geq 6 $ . I have also interested in What will be the maximum dimension of $ M $ when $ n \geq 6 $ ( except $ n = 8 $).



The upper bounds on rank $ 2 $ real matrices
A skew-symmetric matrix of rank 2 is of the form $(uv^T-vu^T)$ for some column vectors $u$ and $v$. Let's denote it $u\wedge v$. Also, if $u, v, w, z$ are linearly independent then $u\wedge v+w\wedge z$ has rank 4. It follows that a subspace of ${\mathfrak{so}}(n)$ consisting of matrices of rank $\leq 2$ is of the form $u\wedge V\_0$ where $V\_0$ is a subspace of ${\mathbb R}^n$, which we can assume lies in the orthogonal complement to ${\mathbb R}u$. In particular, the maximum dimension is $(n-1)$.


