Simultaneous cyclic of $A$ and $A^2$ I'm wondering that whether following statements are right or not, please help me:
a) If a vector space $V$ is $A$-cyclic, then $V$ is $A^2$-cyclic. 
b) If a vector space $V$ is $A^2$-cyclic, then $V$ is $A$-cyclic.
Vector space $V$ is said to be $A$-cyclic if there exists a vector $v \in V$ such that $\{v, Av, A^2v, \dots, A^{n-1}v\}$ is a linearly independent set. One noticeable property of $A$ is that the minimal and the characteristic polynomials of $A$ are the same. 
Thank you so much.
 A: Let
$$A = \operatorname{sip}_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$$
For a vector $v = \begin{bmatrix} v_1 & v_2 \end{bmatrix}^T$, we see that
$$Av = \begin{bmatrix} v_2 \\ v_1 \end{bmatrix},$$
so $\{v, Av\}$ is linearly independent whenever $v_1 \ne \pm v_2$. But, $A^2 = {\rm I}$, so
$$\{v, A^2v\} = \{v, v\}.$$
Technically, $\{v, v\}$ is linearly independent set with only one element, but I don't think that this is acceptable here. Nevertheless, if we define $A = 2\operatorname{sip}_2$, then $A^2 = 4{\rm I}$ and the above will become
$$\{v, A^2v\} = \{v, 4v\},$$
which is a set of $2$ clearly dependent vectors whenever $v \ne 0$. So, a) does not hold.
I expect b) to be true, but I don't know how to prove it at the moment.
A: Hint for statement (b). Let $F$ be the scalar field for $V$. By Cayley-Hamilton theorem, there exists a coefficient matrix $C=(c_{ij})\in M_n(F)$ such that $A^{2(i-1)}=\sum_{j=1}^n c_{ij}A^{j-1}$ for each $i\in\{1,\ldots,n\}$. It follows that for any vector $x\in F^n$, if we define $y^T=x^TC$, then
$$
\sum_{j=1}^n x_jA^{2(j-1)} = \sum_{j=1}^n y_jA^{j-1}.
$$
Given that $V$ is $A^2$-cyclic, is $C$ singular?
