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Suppose I randomly generate a $k \times n$ matrix with $k < n$ over $\mathbb{Z}_q$. or Think of it as $k$ vectors over $\mathbb{Z}_q^n$. We then choose a $k$-sized subset of indices $S \subset \mathbb{Z}_n + 1$ and build a $k \times k$ matrix out of selected indices. Suppose $M(S)$ is a matrix built like that over indices $S$. You can choose any subset but the selected indices must be consistent across all vectors.

For example, is the original matrix is $A_{3 \times n}$ and $S = \{1,3,4\}$, The result is $M(S) = \begin{pmatrix}a_{11} & a_{13} &a_{14} \\ a_{21} & a_{23} &a_{24} \\ a_{31} & a_{33} &a_{34}\end{pmatrix}$

I need all such $k \times k$ matrices to be of full rank. That is $\forall S, S \subset \mathbb{Z}_n + 1$ , $|S| = k \to \rho (M(S)) = k$.

What is the probability that any random $k$ vectors have such property?

Manish Adhikari
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 25 '25 at 16:18
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    Welcome to MSE! I think what you're asking is clear enough. I think it could also be phrased as "how many $n$-tuples of vectors in $\Bbb Z_q^k$ are there such that any $k$ of them are linearly independent" (assuming $q$ is prime and $\Bbb Z_q$ denotes the integers modulo $q$ - is that correct?). The case $k = n$ is then answered by this, and more generally you can get a yucky formula using inclusion-exclusion. Are you more interested in formulas or asymptotics, or something else? Do you have a particular application in mind? – Izaak van Dongen Apr 25 '25 at 17:32
  • @IzaakvanDongen Thanks for the link. Asymptotics will do fine. Actually, I was playing with cryptography and thought of a way to build threshold ZK-proofs and threshold signatures with such a matrix. A construction of such a scheme would require a matrix with this property, so I just got curious about the probability that a randomly generated matrix would have this property. It is still a pure math question, so I asked here instead of crypto SE. – Manish Adhikari Apr 25 '25 at 17:49

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