Show that if vector $\overline{v}$ is already at projection subspace $U$ that $\text{proj}_u(\overline{v})=\overline{v}$

inner-productslinear algebraprojection

Problem

Let's assume that $V$ is finite-dimensional inner product space and $U$ is its subspace. Also, assume that $v \in V$. Show that $\overline{v} \in U \iff \text{proj}_U(\overline{v})=\overline{v}$

Attempt to show

I think I have an intuitive understanding of whats going on. If we project $\overline{v}$ onto space $U$ and $\overline{v}$ is already in that space then it would be the same as not projecting anything (since its already "projected").

The projection itself can be written as

$$ \text{proj}_U(\overline{v}) = \frac{\langle \overline{v} , \overline{u}_1 \rangle}
{\langle \overline{u}_1 , \overline{u}_1 \rangle} \overline{u}_1 + \frac{\langle \overline{v} , \overline{u}_2 \rangle}
{\langle \overline{u}_2 , \overline{u}_2 \rangle} \overline{u}_2 + \dots + \frac{\langle \overline{v} , \overline{u}_n
\rangle}{\langle \overline{u}_n , \overline{u}_n \rangle} \overline{u}_n $$

$$ \iff \text{proj}_U(\overline{v}) = \text{proj}_{\overline{u}_1}(\overline{v})+\text{proj}_{\overline{u}_2}(\overline{v})+\dots +\text{proj}_{\overline{u}_n}(\overline{v}) $$

when $(\overline{u}_1,\dots \overline{u}_n)$ defines orthogonal basis in $U$. Each of these projects to the basis vectors. Sum of them results in the projection to space $U$. The result of this projection should be $\overline{v}$ iff the vector $v$ is already at space $U$.

The only problem is that I cannot see the connection here proof wise. I think I have the intuition right (if not please correct me). You could easily compute examples of this but these don't actually show anything but that it works with certain spaces and certain vectors.

Maybe someone could hint me in the right direction or provide something? I've been trying to figure this out for quite a while now with little to no progression.

Best Answer

Hint: if $v\in U$ and $\{u_1,\dots, u_n\}$ is an orthogonal basis of $U$, what are the coordinates of $v$ w.r.t. that basis?

Let $v\in U$ and $\{u_1,\dots, u_n\}$ an orthogonal basis of $U$. I'm going to show that $$ v = \frac{\langle v , u_1 \rangle}{\langle u_1 , u_1 \rangle} u_1 + \dots + \frac{\langle v , u_n \rangle}{\langle u_n ,u_n \rangle} u_n. $$ To see it, just note that if $v=c_1u_1+\cdots+c_nu_n$, then $$\langle v,u_i\rangle=\langle \sum_i c_iu_i, u_i\rangle = c_i\langle u_i,u_i\rangle.$$

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