I want to verify following proposition and construct a formal proof whenever it is true. This question is related to this one.
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 $\mathbb{F}^n_p$. For which values of $\overrightarrow{v}\not = \overrightarrow{0}$ such that $\mathbf{A}^{t}\overrightarrow{v} \not =\overrightarrow{v}$ for all integers $1\leq t < \mathrm{ord}(\mathbf{A})$?
EDIT: The following examples are in $\mathbb{F}^{3}_{2}$. The matrices are over $\mathbb{F}_{2}$.
For example, if $\mathbf{A} = \Bigl[ \begin{smallmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \Bigr]$ and $\overrightarrow{v}=(1,0,0)^{T}$ then $\mathbf{A}^{i} \overrightarrow{v}\not = \overrightarrow{v}$ for all $1 \leq i < \mathrm{ord}(\mathbf{A})=7$.
Another example, if $\mathbf{A} = \Bigl[ \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \Bigr]$ and $\overrightarrow{v}=(0,1,0)^{T}$ then $\mathbf{A}^{i} \overrightarrow{v}\not = \overrightarrow{v}$ for all $1 \leq i < \mathrm{ord}(\mathbf{A})=2$
However, if we take $\mathbf{A} = \Bigl[ \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \Bigr]$ and $\overrightarrow{v}=(1,0,0)^{T}$, the condition does not hold, since $\mathbf{A}^{1}\overrightarrow{v}=\overrightarrow{v}$ yet $\mathrm{ord}(\mathbf{A})=2$.
I suspect that the non zero vector $\overrightarrow{v}$ must not lies in the eigenspace $E_{\mathbf{A}^{k}}(1)$, i.e. the eigenspace of $\mathbf{A}^{k}$ associated with eigenvalue $1$.
In other words the non zero vector $\overrightarrow{v}$ must not be an eigenvector that correspond to eigenvalue $\lambda$ where $\lambda$ is the $i$-th root of unity for $1 \leq i < \mathrm{ord}(\mathbf{A})$.
Is my argument correct?