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Let $\mathbb{F}_n$ denote the finite field with $n$ elements. Suppose that the (non-tall) matrix ${\bf M} \in \mathbb{F}_n^{r \times n}$, where $r \leq n$, has rank $r$ and for any $k \leq r$, the submatrix of $\bf M$ consisting of the first $k$ rows satisfies the property that every of its $k\times k$ submatrices is invertible. Is this a known property?

Unfortunately, asking people I know in real life and trying to find out more about this online hasn't helped. I do suspect that this is a sufficient condition to show that $M_{ij} = \phi_i(z_j)$ where $\phi_0, \phi_1, \dots, \phi_{r-1}$ is a basis of the polynomial space of degree at most $r-1$ and $z_1, \dots, z_n$ are distinct entries of $\mathbb{F}_n$.

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  • $\begingroup$ I don't think your sufficient condition claim can be quite right as stated. If $\mathbf M$ is a matrix with this property, multiplying each column by a nonzero scalar preserves this property, but your characterization does not seem to be preserved by this property. If you instead write $M_{ij} =c_j \phi_i(z_j)$ then it's true at least for $r=2$. $\endgroup$ Commented Feb 14 at 21:04
  • $\begingroup$ @WillSawin Yes you're right, that should be the proper characterization. And if $r=2$ then yeah it should follow from the fact you can normalize the first row to be all $1$'s, and since every $2\times 2$ matrix has to be invertible then the second row has to be all distinct values and so $\phi_0(x) = 0$ and $\phi_1(x) = x$. It's when $r=3$ that things get trickier. If we consider $\textbf{M} \in \mathbb{F}_3^{3\times n}$ with the property, if $n=3$ then we can show the sufficient condition is true, and $n=4,5$ can be shown by numerical experiments. $\endgroup$ Commented Feb 14 at 21:33
  • $\begingroup$ I can't edit my comment but I meant that $\phi_0(x) = 1$. $\endgroup$ Commented Feb 14 at 21:40

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