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Are the eigenvalues of the matrices $AB$ and $BA$ identical? If yes, why?

From the examples that I have tried I think they are identical but I just can't come up with a formal proof for this.

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  • $\begingroup$ You can find more reasons here. $\endgroup$ – Algebraic Pavel Mar 4 '15 at 13:00
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Hint:

If $\lambda $ is an eigenvalue of $AB$ that is $\exists x\neq 0$ and

$$ AB(x)=\lambda x $$ $BAB(x)=BA(Bx)=\lambda (Bx)$

Then $Bx$ is eigenvector for eigenvalue $\lambda$.

Only case which we have to worry about is when $Bx=0$. For that case see that $\lambda x =0$ which contradicts the very starting assumption (except when $\lambda =0$).

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