Let $e = (e_1, \ldots, e_d)$ be an orthonormal basis of $E$, $p$ an integer such that $1 \leqslant p \leqslant d$ and $\mathcal{I}_p = \{\alpha = (i_1, \ldots, i_p) \in \mathbb{N}^p \mid 1 \leqslant i_1 < \cdots < i_p \leqslant d\}$. For all $\alpha = (i_1, \ldots, i_p) \in \mathcal{I}_p$, we denote $e_{\alpha} = (e_{i_1}, \ldots, e_{i_p}) \in E^p$ and for all $\omega$ and $\omega^{\prime}$ elements of $\mathscr{A}_p(E, \mathbb{R})$ $$\langle\omega, \omega^{\prime}\rangle = \sum_{\alpha \in \mathcal{I}_p} \omega(e_{\alpha}) \omega^{\prime}(e_{\alpha}).$$ Show that the inner product $(\omega, \omega^{\prime}) \mapsto \langle\omega, \omega^{\prime}\rangle$ defined above depends only on the inner product on $E$ and not on the choice of the orthonormal basis $e$.
Let $e = (e_1, \ldots, e_d)$ be an orthonormal basis of $E$, $p$ an integer such that $1 \leqslant p \leqslant d$ and $\mathcal{I}_p = \{\alpha = (i_1, \ldots, i_p) \in \mathbb{N}^p \mid 1 \leqslant i_1 < \cdots < i_p \leqslant d\}$. For all $\alpha = (i_1, \ldots, i_p) \in \mathcal{I}_p$, we denote $e_{\alpha} = (e_{i_1}, \ldots, e_{i_p}) \in E^p$ and for all $\omega$ and $\omega^{\prime}$ elements of $\mathscr{A}_p(E, \mathbb{R})$
$$\langle\omega, \omega^{\prime}\rangle = \sum_{\alpha \in \mathcal{I}_p} \omega(e_{\alpha}) \omega^{\prime}(e_{\alpha}).$$
Show that the inner product $(\omega, \omega^{\prime}) \mapsto \langle\omega, \omega^{\prime}\rangle$ defined above depends only on the inner product on $E$ and not on the choice of the orthonormal basis $e$.