Show that similar matrices have the same characteristic polynomial I am given that $A$ is a square matrix and $B=C^{-1}AC.$ I'm trying to show that $A$ and $B$ have the same characteristic polynomial. So far I have said the following: $$p_A(\lambda)=\det(A-\lambda I)$$ $$p_B(\lambda)= \det(B-\lambda I)=\det(C^{-1}AC-\lambda C^{-1}C)= \det(C^{-1}(A-\lambda I)C) $$ I feel like I'm pretty much there but I don't know how to finish it off. Can anyone help?

Last step (you were so close!): the determinant is multiplicative, so

$$\det\left(C^{-1}(A-\lambda I)C\right)=\color{red}{\left(\det C\right)^{-1}}\cdot\det(A-\lambda I)\cdot \color{red}{\det C}=\det(A-\lambda I)$$