Let $E$ be a linear subspace of ${\bigwedge}^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains a nonzero element of the form $X\wedge Y$, with $X, Y\in{\mathbb R}^n$?


When $n=3$, dimension $1$ is enough. When $n=4$ we would need dimension $4$. For general $n$, it is easy to see $E$ having dimension $\frac{(n-1)(n-2)}{2}+1$ is sufficient, but I don't know if that is optimal.



A question about subspace in ${\bigwedge}^2({\mathbb R}^n)$
Partial answer: the minimal dimension is at least
${n-2 \choose 2} + 1$, with equality if $n-1$ is a power of $2$.
For example, if $n=5$ the minimum is $4$, curiously the same as for $n=4$,
and less than the "easy" bound of ${5-1 \choose 2} + 1 = 7$.


Let $N = {n \choose 2}$, which is the dimension of the alternating square of
an $n$-dimensional vector space $V$. Then the pure tensors $X \wedge Y$
constitute a homogeneous subset of dimension $2n-3$; projectively this is
the Plücker embedding in $(N-1)$-dimensional projective space
of the Grassmanian ${\rm Gr}(2,n)$ of $2$-planes in $V$,
which has dimension $2n-4$.
Thus a general linear space of codimension less than $2n-4$
will miss ${\rm Gr}(2,n)$ for lack of sufficient degrees of freedom.
This gives the lower bound ${n-2 \choose 2} + 1 = N-(2n-4)$.


Over an algebraically closed field this necessary condition is also sufficient,
and the general linear subspace of codimension $2n-4$ meets the Plücker variety
in $d\_n$ points counted with multiplicity, where $d\_n$ is the degree of
the Plücker variety. It is known that $d\_n$ is the Catalan number
$C\_{n-2} = \frac1{n-1}{2n-4 \choose n-2}$.


The field of real numbers is not algebraically closed,
but every polynomial of odd degree has a root.
Thus if $d\_n$ is odd we are still guaranteed a real intersection.
This happens when $n-1$ is a power of $2$, i.e. $n=3,5,9,17,\ldots$.
We've now proven that the bound ${n-2 \choose 2} + 1$ is attained for such $n$.


For $n=4$ it is well-known that the real Grassmannian ${\rm Gr}(2,4)$
is a quadric of signature $(3,3)$, so as **Yuval** found it takes
a subspace of dimension at least $4$ to guarantee a real intersection.
For $n \geq 6$ that are not of the form $2^m + 1$, I do not know
by how much the real answer exceeds the lower bound ${n-2 \choose 2} + 1$.


