Let $E$ be a linear subspace of ${\bigwedge}^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains an 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.

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.

Let $E$ be a linear subspace of $\wedge^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains an 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.

Let $E$ be a linear subspace of ${\bigwedge}^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains an 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.