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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?

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  • $\begingroup$ Your argument is not correct. Take $v=(0,1,-1)$ for your second example. $\endgroup$ – Omnomnomnom Jan 4 '16 at 10:31
  • $\begingroup$ I am sorry Omnomnomnom, what do you mean with $\overrightarrow{v}=(0,1,-1)$? Do you mean $\overrightarrow{v}=(0,1,1)$? Since we have $-1 = 1$ in $\mathbb{F}_{2}$ And of course, if you mean $\overrightarrow{v}=(0,1,1)$ my argument STILL correct because $\overrightarrow{v}=(0,1,1)$ is the eigenvector that correspond with eigenvalue $1$. Therefore, we cannot choose this vector, i.e. $\overrightarrow{v}=(0,1,1)$ does not qualify as a desired vector. $\endgroup$ – Iqazra Jan 4 '16 at 15:15
  • $\begingroup$ By simple calculation, the characteristic polynomial of the second example is $x^3+x^2+x+1 = (x+1)^3$. The eigenvalue is $1$ with algebraic multiplicity $3$. The corresponding eigenspace is $E_{\mathbf{A}}(1) = \mathrm{span}\{(1,0,0)^{T},(0,1,1)^{T}\}$. Therefore, we cannot take $\overrightarrow{v}=(0,1,-1)=(0,1,1)$ in the second example as a counterexample. $\endgroup$ – Iqazra Jan 4 '16 at 15:22
  • $\begingroup$ *I forgot to put period after $\mathbb{F}_2$ in my first comment, it should be "...$-1=1$ in $\mathbb{F}_{2}$. And ...". $\endgroup$ – Iqazra Jan 4 '16 at 15:30
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    $\begingroup$ Sorry, my bad, I forgot to mention that in my example the matrices are over $\mathbb{F}_{2}$. The vector space is $\mathbb{F}^{3}_{2}$. Thank you Morgan Rodgers. $\endgroup$ – Iqazra Jan 4 '16 at 21:55

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