I wanna show that for a finite dimensional vector space $V$ (over a field $K$) and sub-spaces $U,W\subseteq V$ it is true that:
If $U^\perp \cap W^\perp = \{0\}$, then $U+W=V$.
Here's the definition of an orthogonal complement of a sub-space $U\subseteq V$ from the script I'm using: $$U^\perp := \{ \ell\in V^* \mid \ell(u)=0 \quad \forall u\in U\},$$ where $V^*:= \hom(V,K)$ denotes the dual space of $V$.
Can someone give me a hint on how to begin with the proof?
This can be proven more generally for closed subvector spaces $U$ and $W$ (we don't need finite dimensions).
Suppose there is a vector $x$ in $U+W$ (which is closed) that is not in $V$, then by the Hahn-Banach separation theorem, there is a linear functional $\ell\in V^\ast$ such that $\ell(U+V)=\{ 0\}$ and $\ell(x)\neq 0$, but then $\{0\}\subseteq \ell(U)\subseteq\ell(U+V)=\{0\}$ therefore $\ell(U)=\{0\}$ and similarly $\ell(V)=\{0\}$. This means that $\ell\in U^\perp\cap W^\perp$, and since $\ell(x)\neq 0$, we have $\ell\neq 0$.