In the realm of linear algebra, vector spaces and their subspaces are fundamental concepts. A fascinating question arises when we consider a vector space $V$ over a finite field $mathbb{F}_p$: how many $k$-dimensional subspaces does $V$ contain? Let’s delve into this problem step-by-step.
First, let’s understand the size of our vector space. If $V$ is an $n$-dimensional vector space over $mathbb{F}_p$, we can think of it as being built from a basis ${v_1, v_2, cdots, v_n}$. Any vector $v$ in $V$ can be uniquely expressed as a linear combination of these basis vectors:
$$ v = a_1 v_1 + a_2 v_2 + cdots + a_n v_n $$
where each coefficient $a_i$ comes from the finite field $mathbb{F}_p$, which has $p$ elements. Since there are $p$ choices for each of the $n$ coefficients, the total number of vectors in $V$ is $p^n$. This means there are $p^n$ elements in our vector space, and $p^n – 1$ non-zero vectors (excluding the zero vector).
Now, let’s consider a simpler case: counting 1-dimensional subspaces. A 1-dimensional subspace, often called a line through the origin, is spanned by a single non-zero vector. If we pick any non-zero vector in $V$, it generates a 1-dimensional subspace. However, different non-zero vectors can generate the same subspace. Specifically, if we take a non-zero vector $v$ and multiply it by any non-zero scalar from $mathbb{F}_p$, we get another non-zero vector that spans the same 1-dimensional subspace. Since there are $(p-1)$ non-zero scalars in $mathbb{F}_p$, each 1-dimensional subspace is spanned by $(p-1)$ non-zero vectors (each being a basis for that subspace if considered alone).
To find the number of 1-dimensional subspaces, we take the total number of non-zero vectors in $V$, which is $p^n – 1$, and divide it by the number of non-zero vectors that span the same 1-dimensional subspace, which is $(p-1)$. This gives us the number of 1-dimensional subspaces:
$$ frac{p^n – 1}{p-1} $$
This initial approach is insightful, but extending it directly to $k$-dimensional subspaces requires a more careful strategy. The initial thought of using combinations $C^k_m (p-1)^k$ where $m = frac{p^n – 1}{p-1}$ is not quite correct because it oversimplifies the process of choosing linearly independent vectors and doesn’t properly account for the bases of a $k$-dimensional subspace.
To count $k$-dimensional subspaces correctly, we need to consider how to construct a basis for a $k$-dimensional subspace. We can build a basis for a $k$-dimensional subspace by selecting $k$ linearly independent vectors from $V$.
Let’s think about choosing the first vector for our basis. We can choose any non-zero vector, and there are $p^n – 1$ options.
For the second vector, we need to choose a vector that is linearly independent from the first one. This means we must avoid choosing a vector that is a scalar multiple of the first vector. The subspace spanned by the first vector has $p$ vectors (including the zero vector). So, we have $p^n – p$ choices for the second vector to ensure it’s linearly independent from the first.
For the third vector, we must choose a vector that is linearly independent from the first two. The subspace spanned by the first two vectors is a 2-dimensional subspace and contains $p^2$ vectors. Thus, we have $p^n – p^2$ choices for the third vector.
Continuing this pattern, for the $k$-th vector, we need to choose a vector that is linearly independent from the first $k-1$ vectors. The subspace spanned by the first $k-1$ vectors is a $(k-1)$-dimensional subspace and contains $p^{k-1}$ vectors. So, we have $p^n – p^{k-1}$ choices for the $k$-th vector.
Therefore, the number of ways to choose an ordered set of $k$ linearly independent vectors in $V$ is:
$$ (p^n – 1)(p^n – p)(p^n – p^2) cdots (p^n – p^{k-1}) $$
However, this counts ordered bases. Different ordered bases can span the same $k$-dimensional subspace. We need to divide by the number of ordered bases for a single $k$-dimensional subspace. A $k$-dimensional subspace itself is a vector space of dimension $k$ over $mathbb{F}_p$. Using the same logic as above, the number of ordered bases for a $k$-dimensional subspace is obtained by replacing $n$ with $k$ in the previous formula:
$$ (p^k – 1)(p^k – p)(p^k – p^2) cdots (p^k – p^{k-1}) $$
Finally, to find the number of $k$-dimensional subspaces, we divide the number of ordered sets of $k$ linearly independent vectors in $V$ by the number of ordered bases of a $k$-dimensional subspace:
$$ frac{(p^n – 1)(p^n – p)(p^n – p^2) cdots (p^n – p^{k-1})}{(p^k – 1)(p^k – p)(p^k – p^2) cdots (p^k – p^{k-1})} $$
This formula gives us the number of $k$-dimensional subspaces in an $n$-dimensional vector space over $mathbb{F}_p$. This quantity is known as the Gaussian binomial coefficient, often denoted as $begin{bmatrix} n k end{bmatrix}_p$ or $binom{n}{k}_p$.
This result highlights a beautiful aspect of linear algebra over finite fields, providing a precise count for the number of subspaces of a given dimension. Understanding how many $k$-dimensional subspaces exist within a vector space is crucial in various areas, including coding theory, combinatorics, and further studies in linear algebra.
