Prove that {$T(v_1), ... T(v_n)$} is a basis for W. Let $T: V \to W$ be linear as well as one-to one and onto. Suppose $dim(V) = n$ and let $\alpha$ = {$v_1, ...v_n$} be a basis for $V$. Is my logic of going about the proof correct, applying the transformation will result in an independent set and thus since we know the dimension of $V$ which is $n$, then by the basis-theorem I know that my independent set spans W. 
 A: Your idea is correct, but lacks mathematical rigor. What you must do is the following:


*

*Show that $T(v_1), \dots, T(v_n)$ are linearly independent. That is, start with some set of scalars $\alpha_1,\dots, \alpha_n$ for which $$\sum_{i=1}^n\alpha_iT(v_i)=0$$
and show that $\forall i: \alpha_i = 0$.

*Show that the span of $T(v_1), \dots, T(v_n)$ is $W$. That is, take any vector $w\in W$ and find such values $\beta_1, \dots, \beta_n$ that $$\sum_{i=1}^n\beta_iT(v_i)=w$$


It is also possible to use some theorems that can help you, like the theorem that a one-to-one linear mapping preservs dimension, but since that theorem is basically the consequence of what you are trying to show, I doubt you're meant to use it. Your statement can very simply be shown using only basic means, so I think that's a better way.
A: If one knows the rank-nullity theorem, one can argue as follows:
$T$ injective $\implies \text{ker }T = \{0\} \implies \dim(\text{ker }T) = 0 \implies \text{rank}(T) = n = \dim(V)$ .
$T$ surjective $\implies \text{rank}(T) = \dim(W)$, so we see $\dim(W) = n = \dim(V)$.
It is then easy to see that if $B$ is a basis for $V$, that $T(B)$ is a basis for $W$, for if $B = \{v_1,\dots,v_n\}$ and we have for any linear combination:
$T\left(\sum_{i = 1}^n \alpha_iv_i\right) = \sum_{i = 1}^n \alpha_iT(v_i) = 0$
the injectivity of $T$ implies $\sum_{i=1}^n\alpha_iv_i = 0$, and the linear independence of $B$ forces
$a_i = 0,\ i = 1,2,\dots n$, that is, $T(B)$ is linearly independent, and spans because:
$|T(B)| = \dim(W)$.
