Let \(T\) be a linear operator on the finite-dimensional space \(V\). Prove that there exists a vector \(\alpha\) in \(V\) with this property. If \(f\) is a polynomial and \(f(T)\alpha = 0\), then \(f(T) = 0\). (Such a vector \(\alpha\) is called a separating vector for the algebra of polynomials in \(T\).) When \(T\) has a cyclic vector, give a direct proof that any cyclic vector is a separating vector for the algebra of polynomials in \(T\).
Solution
1. A separating vector always exists.
Let \(m_T\) be the minimal polynomial of \(T\). The preceding consequence of the cyclic decomposition theorem gives a vector \(\alpha\in V\) whose \(T\)-annihilator is exactly \(m_T\) [@HK71]. Thus, if \[ f(T)\alpha=0, \] then \(f\) belongs to the annihilator ideal generated by \(m_T\), so \[ m_T\mid f. \] Write \(f=m_Tq\). Since \(m_T(T)=0\) on all of \(V\), \[ f(T)=m_T(T)q(T)=0. \] Hence this \(\alpha\) is a separating vector for the algebra of polynomials in \(T\).
2. Every cyclic vector is separating, directly.
Now suppose \(\alpha\) is cyclic and \(f(T)\alpha=0\). Every vector \(v\in V\) has the form \[ v=g(T)\alpha \] for some polynomial \(g\), because \(Z(\alpha;T)=V\). Polynomial expressions in \(T\) commute, so \[ f(T)v=f(T)g(T)\alpha=g(T)f(T)\alpha=0. \] Thus \(f(T)\) annihilates every vector in \(V\), hence \[ f(T)=0. \] Therefore every cyclic vector is separating.