Let $v_1,\ldots,v_k\in \mathbb{F}_p^n$ be a set of vectors, where $p$ is a prime. Assume further that the components of each vector can be represented by integers smaller than some integer $k$. Is there a way to check linear independence, so that it will be guaranteed for all primes larger than some concrete computable limit? For example, if we have linear independence over $\mathbb{Q}$, then my understanding is that this will hold for large enough $p$, but I don't know of any way of determining if a prime is large enough.
To simplify this problem maybe a bit, we could assume that all the components of the $v_i$'s are one of $-1,0,1$.
Any pointers on material regarding this would also be appreciated.
Somewhat better than the answer by Magdiragdag, which depends on choosing complementary vectors, you can proceed as follows. Take the matrix whose columns are your integer vectors, and compute its Smith normal form over$~\Bbb Z$. If you get any zero entries on the main diagonal, then your vectors are linearly dependent over$~\Bbb Z$ and will remain so when reducing modulo any prime. If you do not get any zeros, then the final (largest) diagonal entry is such that its prime divisors$~p$ are precisely the primes for which reducing the vectors modulo$~p$ gives a linearly dependent family. (Note that if $k\neq n$, then the final diagonal entry is not the one in the lower right-hand corner).
That the above is true can be seen from the fact that the Smith normal form computation amounts to performing a base change over$~\Bbb Z$ (invertible by definition) such that the subgroup spanned by your given vectors is spanned by multiples (given by the diagonal entries) of the new basis vectors.