Show that if $\lambda$ is an eigenvalue of matrix $A$ and $B$, then it is an eigenvalue of $B^{-1}AB$

Question

I was not too sure how to complete this proof. I thought maybe it was related to $B^{-1}AB$ being the diagonalization of another matrix $D$. I tried approaching it with the determinant and the definition ($A=\lambda x$, $B=\lambda y$) but to no avail.

Answer 1

You only need it an eigenvalue of $A $. If $Ax=\lambda x $ and $B $ is invertible, let $v=B^{-1}x $. Then $$ B^{-1}ABv=B^{-1}Ax=\lambda v. $$

Answer 2

Since $A$ and $B^{-1}AB$ are similar, they have the same characteristic polynomial, hence the same eigenvalues.