TITLE: Are there large integer matrices with entries computable in polynomial time, such that all minors are nonzero?
QUESTION [5 upvotes]: Is there a sequence of matrices $(A_n\in M_{2^n\times2^n}(\mathbb{Z}))_{n\in\mathbb{N}}$ such that the $(i,j)$th entry of $A_n$ is computable in polynomial time, such that all minors of each $A_n$ are nonzero?
The last condition is easy to satisfy without the entries being computable in polynomial time, by using Vandermonde matrices. But the entries of a $2^n\times2^n$ Vandermonde matrix are too large to write down in polynomial time (since a row of powers of $k$ will end with $k^{2^n-1}$, which takes $(2^n-1)\log(k)>poly(n)$ digits to write down).
I'm also interested in the same question where rational, rather than just integer, entries are allowed.

REPLY [2 votes]: A $2^n \times 2^n$ Hadamard matrix, see https://en.wikipedia.org/wiki/Hadamard_matrix, has entries $\pm 1$ and is equal to its transposed inverse (up to a scalar factor $2^n$). Thus all its minors are nonzero.
The Wikipedia article contains Sylvester's contruction of a $2^n \times 2^n$ Hadamard matrix. With the information in that article it is easy to see that an entry of such a matrix can be computed in time $O(n)$.