5
$\begingroup$

Let $p$ be an odd prime and $\mathbb Z_p$ be the prime field of order $p$. Consider the matrix ring $R=M_n(\mathbb Z_p)$. Is there any method to count the solutions of the equation (in the ring $R$)

$$X^2=I.$$ Where $I$ is the identity matrix?

$\endgroup$

3 Answers 3

1
$\begingroup$

Denote $k:=Z_p$. If $p\ne2$, the matrix $X$ is in one-to-one correspondence with a decomposition $k^n=E_+ \oplus E_-$, where $E_\pm$ is the eigenspace associated with the eigenvalue $\pm1$.

Given the dimension $m$ of $E_+$ ($n-m$ for $E_-$), these decompositions are in one-to-one correspondence with the bases of $k^n$, quotiented by the action of $GL_m\times GL_{n-m}$. Therefore, their number is $$\frac{|GL_n(Z_p)|}{|GL_m(Z_p)|\cdot|GL_{n-m}(Z_p)|}=p^{m(n-m)}\frac{(p-1)\cdots(p^n-1)}{(p-1)\cdots(p^m-1)(p-1)\cdots(p^{n-m}-1)}.$$ Summing up over all the possible $m$'s, the number of solutions to $X^2=I$ is $$\sum_{m=0}^np^{m(n-m)}\frac{(p^{m+1}-1)\cdots(p^n-1)}{(p-1)\cdots(p^{n-m}-1)}.$$

$\endgroup$
1
  • $\begingroup$ Your formula is precisely the same as my formula, except you wrote $r=n-m$. $\endgroup$ Dec 4, 2012 at 0:18
2
$\begingroup$

Because $p$ is different from $2$, the solutions are the same as direct sum decompositions $\mathbb{Z}_p^{\oplus n} = E_{+1} \oplus E_{-1}$ as $\mathbb{Z}_p$-vector spaces. You can index these by the dimensions, say $r$ and $n-r$. For each, the number of solutions is $$ \frac{(p^n-1)(p^n-p) \cdot \dots \cdot (p^n-p^{r-1})}{(p^r-1)(p^r-p) \cdot \dots \cdot (p^r-p^{r-1})} \cdot p^{r(n-r)} $$ So the final answer is the sum over $r$ from $0$ to $n$ of this factor.

$\endgroup$
1
  • $\begingroup$ Didn't you count twice the powers of $p$ ? $\endgroup$ Dec 3, 2012 at 17:02
2
$\begingroup$

There are $\displaystyle \sum_{i=0}^n \frac{N(n)}{N(i)N(n-i)}$ solutions to your problem, where $N(r)$ is the number of elements in $GL_r(\mathbb{F}_p)$.

Any such $X$ is invertible, and $X\in GL_n(\mathbb{F}_p)$ is a solution if and only if it is conjugate in $GL_n(\mathbb{F}_p)$ to a diagonal matrix with the first $i$ entries equal to $1$ and last $n-i$ entries equal to $-1$. This explains the number of summands. Now $g\in GL_n(\mathbb{F}_p)$ commutes with such diagonal matrix if and only if it preserves both eigenspaces. Hence, there are $\frac{N(n)}{N(i)N(n-i)}$ elements in the conjugacy class.

The formula can be made even more explicit by substituting a formula for $N(r)$.

$\endgroup$
1
  • $\begingroup$ Matrices conjugate to a diagonal matrix with $i$ 1's and $n-i$ $-1$'s on the diagonal form a class of matrices with a fixed rational canonical form. For the general problem of enumerating matrices in $R$ with a fixed rational canonical form, see Theorems 1.10.4 and 1.10.7 of Enumerative Combinatorics, vol. 1, 2nd ed. $\endgroup$ Dec 4, 2012 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.