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
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
- $AB,BA$ 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?
- 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.