We complete, if necessary, $(e_1, e_2, \ldots, e_d)$ to a basis $(e_1, e_2, \ldots, e_n)$ of $E$. Let $\Phi$ be the $d$-th coordinate form which associates to any vector $x$ of $E$ its coordinate along $e_d$. Let $\Psi$ be the linear map from $E$ to $\mathbb{K}^d$ defined, for all $x \in E$, by $$\Psi(x) = \left(\Phi\left(f^i(x)\right)\right)_{0 \leqslant i \leqslant d-1} = \left(\Phi(x), \Phi(f(x)) \ldots, \Phi\left(f^{d-1}(x)\right)\right)$$ Show that $\Psi$ induces an isomorphism between $E_1$ and $\mathbb{K}^d$.
We complete, if necessary, $(e_1, e_2, \ldots, e_d)$ to a basis $(e_1, e_2, \ldots, e_n)$ of $E$. Let $\Phi$ be the $d$-th coordinate form which associates to any vector $x$ of $E$ its coordinate along $e_d$. Let $\Psi$ be the linear map from $E$ to $\mathbb{K}^d$ defined, for all $x \in E$, by
$$\Psi(x) = \left(\Phi\left(f^i(x)\right)\right)_{0 \leqslant i \leqslant d-1} = \left(\Phi(x), \Phi(f(x)) \ldots, \Phi\left(f^{d-1}(x)\right)\right)$$
Show that $\Psi$ induces an isomorphism between $E_1$ and $\mathbb{K}^d$.