Let us have $N$ linear independent multivariate polynomials $p_j(x_1,..,x_m)$ of degree $K$. Consider now their span: since they are linearly independent, this will never be equal to constant $0$. What I am asking is: how many different points $\vec x_{\alpha}=(x_{\alpha,1},...,x_{\alpha,m})$ should we pick to have the guarantee that any (non-trivial) linear combination of the polynomials $p_j$ is not equal $0$ at least in one of these points?
My question comes from the following framework of a polynomial matrix, where each of its $R$ columns is given by the polynomials $p_j$ evaluated at the same point $\vec x_{\alpha}$ for $\alpha=1,...,R$.
\begin{equation*} \begin{pmatrix} p_1(\vec x_{\alpha_1}) & p_1(\vec x_{\alpha_2}) & \dots & p_1(\vec x_{\alpha_R}) \\ p_2(\vec x_{\alpha_1}) & p_2(\vec x_{\alpha_2}) & \dots & p_2(\vec x_{\alpha_R}) \\ \vdots & \vdots & \dots & \vdots \\ p_N(\vec x_{\alpha_1})& p_N(\vec x_{\alpha_2}) & \dots & p_N(\vec x_{\alpha_R}) \end{pmatrix} \end{equation*}
I want the matrix to have full row rank, meaning that the span of the rows must not give $\vec 0$. This indeed translates in having any linear combination of the polynomials not being $0$ in at least one $\vec x_{\alpha}$.
By knowing that the polynomials have at most degree $K$, is there a minimum number of points $\vec x_{\alpha}$ that can be freely chosen, which guarantees that the matrix has full row rank?
And bonus question: What if I can pick $\vec x_{\alpha}$, but only as vectors of positive integers?