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

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?

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.
 A: A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n \times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X \in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
A: 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.
A: 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.
