Use the primary decomposition theorem and the result of Exercise 13 to prove the following. If \(T\) is any linear operator on a finite-dimensional vector space \(V\), then there is a vector \(\alpha\) in \(V\) with \(T\)-annihilator equal to the minimal polynomial for \(T\).
Solution
Factor the minimal polynomial as \[ m_T=p_1^{r_1}\cdots p_k^{r_k}, \] where the \(p_i\) are distinct monic irreducible polynomials. By the primary decomposition theorem, \[ V=W_1\oplus\cdots\oplus W_k, \qquad W_i=\ker p_i(T)^{r_i}, \] and the restriction \(T_i=T|_{W_i}\) has minimal polynomial \(p_i^{r_i}\) [@HK71].
1. Choose one vector realizing each primary power.
Exercise 13, applied to \(T_i\) on \(W_i\), gives a vector \(\alpha_i\in W_i\) whose \(T_i\)-annihilator is exactly \[ p_i^{r_i}. \] Set \[ \alpha=\alpha_1+\cdots+\alpha_k. \]
2. The annihilator of \(\alpha\) is the least common multiple of the component annihilators.
For any polynomial \(f\), \[ f(T)\alpha =f(T)\alpha_1+\cdots+f(T)\alpha_k, \] and the \(i\)th summand lies in \(W_i\) because each \(W_i\) is \(T\)-invariant. The sum \(V=W_1\oplus\cdots\oplus W_k\) is direct, so \[ f(T)\alpha=0 \] if and only if \[ f(T_i)\alpha_i=0 \] for every \(i\). By the choice of \(\alpha_i\), this is equivalent to \[ p_i^{r_i}\mid f \] for every \(i\). Since the distinct irreducibles are pairwise relatively prime, that is equivalent to \[ m_T=p_1^{r_1}\cdots p_k^{r_k}\mid f. \] Thus the annihilator ideal of \(\alpha\) is generated by \(m_T\), and the monic \(T\)-annihilator of \(\alpha\) is \[ \boxed{m_T}. \]