$B^2A - A$ is invertible $\Rightarrow $ $AB - A$ is invertible I'm trying to solve the problem in the title. So far I have concluded:
1) $det(B^2A - A) \neq 0 \Rightarrow det(A)\neq0$ and $det(B^2 - I)\neq0 $
2)Suppose $det(AB - A) = 0$  $\Rightarrow det(A)=0$ or $det(B - I) = 0 $
By (1): $det(A)\neq0$ and so I'd like to show:  $det(B^2 - I)\neq0 \Rightarrow det(B - I)\neq0$
Suggestions?

Hint: $B^2-I=(B+I)(B-I)$. Now take the determinant.
