# Bounding the multiplicative order of matrices in $\mathbb M_n(\mathbb Z)$

Let $\mathbb N$ be the set of positive integers.

Prove that:

$\forall n\in\mathbb{N}:\exists r\in\mathbb{N}$ (let's say $r=r(n)$ as a function of $n$) such that:

If $M\in \mathbb{M}_n(\mathbb{Z})$ and $m$ is a positive integer and $M^m=I_n$, then we have $M^r=I_n$.

(Paraphrase: for all integer $n\times n$ matrices $M$ such that a power of $M$ is identity, we have $M^r$ is identity.)

• Are you sure that is the correct statement? It seems to me that as you are given $M^m=I_n$ you can immediately say $\exists r\in \mathbb N$ so the $M^r=I_n$ - just take $r=m$. – Dom Oct 1 '14 at 13:06
• What are the matrix components? Complex numbers, real numbers, rational numbers, integers? – Daniel Fischer Oct 1 '14 at 13:08
• Depending on how you defined $\mathbb N$ and the matrix exponents $r=0$ could be a solution too. – flawr Oct 1 '14 at 13:08
• If we are working over a finite field, then this is evident, as there are only a finite number of matrices with a given rank. – awllower Oct 1 '14 at 13:22
• The point of the question is that there's a single $r$ that works for every $n\times n$ matrix $M$ that has some power equal to the identity. If $M^m=I$ then the eigenvalues are roots of unity, so the characteristic polynomial involves cyclotomic polynomials, but also the characteristic polynomial is of degree $n$, so there a limit on which cyclotomic polynomials it can involve. But as others have mentioned, it would be a good idea to tell us where the entries come from. – Gerry Myerson Oct 1 '14 at 13:28

This is obviously wrong, already for $n=2$. You've got real rotation matrices $R$ of any finite order $m$, but not one order $r$ such that $R^r=I_2$ for all such $R$ (with varying $m$) at once.
On the other hand, now that the question has changed and entries are forced to be integers, the situation changes. In fact rational coefficients would suffice, which is what I will use. Any $n\times n$ rational matrix satisfies a polynomial$~P$ of degree$~n$ with rational coefficients, by the Cayley-Hamilton theorem. If $M^m=I_n$, then the minimal polynomial $\mu$ of$~M$ over$~\Bbb Q$ is a product of distinct cyclotomic polynomials$~\Phi_d$ with $d\mid m$, since (1) $\mu$ divides $X^m-1$ by assumption; (2) $X^m-1=\prod_{d\mid m}\Phi_d$; and (3) each$~\Phi_d$ is irreducible in$~\Bbb Q[X]$. Since $\mu$ divides $P$, the degree $\phi(d)$ of each factor $\Phi_d$ must certainly be at most$~n$ (in fact their sum must be so). The main point is that for given $n$ there are only finitely many such cyclotomic polynomials. Assuming that, we can take $r$ to be the least common multiple of all possible $d$, and $M^r=I_n$ is assured, since all factors of$~\mu$, and therefore $\mu$ itself, divide$~X^r-1$.
Now $\phi(m)$ is the product over all primes$~p$ that divide$~m$ of $(p-1)p^{e-1}$, where $e$ is the (positive) multiplicity of $p$ in the factorisation of$~m$. With the bound $n$ on $\phi(m)$ imposed, only primes $p\leq n+1$ can participate (a finite number), and for each there is a bound on the possible multiplicity$~e$. This leaves only finitely many possible factorisations, allowing only finitely many different values $m$. QED.
• Yeah I have misread the $\mathbb{Z}$ part. So many thanks!!! – Learning Oct 1 '14 at 13:46
• What do you mean by "Since $M$ divides $P$"? – Learning Oct 1 '14 at 16:16
• Oops, I used $M$ both for the matrix and for its minimal polynomial. In "$M$ divides $P$" it is the latter meaning; this is what the Cauley-Hamilton theorem (over fields) says. But I'll replace the minimal polynomial by $\mu$ now. – Marc van Leeuwen Oct 1 '14 at 16:53