Similar matrices have the same eigenvalues and the same characteristic polynomials.

Theorem: If $A$ and $B$ are similar ($B = P^{-1}AP$), then they have the same eigenvalues and the same characteristic polynomial.

Proof: The characteristic polynomial of $B$ is:

$$\det(B - \lambda I) = \det(P^{-1}AP - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(P^{-1})\det(A - \lambda I)\det(P)$$

Since $\det(P^{-1})\det(P) = 1$, we get:

$$\det(B - \lambda I) = \det(A - \lambda I)$$

So $A$ and $B$ have the same characteristic polynomial, hence the same eigenvalues.

Note: The converse is false. Having the same eigenvalues does NOT imply similarity. For example, $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ both have eigenvalue 1 (with multiplicity 2) but are not similar (one is diagonalizable, the other is not).