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

Question

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 $W$?

Answer 1

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 $$ \text{proj}_W(\bar{v})=\bar v $$