Let \(T\) be a linear operator on a finite-dimensional vector space over the field of complex numbers. Prove that \(T\) is diagonalizable if and only if \(T\) is annihilated by some polynomial over \(C\) which has distinct roots.
Solution
If \(T\) is diagonalizable, let \(c_1,\ldots,c_r\) be its distinct characteristic values. Then in an eigenbasis the polynomial \[ f(x)=\prod_{j=1}^r(x-c_j) \] vanishes on every diagonal entry, so \(f(T)=0\). Its roots are distinct.
Conversely, suppose \(f(T)=0\) for a polynomial \(f\in\mathbb C[x]\) with distinct roots. The minimal polynomial \(m_T\) divides every annihilating polynomial, hence \(m_T\mid f\). Because \(f\) has no repeated root, neither does \(m_T\); and because the base field is \(\mathbb C\), \(m_T\) splits into linear factors. The minimal-polynomial criterion therefore implies that \(T\) is diagonalizable.