- Show that if the minimal polynomial of a linear map \(T\) is irreducible, then every \(T\dash\)invariant subspace has a \(T\dash\)invariant complement.
Solution
Let \(V\) be a finite-dimensional vector space over a field \(F\), let \(T\in\operatorname{End}_F(V)\), and suppose its minimal polynomial \(m_T(x)\) is irreducible.
1. The algebra\[ K=F[x]/(m_T) \]is a field, and \(V\) is naturally a vector space over \(K\).
Because \(m_T\) is irreducible, the ideal \((m_T)\) is maximal in the PID \(F[x]\), so \(K\) is a field. Define \[ [f(x)]\cdot v=f(T)v. \] This is well defined because \(m_T(T)=0\), so two polynomials congruent modulo \(m_T\) induce the same endomorphism of \(V\). Thus the \(F[x]\)-module structure defined by \(x\cdot v=T(v)\) factors through the field \(K\).
2. A subspace \(W\subseteq V\) is \(T\)-invariant if and only if it is a \(K\)-linear subspace.
If \(W\) is \(K\)-linear, it is stable under the class of \(x\), hence under \(T\).
Conversely, if \(W\) is \(T\)-invariant, then it is stable under every polynomial in \(T\), so it is stable under the action of every class \([f(x)]\in K\). Hence \(W\) is a \(K\)-subspace.
3. Every \(T\)-invariant subspace has a \(T\)-invariant complement.
Let \(W\subseteq V\) be \(T\)-invariant. By 2, \(W\) is a \(K\)-subspace of the \(K\)-vector space \(V\). Extend a \(K\)-basis of \(W\) to a \(K\)-basis of \(V\), and let \(U\) be the \(K\)-span of the added basis vectors. Then \[ V=W\oplus U \] as \(K\)-vector spaces. Since \(U\) is a \(K\)-subspace, 2 shows that \(U\) is \(T\)-invariant. Thus \(U\) is the required invariant complement.