# Finding cyclic vectors for a special matrix

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.

• It is not the vector. It is a vector. – Jack Dec 11 '16 at 13:05
• @Jack Thanks, the question is edited by your hint. – Amin235 Dec 11 '16 at 13:18

• Can I ask you to obtain a cyclic vector for my example in the question with using method in lemma $9$ in the section $5$ of article that you introduced. Excuse me for this request. It is a bit difficult to me to understand lemma $9$ in section $5$ for computing a cyclic vector. Thanks again. – Amin235 Dec 11 '16 at 15:27