Proving AB & BA Have Same Characteristic Polynomial - Simple Ways

[td21]

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!

 

[Simon_Tyler]

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)

 

[feynman137]

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|<epsilon, A-lambda I is invertible. Therefore, (A-lambda I)B and B(A-lambda I) must have the same characteristic equation. Also, det(xI-(A-lambda I)B)=det(xI-B(A-lambda 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.