Why Define Inner Products for Complex Vector Spaces Using Complex Conjugation?

[qbslug]

What is the motivation behind defining the inner product for a vector space over a complex field as
<v,u> = v1*u1 + v2*u2 + v3*u3
where * means complex conjugate
as opposed to just
<v,u> = v1u1 + v2u2 + v3u3
They both give you back a scalar. The only reason I can see is the special case for <v,v> in which you get a real number but what does that matter.

 

[Hurkyl]

What is the motivation behind defining the inner product for a vector space over a complex field as
<v,u> = v1*u1 + v2*u2 + v3*u3
where * means complex conjugate

Because it's useful.

One thing to note is that if you "forget" the complex structure and view such a vector space as a vector space of the reals, the complex inner product gives you the same value as the real dot product

 

[AlephZero]

The only reason I can see is the special case for <v,v> in which you get a real number but what does that matter.

It matters a lot in terms of "usefulness". For example, real numbers can be ordered with the relations "greater than" and "less than", but complex numbers can not. <v,v> represents the "length" of v, so if <v,v> was a complex quantity, general results like the triangle inequality would not apply to it.

 

[qbslug]

Ok thanks. This is the only axiom of inner products that bothers me. So we could define the inner product of a complex vector space as
<v,u> = v1u1 + v2u2 + v3u3
with no complex conjugates but we would lose some nice properties that are convenient such as length?

 

[batboio]

One reason that comes to mind is that by defining the inner product as <v,u> = v1*u1 + v2*u2 + v3*u3 you get a real number for <v,v> as you said and that is needed for a prehilbert space (where <v,v>=0 <=> v=0 and otherwise <v,v> > 0), which is basically the generalization of the Euclidean space for a complex field.