Question about $T-$cyclic subspace generated

Question

$\textbf{Definition:}$ Let $V$ be a finite dimensional vector space over a field $F$ and let $T:V \to V$ be a linear operator. If $v$ is a vector in $V$, the $T-$cyclic subspace generated by $v$ is the subspace $Z(v,T) = \displaystyle \bigcap_{W \in C(v,T)}W$, where $C(v,T) = \{W \subset V: W$ is $T-$ invariant subspace of $V$ and $v \in W$$\}$.

$\textbf{Exercise:}$ Prove that $Z(v,T)= \{g(T)(v) \in V: g \in F[x]\}$.

$\textbf{My attempt:}$

It's clear that $\{g(T)(v) \in V: g \in F[x]\}$ is a subspace of $V$. Since polynomials commute, this subspace is $T-$ invariant. Futhermore, if we take $g = 1$ then $v$ is there. Therefore, $Z(v,T) \subset \{g(T)(v) \in V: g \in F[x]\}$.

For otherwise, let $w \in \{g(T)(v) \in V: g \in F[x]\}$, then exists $p \in F[x] $ s.t $w = p(T)(v)$.

Now, since $v \in Z(v,T)$ and $Z(v,T)$ is $T-$ invariant, we have $p(T)(v) = w \in Z(v,T)$.

Therefore $Z(v,T)= \{g(T)(v) \in V: g \in F[x]\}$.

Is that correct?

Answer 1

Your proof is basically correct, but confusing. Denote $U = \{g(T)(v) \in V: g \in F[x]\}$. I would break the proof down into 2 steps:

1. $Z(v,T) \subset U$. This is what you have shown in the first paragraph of your proof, and this part of your proof is understandable. To reiterate, this is a consequence of the fact that $U$ is $T$-invariant and contains $v$, which means that $U$ is one of the sets in the intersection over $C(v,T)$.

2. $U \subset Z(v,T)$. This is true because $U \subset W$ for every $W \in C(v,T)$: for any $W \in C(v,T)$, $T$-invariance implies that $p(T)v \in W$ for any polynomial $p$.