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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
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1 Answer 1

(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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