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 $H$. Deduce that $A$ is $H$-singular if and only if the matrix $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0 \end{pmatrix} \in \mathcal{M}_{n+1}(\mathbb{R})$ is singular.
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 $H$.
Deduce that $A$ is $H$-singular if and only if the matrix $A_{N} = \begin{pmatrix} A & N \\ N^{\top} & 0 \end{pmatrix} \in \mathcal{M}_{n+1}(\mathbb{R})$ is singular.