Showing $P^2=P$ for $P(v)=\frac{\langle v,w\rangle }{||w||^2}w$

Question

My book asserts that for fixed $w$ where $w\neq 0$ that $P^2=P$ for $P(v)=\frac{\langle v,w\rangle }{||w||^2}w$

My book has a general corralary that $v\to P(v)$ is a bounded linear transformation and the fact that $P^2=P$ implies it is a projection. I'm not sure how they made the assertation. Any ideas?

Answer 1

We have

$$\require{cancel}P^2(v)=P(P(v))=P\left(\frac{\langle v,w\rangle }{||w||^2}w\right)=\frac{\langle v,w\rangle }{||w||^2}P\left(w\right)=\frac{\langle v,w\rangle }{||w||^2}\cancelto{=1}{\frac{\langle w,w\rangle }{||w||^2}}w=P(v)$$

Moreover, we have by the Cauchy-Schwarz inequality $$||P(v)||=\left|\frac{\langle v,w\rangle }{||w||^2}\right|\cdot||w||\le \frac{||v||||w||}{||w||^2}\cdot||w||=||v||$$ hence $P$ is bounded and $$||P||\le1$$ and if $v\in \operatorname{Im}(P)$ then $P(v)=v$ and then $||P(v)||=||v||$ so $$||P||\ge 1$$ and we conclude that $$||P||=1$$

Answer 2

Hint: $$P^2(v)=P(P(v))=\frac{\langle P(v),w\rangle}{\|w\|^2}w$$