Let $n$ be the index of $u$, that is, the integer such that $u^{n-1} \neq 0$ and $u^n = 0$. Prove that there exists a vector $v$ such that $v$ is an eigenvector of $h$ and $u^{n-1}(v) \neq 0$.
Let $n$ be the index of $u$, that is, the integer such that $u^{n-1} \neq 0$ and $u^n = 0$. Prove that there exists a vector $v$ such that $v$ is an eigenvector of $h$ and $u^{n-1}(v) \neq 0$.