Let \(F\) be a subfield of the complex numbers, \(V\) a finite-dimensional vector space over \(F\), and \(T\) a semi-simple linear operator on \(V\). If \(f\) is any polynomial over \(F\), prove that \(f(T)\) is semi-simple.
Solution
Because \(T\) is semisimple, its minimal polynomial is squarefree [@HK71]. Over the given subfield of \(\mathbb C\), write \[ m_T=p_1\cdots p_r \] with the \(p_i\) distinct monic irreducibles. The primary decomposition gives \[ V=V_1\oplus\cdots\oplus V_r, \qquad V_i=\ker p_i(T), \] and on each \(V_i\) the algebra \(F[T|_{V_i}]\) is a field isomorphic to \(F[x]/(p_i)\).
Let \(S=f(T)\). On \(V_i\), the operator \(S\) is multiplication by the class of \(f\) in the field \(F[x]/(p_i)\). Therefore the minimal polynomial of \(S|_{V_i}\) is the irreducible minimal polynomial over \(F\) of that field element, hence is squarefree. The minimal polynomial of \(S\) is the least common multiple of these finitely many irreducible polynomials, so it is also squarefree. By the semisimplicity criterion [@HK71], \(S=f(T)\) is semisimple.