# Is there a vector space that cannot be an inner product space?

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.

• The easy answer is, no. But I think you want a non-trivial inner product. Nov 29, 2012 at 17:49
• @rschwieb I meant $\langle a,b\rangle =0$ for every $a,b$ is trivial. (My above comment is a joke only :-) ) So yes, what is interesting the question of the existence of positive definite inner product. As I see, the answer of Christian is complete. Nov 29, 2012 at 21:47
• I'm sorry, in all lectures I have attended so far, the inner product is positive by definition, which is why I did not specify it.
– Huy
Nov 30, 2012 at 8:10
• @vesszabo Sorry, I should have kept my mutterings to myself! I didn't really have anything interesting to say either way :) Nov 30, 2012 at 14:06

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.

• Perhaps you could expand on what convergence w.r.t. (the metric generated by) your inner product means? Nov 29, 2012 at 19:36
• "symmetric bilinear form" should probably be "non-degenerate symmetric bilinear (or sesquilinear) form". Nov 29, 2012 at 19:41
• @kahen: See my edit. Nov 30, 2012 at 9:16
• @vesszabo: As Christian explain, given the axiom of choice, any arbitrary vector space can admit an inner product. In some sense just having a vector space is too floppy. It is more interesting to ask your question for a more rigid class of spaces, the topological vector spaces. In which case the answer is no: you cannot always find a inner product compatible with the given topology. There exists TVSs which are not normable or metrizable. Nov 30, 2012 at 9:30
• @WillieWong It's a valuable remark. Indeed, a vector space structure alone is (usually) not "too much" to work with it. Dec 1, 2012 at 11:04

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.

• I'm not sure, but can one not just use the well-ordering theorem?
– Huy
Nov 29, 2012 at 18:45
• You can't get an ordering compatible with the ring structure. Nov 29, 2012 at 18:56
• @ZhenLin How about the ring of integers? Nov 29, 2012 at 23:04
• The ring of integers isn't a field. Nov 29, 2012 at 23:22

Christian Blatter's answer shows that, assuming the axiom of choice, every vector space can be equipped with an inner product.

Without the axiom of choice, this can fail. As I show in Inner product on $C(\mathbb R)$, it is consistent with ZF+DC that the vector space $$C(\mathbb{R})$$ of continuous functions on $$\mathbb{R}$$ does not admit any inner product, nor even any norm.

The idea is that $$C(\mathbb{R})$$ already has a "nice" topology which is not compatible with an inner product, and under appropriate axioms consistent with ZF+DC, there are "automatic continuity" results saying that it cannot have any other "nice" topology.