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This question already has an answer here:

Let $A$ and $B$ be $n$ x $n$ matrices over reals. Show that $I - BA$ is invertible if $I - AB$ is invertible. Deduce that $AB$ and $BA$ have the same eigenvalues.

I know how to prove that $AB$ and $BA$ have same eigen values, when either $A$ or $B$ is non-singular.

But here, that condition is not mentioned.

Please help me.

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marked as duplicate by Robert Israel matrices Aug 6 '17 at 18:20

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.

  • $\begingroup$ Use the determinant $\endgroup$ – JohnColtraneisJC Aug 6 '17 at 18:22
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Note that $$ B^{-1}(I-BA)B=I-AB. $$ Hence $\det(I-BA)=\det(I-AB)$, so that $I-BA$ is invertible if and only if $I-AB$ is invertible. Actually, the question has been answered here already. The second part is a duplicate as well, see here, and here.

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