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Definition: Suppose $p=x^k+\sum_{i=0}^{k-1}a_i\cdot x^i\in F[x]$ is a monic polynomial. Then companion matrix of $p$ is $$\begin{bmatrix} & & & -a_0\\ 1& & &-a_1\\ &\ddots & & \vdots \\ & & 1& -a_{k-1}\\ \end{bmatrix}\in M_{k\times k}(F)$$

If $U$ is a linear operator on the finite-dimensional space $W$, then $U$ has a cyclic vector if and only if there is some ordered basis for $W$ in which $U$ is represented by the companion matrix of the minimal polynomial for $U$.

Proof: We have just observed that if $U$ has a cyclic vector, then there is such an ordered basis for $W$. Conversely, if we have some ordered basis $\{\alpha_1,…,\alpha_k\}$ for $W$ in which $U$ is represented by the companion matrix of its minimal polynomial, it is obvious that $\alpha_1$ is a cyclic vector for $U$.

Que: Let $\dim (W)=k$ and $U:W\to W$ be a linear operator. Let $m$ be minimal polynomial of $U$. Then $\deg (m)\leq k$. If $\deg (m)\lt k$, then matrix representation of $U$ with respect to some basis equal to companion matrix of $m$ don’t make sense, or is it?

user264745
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  • What would the size of such a matrix be? Would that make sense? – blargoner Feb 09 '23 at 01:37
  • @blargoner yes. Two different size matrix are equal. Can we fix definition of companion matrix so that “matrix representation of $U$ with respect to some basis equal to companion matrix of $m$” make sense? – user264745 Feb 09 '23 at 07:58

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Indeed, it is well known, and quite obvious (look what $Q[M](e_1)$ is when $\deg(Q)\leq{k}$), that the minimal polynomial of a companion matrix$~M$ of $P$ is $P$ itself; in particular its degree equals the dimension of the vector space acted upon. Since the minimal polynomial is a property of a linear operator, it must be the same for any matrix of the operator, so by contrapositive, if the degree of the minimal polynomial of a linear operator is strictly less than the dimension of the space, it cannot have a companion matrix on any base.

The converse is also true: if the degree of the minimal polynomial of a linear operator equals the dimension of the space (so that the minimal polynomial itself coincides with the characteristic polynomial), then there exists a cyclic vector and so a basis for which the matrix it the companion matrix of the minimal polynomial. There are several ways to characterise this situation.

Marc van Leeuwen
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  • Thank you for the answer. To be honest, I find your answer bit hard to follow. Is it really obvious that “minimal polynomial of a companion matrix $A$ of $p$ is $p$ itself”? Following is my proof of that claim: Let $U:F^k\to F^k$ defined by $[U]B=A= \begin{bmatrix} & & & -a_0\ 1& & & -a_1\ &\ddots & & \vdots \ & &1 & -a{k-1}\ \end{bmatrix}$, where $B={e_1,…,e_k}$ is standard basis of $F^k$. Since $U(e_i)=e_{i+1}$, $\forall 1\leq i\leq k-1$, we have $B={e_1,U(e_1),…,U^{k-1}(e_1)}$. So $F^k=\text{span}(B)=\text{span}({U^i(e_1)\mid i\geq 0})=Z(e_1;U)$. – user264745 Feb 12 '23 at 18:51
  • Since $U(e_k)=U^k(e_1)=-a_0e_1-…-a_{k-1}U^{k-1}(e_1)$ and by linear independence of $B$, we have $p$ is $U$-annihilator of $e_1$. By theorem 1 section 7.1, $p$ is minimal polynomial of $U$. Thus $p$ is minimal polynomial of $A$. – user264745 Feb 12 '23 at 18:51
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    I added an answer at that other question, I hope this helps. – Marc van Leeuwen Feb 13 '23 at 07:21