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xEnOnn
How I show that the eigenvalues of matrix A times matrix B is and BA are equal? That's to show eigenvals(AB)=eigenvals(BA).
http://en.wikipedia.org/wiki/Characteristic_polynomial think it might be useful
From this, \[\det(AB-\lambda I)=0\] Since AB and BA are different matrices, I still find it strange that their eigenvalues would turn out the same. How is this so?
Suppose \[ABx=\lambda x\] then multiply by B \[BABx=\lambda Bx\] so \[BA(Bx)=\lambda (Bx)\] thus BA has the same eigenvalues as AB similarly AB has the same eigenvalues as BA.
wow...this is smart... thanks a lot Zarkon!! :D