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Every finite dimensional vector space can made into an inner product space with the same dimension.

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What's your definition of an inner product space? –  azimut Apr 6 '13 at 19:58
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Over a field $F$, up to vector space isomorphism, there is only one $n$-dimensional space. –  André Nicolas Apr 6 '13 at 20:05
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Strictly equivalent to @AndréNicolas's comment: fix a basis of the vector space and use the canonical inner product on the coordinates. –  Did Apr 6 '13 at 21:04

2 Answers 2

The definitions of inner product space that I have seen always require that vectors have a nonnegative inner product with themselves. Since inner products live in the base field, this requirement can only be meaningful if that field is of characteristic $0$. So your statement would be false over fields of prime characteristic. Also over fields larger than $\Bbb C$, like $\Bbb C(X)$, I believe it would be hard to arrange that the inner products of vectors with themselves lie in an ordered subfield like $\Bbb R$.

If (as is usual) you consider inner product spaces only over the fields $k=\Bbb R$ or $k=\Bbb C$, then indeed every finite dimensional vector space can be made into an inner product space, by transport via a vector space isomorphism with $k^n$ of the standard inner product on the latter.

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@StefanSmith: The rational functions (formal quotients of polynomials) with complex coefficients. It is the field of fractions of $\Bbb C[X]$. And about the subfield of $\Bbb C$, basically yes, but if you subfield is not stable under complex conjugation, you might be in trouble still. –  Marc van Leeuwen Apr 6 '13 at 22:02
    
Thanks again. Two more questions, if you don't mind: what do you mean by "stable"? and is it possible for $\mathbf{K}$ to be an ordered field that is not a subfield of $\mathbf{R}$? –  Stefan Smith Apr 7 '13 at 2:48
    
@StefanSmith: Stable here means that if you take the complex conjugate of an element of the subfield, then it must be in the subfield again. If $\alpha$ is a non-real cube root of $2$, then $K=\Bbb Q[\alpha]$ is not stable under complex conjugation, and in particular $\alpha\bar\alpha\notin F$, which forbids using the usual complex inner product. However here $K$ itself can be ordered, which also answers your second question. And ther are also ordered fields much larger than $\Bbb R$, like the surreal numbers. –  Marc van Leeuwen Apr 7 '13 at 4:42
    
Thanks. Maybe my second question was unclear. I know that an ordered field does not necessarily have to be a subfield of $\mathbb{R}$. What I meant was, is there an ordered field that is not a subfield of $\mathbb{R}$ that can be used as the field for an inner product space? –  Stefan Smith Apr 7 '13 at 14:43
    
@StefanSmith: I can see no reason why another ordered field than $\Bbb R$, larger or smaller, cannot be used to define an inner product space. But as I mentioned, I haven't seen it actually done. –  Marc van Leeuwen Apr 7 '13 at 21:01

I think it depends on what field you are using for your vector space. If it is $\mathbf{R}$ or $\mathbb{C}$, the answer is definitely "yes" (see the comments, which are correct). I am pretty sure it is "yes" if your field is a subfield of $\mathbb{C}$ that is closed (the word "stable" also seems to be standard) under complex conjugation, such as the algebraic numbers. Otherwise, e.g. if your field is $\mathbf{F}_2$, I don't know. I consulted Wikipedia's article on inner product spaces and they only dealt with the case where the field was the reals or the complex numbers.

EDIT: Marc's answer is better than mine. See his comment regaring subfields of $\mathbf{C}$. Some such subfields are not stable under complex conjugation and cannot be used as a field for an inner product space. I am pretty sure that if $\mathbb{K}$ is any ordered field (which may or may not be a subfield of $\mathbb{R}$) you can use it as the field for an inner product space.

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