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I have strong computational evidence to think that the answer is $q^{n(n-1)}$, although a proof eludes me. Any ideas?

rschwieb
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Jared
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  • An idea I haven't fleshed out: a matrix $A$ is nilpotent if and only if $\operatorname{tr}(A^i) = 0$ for $i = 1, \dots, n$. These are $n$, hopefully independent equations on a space of dimension $q^{n^2}$, voila. – Ryan Reich Dec 13 '13 at 17:20
  • @RyanReich I was staring at a similar statement on wikipedia, where it said that that characterization of nilpotentce only holds for "sufficiently large characteristic," which they didn't elaborate on. Do you happen to know the limitation? – rschwieb Dec 13 '13 at 17:28
  • @rschwieb: We need that $n!$ is invertible in order to use Newton's identities (this is also explained at Wikipedia). – Martin Brandenburg Dec 13 '13 at 18:08
  • @MartinBrandenburg Great. Consider sharpening the statement in the middle of the paragraph I was looking at with an appropriate pointer or explanation so that other readers don't have to sift through an entire article as you bothered doing. – rschwieb Dec 13 '13 at 18:24
  • @RyanReich Replace $\mathrm{Tr}(A^i)$ with the coefficient of $x^i$ in the characteristic polynomial of $i$, and now you get a statement which is right in every characteristic. But that doesn't help explain why these equations should be independent. – David E Speyer Oct 25 '18 at 17:04

1 Answers1

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There is a nice proof based on the two following lemmas. Let $V$ be a vector space of dimension $n$.

Lemma 1 (Fitting). For all $u \in \mathcal{L}(V)$, there exists a unique decomposition $V=W_N \oplus W_I$ such that :

  • $u(W_N) \subset W_N$ and $u_{|_{W_N}}$ is nilpotent,
  • $u(W_I) \subset W_I$ and $u_{|_{W_I}}$ is inversible.

Sketch of proof. This a very special case of the Jordan normal form theorem.

Lemma 2. For all $k \leq n$ there is a one-to-one correspondence between the decompositions $W_1 \oplus W_2$ of $V$ with $\dim W_1=n-\dim W_2 = k$ and the quotient $$GL(n)/(GL(k)\times GL(n-k)).$$

Sketch of proof. Consider the natural group action of $GL(n)$ on the set of such decompositions.

Let $N_k$ denote the number of nilpotent matrices in $M_k(\Bbb F_q)$. Using the two lemmas, we see that $$ q^{n^2} = \#M_n(\Bbb F_q) = \sum_{k=0}^n (N_k\cdot\#GL_{n-k})\cdot\dfrac{\#GL_n}{\#GL_k\cdot \#GL_{n-k}} = \#GL_n\sum_{k=0}^n \frac{N_k}{\#GL_k} $$ which yields for all $n \geq 2$, $$\dfrac{N_n}{\#GL_n(\Bbb F_q)} = \dfrac{q^{n^2}}{\#GL_n(\Bbb F_q)} - \dfrac{q^{(n-1)^2}}{\#GL_{n-1}(\Bbb F_q)}.$$ The result follows after simplification using the well-known fact that $$\#GL_k(\Bbb F_q) = (q^k-1)(q^k-q)\dots(q^k-q^{k-1}).$$

Siméon
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  • Don't you mean $# \mathrm{GL}_{n-k}(\mathbb{F}_q)$ in the sum? I also don't understand $# \mathrm{GL}_k(\mathbb{F}_q)$ in the denominator. – Martin Brandenburg Dec 13 '13 at 22:57
  • @MartinBrandenburg: Actually the sum is $\sum_{k=0}^n(N_k\cdot#GL_{n-k})\cdot\dfrac{#GL_n}{#GL_k\cdot #GL_{n-k}}$. The second part is the number of couples $(V,W)$ of subspaces with $V\oplus W= \Bbb F_q^n$ and $\dim V = k$. – Siméon Dec 13 '13 at 23:35
  • Ah, yes. Thank you! – Martin Brandenburg Dec 14 '13 at 00:05
  • The edit makes it very clear. Great answer! I wonder if there is a more direct proof which enumerates the nilpotent matrices. – Martin Brandenburg Dec 14 '13 at 09:01
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    @MartinBrandenburg: The proof given in this paper of Gerstenhaber might be more explicit. You can also check the original proof of Fine and Herstein. – Siméon Dec 14 '13 at 10:18
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    For the Lemma of Fitting you don't need Jordan normal form (which works only for alg.cl. fields anyway). See http://planetmath.org/fittingslemma for a complete and easy proof. – Martin Brandenburg Dec 14 '13 at 11:19
  • I really like your proof, it is more simple than the ones by Gerstenhaber, Fine-Herstein and Reiner. Back in the 50s this would have been worth a publication ... – Martin Brandenburg Dec 14 '13 at 11:31
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    @MartinBrandenburg: I'm not the author of this proof. It is a slight modification of the argument given in the paper (in french) by DE SEGUINS PAZZIS, Clément. Dénombrements matriciels sur des corps finis. RMS: Revue de la filière mathématiques, 2006, vol. 116, no 4. – Siméon Dec 14 '13 at 12:14