TITLE: Probability of sampling linearly independent vectors
QUESTION [5 upvotes]: Let $q$ be a prime so that $\mathbb{Z}_q$  is a field. I would like to sample $n$ vectors independently and uniformly from $\mathbb{Z}^n_q$, to get a set $V = \{ v_i \}_{i=1}^n$.
What is the probability that $V$ is a basis for $\mathbb{Z}^n_q$? I had the following proof idea, by induction. The problem reduces to what is the probability of sampling $n$ linearly independent vectors, and as such, we proceed by induction on the number of sampled vectors $k$. For $k = 1$, we only require that we are not sampling the zero vector, and as such we have that $\Pr[\{v_1\} \text{is LI} ] = 1 - q^{-n}$. Now let us suppose that we have $k$ linearly independent vectors $\{ v_i \}_{i=1}^k$. We sample a new vector $v_{k+1}$. In order for it to be linearly independent, it must be that it is not a linear combination of the others, and as such there are at most $q^k$ possible 'forbidden' values. Namely these correspond to the possible coefficients $\alpha_i$ of the   linear combinations $\sum_{i=1}^k \alpha_i v_i$. So we have that $\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] \geq 1 - q^{k - n}$. As such we should have that
$$\Pr[\{v_i\}_{i=1}^n \text{is LI}] = \Pr[\{v_i\}_{i=1}^n \text{is LI} | \{v_i\}_{i=1}^{n-1} \text{is LI}] \Pr[\{v_i\}_{i=1}^{n-1} \text{is LI}] \geq \prod_{i=1}^n (1 - q^{i - n})$$
Do you think this analysis is correct? What I am also interested in is the following generalizations, on which I, unfortunately, had less success.

Suppose we set $n=3k$ and that instead of sampling the vectors independently we sample three matrices $M, M_0, M_1 \in \mathbb{Z}^{3k\times k}_q$ such that each matrix is of full rank. What would be an upper bound on the probability of the columns of said matrices spanning $\mathbb{Z}_q^{3k}$? I have found in the literature (Lemma 8) a claim that it should be at least $1 - \frac{2k}{q}$ but have not found any good argument
Let us relax the very first statement, and admit that $q$ could be composite. Does this change the argument significantly? I am interested in this result both in the vector case and in the three matrices sample case.

Thank you, I hope the formatting is clear enough :)
For context, I am investigating whether the construction in the linked paper would hold in nonprime order groups.

REPLY [2 votes]: Your analysis is correct. There is possibly a typo: $i$ should be $k$. Moreover, I think it is equality. More precisely, $\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}$ if and only if $v_{k+1}$ is not a linear combination of $v_1,\ldots,v_k$. The only thing that you need to make sure is that two different linear combinations of $\{v_i\}_{i=1}^{k}$ give rise to two different vectors, which would be true since $\{v_i\}_{i=1}^{k}$ is LI. Thus $\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] = \frac{q^n-q^k}{q^n}$.
If it is not a field (that is $q$ is composite) then the equality may not hold as two different linear combinations can give you the same vector. To see this take $q=6$, then $v_1=(1,0,0,0,0,0)', v_2=(5,2,0,0,0,0)$, note that here $v_1+v_2 = 4v_1+4v_2=(0,2,0,0,0,0)$.
The inequality that you wrote still holds (after the typo correction:)).
Matrix Case: write $M_0|M_1|M_2$ for the augmented $3k\times 3k$ matrix.
Now $M_0|M_1|M_2 \text{ is LD}$ if one of the columns in $M_1$ is linear combination of columns in $M_0$ or one of the column in $M_2$ is linear combination of columns in $M_0$ and $M_1$. Now probability that a fixed column in $M_1$ is linear combination of columns in $M_0$ is at most $\frac{q^k}{q^{3k}}\leq \frac{1}{q}$, and probability that a fixed column in $M_2$ is linear combination of columns in $M_0$ and $M_1$ is at most $\frac{q^{2k}}{q^{3k}}\leq \frac{1}{q}$. Therefore using union bound
$$
\mathbb{P}(M_0|M_1|M_2 \text{ is LD}) \leq  \frac{2k}{q}.
$$
You can see that this bound holds even if $q$ is not a prime.