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I am stuck with this exercise. Can anyone give me some hints on how to proceed?

Let $A$ be an integral hyperbolic matrix with determinant $\pm 1$ and $v_1, v_2 \in \mathbb{R}^2 \setminus \{ (0, 0) \}$

Show that the equation $(A^T)^n v_1 - v_2 = 0$ has at most one solution $n \in \mathbb{Z}$.

Perhaps I'm wrong but shouldn't $v_1, v_2$ be eigenvectors here?

Kajice
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  • What is $(^tA)$, is that a notation for the solution of $\dot x=Ax$, or ...? Are the $v_k$ vectors or numbers? – Lutz Lehmann Apr 21 '23 at 07:32
  • I assume $^tA$ is the transpose of A. $v_k$ are vectors, I made a typo, I will correct it. – Kajice Apr 21 '23 at 07:47
  • Then $A^\top$ is a more usual notation than ${}^t!!A$. But why use that, the spectrum is unchanged for the transposed matrix, so as general statement it makes no difference which version is used. – Lutz Lehmann Apr 21 '23 at 07:51
  • Yes I was also confused by that. I just copied the exercise exactly how I was given it. – Kajice Apr 21 '23 at 07:55
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    If $X^mv=X^nv$, then $v$ is a kernel vector, which is excluded, or it is an eigenvector to a unit-root eigenvalue which does not exist. – Lutz Lehmann Apr 21 '23 at 07:58

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