Generate an integer matrix such that all submatrices are non-singular

Question

I need to generate an $\infty \times N$ integer matrix with a few properties.

• The top $N$ rows (and $N$ columns) should be the identity matrix.
• Any square submatrix (meaning the result after removing any number of rows and columns) should have a non-zero determinant. There are a few submatrices of the identity matrix where this does not hold, but this rule should apply in all other cases.

I found cauchy matrices, but they contain reals, where I'm working with integers.

I also played around a bit with a symmetric pascal matrix, and I have a good feeling about it, but I'm not sure how to prove that it fulfills the second property. Here's an example with $N=3$

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \\ 1 & 4 & 10 \\ \vdots & \vdots & \vdots \\ \end{bmatrix} $$

Answer 1

Try (after the initial $I$) a Vandermonde matrix, where $a_{ij} = \alpha_i^{j-1}$, $\alpha_i$ being all distinct and positive.

Answer 2

Here is a route to an integer matrix through irrational numbers. This is only a heuristic.

Take $n^2$ (distinct!) prime numbers. Take their square roots as the entries of an $n\times n$ matrix. Then every submatrix would be non-singular as the field generated by every subset of these numbers is distinct.

Now comes the part I am not able to justify rigorously: By choosing the primes sufficiently far apart it should be possible to get submatrices with determinant values huge in absolute value. Now replace all the square roots by integers close to them and get (by continuity) all the determinant less huge and yet non-zero.

Perhaps choosing $n^2$ algebraically independent huge transcendental numbers, and then replacing by their integer parts might work.