# Dual space

1. Nov 23, 2008

### boombaby

Let V be a finite dimensional vector space over the field K, V* be the dual space, the set of all linear functions on V. Now define a h:V*->K^n by h(f)=(f(e_1),f(e_2),...f(e_n)), where e_i is the basis of V. It is known that h(xf+yg)=xh(f)+yh(g), where x,y is in K and f,g in V*. It is told that h is bijective. I understand h is injective, but do not understand why it is surjective? it might be quite simple, all the books just says it is OBVIOUS, but I just don't get it....
More precisely, given x=(x_1, x_2,...,x_n) in K^n, why there is a f in V* such that h(f)=x, or equivalently, f(e_i)=x_i ?
Thanks!

2. Nov 23, 2008

### Hurkyl

Staff Emeritus
Aren't expressions exactly like that how you normally go about specifying a linear transformation?

3. Nov 23, 2008

### boombaby

yea, thanks. I think get it now...in case I made something wrong, check it plz, thanks
since I'm not having much knowledge on linear transformation. This is how I'm thinking:
there's a isomorphism g:V->K^n such that g(e_1)=(1,0,...,0), g(e_2)=(0,1,0...0)....g(e_n)=(0,0,...,1).
given (a_1,a_2,...,a_n) in K^n, there's a linear function f:K^n->K such that f(g(e_1))=a_1,...f(g(e_n))=a_n and f has the form f((x_1,x_2,...,x_n))=a_1*x_1+....a_n*x_n. This is done by solving a system of linear equations.
f(g(v)):V->K is the linear function I need. (linearity is easy to check)

Is this the right way to prove it? Is it possible to have a more explicit form of f?
Thanks

Last edited: Nov 24, 2008
4. Nov 23, 2008

### morphism

That's explicit enough, isn't it? The reason textbooks don't elaborate on why h is bijection is because it really is obvious! Any linear map defined on V is completely determined by its action on a basis; and vice versa, once you define the values of a function on the basis of V, you can extend it (uniquely) to a linear map.

5. Nov 24, 2008

### boombaby

thanks. it's more clear now