I am asked to find how many there are $k$-dimensional subspaces in vector space $V$ over $\mathbb F_p$, $\dim V = n$.
My attempt:
1) Let's find a total number of elements in $V$: assume that $\{v_1, v_2, \cdots, v_n\}$ is a basis in $V$. Then, for every $v \in V$ we can write down
$$ v = a_1 v_1 + a_2 v_2 + \cdots + a_n v_n $$
and since the coordinates ($a_1, \cdots, a_n$) are from $\mathbb F_p$ there are $p^n$ vectors in $V$; $p-1$ without the zero vector.
2) Let's look at a situation where $k=1$. Let's call this 1-dimensional space $V'$.
$$\forall v' \in V'. v' = a_1 v_1$$ where $v_1$ is a basis in $V'$. We know that if there are two non-zero vectors $u \in V'_1$ and $v \in V'_2$, they are not linear dependant. So, every 1-dimensional subspace has $(p-1)$ basises. Therefore, there are $\frac{p^n – 1}{p-1}$ possible 1-dimensional subspaces in $V$
3) k-dimensional subspace is defined by the set of it's basises. Since basis can not contain zero vectors we can write down the formula for selecting $k$ linear independent vectors: $C^k_m (p-1)^k$, where $m = \frac{p^n – 1}{p-1}$. Here we first choose $k$ 1-dimentioanl subspaces and then we choose one of $(p-1)$ non-zero vectors from each of the subspaces.
4) .. unfortunately, this is where I am stuck. My intuition says that the answer may be
$\frac{p^n – 1}{(p-1)^k}$, but this might be completely wrong and I don't know how to go about finishing the problem.
Thanks in advance.
Best Answer
Here is a hint: Find a formula for the number of possibilities to choose $k$ linearly indepenent vectors in $\mathbb{F}_p ^n$ (where order matters). Each of these choices serves as a basis for a $k$-dimensional subspace, but for each subspace there are several bases, so you have to divide by the number of bases for each subspace - and computing this number involves essentially the same formula as above.