Quick question: Can I define some inner product on any arbitrary vector space such that it becomes an inner product space? If yes, how can I prove this? If no, what would be a counter example? Thanks a lot in advance.
Is there a vector space that cannot be an inner product space?
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0@vesszabo Sorry, I should have kept my mutterings to myself! I didn't really have anything interesting to say either way :) – 2012-11-30
3 Answers
How about vector spaces over finite fields? Finite fields don't have an ordered subfield, and thus one cannot meaningfully define a positive-definite inner product on vector spaces over them.
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5The ring of integers isn't a _field_. – 2012-11-29
I'm assuming the ground field is ${\mathbb R}$ or ${\mathbb C}$, because otherwise it's not clear what an "inner product space" is.
Now any vector space $X$ over ${\mathbb R}$ or ${\mathbb C}$ has a so-called Hamel basis. This is a family $(e_\iota)_{\iota\in I}$ of vectors $e_\iota\in X$ such that any $x\in X$ can be written uniquely in the form $x=\sum_{\iota\in I} \xi_\iota\ e_\iota$, where only finitely many $\xi_\iota$ are $\ne 0$. Unfortunately you need the axiom of choice to obtain such a basis, if $X$ is not finitely generated.
Defining $\langle x, y\rangle :=\sum_{\iota\in I} \xi_\iota\ \bar\eta_\iota$ gives a bilinear "scalar product" on $X$ such that $\langle x, x\rangle>0$ for any $x\ne0$. Note that in computing $\langle x,y\rangle$ no question of convergence arises.
It follows that $\langle\ ,\ \rangle$ is an inner product on $X$, and adopting the norm $\|x\|^2:=\langle x,x\rangle$ turns $X$ into a metric space in the usual way.
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0@WillieWong It's $a$ valua$b$le remark. Indeed, a vector space structure alone is (usually) not "too much" to work with it. – 2012-12-01