PROVE: Two vector spaces V and W are isomorphic iff dimV = dimW (say n) I have proved (=>) but showing (<=) is hard for me. 
I think it should start with letting {$v_1, ... ,v_n$} be a basis of V and {$w_1, ... ,w_n$} be a basis of W then suppose there exists a linear transformation T such that $T:V->W$ defined by $T(v_i)=w_i$ . 
Then I have a feeling that it's an isomorphism. How do I show that? 
 A: Just to be sure that there are no misunderstandings, I will add an answer. Suppose that $\{v_1,\ldots,v_n\}$ is a basis for $V$, and that $\{w_1,\ldots,w_n\}$ is a basis for $W$. Define a map $T:V\to W$ by the linear extension of the assignment $v_i\to w_i$ for all $i\in\{1,\ldots,n\}$. To prove that $T$ is injective, it suffices to show that the $\ker(T)=\{0\}$. So suppose that $v=\sum_{i=1}\lambda_iv_i\in\ker(T)$, then
\begin{equation}
\begin{split}
0&=T\left(\sum_{i=1}^n\lambda_iv_i\right) \\
&=\sum_{i=1}^n\lambda_iT(v_i) \\
&=\sum_{i=1}^n\lambda_iw_i,
\end{split}
\end{equation}
which implies that $\lambda_i=0$ for all $i\in\{1,\ldots,n\}$, since $\{w_1,\ldots,w_n\}$ is a basis. Surjectivitiy of $T$ is of course trivial, for if $w=\sum_{i=1}^n\lambda_iw_i\in W$ is given, then $\sum_{i=1}^n\lambda_iv_i\in V$ is such that $T\left(\sum_{i=1}^n\lambda_iv_i\right)=\sum_{i=1}^n\lambda_iw_i$.
A: Let $V$ and $W$ be vector spaces with $\dim V=\dim W$, say $n$. Then $V$ and $W$ have bases $(v_1,\ldots,v_n)$ and $(w_1,\ldots,w_n)$, and the linear map $f$ defined by $f(v_i)=w_i$ is an isomorphism.
