# $I_m -AB$ ivertible if and only if $I_n-BA$ invertible

Let $A$ and $B$ be $m\times n$ and $n\times m$ matrices respectively.

1. Prove that if $\lambda$ is a non-zero eigenvalue of $AB$ then it is also an eigenvalue of $BA$
2. Prove that $I_m-AB$ is invertible if and only if $I_n-BA$ is invertible.

Part (1) is easy:

$$ABx=\lambda x$$

By definition, $x\ne 0$, and by assumption $\lambda \ne 0$. So we have $Bx\ne 0$

Now, $$B(AB)x=(BA)Bx=\lambda Bx$$ and so $Bx$ and a non-zero vector with eigenvalue $\lambda$.

My problem is I have no idea how to use this to do (2). Any help is much appreciated. Thanks!

As hinted in Carmichael's comment, we have that the following are equivalent:

(1) $(I_n-BA)$ is invertible.

(2) $1$ is not an eigenvalue of $BA$.

(3) $1$ is not an eigenvalue of $AB$.

(4) $(I_m-AB)$ is invertible.

You've proven that $(2) \iff (3)$, so just note that

$$(I-M)v \iff Mv=Iv=v$$

for any matrix $M$.

For part (2):

Let $𝐼_{𝑚}−𝐴𝐵$ be invertible and let’s consider the following matrix expression $(𝐼_{𝑛}−𝐵𝐴)𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴$ which you can verify simplifies to $BA$.

If that is true, then what is $(𝐼_{𝑛}−𝐵𝐴)[𝐵(𝐼_{𝑚}−𝐴𝐵)^{−1}𝐴+I_{n}]$?

You can easily construct a similar argument if you first let $𝐼_{𝑛}−𝐵𝐴$ be invertible.