Show that for a projection matrix, $P$, $~~v \notin \text{range}~ P \implies P(P-I)v=0$.

Question

$P^2=P$.

I have a statement: if $v \notin \text{range}~ P \implies P(P-I)v=0$.

I understand why the $0$ happens.

I don’t understand how we arrive to $P(P-I)v$ once we know that $v \notin \text{range}~ P$. Range denotes column space of $P$.

Answer 1

$P^2 -P$ is a zero operator, and so $(P^2 -P)(v) =0$ for all $v \in V.$ So, it's irrelevant whether $v \in range(P)$ or not.

Answer 2

If $P^2=P,$ then for all(!) $v$ we have

$$P(Pv-v)=P^2v-Pv=Pv-Pv=0.$$