Consider matrices $M$ of size $L\times L$ over a finite field $\mathbb{Z}_p$, for simplicity focus on $p$ prime. The size $L$ is even. We want to find the order of a specific class of matrices, namely we want to find the smallest non-zero integer $n$ such that $$M^n=1,$$ where 1 is here the identity matrix of size $L\times L$. The matrix $M$ has a specific block structure, which originates from a specific linear cellular automaton. We have $$M=AB,$$ where $$A= \begin{pmatrix} 1 & 1 & & && & \\ 1 & -1 && & & & \\ & & 1 & 1 && &\\ & & 1 & -1 && & \\ &&&& \ddots & & \\ & & & &&1 & 1\\ & & & &&1 & -1\\ \end{pmatrix}$$ and $$ B= \begin{pmatrix} -1 && & & && 1&\\ & 1 & 1 && && \\ & 1 & -1 && && \\ &&& \ddots & & & \\ & & &&1 & 1&\\ & & &&1 & -1&\\ 1 && & & && 1\\ \end{pmatrix}$$

You can see $B$ as $CAC^{-1}$ where $C$ is a cyclic shift over $L$ variables. Separately $A$ and $B$ have simple properties, but their alternating product becomes complicated.

I did some numerical testing, choosing first $\mathbb{Z}_3$. The interesting thing is (which also motivates me to look deeper into this) that the order $n$ seems to have a complicated behaviour as a function of $L$ and it is not clear to me what is the precise source of this. Sometimes $n$ is very big, seemingly exponentially growing with $L$. For example $n(L=46)=354292=2^2\cdot 23\cdot 3851$. Or $n(L=58)=9565940=2^2\cdot 5\cdot 29\cdot 16493$. But if $L$ is divisible by 6 then we get much lower numbers, $n(60)=120$ for example. I understand that it is natural to see the prime $p$ somehow reflected in the function $n(L)$, but what I don't understand is how the big primes mentioned above can enter the game.

Update: Further numerical experiments showed that the smallest orbit happens when $L=2p^m$, in which case $n=2L$, we have simple linear growth. It gets complicated and quickly growing for most other $L$.

The origin of the problem is the following cellular automaton. Consider $L$ variables $s_j\in \mathbb{Z}_p$, with $j=1,2,\dots,L$. We define a dynamics on the model, such that we perform two operations cyclically. In each operation we group two neighbouring variables into a pair, and we alternate the ways how we make the pairing. For each pairing we perform the update $$(s_j,s_{j+1})\to (s_j+s_{j+1},s_j-s_{j+1})$$ What changes is that at every second update $j$ is even or odd. If you represent this linear operation with matrices, you get $A$ and $B$ and we want to iterate the product $M=AB$.


2 Answers 2


Big primes enter the picture as follows: Compute the characteristic polynomial $P_L$ of your square matrix $M$ of size $L$. Factor $P_L$ over your working prime $p$. This gives you the eigenvalues of the involved Jordan-blocs over $\mathbb F_p$. These eigenvalues are elements of field extensions of degree at most $L$ of your groundfield $\mathbb F_p$. You can thus get large primes dividing the order of $M$ if they are divisors of $p^k-1$ for $k$ the degree of (at least) one irreducible polynomial (over $\mathbb F_p$) dividing $P_L$.

A fast way for computing the order of $M$ is thus to compute the characteristic polynomial $P_L$ of $M$, factor it over $\mathbb F_p$ and check then if prime-divisors of $p^k-1$ (for $k$ the degree of an involved irreducible polynomial) divide the order.

The order of $M$ over $\mathbb F_p$ is necessarily a divisor of $U=p^{L-1}\prod_{k=1}^{L}(p^k-1)$. One can of course remove a factor $p^k-1$ from this product if $P_L$ mod $p$ has no irreducible divisor modulo $p$ of degree $k$. One can further reduce the size of $U$ by taking greatest common divisors of all involved factors. The exponent $L-1$ of $p$ in $U$ can of course be replaced by $\mu-1$ where $\mu$ is largest occuring multiplicity among the irreducible divisors of $P_L \pmod p$. (There seem indeed to be often multiplicities in $P_L$: This leads to possibly non-trivial Jordan-blocks in $M$ which contribute a factor $p^d$ to the order of $M$ where $d+1$ is the dimension of the largest involved Jordan bloc.)

The computational bottleneck of this approach is the factorization of $p^k-1$. The rest is computationally easy using fast exponentiation. Concretetely, You have to check if $M^{U/q}$ is still the identity for $q$ any prime divisor of $U$. If this is the case, replace $U$ by $U/q$ (and recheck if you can remove one more power of $q$ from $U$). Going through all prime divisors of $U$ gives you at the end the exact order of $M$ over $\mathbb F_p$.

Experimental observation The characteristic polynomial $P_L$ of the square matrix $M=AB$ of even size $L$ seems to be given by $$P_L=(x^2+4)Q^2_{L/2}$$ where $Q_1=1,\quad Q_2=x+2$ and $Q_n=(x+2)Q_{n-1}-xQ_{n-2}$ for $n\geq 3$. (The polynomials $Q_n$ are 'almost' orthogonal polynomials and seem to satisfy the divisibility property $P_a\vert P_b$ if $a\vert b$.) This formula holds for $L\leq 60$ even.

  • $\begingroup$ Thank you, this is useful! But there are indeed many problems from a practical point of view. For example, would you expect that the factorization of the char.pol. over the finite field can be done analytically, for every $L$? I don't expect this to happen.... (I guess my $L$ is your $n$) But at least now I know how the large primes can enter the game... Perhaps the nice result for $L=2p^m$ could be proven analytically, without a case by case computation... $\endgroup$ Commented Dec 10, 2021 at 13:12
  • $\begingroup$ I have changed my notations to yours and I have added a small experimental observation on the involved characteristic polynomials. $\endgroup$ Commented Dec 10, 2021 at 20:22
  • $\begingroup$ Thanks, interesting! For the characteristic polynomial did you work here above the real numbers? I don't see the appearance of the field characteristic. But if the divisibility holds with integer coefficient polynomials, then I understand that it also holds over the finite field case. In any case, if we prove this relation and the divisibility, it could lead to a proof of at least the case $L=2\cdot p^m$. That is already useful. $\endgroup$ Commented Dec 11, 2021 at 4:18
  • $\begingroup$ Divisibility of $P_{ab}$ by $P_a$ is obviously true (independently of the conjectural truth of the recursion relations for $Q_n$) by the Fourier analysis argument made by Denis Serre (easy proof: Repeat coordinates $(v_1,\ldots, v_{2a})$ of an eigenvector for $M_a$ in order to get an eigenvector of the same eigenvalue for $M_{ka}$. This factorization holds thus over ℤ[𝑥]. Irreducible polynomials over ℤ decay further over primes according to the ramification/intertia stuff of classical algebraic number theory. $\endgroup$ Commented Dec 17, 2021 at 14:22
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    $\begingroup$ @Balazs Pozsgay : I could not find your e-mail so easily. Try [email protected] (this should work, my University had recently a namechange and the administrative people at the top considered it a great idea to disable the old adresses) $\endgroup$ Commented Dec 23, 2021 at 12:30

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity in a suitable extension of ${\mathbb F}_p$.

The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$ (which turns out to be a discrete Fourier analysis). Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.

In other words, defining the vectors $v_\omega=(1,\omega,\ldots,\omega^{\ell-1})$, we see that the subspaces defined by $y,z\in{\rm Span}(v_\omega)$ are stable under both $A$ and $B$, and we are able to compute the spectrum of $AB$ by studying its restriction to these $2$-dimensional spaces. Notice that the action on the spaces associated with $v_{\omega}$ and $v_{\omega^{-1}}$ are conjugate to each other.

  • $\begingroup$ Thanks! I understand the Fourier transform method, that would have been the natural idea if we are above the complex numbers. Here it is not so clear to me, what is where. For example the $\omega$ root of unity is a complex number, right? Or do you treat it in some field extension? And what about $\lambda_\omega, \mu_\omega$? Where are they? From a practical point of view, I can regard them as complex numbers, and then for their orders I just compute multiplication in $\mathbb{C}$ and observe when is a specific power an integer and a multiple of $p$. This is what I would do now... $\endgroup$ Commented Dec 13, 2021 at 9:37
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    $\begingroup$ @BalázsPozsgay No, there are not complex numbers. They belong to some extension, and this is why I assume that $p$ does not dvide $L$. $\endgroup$ Commented Dec 13, 2021 at 11:02

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