I came across this result, and was trying to understand why it is true: Suppose $V$ is a finite-dimensional vector space and $U \subset V$, and has a bilinear form $b: V \times V \xrightarrow{} k$, then we define the orthogonal complement $U^\perp$ as:
$$ U^\perp = \{v \in V | b(v, w) = 0 \forall w \in U\} $$
As such, we have $dim(U) \leq dim(U^\perp)$.
Is there any reason as to how to prove this result? The only result to do with dimensions of $U^\perp$ that I know of is that $dim(U) + dim(U^{\perp}) = dim(V)$, but it doesn't seem to help in any way.