The question is this:
Let $A$ be an orthonormal subset of vectors in an infinite dimensional vector space $V$. Suppose for every $0\neq y \in V-A$ there is a $v \in A$ so that $\langle v,y \rangle \neq 0$. Show that $A$ is then a maximal orthonormal subset of $V$.
So in trying to prove this I have run into a number of "uncertainties":
- I suppose such a subset $A$ need not be finite? A maximal orthonormal subset need not be finite? Could it even be that $A$ is not countable?
- Let's say we have an orthonormal subset $A$, and we can find a normal vector $w \in V-A$ so that $A \cup \{w\}$ is linearly independent. Can we say \begin{equation} u=w-\sum_{v \in A}\langle v,w \rangle v \end{equation} is orthogonal to every vector in $A$ (in other words apply Gram-Schmidt to turn $A \cup {w}$ into an orthonormal subset). Does this depend at all on whether $A$ is finite/countable, or does Gram-Schmidt "work" regardless.
Any confirmation/counter-examples would be much appreciated...also any suggestions for another approach to proving this would be much appreciated. The question is (part of) a question from "The linear algebra a beginning graduate student ought to know" by Golan (the converse is straightforward).
If for every y in V -A there is at least one element v of A such that $\langle v,y \rangle \neq 0$ then then there is no vector y in V -A which is orthogonal to every vector in A.
Therefore A is maximal.