Let $E$ be a Euclidean space of dimension $N$. Let $u$ be a self-adjoint endomorphism of $E$ such that for all $x \in E$, $q_u(x) = (u(x) \mid x) \geq 0$. We assume that 0 is a simple eigenvalue of $u$ and we denote by $\lambda_2$ the smallest nonzero eigenvalue of $u$. We denote by $p : E \rightarrow E$ the orthogonal projection onto the vector line $\ker(u)$. Show that for all $x \in E$, $q_u(x - p(x)) \geq \lambda_2 \|x - p(x)\|^2$.
Let $E$ be a Euclidean space of dimension $N$. Let $u$ be a self-adjoint endomorphism of $E$ such that for all $x \in E$, $q_u(x) = (u(x) \mid x) \geq 0$. We assume that 0 is a simple eigenvalue of $u$ and we denote by $\lambda_2$ the smallest nonzero eigenvalue of $u$. We denote by $p : E \rightarrow E$ the orthogonal projection onto the vector line $\ker(u)$.\\
Show that for all $x \in E$, $q_u(x - p(x)) \geq \lambda_2 \|x - p(x)\|^2$.