This questsion is an extension of a previously asked question that I think needs a more rigorous proof (please see the link and the answer by user Aaron).
The question is, given a little bit of different notation (in accordance with control theory), for $b\in\mathbb{R}^{n}$ and $A\in\mathbb{R}^{n\times n}$, what is a sufficient condition for
\begin{align} b,Ab,A^{2}b,\dots,A^{n-1}b \end{align}
all being linearly independent? Furthermore, if we assume that $A$ has $n$ distinct eigenvalues $\lambda_{i}$, $i=1,\dots,n$, is this condition sufficient for the above to hold and if so how do we prove it?
In the above linked question Aaron's answer was:
If the characteristic polynomial of $A$ has no repeated roots, then we have a basis $\{v_{i}\;\vert\;1\leq i\leq n\}$ of eigenvectors for $\mathbb{C}^{n}$, and we can express $x=\sum \alpha_{i}v_{i}$ in terms of the basis. If all of the $\alpha_{i}$ are non-zero, then all of your vectors will be linearly independent. In fact, with this condition on $A$, this is a necessary and sufficient condition on $x$.
If $A$ has an eigenvalue with geometric multiplicity greater than 1, then no $x$ will work.
If $A$ has repeated eigenvalues but all eigenvalues have geometric multiplicity of 1, then it is still possible to find such an $x$, but things are a bit more complicated.
I would be very happy if someone could fill in the blanks for his statements, it is not immediately evident that a $b$ with nonzero entries leads to all vectors $b,Ab,\dots,A^{n-1}b$ being linearly independent.
No assumption is made on the field $k$ and $A\in M_n(k)$.
For $b\in k^n$, the $A^0 b,\ldots, A^{n-1}b$ are linearly dependent iff there is some non-zero polynomial $h\in k[x]$ of degree $<n$ such that $h(A)b=0$.
With $p\in k[x]$ the minimal polynomial of $A$ and $g=\gcd(p,h)$ then $h(A)b=0,p(A)b=0$ implies $g(A)b=0$.
Thus the necessary and sufficient condition for the $A^0 b,\ldots, A^{n-1}b$ to be linearly independent is that $\deg(p)=n$ and $f(A)b\ne 0$ for each $f\in k[x]$ strictly dividing $p$.
When the $\deg(p)=n$ condition is satisfied, factorize in irreducibles $p=\prod_j q_j^{e_j}\in k[x]$. Taking $b_j \in \ker(q_j(A)^{e_j}) ,\not \in \ker(q_j(A)^{e_j-1})$ then $b = \sum_j b_j$ works.
(take $f$ such that $p = q_j f$, then $0=f(A)b = f(A)b_j=\gcd(f,q_j^{e_j})(A)b_j=q_j(A)^{e_j-1}b_j$ is a contradiction)