Basis if and only if $\varphi$ is an isomorphism Let $V$ be a finite dimensional vector space. I am aware of the theorem stating that if $\varphi \colon V \rightarrow V$ is an automorphism and $\mathcal{A}=(a_1, \ldots, a_n)$ is a basis of V then $\varphi(\mathcal{A})=(\varphi(a_1), \ldots, \varphi(a_n))$ is another basis of $V$. However, suppose we have some other set $\mathcal{B}=(b_1, \ldots, b_n)$ and we want to prove that it is also a basis of V. I know how to prove it working from definitions but i was wondering if you could construct a linear map $f$ 
$$
f(x_1 a_1 + \ldots + x_n a_n) = y_1 b_1 + y_n b_n
$$
such that if $\mathcal{A}$ is a basis then $\mathcal{B}$ is another basis if and only if $f$ is an isomorphism. That way we could reduce a problem of showing that $\mathcal{B}$ is a basis to show that a map $f$ is an isomorphism.

Yes, you are right.  If you can show that $b_1,\ldots,b_n$ is the image of a basis $a_1,\ldots,a_n$ by an isomorphism $f$, then you have shown that $\left\{b_i\right\}_{i=1,\ldots,n}$ is a basis.
This happens because since $f$ is an isomorphism, the image of $f$ has dimension $n$. But the image of $f$ is generated by $f(a_1),\ldots,f(a_n)$, hence the vectors $b_i=f(a_i)$ generate a space of dimension $n$. There are $n$ such vectors, and so they form a basis for the space they generate (which is $V$ itself).
