We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$. Show that $A$ is $H$-singular if and only if there exist a non-zero vector $X$ in $H$ and a real number $\lambda$ such that $AX = \lambda N$.
We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. We assume $n \geqslant 2$. Let $F = H$ be a hyperplane of $E_{n}$ and let $N \in E_{n}$ be a unit vector normal to $H$.
Show that $A$ is $H$-singular if and only if there exist a non-zero vector $X$ in $H$ and a real number $\lambda$ such that $AX = \lambda N$.