Show that $||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$ for orthogoonal projection

Question

I'm working on some practice problems from Noble & Daniel's Applied Linear Algebra (3rd), specifically here looking for help with question 5 from section 5.8 on pg. 232.

Suppose that $P_0$ is the orthogonal projection onto $V_0$. Show that for every v in V,

$||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$

This whole section about orthogonal projections and bases has been a bit confusing for me. I'm not sure where to begin on this question. I know that $||v - P_0v||$ is orthogonal to each of the $v_i$ in v but and I thought that might be useful but I'm not sure how to leverage it.

Answer 1

It's a calculation. Assume a real vector space, and note that $v=P_0v+w$ where $P_0v\perp w$ so the Pythagorean theorem applies to show that

$\|v\|^2=\|P_0v+w\|^2=\|P_0v\|^2+\|w\|^2=\|P_0v\|^2+\|v-P_0v\|^2.$

Answer 2

In general, $||v+w||^2 = <v+w,v+w> = \ldots = ||v||^2 + ||w||^2 + 2<v,w>$. Now if $v \bot w$, you have that $<v,w>=0$, and you are done.