6
$\begingroup$

This question already has an answer here:

Prove that if $I-BA$ is invertible then $I-AB$ is invertible.

Though I have found this question already posted and also it has some answers like

Use:$(I−BA)(I+B(I−AB)^{−1}A)=I(I−BA)(I+B(I−AB)^{−1}A)=I $

I have done it like this.

$I-BA$ is invertible $\implies 0$ is not an eigen value of $I-BA\implies 1$ is not an eigen value of $BA\implies 1$ is not an eigen value of $AB\implies 0$ is not an eigen value of $I-AB\implies I-AB$ is invertible.

I have used the facts the

  1. $AB,BA$ have same non-zero eigen values

Proof:Let $c\neq 0$ be a eigen value of $AB$ corresponding to eigen vector $\alpha$.Then $A(B\alpha)=c\alpha$

Now $(BA)(B\alpha)=B(AB\alpha)=c(B\alpha)\implies c$ is an eigen vector of $BA$ corresponding to $B\alpha$.Also $B\alpha\neq 0$ otherwise $c=0$ .

Similarly every eigen value of $BA$ is an eigen value of $AB$.

How to show that they have same eigen values for if $A,B$ are $n\times n$ matrices?

and

  1. If $c$ is an eigen value of a matrix $M$, then $1-c$ is an eigen value of $I-M$.

Please check whether my answer is correct/not.

$\endgroup$

marked as duplicate by Brahadeesh, stressed out, Lord Shark the Unknown, onurcanbektas, Davide Giraudo Jan 4 at 15:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

5
$\begingroup$

In fact, what you call the "use" of this fact is sufficient for a proof.

However, your steps are indeed (mostly) correct. As long as you and the reader can agree to take for granted that $AB$ and $BA$ share their non-zero eigenvalues (which is a non-trivial statement to make), then this proof is valid. Note that, depending on your professor, this might not be sufficient detail for a proof on a test.

Note that $AB$ and $BA$ do not necessarily share all of their eigenvalues unless they happen to be square. It is possible for $AB$ to be invertible without $BA$ being invertible.

$\endgroup$
  • $\begingroup$ Yes I have proved that $AB$ and $BA$ have same non-zero eigen values; :I have added a proof also $\endgroup$ – Learnmore Aug 22 '16 at 14:14
  • $\begingroup$ But how can I show that if $A,B$ are $n\times n$ matrices then they have the same eigen values $\endgroup$ – Learnmore Aug 22 '16 at 14:15
  • 2
    $\begingroup$ @BobWilson it suffices to note that, in this case, $$ \det(AB) = \det(A)\det(B) = \det(BA) $$ $\endgroup$ – Omnomnomnom Aug 22 '16 at 14:20
  • $\begingroup$ @BobWilson in the future, rather than adding a question to your original post (which has already been answered), it is better to start a new post. $\endgroup$ – Omnomnomnom Aug 22 '16 at 14:22
  • $\begingroup$ Yes that's true but it would add another question to the already existing bunch of questions ;so I thought it to be a better idea $\endgroup$ – Learnmore Aug 22 '16 at 14:23

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