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Can we define any modular eigenvalue for an integer matrix $A\in \mathbb{Z}^{m\times m}$?

If the answer is yes, is there any way to find these modular eigenvalues of a matrix (probably based on roots of unity or something else)? Then I want to write a matrix's characteristic polynomial as: $$p_A(\lambda)\equiv\prod_i (\lambda-y_i)$$ In a way that $\forall\lambda\in\mathbb{Z}\,\,,p_A(\lambda)\equiv\text{det}(\lambda I-A).$


Note: I'm trying to find determinant of $A^{2^m-1}\pm I$ modulo $2^{m+1}$, where $A$ is the following matrix: $$ A=\left(\begin{array}{cccc} a_1 & \ldots & a_{m-1}& a_m\\ 1 & \ldots & 0 & 0\\ \vdots & \ddots & \vdots &\vdots\\ 0 & \ldots &1 & 0 \end{array}\right) $$ with primitive polynomial modulo 2.


EDIT: To this aim I want to use the results of this answer. The results show that if $p_A(\lambda)=\prod_i (\lambda-y_i)$ then $p_{A^n}(\lambda)=\prod_i (\lambda-y_i^n)$. Therefore I want to compute modular eigenvalues of $A$. However for example if $p_A(x)=x^2+x+1$ the roots are not integers.

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The definition of the characteristic polynomial is the same, with the obvious caveat that $$p_A(\lambda)\,\equiv\,det(\lambda\boldsymbol{I}\,-\,A)=\,\Pi_i(\lambda\,-\,\lambda_i)\,\mod\,2^{m+1}$$ where $2^{m+1}>\lambda_i\in\mathbb{Z}$ are the eigenvalues.

For you particular matrix, you may try to find a recurrence that will allow to get $A^{2^m-1}$ or diagonalize it first, exponentiate, add and change back to the original base.

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  • $\begingroup$ Would you please explain it more? How can I diagonalize $A$? $\endgroup$ – SMA.D Sep 15 '15 at 15:57
  • $\begingroup$ All algebraic operations needed are the same as usual. The meaning of the numbers obtained change though: they are all modulo $2^{m+1}$. Where exactly are you stuck? $\endgroup$ – MASL Sep 15 '15 at 16:20
  • $\begingroup$ My problem is that in some cases the eigenvalues are not integers, for example if $p_A(x)=x^2+x+1$ the roots are complex. $\endgroup$ – SMA.D Sep 15 '15 at 16:50
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    $\begingroup$ Well, nobody said that the matrix $A$ of yours is indeed diagonalizable in $\mathbb{Z}$. If you'd come up with a characteristic polynomial like that, you'd conclude it isn't diagonalizable. For example, all symmetric real matrix is diagonalizable in $\mathbb{R}$, but not all real matrix is. All real matrix is diagonalizable in $\mathbb{C}$ however. $\endgroup$ – MASL Sep 15 '15 at 17:16

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