We assume $n \geq 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_1, N_2)$ a basis of $F^{\perp}$ and we set $N = \left(\begin{array}{ll} N_{1} & N_{2} \end{array}\right) \in \mathcal{M}_{n,2}(\mathbb{R})$. Deduce that $A$ is $F$-singular if and only if the matrix $$A_{N} = \left(\begin{array}{ccc} A & N_{1} & N_{2} \\ N_{1}^{\top} & 0 & 0 \\ N_{2}^{\top} & 0 & 0 \end{array}\right) = \left(\begin{array}{cc} A & N \\ N^{\top} & 0_{2} \end{array}\right) \in \mathcal{M}_{n+2}(\mathbb{R})$$ is singular.
We assume $n \geq 3$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-2$. We consider $(N_1, N_2)$ a basis of $F^{\perp}$ and we set $N = \left(\begin{array}{ll} N_{1} & N_{2} \end{array}\right) \in \mathcal{M}_{n,2}(\mathbb{R})$.
Deduce that $A$ is $F$-singular if and only if the matrix
$$A_{N} = \left(\begin{array}{ccc} A & N_{1} & N_{2} \\ N_{1}^{\top} & 0 & 0 \\ N_{2}^{\top} & 0 & 0 \end{array}\right) = \left(\begin{array}{cc} A & N \\ N^{\top} & 0_{2} \end{array}\right) \in \mathcal{M}_{n+2}(\mathbb{R})$$
is singular.