# Show that $\det(A-\lambda I)=\det(B-\lambda I)$

Alright, I've been trying to work this linear algebra problem out for a bit and I don't seem to be getting anywhere. The problem is this:

Assume that $A=M^{-1}BM$. Show that $\det(A-\lambda I)=\det(B-\lambda I)$.

So my instincts tell me that this has something to do with the fact that $A^n$ can be expressed as $M^{-1}BM$

-
This rather has something to do with the fact that det(UV)=det(U)det(V) for suitable matrices U and V... – Did Oct 5 '11 at 7:20
Hint: $\lambda I = M^{-1}(\lambda I) M$. – Ted Oct 5 '11 at 7:23

A small hint: In the expression $\det(A-\lambda I)$ you can use the given fact that $A=M^{-1}BM$ of course, but also that $I=M^{-1}IM$.