Exercise 7.3.15

Answers

1.: Let $W$ be the $T$ -cyclic subspace generated by $x$ . And let $p (t)$ be the minimal polynomial of $T_{W}$ . We know that $p (T_{W}) = T_{0}$ . If $q (t)$ is a polynomial such that $q (T) (x) = 0$ , we know that $q (T) (T^{k} (x)) = T^{k} (q (T) (x)) = 0$ for all $k$ . So $p (t)$ must divide $q (t)$ by Theorem 7.12. Hence $p (t)$ is the unique $T$ -annihilator of $x$ .
2.: The $T$ -annihilator of $x$ is the minimal polynomial of $T_{W}$ by the previous argument. Hence it divides the characteristic polynomial of $T$ , which divides any polynomial for which $g (T) = T_{0}$ , by Theorem 7.12.
3.: This comes from the proof in Exercise 7.3.15(c).
4.: By the result in the previous question, the dimension of the $T$ -cyclic subspace generated by $x$ is equal to the degree of the $T$ -annihilator of $x$ . If the dimension of the $T$ -cyclic subspace generated by $x$ has dimension $1$ , then $T (x)$ must be a multiple of $x$ . Hence $x$ is an eigenvector. Conversely, if $x$ is an eigenvector, then $T (x) = λx$ for some $λ$ . This means the dimension of the $T$ -cyclic subspace generated by $x$ is $1$ .

jephianlin

2011-06-27 00:00