- Show that if a matrix is diagonalizable, its minimal polynomial is squarefree.
Solution
Suppose \(A\) is diagonalizable over \(F\). Then for some \(P\in\operatorname{GL}_n(F)\), \[ P^{-1}AP=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \] Let \(\mu_1,\dots,\mu_r\) be the distinct eigenvalues of \(A\) and put \[ q(x)=\prod_{j=1}^r(x-\mu_j). \] Then \(q\) is squarefree and \[ q(A)=P\,\operatorname{diag}(q(\lambda_1),\dots,q(\lambda_n))P^{-1}=0. \] Hence the minimal polynomial \(m_A\) divides \(q\).
Conversely, if \(\mu_j\) is an eigenvalue with eigenvector \(v\ne0\), then \[ 0=m_A(A)v=m_A(\mu_j)v, \] so \(m_A(\mu_j)=0\). Thus every factor \((x-\mu_j)\) divides \(m_A\). Since these factors are pairwise coprime, \[ q\mid m_A. \] Therefore \(m_A=q\), a product of distinct linear factors. In particular, \(m_A\) is squarefree.