In this question, $u$ denotes an endomorphism of $\mathbb{R}^n$ that is self-adjoint positive definite. We propose to prove that there exists a unique endomorphism $v$ of $\mathbb{R}^n$ that is self-adjoint, positive definite, such that $v^2 = u$. Let $v$ be an endomorphism of $\mathbb{R}^n$, self-adjoint positive definite and satisfying $v^2 = u$, and let $\lambda$ be an eigenvalue of $u$. Show that $v$ induces an endomorphism of $\operatorname{Ker}(u - \lambda \mathrm{Id})$ which we shall determine.
In this question, $u$ denotes an endomorphism of $\mathbb{R}^n$ that is self-adjoint positive definite. We propose to prove that there exists a unique endomorphism $v$ of $\mathbb{R}^n$ that is self-adjoint, positive definite, such that $v^2 = u$.
Let $v$ be an endomorphism of $\mathbb{R}^n$, self-adjoint positive definite and satisfying $v^2 = u$, and let $\lambda$ be an eigenvalue of $u$. Show that $v$ induces an endomorphism of $\operatorname{Ker}(u - \lambda \mathrm{Id})$ which we shall determine.