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> Let $E$ be a subset of all $4\times 4$ real skew symmetric matrices with the property that for any $A,B \in E,\ \operatorname{rank}(A-B)\leq 2$, then what can be said about the maximal dimension of $\newcommand{\span}{\operatorname{span}}\span E$?
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For less calculations, I considered $\Lambda^2(\mathbb R^4)$ [space of all skew symmetric bilinear forms] as the space of $4\times 4$ real skew symmetric matrices. What I have done so far is that "if $e,e+m\wedge n\in E$ with $\{m,n\}$ is linearly independent, then any element of $E$ should be of the form $e+m\wedge x+n\wedge y$ for some $x,y\in \mathbb R^4.$ **Also I am getting after so many computations that $\dim \span E\leq 4.$**" So I am trying to prove that "if $\{e,e+m\wedge n, e+m\wedge x+n\wedge y,e+m\wedge x'+n\wedge y',e+m\wedge x''+n\wedge y''\}=E$ where $\{e,e+m\wedge n, e+m\wedge x+n\wedge y,e+m\wedge x'+n\wedge y'\}:=E'(say)$ is linearly independent, then the remaining element $e+m\wedge x''+n\wedge y''$ should lie in $\span E'.$ In other words, we can also prove that $\dim \span\{m,n,x,y,x',y',x'',y''\}\leq 3.$ "


Basically I want to prove that maximal dimension of $E$ should be $4$.


Any help in this direction is welcome. Thank you.



Maximal dimension of linear span of a subset of all $4\times 4$ real skew symmetric matrices
Yes, such a subset $E$ must be contained in a $4$-dimensional subspace of
$W := \Lambda^2({\bf R}^4)$.


The key fact here is that the 6-dimensional space $W$
carries a symmetric bilinear pairing $(\cdot,\cdot) : W \times W \to {\bf R}$
such that $(\omega,\omega) = 0$ if and only if $\omega$ has rank at most $2$.
Indeed if we fix nonzero $\delta$ in the one-dimensional space
$\Lambda^4({\bf R}^4)$ then we can define $(\omega,\omega')$ by
$\omega \wedge \omega' = (\omega,\omega') \delta$.
Moreover, the pairing is nondegenerate
(though not definite $-$ indeed its signature is $(3,3)$).
Thus the maximal dimension of an isotropic subspace $U \subset W$
(that is, a subspace $U$ such that $(u,u') = 0$ for all $u,u' \in U$)
is at most $\frac12 \dim W = 3$.


Now suppose $E$ contains five linearly independent elements
$\omega\_0, \omega\_1, \omega\_2, \omega\_3, \omega\_4$,
and let $\psi\_i = \omega\_i - \omega\_0$ for $i=1,2,3,4$.
By hypothesis $(\psi\_i,\psi\_i) = 0$ for each $i$, and also
$(\psi\_i - \psi\_j, \psi\_i - \psi\_j) = 0$ for $i,j \in \{1,2,3,4\}$
because $\psi\_i - \psi\_j = \omega\_i - \omega\_j$
is also assumed to have rank at most $2$.
By linearity it follows that $(\psi\_i, \psi\_j) = 0$ for all
$i,j \in \{1,2,3,4\}$, and then that $(\psi,\psi') = 0$
for all $\psi,\psi'$ in the $4$-dimensional space generated by the $\psi\_i$.
We have thus found an isotropic subspace of dimension $4$ $-$ contradiction.
**QED**


(It is known, and not too hard to check,
that a maximal isotropic space is either
$\{x\_0 \wedge x : x \in {\bf R}^4\}$ for some nonzero
$x\_0 \in {\bf R}^4$, or $\{x \wedge y : x,y \in H\}$
for some hyperplane $H \subset {\bf R}^4$.)


