I want to verify the following proposition and construct a formal proof whenever it is true.

Suppose $\mathbf{A}$ is an $n \times n$ matrix over $\mathbb{F}_p$ such that $\mathrm{ord}(\mathbf{A})= \ell$, i.e. $\ell$ is the least positive integer such that $\mathbf{A}^\ell = \mathbf{I}$, where $\mathbf{I}$ denotes the identity matrix.

Let $\overrightarrow{v}$ be a non zero vector in $\mathbb{F}^n_p$. If $\mathbf{A}^i \overrightarrow{v}= \mathbf{A}^j \overrightarrow{v}$ for two distinct to positive integer $i < j$, then $i \equiv j \pmod \ell$.

If not all non zero vectors satisfy the condition, then determine all non zero vectors that satisfy the condition.

I suspect that the condition holds for all non zero vectors. My attempt to prove the proposition is as follows. Let $\overrightarrow{v}$ be a non zero vector that satisfy $\mathbf{A}^i \overrightarrow{v}= \mathbf{A}^j \overrightarrow{v}$ for two distinct to positive integer $i < j$, since $\mathbf{A}$ is invertible, we have $\overrightarrow{v}= \mathbf{A}^{(j-i)} \overrightarrow{v}$.

What are the next steps? We do not have a cancellation law for vectors. (If such thing exists, then we have $\mathbf{I}=\mathbf{A}^{(j-i)}$, and thus $i \equiv j \pmod \ell$).

Hence, I tried the following reasoning. The condition $\overrightarrow{v}= \mathbf{A}^{(j-i)} \overrightarrow{v}$ yields $(\mathbf{A}^{(j-i)}-\mathbf{I})\overrightarrow{v}= \overrightarrow{0}$. This yields an equation: $(\mathbf{A}^{t}-\mathbf{I})\overrightarrow{v}= \overrightarrow{0}$ for $t \in \mathbb{N}$. If for any $t \not \equiv 0 \pmod \ell$ the matrix $\mathbf{A}^{t}-\mathbf{I}$ is invertible, then $\mathrm{ker}(\mathbf{A}^t-\mathbf{I})=\overrightarrow{0}$. Thus $\mathbf{A}^i \overrightarrow{v}\not = \mathbf{A}^j \overrightarrow{v}$ if $i \not \equiv j \pmod \ell$. I got stuck here. I have no idea to prove the invertibility.

ADDENDUM: I have already checked using SAGE mathematical software that the condition is true in $GL(n,2)$ for $n=3,4,5$, where the matrices $\mathbf{A}$ have irreducible characteristic polynomials.

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    $\begingroup$ The condition certainly does not hold for all nonzero vectors. Try $\mathbf{A} = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}\right)$, $\ell = 2$ and $\overrightarrow{v} = \left(1,0,0\right)^T$. Then, $\mathbf{A}^1 \overrightarrow{v} = \mathbf{A}^2 \overrightarrow{v}$. $\endgroup$ – darij grinberg Jan 3 '16 at 15:18
  • $\begingroup$ You need to make sure that $v$ is not taken from the wrong eigenspaces. $\endgroup$ – Omnomnomnom Jan 3 '16 at 21:27
  • $\begingroup$ What does "wrong eigenspaces" mean? $\endgroup$ – Iqazra Jan 4 '16 at 2:11
  • $\begingroup$ Thank you darij grinberg, then my question is: is there any $\overrightarrow{v}\not =\overrightarrow{0}$ that satisfy the condition? $\endgroup$ – Iqazra Jan 4 '16 at 2:51
  • $\begingroup$ I see that the vector $\overrightarrow{v}=(1,0,0)^{T}$ in darij grinberg counterexample is also the eigenvector of the matrix $\mathbf{A}$ associated with eigenvalue $1$. So my question is: is it sufficient to take $\overrightarrow{v}$ such that $\overrightarrow{v} \not \in E_{1}$ where $E_{1}$ denotes the eigenspace of $\mathbf{A}$ associated with eigenvalue $1$? In accordance to omnomnom hint, does this mean the "wrong eigenspace" is $E_{1}$? $\endgroup$ – Iqazra Jan 4 '16 at 6:06

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