1. ## dimensions

Let F = F5 (prime field). How many subspaces of each dimension does the space F^3 contain?

2. Originally Posted by jin_nzzang
Let F = F5 (prime field). How many subspaces of each dimension does the space F^3 contain?
Let V be a n-dimensional vector space over a finite field F with p elements.

A k-dimensional space is spanned by k independent vectors. As shown in my previous reply to your post (link), there are $(p^n - 1)(p^n - p)...(p^n - p^{k-1})$ choices for picking k-independent vectors from an n-dimensional space. Among them, we should consider the cases for which k-independent vectors span the same subspace. Two sets of k-independent vectors span the same subspace W of V if and only if they are both bases of the k-dimensional space W. The number of distinct bases for the k-dimensional space W is $(p^k - 1)(p^k - p)...(p^k - p^{k-1}).$ We divide $(p^n - 1)(p^n - p)...(p^n - p^{k-1})$ by $(p^k - 1)(p^k - p)...(p^k - p^{k-1})$ in order to get the number of subspaces of an n-dimensional vector space V of dimension k ( $n \geq k$ ).

$\frac{(p^n - 1)(p^n - p)...(p^n - p^{k-1})}{(p^k - 1)(p^k - p)...(p^k - p^{k-1})}$ is the required number of subspaces of an n-dimensional vector space V of dimension k.

For your problem, $V=F^3$, n=3 and p = 5. You can compute the number of subspaces for k=1, 2 using the above formula.

Below is a simple example that you can easily check.

Let $W = F^2$ where $F=F_2$.
There are six distinct basis for W.

{(1,0)(0,1)}, {(1,0),(1,1)}, {(0,1),(1,0)}, {(0,1),(1,1)} {(1,1)},{(0,1)}, {(1,1), (1,0)}

Note that the above is an ordered basis, which implies that {(1,0)(0,1)} and {(0,1),(1,0)} are not the same.

Now consider $V = F^3$ and 2-dimensional subspace W of V. Assume W is spanned by two independent vectors ${w_1, w_2}$. The number of distinct choices of basis for $w_1, w_2$ is $(2^3-1)(2^3-2) = 42$, since V contains 8 vectors including 0. This number should be divided by 6, because each 6 distinct basis (shown in the above example) basically span the same subspace. Thus, the number of subspace W of dimension 2 in V of dimension 3 is 42/6=7 (They are both vector spaces over $F=F_2$).