Let $A\in M_n$ be a non-derogatory matrix (in other words, its minimal and characteristic polynomials coincide). We call vector $\vec{v} \in \mathbb{R}^n$ is cyclic if $\{A^i \vec{v}\}_{i = 0}^{n – 1}$ be linearly independent. The following theorem ensure that there are such cyclic vectors.
Theorem: Let $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same.(section 7.1 in Linear algebra by Hoffman-Kunze)
My question: Is there a method for obtaining the cyclic vectors when we have a matrix that it's minimal and characteristic polynomials coincide or should choose a random vector and test it, is cyclic or not?
Example: Consider the following matrix
$$A=\left( \begin {array}{ccc} -3&1&-1\\ -7&5&-1
\\-6&6&-2\end {array} \right)$$
The minimal and characteristic polynomials matrix $A$ is as follows
$$
{x}^{3}-12\,x-16
$$
If we choose vector $$v= {\left[ \begin {array}{ccc} 1&2&1\end {array} \right]}^T $$
then we have
$$
F=\{A^i \vec{v}\}_{i = 0}^{3 – 1}= \left( \begin {array}{ccc} 1&-2&4\\ 2&2&20
\\ 1&4&16\end {array} \right)
$$
we can check that $det(F)=0$ and it means vector $v$ is not cyclic. But If we choose vector $$v= {\left[ \begin {array}{ccc} 1&3&1\end {array} \right]}^T $$
then we have
$$
F=\{A^i \vec{v}\}_{i = 0}^{3 – 1}=\left( \begin {array}{ccc} 1&-1&0\\3&7&32
\\ 1&10&28\end {array} \right)
$$
we can check that $det(F)=-72$ then we conclude that vector $v$ is cyclic.
Best Answer
The following research paper
ON THE COMPUTATION OF MINIMAL POLYNOMIAL, CYCLIC VECTORS AND THE FROBENIUS FORM (section 5: Searching for a cyclic vector)
should give an answer to your question.