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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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 $\lambda =0$).