Why must inner product spaces be over the field of real or complex numbers?

[cavalier]

Friedberg's Linear Algebra states in one of the exercises that an inner product space must be over the field of real or complex numbers. After looking at the definition for while, I am still having trouble seeing why this must be so. The definition of a inner product space is given as follows.

Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F, denoted <x,y>, such that for all x, y, and z in V and all c in F, the following hold:
(a) <x+z, y>=<x,y> + <z,y>
(b) <cx,y>=c<x,y>
(c) <x,y>=$\overline{<y,x>}$
(d) <x,x> > 0 if x$\neq$0.

I can't convince myself that I could not contrive some vector space and some inner product such that the resulting inner product space would not use the whole number line.

 

[micromass]

The short answer is that it must not be over the field of real or complex numbers. It can be defined over more general structures as well (for example, over $C^*$-algebras). This yields the notion of $C^*$-modules

However, there are at least two reasons why we limit ourself to real or complex numbers:
1) The theory is nicer and easier of reals or complex numbers. For example, the theorem of Pythagoras mustn't hold in the more general case.

2) All the applications occur when working over the real or complex numbers. So looking at others structures is just not as interesting.

 

[Sina]

fields have multplicative inverses and in this way for instance you can normalize your vectors. If the norm of your vector is |v| which is in your field then it has an inverse 1/|v|. So v/|v| exists and you get a normalized vector. So the inner product comes into the role by the fact that it gives the norm of your vectors as a value in your field.

The concept of normalization is of outmost importance in many applied areas first and mostly quantum mechanics. Existance of othonormal bases is also of great use.

 

[lavinia]

Friedberg's Linear Algebra states in one of the exercises that an inner product space must be over the field of real or complex numbers. After looking at the definition for while, I am still having trouble seeing why this must be so. The definition of a inner product space is given as follows.

Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F, denoted <x,y>, such that for all x, y, and z in V and all c in F, the following hold:
(a) <x+z, y>=<x,y> + <z,y>
(b) <cx,y>=c<x,y>
(c) <x,y>=$\overline{<y,x>}$
(d) <x,x> > 0 if x$\neq$0.

I can't convince myself that I could not contrive some vector space and some inner product such that the resulting inner product space would not use the whole number line.

You can definitely do this. But inner products are generally used to define angles and lengths. For this some concept of a continuum is natural.

However symmetric bilinear forms appear all over in mathematics and in the scalars do not even have to be in a field.

 

[Deveno]

the positive-definite property of an inner product only makes sense over fields of characteristic 0. conjugate-symmetry only makes sense for fields that possesses an automorphism that fixes an ordered subfield, that is also an involution.

this limits the possible choices for our underlying field. more choices than R and C are indeed possible, but not commonly used (one of the reasons C is so popular is that C is algebraically complete, so for a linear transformation of a complex vector space, we are guaranteed eigenvalues).