Show that $V = U^\perp \bigoplus U$

Question

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 $V = U^\perp + U$?

Answer 1

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.

Answer 2

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.