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.
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 ${n-2 \choose 2} + 1$.
One more exact answer: for $n=8$ the minimal dimension is $22$, again attaining the "easy" bound ${n-1 \choose 2} + 1$ and the same as the value for the next dimension $n=9$.
First to explain why ${n-1 \choose 2} + 1$ is enough, not just for real vector spaces but for an $n$-dimensional vector space $V$ over any field. Fix nonzero $X_0 \in V$. Then $X_0 \wedge V = \{ X_0 \wedge Y : Y \in V \}$ is a linear subspace of dimension $n-1$ in $\bigwedge^2 V$. Thus if $E \subset \bigwedge^2 V$ is a linear subspace of codimension $n-2$ then it must have nonzero intersection with $X_0 \wedge V$.
Now take $n=8$ and identify $V$ with the Cayley octonions. Let $V_0 \subset V$ consist of the "purely imaginary" octonions, so $V$ is the orthogonal direct sum of $\bf R$ with $V_0$. Write $\bigwedge^2 V = ({\bf R} \wedge V_0) \oplus \bigwedge^2 V_0$, and let $E$ be the kernel of the homomorphism $h: \bigwedge^2 V \to V_0$ that takes $1 \wedge X$ to $X$ and $X \wedge Y$ to the imaginary part of the octonion $XY$, for any $X,Y \in V_0$. [This is well-defined because $XY + Y\!X \in \bf R$ for all $X,Y \in V_0$, so $h(Y\wedge X) = - h(X \wedge Y)$.] Then $E$ has dimension $21$. I claim that $E$ contains no nonzero pure tensors. Indeed a pure tensor in $\bigwedge^2 V$ has the form $1 \wedge X$, $X \wedge Y$, or $(1+X) \wedge Y$ for some $X,Y \in V_0$ which are linearly independent (so in particular nonzero). Certainly $h(1\wedge X) = X$ is nonzero. So is $h(X \wedge Y)$, because if $XY \in \bf R$ for $X,Y \in V_0$ then $X$ and $Y$ are proportional. Finally $h\bigl((1+X) \wedge Y\bigr) = Y + h(X \wedge Y)$ cannot be zero because the imaginary part of $XY$ is orthogonal to $Y$.
It also follows that for $n=6,7$ the minimal $\dim E$ is at least ${n \choose 2} - 6 = 9, 15$, and thus exceeds the lower bound ${n-2 \choose 2} + 1 = 7, 11$ coming from the dimension of the Grassmannian.
Replacing the Cayley octonions by the Hamilton quaternions recovers the answer of $4$ for $n=4$.
