subspaces of $\mathbb{F}_5^3$

Question

Let $F=\mathbb{F}_5$. How many subspaces of each dimension does the space $F^3$ contains.

So, $\mathbb{F}_5=\{\bar0,\bar1,\bar2,\bar3,\bar4\}$. But I don't know how to proceed.

Answer 1

There is obviously one subspace of dimension $0$ and one of dimension $3$.

Each subspace of dimension $1$ contains $(0,0,0)$ and four other elements, which are of the form $\lambda (a,b,c)$, where $(a,b,c)$ is some non-zero vector and $\lambda \in \{ 1, 2, 3, 4 \}$. These $1$-dimensional subspaces partition the set of non-zero vectors into even chunks of $4$, since if two subspaces contain the same non-zero vector, then they are equal. It follows that there are $$\frac{5^3-1}{4} = \frac{124}{4} = 31$$ subspaces of dimension $1$.

Likewise, there are $\dfrac{5^2-1}{4} = \dfrac{24}{4} = 6$ subspaces of $\mathbb{F}_5^2$ of dimension $1$. (It's the same proof; more generally, there are $\dfrac{p^k-1}{p-1}$ subspaces of $\mathbb{F}_p^k$ of dimension $1$ for any prime $p$ and any $k \ge 1$.)

Since every $2$-dimensional subspace of $\mathbb{F}_5^3$ is isomorphic to $\mathbb{F}_5^2$ and is spanned by two $1$-dimensional subspaces, this means that you can count the number of $2$-dimensional subspaces of $\mathbb{F}_3^3$ by choosing two spanning $1$-dimensional subspaces of the $2$-dimensional subspace, and then accounting for the fact that you overcounted every time you chose two $1$-dimensional subspaces spanning the same $2$-dimensional subspace. That is, there are: $$\dfrac{\binom{31}{2}}{\binom{6}{2}} = 31$$ subspaces of dimension $2$.

[My linear algebra is rusty; you likely could have deduced this using the fact that a $2$-dimensional subspace has codimension $1$, so that there are just as many $2$-dimensional subspaces as there are $1$-dimensional ones.]