Orthogonal vectorspace properties I got V as an euclidean space, U $\subset$ V as subspace and v $\in$ V$\setminus$U.
I want to show that there exists exactly one u $\in$ U so that (v-u) $\in$ U$^{\bot}$.
U$^{\bot}$ is the vector space which includes every vector x that has the property $\langle$x,u$\rangle$= $0$ , with u $\in$ U.
In my case there is $\langle$(v-u),u$\rangle$= $0$. That's the same as $\langle$v,u$\rangle$$-$$\langle$u,u$\rangle$= $0$ and $\langle$v,u$\rangle$=$\langle$u,u$\rangle$
But v can't be u and u can't be v, because of v $\in$ V$\setminus$U and u $\in$ U.
How can I find the "fitting" u for each v?
Of course u = $0$$_v$ is not the answer I am searching.
 A: Let $P$ be the orthogonal projection onto $U$. The desired $u \in U$ is equal to $Pv$.
Indeed $v = Pv + (v - Pv)$ and $v - Pv \perp U$:
$$\langle v - Pv, x\rangle = \langle v, x\rangle - \langle Pv, x\rangle = \langle v, x\rangle - \langle v, Px\rangle = \langle v, x\rangle - \langle v, x\rangle = 0, \forall x \in U$$
To show uniqueness, assume that $v = a + a' = b + b'$ where $a,b \in U$, and  $a', b' \in U^\perp$.
Then we have
$$\underbrace{a - b}_{\in U} = \underbrace{b' - a'}_{\in U^\perp}$$
so $a - b = 0$ and $b' - a'=  0$.

Without assuming that we know that $P^T = P$, we can proceed like this: let $\{e_1, \ldots, e_k\}$ be an orthonormal basis for $U$ and let $\{e_1, \ldots, e_n\}$ be its extension to an orthonormal basis for $V$. Then the projection $P$ is given by $$Px = \sum_{i=1}^k \langle x, e_i\rangle e_i$$
Now it is clear that $$v - Pv = \sum_{i=1}^n \langle v, e_i\rangle e_i 
- \sum_{i=1}^k \langle v, e_i\rangle e_i = \sum_{i=k+1}^n\langle v, e_{i}\rangle e_i \in U^\perp$$
because $e_j \perp U$ for all $k+1 \le j \le n$.
