We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$. Show that if $X \in \mathcal{M}_{p,1}(\mathbb{R})$ is non-zero then $X^{\top}N'^{\top}AN'X > 0$.
We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Show that if $X \in \mathcal{M}_{p,1}(\mathbb{R})$ is non-zero then $X^{\top}N'^{\top}AN'X > 0$.