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$F$ is a field of order $q$.

We are essentially counting the total number of possible ordered bases in the vector space $F^d$.

The first vector has $q^d - 1$ possibilities because we need to exclude the zero vector.

The second vector cannot be a scalar multiple of the first vector, and there are $q$ scalars in $F$, so once we have chosen the first vector, there are $q^d-q$ possible ways to construct the second vector such that $v_1$ and $v_2$ are linearly independent.

But what about the third vector? Why are there $q^d-q^2$ possibilities?

Sid Caroline
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    Because it can be any vector except the $q^2$ linear combinations of the previous $2$ (they result in exactly $q^2$ vectors because the previous $2$ vector are linearly independent). –  Jun 09 '17 at 07:40
  • Thanks! I think I get it. $q^2$ is the number of ways to have two scalars in $F$ – Sid Caroline Jun 09 '17 at 07:53

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