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).

Question

anonymous · Answer

http://en.wikipedia.org/wiki/Characteristic_polynomial think it might be useful

anonymous · Answer

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?

zarkon · Answer

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.

anonymous · Answer

wow...this is smart... thanks a lot Zarkon!! :D