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I need to show that if $A$ and $B$ have a cyclic vector and have equal characteristic polynomial, then they are similar.

Here I know that given $T: V \rightarrow V$ with $n=\operatorname{dim} V,$ a $T$ -cyclic vector $v \in V$ satisfies $B=\left(v, T(v), \ldots, T^{n-1}(v)\right)$ is an ordered basis of $V$, and $[T]_{B}=L_{\Delta_{T}}=L_{\psi_{T}}$, where $L_{f}$ is the companion matrix defined by the polynomial $f$. But I'm not seeing how I can use this information to prove what I need.

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You have all of the pieces, you just need to put them together.

$A$ and $B$ have a cyclic vector and have common characteristic polynomial $f$. Let $v$ denote an $A$-cyclic vector, and let $w$ denote a $B$-cyclic vector. Let $\mathcal A = (v,Av,\dots,A^{n-1}v)$, and let $\mathcal B = (w,Bw,\dots,B^{n-1}w)$. We note that $$ [A]_{\mathcal A} = [B]_{\mathcal B} = L_f. $$ In other words, $A$ is similar to $L_f$ and $B$ is similar to $L_f$. Because matrix similarity is an equivalence relation, this means that $A$ is similar to $B$.

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