Let \(T\), \(V\), and \(F\) be as in Exercise 5. Suppose \(\alpha\) and \(\beta\) are vectors in \(V\) which have the same \(T\)-annihilator. Prove that, for any polynomial \(f\), the vectors \(f\alpha\) and \(f\beta\) have the same \(T\)-annihilator.
Solution
In the section’s notation, \(f\alpha\) means \(f(T)\alpha\) [@HK71]. Let \[ I=\operatorname{Ann}_T(\alpha)=\operatorname{Ann}_T(\beta) =\{h\in F[x]:h(T)\alpha=0\} =\{h\in F[x]:h(T)\beta=0\}. \]
1. Compute the annihilator of \(f(T)\alpha\) from \(I\).
For any polynomial \(g\), \[ \begin{aligned} g\in\operatorname{Ann}_T(f(T)\alpha) &\iff g(T)f(T)\alpha=0\\ &\iff (gf)(T)\alpha=0\\ &\iff gf\in I. \end{aligned} \] Therefore \[ \operatorname{Ann}_T(f(T)\alpha) =\{g\in F[x]:gf\in I\}. \]
2. The same description holds for \(f(T)\beta\).
Using \(\operatorname{Ann}_T(\beta)=I\) in exactly the same calculation gives \[ \operatorname{Ann}_T(f(T)\beta) =\{g\in F[x]:gf\in I\}. \] Hence \[ \boxed{ \operatorname{Ann}_T(f(T)\alpha) =\operatorname{Ann}_T(f(T)\beta).} \] Thus applying any polynomial in \(T\) preserves equality of the two \(T\)-annihilators.