I'm studying linear algebra using Axler's text and am stuck on 6.17.
The problem is: Prove that if $P \in \mathcal{L}(V)$ is such that $P^2 = P$ and every vector in $\operatorname{null} P$ is orthogonal to every vector in
$\operatorname{range} P$, then $P$ is an orthogonal projection.
My proof so far: Suppose $P \in \mathcal{L}(V)$ is such that $P^2 = P$ and every vector in $\operatorname{null} P$ is orthogonal to every vector in
$\operatorname{range} P$. Let $v \in P$. We have a decomposition $V = \operatorname{range} T \oplus \operatorname{null} T
= \operatorname{range} T \oplus (\operatorname{range} T)^\perp$. Then each vector $v \in V$ can be uniquely represented in the form
$v = u + w$ where $u \in \operatorname{null} P$ and $w \in \operatorname{range} P$.
We have $Pv=P(u + w)=Pu+Pw=Pw=P^2w$.
I would appreciate some hints on how to finish my proof. I have the idea that from here, we can derive that $P(u+w)=w$ which would complete the proof (please correct me if I'm wrong).
To begin with, a hint on how $P^2 = P$ is useful would be very helpful.
Best Answer
Hint:
You know that $w$ is in the range of $P$, and so $w = Px$ for some $x \in V$.
Try using this and the fact that $P^2 = P$.