If $I-AB$ is invertible, then is $I-BA$ invertible?

Question

If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?

This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.

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My thoughts.

If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.

However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?

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EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. $$ So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.

Answer 1

Hint: Note that the two matrices have the same determinant. There are several proofs of this, such as those given here.

Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.

Answer 2

Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.