If $A^3B-B$ is invertible then $BA-B$ is invertible.

Question

I assume this is a simple proof but i'm stuck here.

I need to prove that if $A^3B-B$ is invertible then $BA-B$ is invertible.

So $A^3B-B=(A^3-I)B$ and then both $(A^3-I)$ and $B$ are invertible.

So I need to show that $BA-B=B(A-I)$ are both invertible.

I thought maybe I could somehow show that if $(A^3-I)$ is invertible then $(A-I)$ is invertible but I guess there's a simpler way to get there. What am I missing?

Thanks!

Answer 1

Note that

$A^3B - B = (A^3 - I)B; \tag 1$

thus the invertibility of $A^3B - B$ implies both $A^3 - I$ and $B$ are invertible. Since

$A^3 - I = (A - I)(A^2 + A + I), \tag 2$

the invertibility of $A^3 - I$ yields the same for $A - I$ and $A^2 + A + I$. Since now $A - I$ and $B$ are invertible, so is their product

$B(A - I) = BA - B. \tag 3$

Answer 2

Hint $$A^3-I = (A-I)(A^2 + A + I).$$