Do similar matrices have same characteristic equations?

Question

Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?

Answer 1

Note that $\det (\lambda I -A) = \det S \det (\lambda I -A) \det S^{-1} = \det (\lambda I - S A S^{-1})$.

Answer 2

Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then \begin{align*}\text{charpoly}(A,t) & = \det(A - tI)\\ & = \det(PBP^{-1} - tI)\\ & = \det(PBP^{-1}-tPP^{-1})\\ & = \det(P(B-tI)P^{-1})\\ & = \det(P)\det(B - tI) \det(P^{-1})\\ & = \det(P)\det(B - tI) \frac{1}{\det(P)}\\ & = \det(B-tI)\\ & = \text{charpoly}(B,t). \end{align*}

This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.