Let \(T\) be a linear operator on \(V\) with minimal polynomial of the form \(p^n\), where \(p\) is irreducible over the scalar field. Show that there is a vector \(\alpha\) in \(V\) such that the \(T\)-annihilator of \(\alpha\) is \(p^n\).
Solution
Let the minimal polynomial of \(T\) be \[ m_T=p^n. \] Because \(p^{n-1}\) is a proper divisor of the minimal polynomial, \[ p(T)^{n-1}\ne0. \] Hence there exists a vector \(\alpha\in V\) such that \[ p(T)^{n-1}\alpha\ne0. \]
Let \(m_\alpha\) be the monic generator of the ideal \[ \{f\in F[x]:f(T)\alpha=0\}, \] the \(T\)-annihilator of \(\alpha\) [@HK71]. Since \(p(T)^n=0\) on all of \(V\), one has \[ m_\alpha\mid p^n. \] The polynomial \(p\) is irreducible, so every monic divisor of \(p^n\) has the form \(p^r\) for some \(0\le r\le n\). Thus \[ m_\alpha=p^r. \] If \(r\le n-1\), then \(m_\alpha\) divides \(p^{n-1}\), so \[ p(T)^{n-1}\alpha=0, \] contrary to the choice of \(\alpha\). Therefore \(r=n\), and \[ \boxed{m_\alpha=p^n}. \] This is the required vector.