Show that if the minimal polynomial of a linear operator \(T: V \to V\) on a finite-dimensional vector space \(V\) over a field \(F\) is irreducible, then every \(T\)-invariant subspace has a \(T\)-invariant complement (i.e. \(V\) is a semisimple \(F[T]\)-module).
Solution
Let \(m_T(x)=p(x)\) be irreducible. Regard \(V\) as an \(F[x]\)-module by letting \(x\) act as \(T\). Since \(p(T)=0\), the action factors through \[ K:=F[x]/(p), \] which is a field. Hence \(V\) is naturally a vector space over \(K\).
A subspace \(W\subseteq V\) is \(T\)-invariant exactly when it is an \(F[x]\)-submodule. Because the \(F[x]\)-action factors through \(K\), such a \(W\) is a \(K\)-subspace. Choose a \(K\)-linear complement \(W'\) so that \[ V=W\oplus W'. \] Since \(W'\) is a \(K\)-subspace, it is stable under the class of \(x\), hence under \(T\). Thus every \(T\)-invariant subspace has a \(T\)-invariant complement.