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Problem For arbitrary matrix $A\in M_n(\mathbb Q)$ such that $\det (xI-A)\in \mathbb Z[x]$ and $\det A=\pm1$, is there any $k\in \mathbb N_+$ such that $A^k\in M_n(\mathbb Z)$?

  • This problem is inspired by the math-overflow answer here. I wonder a generalisation of $(2)\implies(0)$ part: if we assume $\det A=\pm 1$, is there any $k\in \mathbb N_+$ such that $A^k\in \mathrm{SL}_n(\mathbb Z)$?

What I have tried.

  1. So far I have done some numerical verification, where the least $k$ is usually a large integer even for $k=3,4$.

  2. I have also tried to prove the periodicity of $\{d_k\}_{k\in \mathbb Z}$ where $d_k$ is the least positive integer such that $d_k\cdot A^k\in M_n(\mathbb Z)$, but have no ideas.

Drinzjeng Triang
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  • Good question. Note that $\begin{pmatrix}1 & 1/2\0 & 0\end{pmatrix}$ is a counter-example if you didn't require $\det A = \pm 1$, so probably having no non-zero eigenvalues plays into this. – SomeCallMeTim Jul 10 '24 at 15:40
  • @SomeCallMeTim Even requiring just $\det A \ne 0$ isn't enough: $\begin{pmatrix}1&1/2\0&2\end{pmatrix}$ has integer characteristic polynomial and determinant 2, but no power of $A$ is integral. – arkeet Jul 11 '24 at 00:46

1 Answers1

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Since the characteristic polynomial has integer coefficients, there are matrices $B \in M_n(\mathbb{Z})$ and $P \in GL_n(\mathbb{Q})$ with $A = PBP^{-1}$ (by the MO answer linked in the question).

Let $m$ be a positive integer such that $mP$ and $mP^{-1}$ have integer entries. Then $\det B = \det A = \pm 1$ is invertible mod $m^2$, so $B$ represents an element $\bar B$ of the finite group $GL_n(\mathbb{Z}/m^2 \mathbb{Z})$ of invertible matrices mod $m^2$. Therefore, there is a positive integer $k$ such that $\bar B^k = \bar I$ in $GL_n(\mathbb{Z}/m^2 \mathbb{Z})$; that is, $B^k = I + m^2 X$ for some integer matrix $X$. Then $$ A^k = PB^k P^{-1} = I + (mP) X (mP^{-1}) $$ is an integer matrix.

arkeet
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