Prove the number of unordered pairs of linearly independent elements

Question

Let $V$ be a vector space over $K$.

Let $K={\mathbb{Z}}/{p\mathbb{Z}}$, and $\dim V=3$.

We know that $V$ has $p^3$ elements.

I need to show that the number of unordered pairs of linearly independent elements of V is: $$ \frac1 2(p^3−1)(p^3−p). $$

Answer 1

As usual you can regard an element of $V$ as a triple $(x, y, z)$ of elements $x, y, z \in V$. The first element of any such unordered pair $(v_1, v_2)$ must be nonzero (otherwise the pair is linearly dependent), but otherwise there are no restrictions, that is, any nonzero element can be the first element of a pair. When specifying a vector in $V$ there are $p$ choices for each of $x, y, z$, so $p^3$ vectors total, and thus our first vector $v_1$ in the pair can be any of $p^3 - 1$ nonzero vectors.

Now, for an arbitrary choice of first vector, how many choices are there for the second? How many pairs of choices are there in total? How do we account for the fact that we want unordered rather than ordered pairs?

A line of reasoning along these lines lets you compute, for example, the number of nondegenerate matrices (whose columns are linearly independent vectors) over a finite field $\mathbb{F}$, that is, the order of the finite group $GL(k, \mathbb{F})$ and its cousins.