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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.

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    Could you explain your notation? It's not familiar to me.2011-10-04
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    (Please post the question in the body as well, not just in the title.) How do you define `proj`? I guess one way would be to define projection of $b$ onto $a$ as $\langle b,a \rangle a$, assuming $a$ is a unit vector. Assuming this definition, it seems quite straightforward.2011-10-04
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    If it makes sense conceptually (by this I imagine you mean geometrically), then you have proved it, as long as the formal version of projection does what it claims to do. For a mechanical verification, let $a_u$ be the unit vector in the direction of $a$. Then the right-hand side is $(b\cdot a_u)a_u$. Straightforward computation shows the left-hand side is the same.2011-10-04

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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 $$\bigg(({\bf b} \cdot {{\bf a} \over ||{\bf a}||^3}) ({\bf a} \cdot {\bf a})\bigg) {{\bf a} \over ||{\bf a}||}$$ $$= ({\bf b} \cdot {{\bf a} \over ||{\bf a}||^3}) ||{\bf a}||^2 {{\bf a} \over ||{\bf a}||}$$ $$ = ({\bf b} \cdot {{\bf a} \over ||{\bf a}||}) {{\bf a} \over ||{\bf a}||}$$ $$= proj_{\bf a}({\bf b}) $$

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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.

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    Furthermore, a linear operator acting on a vector space satisfying the condition $P^2 = P$ is how a projection is (usually) defined.2011-10-04
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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 $\pi(\pi (u)) = \pi (u)$.