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Given a linearly independent set of vectors in some vector space, is it always possible to construct an inner product so that the vectors are orthogonal?

I know I can construct an appropriate inner product for finite dimensional vector spaces by simply inverting the matrix consisting of the vectors in the set and their orthogonal compliment. But what about for infinite dimensional vector spaces?

Alex Pavellas
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  • A very interesting problem. I'd also recommend adding a functional-analysis tag, as this is infinite dimensional. – Marcus M Jul 19 '15 at 18:54
  • Without further restrictions [assuming we're speaking of vector spaces over $\mathbb{R}$ or $\mathbb{C}$, possibly $\mathbb{Q}$], yes, it's possible. Just extend to a basis and define the inner product so that the basis vectors are an orthonormal set. – Daniel Fischer Jul 19 '15 at 19:00
  • Usually not for if you could you would have a (pre)Hilbert space. But for Banach spaces like ${L^p}(R^n)$ with $p$ not equal to 2 there is no such inner product. – Urgje Jul 19 '15 at 19:02
  • This appears to be related. http://math.stackexchange.com/questions/632690/orthogonal-basis-for-infinite-dimensional-vector-spaces – ZenoCosini Jul 19 '15 at 20:32
  • Such linearly independent set might be uncountable. However, the induced inner product space might be separable ... I see no canonical way. You probably need some AC argument anyway. – user251257 Jul 19 '15 at 21:23
  • Another point worth making: in order for the inner product to be continuous with respect to some existing norm, we would need the vectors to have a bounded length. – Ben Grossmann Jul 20 '15 at 16:35

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