Let $\alpha$ be an endomorphism of a finite-dimensional real vector space $U$ and let $\beta$ be another endomorphism of $U$ that commutes with $\alpha$. If $\lambda$ is an eigenvalue of $\alpha$, show that $\beta$ maps the kernel of $\alpha-\lambda \iota$ into itself, where $\iota$ is the identity map. Suppose now that $\alpha$ is diagonalizable with $n$ distinct real eigenvalues where $n=\operatorname{dim} U$. Prove that if there exists an endomorphism $\beta$ of $U$ such that $\alpha=\beta^{2}$, then $\lambda \geqslant 0$ for all eigenvalues $\lambda$ of $\alpha$.