Let $X$ be a Hilbert space and let $f:X\to\mathbb{R}$.
Let $M=\{x\in X:f(x)=0\}$ be the nullspace of $f$.
Let $M^\perp=\{x\in X:(x,y)=0\text{ for all }y\in M\}$ be the orthogonal complement of $M$.
Show that $M^\perp$ is at most one-dimensional, i.e. that if $x$ and $y$ are members of $M^\perp$ then there exist scalars $a$ and $b$, not both zero, such that $ax+by=0$.