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 $V = U^\perp + U$?

The essential steps should be the following:
You can take the orthogonal projection $P:V \rightarrow U$, that projects each point $v\in V$ onto the closest point in $u\in U$. (This can be done either by this definition or more easy if you take a basis of $U$ and just project each point $v$ on this basis)
If you have that, then you can show that $P$ is linear, $Pv \in U$, $v-Pv\in U^\bot$ and obviously $v=Pv + (v-Pv)$ which is all you need.
