Let $V$ be finite dimensional v.s. and $0 \ne T\in \mathscr L(V)$ , then $\exists$ $S \in \mathscr L(V)$ such that $0 \ne T \circ S$ is idempotent If $V$ is a finite dimensional vector space and $T \ne0$ is a linear operator on $V$ , then how may we prove that there is a linear operator $S$ on $V$ such that $T\circ S$ is non-zero and idempotent?

Hint: Suppose $(c,v)$ is an eigenpair of $T$. Consider $T \circ P_v$, where $P_v$ is the orthogonal projector onto the span of $v$. When $c=1$ or $c=0$, I claim this is idempotent. How do you fix it otherwise?
