Solution: Suppose \( A \) and \( B \) are similar matrices. Then there exists an invertible matrix \( P \) such that \( B = P^{-1}AP \). Let \( f(t) \) be the minimal polynomial of \( A \). Then \( f(A) = 0 \). We need to show that \( f(B) = 0 \). Let \[ f(t) = a_nt^n + a_{n-1}t^{n-1} + \cdots + a_1t + a_0. \] Since \( B = P^{-1}AP \), we have \[ f(B) = f(P^{-1}AP). \] Thus, \begin{align*} f(P^{-1} A P) & = a_n(P^{-1} A P)^n + a_{n-1}(P^{-1} A P)^{n-1} + \cdots + a_1(P^{-1} A P) + a_0I \\ & = P^{-1} a_n A^n P + P^{-1} a_{n-1} A^{n-1} P + \cdots + P^{-1} a_1 A P + P^{-1} a_0 P \\ & = P^{-1} (a_n A^n + a_{n-1} A^{n-1} + \cdots + a_1 A + a_0I)P \\ & = P^{-1} f(A) P = P^{-1} 0 P = 0. \end{align*} Thus, \( f(B) = 0 \) and hence the minimal polynomial of \( A \) is the same as the minimal polynomial of \( B \).