Suppose $V$ is a finite-dimensional vector space. Take a linear operator $T \in L(V)$.
Now suppose that we know for every $v \in V$, the set of vectors $\{v, T(v)\, …, T^{k}(v)\}$ is linearly dependent. I want to show that this would imply that the set of linear operators $\{I, T, …, T^{k}\}$ is also linearly dependent.
Here are my thoughts so far:
Define $A_v = \{ p(x) \in F[x]:p(T)(v)=0\}$ where $F$ is the scalar field of $V$. Since $A_v$ is an ideal of $F[x],$ there exists a unique monic polynomial $g_v$, such that $g_v$ generates $A_v$:
$$\langle g_v\rangle=A_v.$$
Now define $G=\{g_v(x): v \in V\}$. If we take $q(x)=\operatorname{lcm}(G)$ (that is, if such a $q$ exists), then it would suffice to show that $\deg(q(x)) \leq k$. If that's proven, then $\{I, T, …, T^{k}\}$ would be linearly dependent.
Edit: This statement actually follows from the Cyclic Decomposition Theorem (Linear Algebra (Ed2), Hoffman, Kunze, p233). As a corollary to this theorem, we have that there exists a vector $\alpha \in V$ such that $g_\alpha$ is the minimal polynomial of $T$ (Again, Hoffman, p237). Now by the hypothesis, for this $\alpha$ we have a polynomial $p(x)$ of degree at most $k$ such that $p(T)(\alpha)=0$. By the definition of $g_\alpha$, $g_\alpha(x)$ divides $p(x)$. We also know that $g_\alpha(x)=m_{T}(x)$. Thus $m_{T}(x)$ divides $p(x)$ and has a degree of at most $k$. Hence, there exists a polynomial of degree less than or equal to $k$ such that its value at $T$ would be zero, which is equivalent to what we are trying to prove.
Although this completes the implication, I'm hoping to find a more elementary proof.
Edit 2: Statement of the Cyclic Decomposition Theorem:
Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let $W_0$ be a proper $T$-admissible subspace of $V$. There exist non-zero vectors $\alpha_1, …, \alpha_r$ in $V$ with respective $T$-annihilators $p_1, …, p_r$ such that
(i) $V=W_0 \bigoplus Z(\alpha_1; T) \bigoplus … \bigoplus Z(\alpha_r; T)$;
(ii) $p_k$ divides $p_{k-1}$, $k=2, …, r$
($Z(\alpha_i; T)$ is the cyclic subspace of $\alpha_i$ (smallest $T$-invariant subspace of $V$ including $\alpha_i$)).
Edit 3: By a $T$-admissible $W$, we mean a $T$-invariant subspace such that for every polynomial $f(x) \in F(x)$, if $f(T)\beta$ is in $W$, then there exists $\gamma \in W$ such that $f(T)\beta=f(T)\gamma$.
Statement of the CDT corollary: Let $T$ be a linear operator on a finite-dimensional vector space $V$. There exists a vector $\alpha$ in $V$ such that the $T$-annihilator of $\alpha$ is the minimal polynomial for $T$.
Best Answer
Given $v\in V$,
$T$ restricts to an endomorphism of $K[T]v$.
Let $f_v$ be the minimal polynomial of $T|_{K[T]v}$.
If $z\in K[T]v\cap K[T]w$ then $f_z$ divides both $f_v$ and $f_w$.
So if $\gcd(f_v,f_w)=1$ then $K[T]v\cap K[T]w=\{0\}$ and hence $f_{v+w} = f_vf_w$.
If $f_v=gh$ then $f_{g(T)v} = h$.
From there you should be able to construct a cyclic vector, that is some $u\in V$ such that $f_u$ is the minimal polynomial of $T$.
From a cyclic vector the result is immediate.