Proof about finite dimensional vector spaces over fields Prove that every finite dimensional vector space $V $of dimension $n$ over a field $F$ is isomorphic to the vector space $F^n$.
Okay, lot's of stuff here. I think most of the reason I can not do this one is because I don't have a clear understanding of everything going on here.
As far as I know, $F^n$ is the set of all $n$-tuples of $F$, or
$F^n$ = $\{(a_1,a_2,...,a_n) | a_i \in F\}$
I have many questions about the rest, though. is $V$ finite ONLY when viewed as a vector space over $F$? What does it really mean for $V$ to be viewed as a vector space over $F$?
To approach this problem, I need to find an isomorphism between V and $F^n$ but it sure helps to know the stuff I posted above to start.
 A: You're right about $F^n$; that's exactly what it means.  The key here is that $V$ is finite dimensional; so here's where you should start:
What does it mean for a vector-space $V$ to be $n$-dimensional?
A: For a vector space $V$ over $\mathbb{F}$, this means that the scalar multiplication of elements of $V$ also lie in $V$ (satisfying the vector space axioms).
To see that it matters what field a vector space is over, take $\mathbb{R}$ as a vector space over itself.  This is pretty trivial of an example.  However, taking $V=\mathbb{R}$ and the field as $\mathbb{Q}$ gives a wildly different vector space. 
As far as establishing an isomorphism, here's a hint: Select a basis for $V$.
A: The meaning of $V$ being a finite dimensional vector space over $F$ is that it is spanned by a finite amount of elements. Thus it is finite iff $F$ is finite. Specifically, if $V$ is an $n$-dimensional $F$-space, then $V$ is of the form $\{\alpha_1e_1+\ldots+\alpha_ne_n\,\,|\,\,\alpha_1,\ldots,\alpha_n\in F\}$, where $e_1,\ldots,e_n$ are the spanning elements. Note that the vector space axioms imply that $\{e_1,\ldots,e_n\}$ is an abelian group. Can you now think of a possible homomorphism from $V$ to $F^n$?
