Tall $m \times n$ matrices whose submatrices with $n$ rows have full rank

Question

I want to find some $m \times n$ (where $m>n$) matrices that have the property that any submatrix with $n$ rows has full rank. The Vandermonde and Cauchy matrices are the only two matrices I know. Can you please give me some other matrices?

P.S. I forgot to mention that the entries in the matrices must be integers, and the base field is infinite field.

Answer 1

Let the entries be $\sqrt{p_i}$ for distinct primes $p_i$.

EDIT: It appears the entries are to be integers. If you take them to be distinct members of a sufficiently rapidly growing sequence, maybe $10^{10^{k!}}$ for $1\le k\le mn$, then no $n\times n$ submatrix can be singular. For any such matrix will have an entry much bigger than all the others, so if you evaluate the determinant by expanding along the row or column containing that entry, the coefficient of that entry would have to be zero for the determinant to vanish; but that coefficient is an $n-1\times n-1$ determinant, and induction takes over.

Answer 2

The Moore Matrix is similar to vandermonde matrix and has a full rank, but it is more generalized.

The Wronskian determinant can be used to generate full-rank matrices.