$B^2A - A$ is invertible $\Rightarrow $ $AB - A$ is invertible

Question

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?

Answer 1

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

Answer 2

$det(AB-A)=det(A)det(B-I)$, $det(A)=0$ implies $det(B^2A-A)=0$, $det(B-I)=0$ implies that $det(B-I)(B+I)=det(B^2-I)=0$.