# The number of $3\times 3$ nilpotent matrices over $\mathbb{F}_q$ using the Orbit-Stabilizer theorem

The Fine-Herstein theorem says that the number of nilpotent $n\times n$ matrices over $\mathbb{F}_q$ is $q^{n^2-n}$. I am trying to verify this for the case $n=3$ using the orbit-stabilizer theorem.

Here is my method:

• Knowing that the minimal polynomial is $x^m$ for $m\leq n$, I determine the possible rational canonical forms:

$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$ $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)$ $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)$

• Let $GL_n(\mathbb{F}_q)$ act by conjugation on each of these matrices, then the sum of the sizes of the three orbits is the number of nilpotent matrices.

We know that $\mid GL_n(\mathbb{F}_q) \mid =(q^n-1)\cdots (q^n-q^{n-1})$, so that all the work goes in to finding the size of the stabilizer.

This method worked like a charm for the $n=2$ case, but I am having trouble finding what an element of the stabilizer looks like in each case for $n=3$. Is this possible to do just by conjugating by a general matrix and setting it equal to the class representative?

• The orbit of each of the above matrices under the conjugation action by the general linear group consists of nilpotent elements. Further, all nilpotent matrices lie in one of these orbits. Thus the sum of the sizes of the orbits is the number of nilpotent matrices. May 25, 2015 at 22:12

## 2 Answers

I looked up this theorem and a proof I found was so nice I wanted to reproduce it here.

A standard fact of linear algebra states that given any linear map $T$ of a finite-dimensional space $V$, we have $V=\operatorname{im}(T^d)\oplus\ker(T^d)$ where $d=\dim V$.

The map $T$ decomposes as $\rm invertible\oplus nilpotent$ acting on this decomposition; moreover every decomposition $V=A\oplus B$ and pair $(X,Y)\in{\rm GL}(A)\times{\rm Nil}(B)$ uniquely determines every endomorphism of $V$. Note ${\rm Nil}(-)$ is the space of nilpotent maps. Therefore, assuming now that $V$ is finite-dim over a finite field $\Bbb F_q$, the total number of endomorphisms of $V=\Bbb F_q^d$ is given by

$$q^{d^2}=\sum_{V=A\oplus B} |{\rm GL}(A)|\cdot|{\rm Nil}(B)|.$$

Write $N(k)$ for the number of nilpotent matrices in $M_{k\times k}(\Bbb F_q)$. We may count the number of decompositions $V=A\oplus B$ with $\dim B=k$ using orbit-stabilizer: ${\rm GL}(\Bbb F_q^n)$ acts transitively on such decompositions and the stabilizer of $\Bbb F_q^{n-k}\oplus\Bbb F_q^k\,$ is ${\rm GL}(\Bbb F_q^{n-k})\times{\rm GL}(\Bbb F_q^k)$. Thus

$$\begin{array}{ll} \displaystyle q^{d^2} & \displaystyle =\sum_{k=0}^d\left(\frac{|{\rm GL}(\Bbb F_q^n)|}{|{\rm GL}(\Bbb F_q^k)\times{\rm GL}(\Bbb F_q^{n-k})|}\right)|{\rm GL}(\Bbb F_q^{n-k})|N(k) \\ & \displaystyle = \sum_{k=0}^d\frac{|{\rm GL}(\Bbb F_q^d)|}{|{\rm GL}(\Bbb F_q^k)|}N(k) \\ & \displaystyle = N(d)+\frac{|{\rm GL}(\Bbb F_q^d)|}{|{\rm GL}(\Bbb F_q^{d-1})|} \sum_{k=0}^{d-1}\frac{|{\rm GL}(\Bbb F_q^{d-1})|}{|{\rm GL}(\Bbb F_q^k)|}N(k) \\ & = \displaystyle N(d)+\frac{|{\rm GL}(\Bbb F_q^d)|}{|{\rm GL}(\Bbb F_q^{d-1})|}q^{(d-1)^2}. \end{array}$$

Counting $|{\rm GL}_n(\Bbb F_q)|=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$ is a classical exercise: to construct an invertible matrix, first choose a nonzero vector, second choose a vector which is not a scalar multiple of the first, third choose a vector which is not in the span of the first two, and so on.

So one may compute

$$\frac{|{\rm GL}(\Bbb F_q^d)|}{|{\rm GL}(\Bbb F_q^{d-1})|}=\frac{q^d-1}{1}\frac{q^{d\phantom{-1}}-q}{q^{d-1}-1}\cdots\frac{q^{d\phantom{-1}}-q^{d-1}}{q^{d-1}-q^{d-2}}=(q^d-1)q^{d-1}.$$

Therefore, solving for $N(d)$ we obtain

$$N(d)=q^{d^2}-(q^d-1)q^{d-1}q^{(d-1)^2}=q^{d(d-1)}.$$

• I got the proof from here. Reading the pdf it's not entirely clear if the author is attributing the theorem to Hall or its proof, although the pdf does cite Fine and Herstein. Hall's text appears to have came before F&H.
– anon
May 26, 2015 at 1:51
• Also, just as a funny note, I have usually heard what they call the Fitting decomposition (in the link) referred to as the Jordan decomposition of a matrix. But then I looked at the French version of the Wiki page, and there it is called la décomposition de Dunford. How very generous of each country to attribute to a mathematician from the other! May 26, 2015 at 4:04
• @ViktorVaughn In the above answer what is the action of $GL_{n}(\Bbb F_q)$ on the given set ? Oct 12, 2021 at 18:29
• @user371231 I think by conjugation. Oct 13, 2021 at 3:52
• @user371231 $g(A,B)=(gA,gB)$.
– anon
Apr 1, 2022 at 20:32

Hint. Try to find the form of the invertible matrices which commute with each of the matrices you listed. There are $(q^3-1)(q^3-q)(q^3-q^2)$, $q^3(q-1)^2$, respectively $q^2(q-1)$ such matrices in each case. Now you get $$1+\dfrac{(q^3-1)(q^3-q)(q^3-q^2)}{q^3(q-1)^2}+\dfrac{(q^3-1)(q^3-q)(q^3-q^2)}{q^2(q-1)}=q^6$$ nilpotent matrices.