Let $A,B\in M_{3x3}(\mathbb Z/p\mathbb Z)$ ($p$ a prime number). Find the probability $P$ that $AB=BA$ that is $P(AB=BA)$

$$A=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} $$

$$B=\begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \\ \end{pmatrix} $$

Please I would really appreciate if you can help me with this problem. Any ideas or suggestions would be highly appreciated :)

  • $\begingroup$ This is a nice problem---what have you tried so far? $\endgroup$ – Travis Willse Oct 14 '14 at 3:47
  • $\begingroup$ The first thing that came to my mind was just to multiply the matrices then, I have the coefficients of $AB$ and $BA$ then I took the first of them : $a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}=b_{11}a_{11}+b_{12}a_{21}+b_{13}a_{31}$ wich gives us $a_{12}b_{21}+a_{13}b_{31}=b_{12}a_{21}+b_{13}a_{31}$ $\endgroup$ – user128422 Oct 14 '14 at 3:55
  • $\begingroup$ but I don´t know exactly in how many ways $a_{12}b_{21}+a_{13}b_{31}=b_{12}a_{21}+b_{13}a_{31}$ $\endgroup$ – user128422 Oct 14 '14 at 4:00
  • $\begingroup$ You can certainly do a case analysis for this equation. If $a_{12} \neq 0$, then you can solve uniquely for $b_{21}$ for any values of the six remaining variables, and if $a_{12} = 0$, you reduce the problem to a simpler equation. But you'd need to do this for every entry, and the equations are interdependent, so a brute-force analysis of the component equations is probably not the best way to proceed. $\endgroup$ – Travis Willse Oct 14 '14 at 4:03
  • 1
    $\begingroup$ This may well be hard. It might be somewhat easier if you restrict to invertible matrices, which makes this a problem in group theory. Anyway, this article looks potentially relevant: projecteuclid.org/euclid.dmj/1077468920 (gated). $\endgroup$ – Travis Willse Oct 14 '14 at 4:19

This is perhaps not very satisfying, but brute-forcing the computation for $p = 2$ in Maple with the below code gives that there are $7456$ commuting (ordered) pairs of matrices in $M_{3 \times 3}(\mathbb{Z} / 2\mathbb{Z})$ (out of $(2^9)^2$ total pairs), so the probability in that case is $$\frac{7456}{(2^9)^2} = \frac{233}{8192} = 0.028442\ldots,$$ which agrees with loup blanc's experiments.

p := 2;
M := [seq(seq(seq(seq(seq(seq(seq(seq(seq(Mod(p, Matrix(3, (i, j) -> m||i||j), integer), m33 = 0..(p-1)), m32 = 0..(p-1)), m31 = 0..(p-1)), m23 = 0..(p-1)), m22 = 0..(p-1)), m21 = 0..(p-1)), m13 = 0..(p-1)), m12 = 0..(p-1)), m11 = 0..(p-1))]:
k := 0;
for i from 1 to nops(M) do
    for j from 1 to nops(M) do
        if op(convert(Mod(p, M[i].M[j] - M[j].M[i], integer[]), set)) = 0 then k := k + 1 end if;
    end do;
end do;

(Incidentally, if someone knows a better way to generate a list of all matrices in $M_{3 \times 3}(\mathbb{Z} / p \mathbb{Z})$ I'd be grateful to learn how.)

  • $\begingroup$ thanks a lot for your time travis I really appreciate it :) $\endgroup$ – user128422 Oct 16 '14 at 3:22
  • $\begingroup$ You're welcome, I hope you found all of this useful. $\endgroup$ – Travis Willse Oct 16 '14 at 4:38

Let $P_p$ be the required probability. The couples $\{(A,uI_3+vA+wA^2)|A\in M_3\}$ are some solutions ; moreover, "generically", they are THE solutions. Then $P_p\approx \dfrac{p^9p^3}{p^{18}}=\dfrac{1}{p^6}$ ; moreover $P_p\geq 1/p^6$. Numerical experiments (for $p=2,3,5,7$) seem to "show" that $1/p^6\leq P_p\leq 2/p^6$.

  • $\begingroup$ They are THE solutions only if the minimal polynomial of $A$ coincides with the characteristic polynomial. $\endgroup$ – Sungjin Kim Oct 15 '14 at 21:41
  • $\begingroup$ @ i707107 , that you write is not a scoop; indeed, this property of cyclic endomorphisms is well-known. I speak about generic matrices; when the underlying field $K$ is infinite, then almost all matrices are cyclic because almost all matrices have distinct eigenvalues in an algebraic extension of $K$ (choose randomly a complex or real matrix and calculate the discriminant of its characteristic polynomial). $\endgroup$ – user91684 Oct 15 '14 at 23:56
  • $\begingroup$ When $K$ is a finite field, "generic" means that the probability that a matrix is cyclic is close to 1. For instance, if $p=101$, then $99/100$ of the matrices of $M_3(\mathbb{Z}/p\mathbb{Z})$ have distinct eigenvalues. $\endgroup$ – user91684 Oct 15 '14 at 23:57

Edit on 10/13/2016

The number of similarity classes $|I|$ of $\mathrm{GL}_3(\mathbb{F}_p)$ below easily follows from this answer. In fact, without having to consider $8$ different types, we have $$ |I|=\sum_{\lambda \in P_3} p^{\ell(\lambda)} $$ where $P_3$ is the set of partition of $3$, and $\ell(\lambda)$ is the number of parts of $\lambda$.

Then $$ |I|=p^3+p^2+p. $$

An Upper bound for $p\geq 5$

Let $A\in M_3(\mathbb{Z}/p\mathbb{Z})$. Denote by $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z} $ and $\mathbb{F}_p[x]$ the polynomial ring over $\mathbb{F}_p$. We write $M^A=\mathbb{F}_p^3$ the finitely generated module over $\mathbb{F}_p[x]$ where $x \cdot v = A \cdot v$ for any $v\in\mathbb{F}_p^3$. It is well-known that the dimension of the space of commuting matrices $C_A = \{ B \in M_3(\mathbb{Z}/p\mathbb{Z}) | AB=BA\}$ depends only on the similarity class of $A$. The similarity class of $A$ falls into one of the following eight types:

  1. $aaa$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-a)}$ where $a\in\mathbb{F}_p$.

  2. $aa^2$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-a)^2}$ where $a\in\mathbb{F}_p$.

  3. $a^3$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)^3}$ where $a\in\mathbb{F}_p$.

  4. $C$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(C(x))}$ where $C(x)\in\mathbb{F}_p[x]$ is a monic cubic irreducible polynomial over $\mathbb{F}_p$.

  5. $abb$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-b)}\oplus \frac{\mathbb{F}_p[x]}{(x-b)}$ where $a, b\in\mathbb{F}_p$ and $a\neq b$.

  6. $ab^2$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-b)^2}$ where $a, b\in\mathbb{F}_p$ and $a\neq b$.

  7. $aB$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(B(x))}$ where $a \in\mathbb{F}_p$ and $B(x)$ is a monic quadratic irreducible polynomial over $\mathbb{F}_p$.

  8. $abc$-type: $M^A \simeq \frac{\mathbb{F}_p[x]}{(x-a)}\oplus \frac{\mathbb{F}_p[x]}{(x-b)}\oplus \frac{\mathbb{F}_p[x]}{(x-c)}$ where $a,b,c\in\mathbb{F}_p$ are distinct.

Let $\{I_1,I_2,\ldots, I_8\}$ be the set of the above eight types similarity classes, and let $I=\cup_{j=1}^8 I_j$ be the set of all distinct similarity classes. Define $G=GL_3(\mathbb{F}_p)$, and $G_A=\{H\in G| AH=HA\}$. By the orbit-stabilizer formula, $$ |\{(A,B)\in M_3(\mathbb{F}_p)^2 | AB=BA\}|=\sum_{A\in I} |G/G_A| |C_A|\leq \sum_{A\in I} |G| \frac{|C_A|}{|C_A|-3p^{\dim C_A -1}}. $$ For the last inequality, let $A\in I_j$ and $\dim C_A = m$. Let $\{I=B_1, \ldots, B_m\}$ be a basis for $C_A$. Consider $$ \det (c_1 B_1 + \cdots + c_m B_m) = 0. \ \ \ (*) $$ For any given $c_2,\ldots, c_m$, the equation $(*)$ is a cubic equation in $c_1$. Therefore, the number of solution to $(*)$ is at most $3p^{m-1}$. We remark that with a lot of effort, we will be able to find explicit formula for $|G_A|$. This is certainly more difficult than finding $|C_A|$.

Note that $$ 1\leq\frac{|C_A|}{|G_A|}\leq\frac{|C_A|}{|C_A|-3p^{\dim C_A -1}}=\frac1{1-\frac3p}=\frac p{p-3}. $$ The number of elements in eacy $I_j$ can be easily found: $$|I_1|= |I_2|= |I_3|=p, \ |I_4|= \frac{p^3-p}3, \ |I_5| =|I_6|=p(p-1), \ |I_7|=\frac{p(p^2-p)}2,$$ $$|I_8|=\frac{p(p-1)(p-2)}6.$$

Therefore, $$ |\{(A,B)\in M_3(\mathbb{F}_p)^2 | AB=BA\}|\leq \frac p{p-3}|G|\left(3p+\frac{p^3-p}3 +2p(p-1)+\frac{p(p^2-p)}2+\frac{p(p-1)(p-2)}6\right) $$ $$ =\frac p{p-3} |G| (p^3+p^2+p). $$

Combining these, we obtain that $$ \frac{ |G|(p^3+p^2+p)}{(p^9)^2}\leq \frac{|\{(A,B)\in M_3(\mathbb{F}_p)^2 | AB=BA\}|}{(p^9)^2}\leq \frac p{p-3}\frac{|G|(p^3+p^3+p)}{(p^9)^2}. $$ Since $|G|=(p^3-1)(p^3-p)(p^3-p^2)$, we have as $p\rightarrow\infty$, $$ \frac{|\{(A,B)\in M_3(\mathbb{F}_p)^2 | AB=BA\}|}{(p^9)^2}\sim \frac1{p^6}. $$


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