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Following the recommendation here to get this question out of the unanswered queue, I've changed this from a proof-verification question into an answer-your-own.

Here's the question again in case someone has their own proof (which may be better than my own below):

How many (unordered) bases does $\Bbb F_q^n$ have as a vector space over $\Bbb F_q$?

Marc van Leeuwen
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    This is precisely the proof I would have given. So I guess I have to say that it looks right to me. – Lubin Jul 07 '14 at 22:07
  • Thanks for the verification @Lubin –  Jul 07 '14 at 22:11
  • Yep, this is perfect. – anon Jul 07 '14 at 22:42
  • The proof is correct. A nice corollary is that $(q^n-1)\dotsc (q^n-q^{n-1})/n!$ is an integer. Can you prove this directly? :) Also, what about counting linearly independent subsets of size $k$? – Martin Brandenburg Jul 08 '14 at 06:06
  • @MartinBrandenburg I arrived at $$\frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{k!}$$ for your second question by a similar approach. Your first question is more challenging though. Simple induction isn't immediately obvious since we don't have that $(n+1)\mid(q^{n+1}-1)q^n$ in general. And even with the second result, inducting on $k$ doesn't seem to help since $k\mid(q^n-q^{k-1})$ isn't true in general. Do you know of a direct method of showing it? And if so, can I get a hint? –  Jul 08 '14 at 09:53
  • What about induction on $k$? To be honest, I haven't tried to prove it. By the way, the linear algebra proof only gives divisibility when $q$ is a prime power. Similar to the proof of $1+q+\dotsc+q^n = \frac{q^{n+1}-1}{q-1}$ using the decomposition of projective $n$-space over $\mathbb{F}_q$ into affine spaces. :) – Martin Brandenburg Jul 08 '14 at 14:03

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We shall first consider how many different ordered bases $\Bbb F_q^n$ has.

Recall that $|GL_n(\Bbb F_q)|=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$. Each element of $GL_n(\Bbb F_q)$ represents a linear map that carries the standard (ordered) basis $\{e_1, e_2, \ldots, e_n\}$ to another ordered basis. We can establish a bijection between the ordered bases of $\Bbb F_q^n$ with the elements of $GL_n(\Bbb F_q)$ by the rule $$\{v_1, \ldots, v_n\}\mapsto\text{the linear extension of the map that carries the standard basis to } \{v_1, \ldots, v_n\}$$

Thus we know how many ordered bases $\Bbb F_q^n$ has. Now we define an equivalence relation on the set of all ordered bases of $\Bbb F_q^n$. We call two ordered bases equivalent if they are permutations of one another. The quotient set under this equivalence relation is clearly in bijection with the set of all unordered bases of $\Bbb F_q^n$. Overmore, each equivalence class has the same number of elements, namely $n!$ since there are $n!$ permutations of any ordered basis. Since the set of unordered bases is in bijection with the quotient set we have that the total number of unordered bases of $\Bbb F_q^n$ is: $$\frac{(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})}{n!}$$ This approach can be mirrored to conclude that the number of linearly independent subsets of order $k$ is: $$\frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{k!}$$

  • There is no need to use the linear group here. The formula in your first line is proven via bases. – Martin Brandenburg Jul 24 '14 at 14:30
  • I thought that was redundant, since its order is usually derived via counting bases. But there are probably other proofs about its order without combinatorics. And those who know those proofs might like this approach more. –  Jul 24 '14 at 14:39