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.

Martin Sleziak
  • 50,316
  • 18
  • 169
  • 342
  • 6,275
  • 5
  • 42
  • 75
  • The easy answer is, no. But I think you want a non-trivial inner product. – vesszabo Nov 29 '12 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. – vesszabo Nov 29 '12 at 21:47
  • 6
    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 '12 at 08:10
  • @vesszabo Sorry, I should have kept my mutterings to myself! I didn't really have anything interesting to say either way :) – rschwieb Nov 30 '12 at 14:06

3 Answers3


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.

Christian Blatter
  • 216,873
  • 13
  • 166
  • 425
  • Perhaps you could expand on what convergence w.r.t. (the metric generated by) your inner product means? – kahen Nov 29 '12 at 19:36
  • "symmetric bilinear form" should probably be "non-degenerate symmetric bilinear (or [sesquilinear](https://en.wikipedia.org/wiki/Sesquilinear_form)) form". – kahen Nov 29 '12 at 19:41
  • @kahen: See my edit. – Christian Blatter Nov 30 '12 at 09:16
  • 24
    @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](http://en.wikipedia.org/wiki/Topological_vector_space). 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](http://en.wikipedia.org/wiki/Metrizable). – Willie Wong Nov 30 '12 at 09:30
  • @WillieWong It's a valuable remark. Indeed, a vector space structure alone is (usually) not "too much" to work with it. – vesszabo Dec 01 '12 at 11:04
  • @WillieWong By "an inner product being compatible with a topology" do you mean that the metric induced by the inner product generates the starting topology? So if you turn any vector space with a topology that is not metrizable, that automatically means that no inner product can be compatible with it? – user56202 Sep 04 '21 at 20:25
  • @user56202 Q1: yes. Q2: inner product spaces are automatically metric spaces with $d(x,y) = \sqrt{\langle x-y,x-y\rangle}$. So yes. – Willie Wong Sep 06 '21 at 15:34
  • @WillieWong Thanks! – user56202 Sep 07 '21 at 23:14

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.

Ilmari Karonen
  • 24,602
  • 3
  • 64
  • 103

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.

Nate Eldredge
  • 90,018
  • 13
  • 119
  • 248