Is the cyclic vector for companion matrix independent of the specific companion matrix under consideration?

Question

Suppose $C_1, C_2 \in M_n(\mathbb R)$ are two matrices in companion form. If $v$ is a cyclic vector for $C_1$, i.e., $\{v, C_1 v, C_1^2 v, \dots, C_1^{n-1}v\}$ is a basis for $\mathbb R^n$, does this imply $v$ is also cyclic for $C_2$? The obvious choice of a cyclic vector for companion matrices is $e_1$ and this is certainly cyclic for all companion matrices. But does this in general hold or there is a counterexample?

Answer 1

The answer is no. Consider $$ A=\pmatrix{0&0&1\\ 1&0&0\\ 0&1&0},\ B=\pmatrix{0&0&0\\ 1&0&\ast\\ 0&1&\ast}, \ C=\pmatrix{0&0&1\\ 1&0&-1\\ 0&1&-1}. $$ Then

• $v=e_3$ is cyclic for $A$ but not for $B$. The problem is that $v\in\operatorname{ran}(B)\subsetneqq\mathbb R^n$.
• $(1,1,1)^T$ is cyclic for $C$ but not for $A$, because it is an eigenvector of $A$.

Answer 2

From a probabilistic point of view, you are right.

Let $C_1,\cdots,C_k\in M_n(\mathbb{R})$ be cyclic matrices. Firstly, randomly choose a vector $v\in\mathbb{R}^n$ (for example, the $(v_i)$ follow independent normal laws).

Then, with probability $1$, $v$ is a common cyclic vector of all the $(C_i)$.