In a generator matrix of a linear block code ,how does increasing linear vectors in a field $F^k$ has $q^k-q^i$ choices?

Question

I am trying to study error control and coding theory by myself. There is an unsolved question which says that the total number of distinct generator matrices of a linear [n,k,d] code over $F=GF(q)$ is $\prod_{i=0}^{k-1}(q^k-q^i)$

Now the hint says that the required number of generator matrices is equal to the number of $k\times k$ non singular matrices over $F$, which is clear to be as I can write generator matrix $(k\times n)$ as $$G=(I_k|A)$$ where $I_k$ is the identity matrix with $k\times k$ dimensions and is the non singular part and rest $A$ is some linear function.

Now comes the main doubt, the hint says the number of such matrices is given as, suppose there are $i<k$ first independent rows in the matrix then adding a linearly independent row has

$$q^k-q^i$$ choices.

But I believe once I have already used $q^k-(i-1)$ choices so now I have $$q^i-i$$ choices of selection, in this way my answer turns out to be $$\prod_{i=0}^{k-1}(q^k-i)$$

Where I am doing wrong?

Answer 1

I think @Phicar comment was enough to answer my questions. If I already have $i$ independent vectors then the total number of linear combinations, i.e., dependent vectors I can make from them is $q^i$, so the total number of independent vectors left are $q^k-q^i$ and this is how the question is being solved to get the requisite answer.