I'll let $\displaystyle T:V\to W$ is the isomorphism, and $\displaystyle \{v_1,\cdots,v_n\}$ a basis for $\displaystyle V$. Note that since $\displaystyle T$ is injective that $\displaystyle \ker T=\{\bold{0}\}$ (why?) and so $\displaystyle \displaystyle \sum_{j=1}^{n}\alpha_j T\left(v_j\right)=\bold{0}\implies T\left(\sum_{j=0}^{n}\alpha_j v_j\right)=\bold{0}\implies \sum_{j=1}^{n}\alpha_j v_j=\bold{0}\implies \alpha_1=\cdots=\alpha_n$ and so $\displaystyle \left\{T(v_1),\cdots,T(v_n)\right\}$ is l.i. Now, since $\displaystyle T$ is surjective we have that for any $\displaystyle w\in W$ there exists some $\displaystyle v\in V$ such that $\displaystyle T(v)=w$. Note though that since $\displaystyle \{v_1,\cdots,v_n\}$ is a basis for there exists $\displaystyle \alpha_1,\cdots,\alpha_n\in F$ such that $\displaystyle \displaystyle \sum_{j=1}^{n}\alpha_j v_j=v$ and so $\displaystyle \displaystyle w=T(v)=T\left(\sum_{j=1}^{n}\alpha_j v_j\right)=\sum_{j=1}^{n}\alpha_j T(v_j)$ and thus $\displaystyle \left\lange\left\{T(v_1),\cdots,T(v_n)\right\}\rig ht\rangle=W$. Thus, $\displaystyle \left\{T(v_1),\cdots,T(v_n)\right\}$ is a basis for $\displaystyle W$