$\textbf{Definition:}$ 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 $Z(v,T) =\{ g(T)(v) \in V: g \in F[x]\}$.
$\textbf{Exercise:}$ Show that $\dim Z(v,T) = 1 \iff v$ is engevector of $T$.
$\textbf{My attempt:}$
$(\Rightarrow)$
Note that If $\dim Z(v, T) = 1$, $v \neq 0$, because if $v = 0$ then $Z(T,v) =\{0\}$.
Now suppose that $\dim Z(v, T) = 1$, then there is a vector $w \in V$ s.t $Z(v,T) = \langle w \rangle$, that is, $g(T)(v) = \alpha w,$ for all polynomial $g$ in $F[x]$ and for some $\alpha \in F$. Thus, if $g = 1$ we have $v = \lambda w$, that is, we can considerer $Z(v,T) = \langle v \rangle$.
Since $T(v) \in Z(v,T)$, there exists $\lambda \in F$ s.t $T(v) = \lambda v$.
$(\Leftarrow)$
Suppose that $v$ is eigenvector of $T$, then there exist $\lambda \in F$ s.t $T(v) = \lambda v$.
If $\lambda = 0$ then $v \in \ker T$ and for all non-constant polynomial $g \in F[x]$ we have $g(T)(v) = 0$. Thus $Z(v,T) = \langle v \rangle$ and $\dim Z(v,T) = 1$.
If $\lambda \neq 0$, my idea is: Take $p,q \in F[x]$ and show that $p(T)(v)$ and $q(T)(v)$ are linearment dependents, but i'm stuck, let's see:
Let's $\alpha_1, \alpha_2 \in F$, s.t
$\alpha_1 p(T)(v) + \alpha_2 q(T)(v) = 0$
Then we have:
$\alpha_1 P(\lambda)(v) + \alpha_2 q(\lambda)(v) = 0$
Then
$\alpha_1P(\lambda) + \alpha_2 q(\lambda) = 0$
How can I prove that $\alpha_1$ or $\alpha_2$ is nonzero??
Here's how I would do it: If $\lambda \neq 0$, then for any polynomial $g \in F[x]$ we have $$g(T)(v) =g(\lambda)v,$$
where $g(\lambda) \in F$, hence $Z(v, T) \subset \langle v \rangle$. Conversely, it can easily be shown $\langle v \rangle \subset Z(v, T)$.
Using your idea: Note that you may assume both $p(\lambda)v$ and $q(\lambda)v$ are non-zero vectors (otherwise, they are trivially linearly dependent). In particular, since $v$ is non-zero, we may assume $p(\lambda)$ and $q(\lambda)$ are nonzero. If they are dependent, we need to show we can find an $\alpha_1$ and $\alpha_2$ not both zero, such that $\alpha_1p(\lambda)v + \alpha_2q(\lambda)v = 0$. Taking $\alpha_1 = q(\lambda)$ and $\alpha_2 = -p(\lambda)$ works.