31
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The easy answer is, no. But I think you want a non-trivial inner product. $\endgroup$ – vesszabo Nov 29 '12 at 17:49
  • $\begingroup$ @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. $\endgroup$ – vesszabo Nov 29 '12 at 21:47
  • 3
    $\begingroup$ 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. $\endgroup$ – Huy Nov 30 '12 at 8:10
  • $\begingroup$ @vesszabo Sorry, I should have kept my mutterings to myself! I didn't really have anything interesting to say either way :) $\endgroup$ – rschwieb Nov 30 '12 at 14:06
21
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm not sure, but can one not just use the well-ordering theorem? $\endgroup$ – Huy Nov 29 '12 at 18:45
  • 9
    $\begingroup$ You can't get an ordering compatible with the ring structure. $\endgroup$ – Zhen Lin Nov 29 '12 at 18:56
  • $\begingroup$ @ZhenLin How about the ring of integers? $\endgroup$ – chaohuang Nov 29 '12 at 23:04
  • 6
    $\begingroup$ The ring of integers isn't a field. $\endgroup$ – Zhen Lin Nov 29 '12 at 23:22
35
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Perhaps you could expand on what convergence w.r.t. (the metric generated by) your inner product means? $\endgroup$ – kahen Nov 29 '12 at 19:36
  • $\begingroup$ "symmetric bilinear form" should probably be "non-degenerate symmetric bilinear (or sesquilinear) form". $\endgroup$ – kahen Nov 29 '12 at 19:41
  • $\begingroup$ @kahen: See my edit. $\endgroup$ – Christian Blatter Nov 30 '12 at 9:16
  • 17
    $\begingroup$ @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. $\endgroup$ – Willie Wong Nov 30 '12 at 9:30
  • $\begingroup$ @WillieWong It's a valuable remark. Indeed, a vector space structure alone is (usually) not "too much" to work with it. $\endgroup$ – vesszabo Dec 1 '12 at 11:04
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.