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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 $ax+by=0$.

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    I take it $f$ is assumed linear.2012-03-07

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