Show that $V = U^\perp \bigoplus U$ If $(V,\langle , \rangle)$ is a Euclidean vector space, $U \subseteq V$ is a subspace of V and $U^\perp := \{v \in V | \langle v,u \rangle = 0, \forall u \in U\}$. Show $V = U^\perp  \bigoplus U$
In the first part I proved that $U^\perp$ is a linear subspace of V already.
I need to prove that $V = U^\perp + U$ and $U^\perp \cap U = \{0\}$.
For $U^\perp \cap U = \{0\}$ I have: $0 \in U^\perp \cap U$ since $U,U^\perp$ are subspaces of V. Also if $u \in U^\perp \cap U \Rightarrow \langle u,u \rangle = 0$, but by the definition of the inner product it follows that $u=0$.
But how can I prove that $V = U^\perp + U$?

Hint: Consider $(u_1, \ldots, u_k)$ an orthogonal basis of $U$, and consider the function
$$P:V \to V:v \mapsto \sum_{i=1}^k \frac{\langle v,u_i\rangle}{\|u_i\|_2^2}u_i$$
Then you should show that $P(v) \in U$ and $v-P(v) \in U^{\perp}$ for every $v \in V$. So you will be able to write $v = P(v)+(v-P(v))$. Note that $P$ is a projection.
