# Prove: if $W$ is subspace of $V$ and vector $v \in W$, then orthogonal projection of vector $v$ onto $W$ is $v$ itself

Let's assume that $$\bar v \in V$$. Let's also assume that $$\bar v \in W$$, where $$W$$ is a subspace of $$V$$.

How to prove that then orthogonal projection $$\operatorname{proj}_{W} (\bar v) = {\left\langle \bar v, \bar w_1 \right\rangle \over \left\langle \bar w_1, \bar w_1 \right\rangle} \bar w_1 + {\left\langle \bar v, \bar w_2 \right\rangle \over \left\langle \bar w_2, \bar w_2 \right\rangle} \bar w_2 + ... + {\left\langle \bar v, \bar w_k \right\rangle \over \left\langle \bar w_k, \bar w_k \right\rangle} \bar w_k = \bar v$$?

I understand that $$\operatorname{proj}_{\bar v} (\bar v)$$ = $${\left\langle \bar v, \bar v \right\rangle \over \left\langle \bar v, \bar v \right\rangle} \bar v = \bar v$$ , but how can I prove that this applies also as an orthogonal projection towards the subpace $$W$$?

• How is $\mathrm{proj}_W$ defined? – Berci Aug 17 at 18:30

Since $$W \oplus W^{\perp} = V$$ and $$\text{proj}_W(\cdot)$$ is the orthogonal projection on $$W$$, it's clear that $$\forall w \in W, \text{proj}_W(w) = w$$ We have $$\bar{v} \in W$$, so $$\text{proj}_W(\bar{v})=\bar v$$