I hope this is not too elementary. 


Let $B: V\ \times V \to W$ be a skew-symmetric bilinear map where $V$, $W$ are 
finite dimensional real vector spaces. Assume that $B (u, v)$ is never zero 
for linearly independent $u$, $v\in V$. 



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> What can we say about the dimensions of $V$ and $W$?
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> 
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An elementary estimate is 



> 
> $$ \dim W \geq \dim V - 1 $$
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In fact, for $u\in V$, $u\neq0$, the linear map $B\_u : V\to W$, $B\_u(v)=B(u,v)$, 
is injective on a codimension $1$ subspace of $V$. Can the above estimate be 
improved? Recall that the image of a bilinear map does not need to be a subspace. 



Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
A better bound is $\dim W\geq 2\dim V-3$. This is obtained as follows. In the projective space $\mathbb{P}(\wedge^2V)$, the set of decomposable bivectors is the Grassmannian $\mathbb{G}(2,V)$, of dimension $2\dim V-4$. The bilinear map $B$ induces a linear map $b:\wedge^2V\rightarrow W$, and the hypothesis is $\mathbb{P}(\mathrm{Ker}\, b)\cap \mathbb{G}(2,V)=\varnothing$. This implies $2\dim V-4<\mathrm{codim}\, \mathrm{Ker}\, b=\dim \mathrm{Im}\, b$, hence the result.


