## what are the simplest way to show that AB and BA have the same characteristic polyno?

polynomial?
i find this but i do not understand;its too complex.

http://www.math.sc.edu/~howard/Classes/700/charAB.pdf

any simpler way/idea?Thanks!
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 If you're happy that Det(AB)=Det(BA), then with I the identity matrix, the characteristic polynomial is p(x) = Det(AB-xI) = Det(IAB-xI2) = Det(B-1BAB-xB-1BI) = Det(BABB-1-xBIB-1) = Det(BA-xI)
 If A is invertible, than it is evident that AB and BA are similar matrices, therefore they have the same characteristic polynomial. Otherwise, we notice that the equation in lambda, det(A-lambda I)=0 has finitely many solutions. We can take epsilon such that, for all lambda 0<|lambda|0 and we are done.
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