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What are the simplest way to show that AB and BA have the same characteristic polyno? 
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#1
Jun1111, 03:59 AM

P: 104

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! 


#2
Jun1111, 05:58 AM

P: 313

If you're happy that Det(AB)=Det(BA), then with I the identity matrix,
the characteristic polynomial is p(x) = Det(ABxI) = Det(IABxI^{2}) = Det(B^{1}BABxB^{1}BI) = Det(BABB^{1}xBIB^{1}) = Det(BAxI) 


#3
Jun1111, 07:16 AM

P: 8

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(Alambda I)=0 has finitely many solutions. We can take epsilon such that, for all lambda 0<lambda<epsilon, Alambda I is invertible. Therefore, (Alambda I)B and B(Alambda I) must have the same characteristic equation. Also, det(xI(Alambda I)B)=det(xIB(Alambda I)). For fixed x, each side of this equation is a polynomial in lambda, hence it is continuous. We can take lambda>0 and we are done.



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