# True or False: Every finite dimensional vector space can made into an inner product space with the same dimension.

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
Over a field $F$, up to vector space isomorphism, there is only one $n$-dimensional space. – André Nicolas Apr 6 '13 at 20:05
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

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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