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Thread: dimensions

  1. #1
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    dimensions

    Let F = F5 (prime field). How many subspaces of each dimension does the space F^3 contain?
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  2. #2
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    Quote Originally Posted by jin_nzzang View Post
    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 $\displaystyle (p^n - 1)(p^n - p)...(p^n - p^{k-1})$ 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 $\displaystyle (p^k - 1)(p^k - p)...(p^k - p^{k-1}).$ We divide $\displaystyle (p^n - 1)(p^n - p)...(p^n - p^{k-1})$ by $\displaystyle (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 ($\displaystyle n \geq k$ ).

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

    For your problem, $\displaystyle V=F^3$, 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 $\displaystyle W = F^2$ where $\displaystyle F=F_2$.
    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 $\displaystyle V = F^3$ and 2-dimensional subspace W of V. Assume W is spanned by two independent vectors $\displaystyle {w_1, w_2}$. The number of distinct choices of basis for $\displaystyle w_1, w_2$ is $\displaystyle (2^3-1)(2^3-2) = 42$, 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 $\displaystyle F=F_2$).
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