How to prove that proj(proj(b onto a) onto a) = proj(b onto a)? How to prove that proj(proj(b onto a) onto a) = proj(b onto a)?
It makes perfect sense conceptually, but I keep going in circles when I try to prove it mathematically. Any help would be appreciated.
 A: If they are vectors in ${\mathbb R}^n$, you can do it analytically too. You have
$$proj_{\bf a}({\bf b}) = ({\bf b} \cdot {{\bf a} \over ||{\bf a}||}) {{\bf a} \over ||{\bf a}||}$$
So if ${\bf c}$ denotes $proj_{\bf a}({\bf b})$ then
$$proj_{\bf a}({\bf c}) = ({\bf c} \cdot {{\bf a} \over ||{\bf a}||}) {{\bf a} \over ||{\bf a}||}$$
$$= \bigg(({\bf b} \cdot {{\bf a} \over ||{\bf a}||}) {{\bf a} \over ||{\bf a}||}\cdot {{\bf a} \over ||{\bf a}||}\bigg) {{\bf a} \over ||{\bf a}||}$$
Factoring out constants this is
$$\bigg(({\bf b} \cdot {{\bf a} \over ||{\bf a}||^3}) ({\bf a} \cdot {\bf a})\bigg) {{\bf a} \over ||{\bf a}||}$$
$$= ({\bf b} \cdot {{\bf a} \over ||{\bf a}||^3}) ||{\bf a}||^2 {{\bf a} \over ||{\bf a}||}$$
$$ = ({\bf b} \cdot {{\bf a} \over ||{\bf a}||}) {{\bf a} \over ||{\bf a}||}$$
$$= proj_{\bf a}({\bf b}) $$
A: Note that a projection $P$ satisfies $P^2 = P$. You are just applying a projection twice, so that's the same thing as applying it once. 
A: Assuming that you're talking about the projection onto a subspace $F$ of a vector space $E$ parellel to a complementary $G$ of $F$, this follows by definition.
Indeed, this parallel projection $\pi : E \longrightarrow F$ is defined as follows: for every vector $u \in F$
$$
\pi (u) = v  \qquad \Longleftrightarrow \qquad v\in F \quad \text{and} \quad u - v \in G \ .
$$
So, who is going to be $ \pi(\pi (u)) $? Well, certainly $\pi (u) \in F$ already and $\pi (u) -\pi (u) = 0 \in G$ too. That is, $\pi (u)$ verifies the conditions of the definition of the projection $ \pi(\pi (u)) $
Hence $\pi(\pi (u)) = \pi (u)$.
