Prove that the minimal polynomial of a square matrix divides its characteristic polynomial.
Solution
Let \(A\in M_n(F)\), let \(m_A(x)\) be its minimal polynomial, and let \(\chi_A(x)\) be its characteristic polynomial.
By the Cayley–Hamilton theorem, \[ \chi_A(A)=0. \] By definition, \(m_A\) is the monic polynomial of least positive degree annihilating \(A\). Divide \(\chi_A\) by \(m_A\): \[ \chi_A=q\,m_A+r, \qquad \deg r<\deg m_A. \] Evaluating at \(A\) gives \[ 0=\chi_A(A)=q(A)m_A(A)+r(A)=r(A). \] The minimality of \(m_A\) forces \(r=0\). Hence \[ m_A\mid\chi_A. \]