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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. – Gerry Myerson Mar 7 '12 at 5:52
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and ensure that they write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Show") to be rude when asking for help; please consider rewriting your post. – Arturo Magidin Mar 7 '12 at 5:54

(Assuming $f$ is linear, of course...)

If $M=X$, then $M^{\perp}=\{\mathbf{0}\}$ and we are done.

Assume then that $M\neq X$. Let $x,y\in M^{\perp}$. If $x=\mathbf{0}$, then $1x+0y = \mathbf{0}$ and we are done. If $y=\mathbf{0}$, then $0x+1y=\mathbf{0}$ and we are done. So we may assume that $x\neq\mathbf{0}$ and $y\neq\mathbf{0}$.

Then $x\notin M$, and $y\notin M$. Hence $f(x)=\alpha\in\mathbb{R}$, $f(y)=\beta\in\mathbb{R}$, and $\alpha\beta\neq 0$.

What can you say about $\beta x-\alpha y$?

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