Let \(T\) be a linear operator on a finite-dimensional vector space \(V\). Suppose that
the minimal polynomial for \(T\) is a power of an irreducible polynomial;
the minimal polynomial is equal to the characteristic polynomial.
Show that no non-trivial \(T\)-invariant subspace has a complementary \(T\)-invariant subspace.
Solution
Let the common minimal and characteristic polynomial be \[ m_T(x)=\chi_T(x)=p(x)^s, \] where \(p\) is monic irreducible. Since the minimal and characteristic polynomials coincide, \(T\) has a cyclic vector \(\alpha\) [@HK71]. Thus \[ V=Z(\alpha;T) \] and the \(T\)-annihilator of \(\alpha\) is \(p^s\).
1. Every \(T\)-invariant subspace is a power-of-\(p\) submodule.
Let \(W\subseteq V\) be \(T\)-invariant and define \[ I_W=\{f\in F[x]:f(T)\alpha\in W\}. \] Because \(W\) is a subspace invariant under \(T\), it is invariant under every polynomial in \(T\). Hence \(I_W\) is an ideal of \(F[x]\). It contains \((p^s)\), because \(p(T)^s\alpha=0\in W\).
Since \(F[x]\) is a principal ideal domain, write \(I_W=(q)\). The containment \((p^s)\subseteq(q)\) means \(q\mid p^s\). Irreducibility of \(p\) gives, after choosing the monic generator, \[ q=p^j \] for some \(0\le j\le s\). Moreover every vector in \(V\) is \(f(T)\alpha\), so by definition of \(I_W\), \[ W=\{f(T)\alpha:f\in I_W\}=p(T)^jV. \] Thus the invariant subspaces are exactly the spaces \(p(T)^jV\).
2. Any two nonzero invariant subspaces have nonzero intersection.
The subspace \(p(T)^jV\) is zero exactly when \(j=s\). Indeed \(p(T)^{s-1}\alpha\ne0\), because the annihilator of \(\alpha\) is exactly \(p^s\).
Take nonzero invariant subspaces \[ W=p(T)^jV, \qquad W'=p(T)^kV, \] so \(j,k<s\). Then \[ p(T)^{\max(j,k)}V\subseteq W\cap W'. \] Because \(\max(j,k)<s\), this contained subspace is nonzero. Hence \[ W\cap W'\ne\{0\}. \]
3. No proper nonzero invariant subspace has an invariant complement.
Let \(0\ne W\ne V\) be \(T\)-invariant. If \[ V=W\oplus W' \] with \(W'\) also invariant, then \(W'\ne0\) because \(W\ne V\). But step 2 gives \(W\cap W'\ne0\), contradicting directness. Therefore no nontrivial \(T\)-invariant subspace admits a complementary \(T\)-invariant subspace.