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This query is inspired by this previous question.

Suppose $A$ is an $n \times n$ matrix whose entries are integers between $-s$ and $s$. Suppose further that $A^k=I$ and moreover $k$ is the smallest positive integer with this property. What sort of bounds can be derived on $k$ in terms of $n$ and $s$?

A related question is considered in the discussion to this answer. The question I am asking is slightly different because I am restricting the integers to lie between $-s$ and $s$.

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robinson
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  • There are matrices, e.g. $$A=\begin{pmatrix} 0 & \cdots & 0\ \vdots & \ddots & \vdots \ 0 & \cdots & 0\end{pmatrix}$$ for which there is no such $k$. A slightly less trivial example you get when $s>0$ and $n>1$ is $$A=\begin{pmatrix} 1 & 1 & 0 & \cdots & 0\ 1 & 0 & 0 & \cdots & 0\ 0 & 0 & 0 & \cdots & 0\ \vdots & \vdots & \vdots & \ddots & \vdots\ 0 & 0 & 0 & \cdots & 0\end{pmatrix}$$ – Zev Chonoles Aug 26 '11 at 03:59
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    Zev, the question assumed that there exists an integer $k$ with the property $A^k=I$. I don't understand your "slightly less trivial" example; the matrix has no powers which equals the identity, since it maps $e_n$ to zero. – robinson Aug 26 '11 at 04:53
  • Anyway, the circular shift matrix $A$ satisfies $A^{n-1}=I$ and no smaller power works, so $k$ can be at least that large. I suspect that may be the point you were trying to make. – robinson Aug 26 '11 at 04:54
  • Indeed, I meant a "slightly less trivial example of a matrix with no power equal to the identity". I didn't have any particular point, just thought I'd make the observation - I'm afraid I didn't have anything nearly as useful to say as "$k$ can be at least as large as $n-1$" :) – Zev Chonoles Aug 26 '11 at 05:18

2 Answers2

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I'm not sure the restriction on the entries makes much difference. If $p$ is prime, the polynomial for roots of unity of order $p^r$ has coefficients in $\lbrace\,-1,0,1\,\rbrace$, so the companion matrix for this polynomial has entries from that same set; then a block matrix built from such companion matrices for various primes will have order the product of the prime powers and entries integers between $-1$ and $1$.

Gerry Myerson
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Let $m(x)$ be the minimal polynomial of $A$, and let $\Phi_k(x)$ be the $k$th cyclotomic polynomial (the minimal polynomial in $\mathbb Z[x]$ for a primitive $k$th root of unity). We know that $\Phi_k(x)$ has degree $\phi(k)$. Since $A$ has integer entries, $m(x)$ has integer coefficients. Since $k$ is minimal, this implies that $\Phi_k(x)$ divides $m(x)$. Since $m(x)$ has degree at most $n$, we conclude $\phi(k) \le n$.

This gives a bound on $k$ depending on $n$ only. For example, using the bound $\phi(k) \ge \sqrt{k}$ (valid for $k>6$), we conclude $k \le \max(6, n^2)$. Using better lower bounds on $\phi(k)$, we can improve the bound.

EDIT: This doesn't quite work if $k$ is composite. See comments below.

Ted
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  • I don't know if it is that easy. In fact, Gerry Myerson tried to use the same argument in this question to conclude $\phi(k)\leq n$; but see the discussion that followed. – Arturo Magidin Aug 26 '11 at 04:54
  • Oops, I guess that was too easy. I was trying to generalize the argument for $n=2$ where I knew 6 was the biggest possible value. Still, as Gerry Myerson mentions in that discussion, given a fixed $n$, there are only finitely many possible $k$. – Ted Aug 26 '11 at 05:16