$U$ is timelike if its orthogonal complement $U^\perp$ is spacelike Consider the bilinear form $\left<x,y\right>_{n,1} = \sum_{j=1}^n x_j y_j - x_{n+1} y_{n+1}$ on $\mathbb{R}^{n+1}$.
A vector $x \in \mathbb{R}^{n+1}$ is said to be timelike if $\left<x,x\right>_{n,1} > 0$ while $x$ is called spacelike if $\left<x,x\right>_{n,1} < 0$.
Furthermore, a subspace $U \subset \mathbb{R}^{n+1}$ is timelike if there exists a timelike vector $u \in U$ and $U$ is called spacelike if every non-zero vector in $U$ is spacelike.
The orthogonal complement is defined as
$U^\perp = \{ v \in \mathbb{R}^{n+1} | \left<v,u\right>_{n,1} = 0 \: \forall u \in U \}$
Now I would like to show that if $U^\perp$ is spacelike then $U$ is timelike.
I already know that the converse holds and one has to use the fact that $(U^\perp)^\perp=U$. But I seem to not be able to prove the statement above, even with this hint.
Could someone help me out? Thanks!

Suppose that $U^{\perp}$ is spacelike. Because $U^{\perp}$ nondegenerate, then the whole space $V = U \oplus U^{\perp}$. Suppose the contrary that $U$ is not timelike (i.e there is no timelike vector in $U$). Then for any $u \in U$, $g(u,u)  \leq 0$. So for any $v \in V$ can be expressed as $v = u + w$ where $u \in U$ and $w \in U^{\perp}$. With this
\begin{align}
g(v,v) &= g(u+w,u+w) \\ &= g(u,u) + g(u,w) + g(w,u) + g(w,w) \\
&=g(u,u) + g(w,w)  \leq 0
\end{align}
which tells us that there is no timelike vector at all in $V$. 
The key here is that by realize that a subspace $W\subset V$ is nondegenerate iff the direct sum of $W$ and $W^{\perp}$ is equal to $V$. You can see the proof in O'neill's Semi-Riemannian Geometry p.59.
