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 $N = \left(\begin{array}{ll} N_{1} & N_{2} \end{array}\right) \in \mathcal{M}_{n,2}(\mathbb{R})$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. Show that there exists $P \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}\left(P^{\top} A^{-1} P\right) = 0$ if and only if there exists $P^{\prime} \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}\left(P^{\prime\top} A P^{\prime}\right) = 0$.
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 $N = \left(\begin{array}{ll} N_{1} & N_{2} \end{array}\right) \in \mathcal{M}_{n,2}(\mathbb{R})$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$.
Show that there exists $P \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}\left(P^{\top} A^{-1} P\right) = 0$ if and only if there exists $P^{\prime} \in \mathcal{G}_{n,2}(\mathbb{R})$ such that $\operatorname{det}\left(P^{\prime\top} A P^{\prime}\right) = 0$.