Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$. Deduce that $\operatorname{det}\left(N^{\prime\top} A N^{\prime}\right) > 0$.
Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$. Deduce that $\operatorname{det}\left(N^{\prime\top} A N^{\prime}\right) > 0$.