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

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 $N=3$

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

Try (after the initial $I$) a Vandermonde matrix, where $a_{ij} = \alpha_i^{j-1}$, $\alpha_i$ being all distinct and positive.
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.
Perhaps choosing $n^2$ algebraically independent huge transcendental numbers, and then replacing by their integer parts might work.