Finding dimension of a submodule

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Let $G= (\mathbb{C}^3, A)$ be the $\mathbb C[x]$-module given by $$ A=\left( \begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{matrix}\right). $$ For a vector $v ∈ \mathbb C^3$ let $L(v) := \{f(x)v \mid f ∈ \mathbb C[x]\}$ be the submodule of $G$ generated by $v$.

I'm trying to find all $v$ such that $\dim L(v)=3$.

I've found that the eigenspace of $A$ is $\mathbb{C}e_1$, not really sure what to do next

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The matrix $A$ is nilpotent (satisfying $A^3 = 0$). Thus,

$$ L(v) = \mathrm{span} \{ v, Av, A^2v \}. $$

Check that if $A^2v \neq 0$, then the set $\{ v, Av, A^2v \}$ is linearly independent (this is a general useful property for a sequence of vectors of the form $(v, Av, \ldots, A^kv)$ where $A^{k+1}v = 0$ which comes in handy when analyzing nilpotent matrices). Thus, $L(v) = \mathbb{C}^3$ if and only if $A^2v \neq 0$. You can write explicitly what this means.