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?
Best Answer
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 $T\left(\sum_{i=1}^n\lambda_iv_i\right)=\sum_{i=1}^n\lambda_iw_i$.