For all $e \in E^p$, we consider $\Omega_p(e) : E^p \rightarrow \mathbb{R}$ defined for all $u \in E^p$ by $$\Omega_p(e)(u) = \operatorname{det}(\operatorname{Gram}(e, u)).$$ (a) Let $M \in \mathcal{M}_p(\mathbb{R})$, $e = (e_1, \ldots, e_p)$ and $e^{\prime} = (e_1^{\prime}, \ldots, e_p^{\prime})$ in $E^p$ satisfying $e_i^{\prime} = \sum_{j=1}^p M_{ij} e_j$ for all $i \in \llbracket 1, p \rrbracket$. Show that $\Omega_p(e^{\prime}) = \operatorname{det}(M) \Omega_p(e)$. (b) Let $e \in E^p$. Show that $\Omega_p(e) \neq 0$ if and only if $e$ is a free family. (c) Verify that $\Omega_p(e)(e) \geqslant 0$ for all families $e \in E^p$.
For all $e \in E^p$, we consider $\Omega_p(e) : E^p \rightarrow \mathbb{R}$ defined for all $u \in E^p$ by
$$\Omega_p(e)(u) = \operatorname{det}(\operatorname{Gram}(e, u)).$$
(a) Let $M \in \mathcal{M}_p(\mathbb{R})$, $e = (e_1, \ldots, e_p)$ and $e^{\prime} = (e_1^{\prime}, \ldots, e_p^{\prime})$ in $E^p$ satisfying $e_i^{\prime} = \sum_{j=1}^p M_{ij} e_j$ for all $i \in \llbracket 1, p \rrbracket$. Show that $\Omega_p(e^{\prime}) = \operatorname{det}(M) \Omega_p(e)$.
(b) Let $e \in E^p$. Show that $\Omega_p(e) \neq 0$ if and only if $e$ is a free family.
(c) Verify that $\Omega_p(e)(e) \geqslant 0$ for all families $e \in E^p$.