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 $f \in \mathrm{O}(E)$. Deduce that $f$ is orthocyclic if and only if $\chi_f = X^n - 1$ or $\chi_f = X^n + 1$.
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 $f \in \mathrm{O}(E)$. Deduce that $f$ is orthocyclic if and only if $\chi_f = X^n - 1$ or $\chi_f = X^n + 1$.