We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix in $\mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$.
Show that a matrix in $\mathcal{M}_{n}(\mathbb{R})$ is singular if and only if it is $E_{n}$-singular.