Prove that the minimal polynomial divides the characteristic polynomial.
Solution
Let \(m_A(t)\) be the minimal polynomial and \(\chi_A(t)\) the characteristic polynomial of \(A\). By Cayley–Hamilton, \[ \chi_A(A)=0. \] Divide \(\chi_A\) by \(m_A\) in \(F[t]\): \[ \chi_A=q\,m_A+r,\qquad \deg r<\deg m_A. \] Evaluating at \(A\) gives \[ 0=\chi_A(A)=q(A)m_A(A)+r(A)=r(A). \] By minimality of \(m_A\), no nonzero polynomial of smaller degree annihilates \(A\). Hence \(r=0\), and therefore \[ \boxed{m_A\mid\chi_A}. \]