Consider a polynomial \(f \in K[x]\), where \(K\) is an algebraically closed field. Suppose that \(f\) has the property that for all matrices \(A \in M_n(K)\) of any size \(n\), if \(f(A) = 0\), then \(A\) is a diagonalizable matrix; then we say that the polynomial \(f\) forces diagonalizability.
Characterize by a simple rule exactly which polynomials in \(K[x]\) force diagonalizability. Prove your answer.
Fix \(m \geq 1\). Is every square matrix \(A\) with entries in \(K\) satisfying \(A^m = I\) diagonalizable? (The answer depends on \(K\).)
Solution
1. A polynomial \(f\in K[x]\) forces diagonalizability if and only if \(f\) is squarefree.
Because \(K\) is algebraically closed, write \[ f(x)=c\prod_{i=1}^r (x-\lambda_i)^{e_i} \] with distinct \(\lambda_i\in K\).
Suppose first that \(f\) is squarefree, so every \(e_i=1\). If \(f(A)=0\), then the minimal polynomial \(m_A\) divides \(f\). Hence \(m_A\) is also a product of distinct linear factors. A matrix over an algebraically closed field is diagonalizable exactly when its minimal polynomial is a product of distinct linear factors. Thus \(A\) is diagonalizable.
Conversely, suppose \(f\) is not squarefree. Then some root \(\lambda\) occurs with multiplicity at least \(2\), so \((x-\lambda)^2\mid f\). Let \[ J=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}. \] Its minimal polynomial is \((x-\lambda)^2\), so \(f(J)=0\), but \(J\) is not diagonalizable. Therefore \(f\) does not force diagonalizability.
2. Fix \(m\ge 1\).Every matrix \(A\) satisfying \(A^m=I\) is diagonalizable if and only if \(\operatorname{char}K\) does not divide \(m\).
The condition \(A^m=I\) is equivalent to \((x^m-1)(A)=0\). By 1, every such \(A\) is diagonalizable exactly when \(x^m-1\) is squarefree.
Its derivative is \(mx^{m-1}\). If \(\operatorname{char}K\nmid m\), then \(m\neq0\) in \(K\), and any common root of \(x^m-1\) and \(mx^{m-1}\) would have to be both nonzero and zero, impossible. Hence \(x^m-1\) is squarefree.
If \(\operatorname{char}K=p\) divides \(m\), write \(m=pr\). Then \[ x^m-1=(x^r)^p-1=(x^r-1)^p \] in characteristic \(p\), so the polynomial has repeated roots. By 1 there is a non-diagonalizable matrix annihilated by \(x^m-1\); explicitly, for any \(m\)th root \(\lambda\) one may take the \(2\times2\) Jordan block \(J_2(\lambda)\) whenever \((x-\lambda)^2\mid x^m-1\).