Proving the Existence of an Isomorphism between F^n and Hom(Hom(F^n,F),F)

[happyg1]

Homework Statement

Here's the problem I've been trying to get my mind around:
Prove that there exists an isomorphism of F^n into Hom(Hom(F^n,F),F).

I'm missing something. Here's what I get to:

F^n is an n-tuple and F is a field. So I can see that there is a set of homomorphisms from F^n into F.
It would be a finite n-tuple mapped into an infinite field, so there would just be a finite number of elements mapped infinitely.

I used n=3 and the real numbers as an example for the sake of trying to understand this:
Define a homomorphism T:F^n \rightarrow F as follows:
x_1=1,x_2=2,x_3=3,x_3=4,x_3=4.1,x_3=...
so x_3 is everything else besides 1 and 2.

Then I get confused trying to define Hom(Hom(F^n,F),F) Am I just mapping everything back into the origional F? And isn't that just what I started with, which is F?(or the reals in my attempted example)

So the whole point is to show that there is an isomorphim from F^n into the Hom((HomF^n,F),F) But it looks to me like F^n is a finite n-tuple and I can't get my mind around how there can be an isomprphism between an infinite field and a finite n-tuple.

What am I missing? Where have I gone wrong?

any clarification will be greatly appreciated.
CC

Homework Statement

Homework Equations

The Attempt at a Solution

 

[StatusX]

I don't really follow what you've tried so far.

Remember, Hom(F^n,F) is the set of linear functions from F^n into F. That is, functions f where x is in F^n and f(x) is in F, and which satisfy f(ax+by)=af(x)+bf(y), where x,y are vectors in F^n and a,b are scalars (in F). Such linear functions from a vector space into the underlying field are usually called linear functionals.

Note that the linearity of the function implies that its value at an arbitrary vector x is completely determined by its values on a basis of F^n. So, for example, if F=R and n=3, there is a bijection between 3-tuples (a,b,c), where a,b,c are real, and linear functionals, given by sending (a,b,c) to the functional f defined by f(1,0,0)=a, f(0,1,0)=b, f(0,0,1)=c, so that f(x,y,z)=ax+by+cz. You can show that the set of such linear functionals forms another vector space over F, and in this case the bijection just mentioned is in fact a vector space isomorphism.

Does this help at all?

 

[happyg1]

Hey,
That makes a whole bunch of sense...but this question is in the section before we study what a basis is. I just went through that section so I see what you are talking about. I don't think I can apply the basis to this problem BUT I'm going to rewrite my definition of the homomorphismF^n \rightarrow F and try again.
THANKYOU!