In this part, we assume that $\mathbb{K} = \mathbb{R}$ and that $E$ is a Euclidean space. The inner product of two vectors $x, y$ of $E$ is denoted $(x \mid y)$ and we denote by $\mathrm{O}(E)$ the group of vector isometries of $E$. We say that an endomorphism $f$ of $E$ is orthocyclic if there exists an orthonormal basis of $E$ in which the matrix of $f$ is of the form $C_Q$ (companion matrix). Let $f \in \mathrm{O}(E)$. Let $f' \in \mathrm{O}(E)$ having the same characteristic polynomial as $f$. Show that there exist orthonormal bases $\mathcal{B}$ and $\mathcal{B}'$ of $E$ for which the matrix of $f$ in $\mathcal{B}$ is equal to the matrix of $f'$ in $\mathcal{B}'$.
In this part, we assume that $\mathbb{K} = \mathbb{R}$ and that $E$ is a Euclidean space. The inner product of two vectors $x, y$ of $E$ is denoted $(x \mid y)$ and we denote by $\mathrm{O}(E)$ the group of vector isometries of $E$. We say that an endomorphism $f$ of $E$ is orthocyclic if there exists an orthonormal basis of $E$ in which the matrix of $f$ is of the form $C_Q$ (companion matrix).
Let $f \in \mathrm{O}(E)$. Let $f' \in \mathrm{O}(E)$ having the same characteristic polynomial as $f$. Show that there exist orthonormal bases $\mathcal{B}$ and $\mathcal{B}'$ of $E$ for which the matrix of $f$ in $\mathcal{B}$ is equal to the matrix of $f'$ in $\mathcal{B}'$.