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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.

2 Answers2

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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.

Vedran Šego
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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?

user1551
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