## xEnOnn Group Title 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). 2 years ago 2 years ago

1. hahd Group Title

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

2. xEnOnn Group Title

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?

3. Zarkon Group Title

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.

4. xEnOnn Group Title

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