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 $(x \mid y)$. 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 $f$ be a nilpotent endomorphism of $E$. Deduce that $f$ is orthocyclic if and only if $$f \text{ has rank } n-1 \quad \text{and} \quad \forall x, y \in (\ker f)^{\perp}, \quad (f(x) \mid f(y)) = (x \mid y).$$
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 $(x \mid y)$. 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 $f$ be a nilpotent endomorphism of $E$. Deduce that $f$ is orthocyclic if and only if
$$f \text{ has rank } n-1 \quad \text{and} \quad \forall x, y \in (\ker f)^{\perp}, \quad (f(x) \mid f(y)) = (x \mid y).$$