True or False: For a basis $\{v_1, \dots, v_n\}$ of $\Bbb R^n$, $\{Av_1, \dots, Av_n\}$ is a basis of $\Bbb R^n$ if $\det(A) \neq 0$.
The answer is true, but I don't know why. My guess is that if $\det(A) \neq 0$, there exist $A^{-1}$ and therefore $A$ has trivial solution ($x=0$). It means nullspace of $A$ has one dimension? I'm stuck in here. If you know the logic which is used in here, please explain to me.