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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.