Show that AB and BA have same eigenvalues

Question

If $A$ and $B$ are $n\times n$ matrices, with $A$ nonsingular prove that $AB$ and $BA$ have the same set of eigenvalues.

Can we begin with $ABX_1=\lambda_1X_1$, somehow show by manipulation that $BAX_1=\lambda_1X_2$?

Answer 1

Similar matrices have the same eigenvalues, and $A^{-1}(AB)A=BA$.

Working along the lines of the proof that similar matrices have the same eigenvalues, if $ABX_1=\lambda_1X_1$ then $$BA(A^{-1}X_1)=A^{-1}(AB)A(A^{-1}X_1)=A^{-1}(ABX_1)=A^{-1}\lambda_1 X_1=\lambda_1A^{-1}X_1$$