$f_w$ must vanish identically on the line $L$ and so $f_w(v_{r+2})= {\langle w,u_{r+2} \rangle} =0$ as well.

To show the ‘moreover’ part, suppose in contradiction that $u_1,\ldots,u_r$ span $u_{r+1}$. We can find, by interpolation, a non zero polynomial $f \in \C[x_1,\ldots,x_d]$ of degree at most $r$ such that $f(v_1)= \ldots = f(v_r) = 0$ and $f(v_{r+1})=1$. More formally, we can translate the line containing the $r+1$ points to the $x_1$-axis and then interpolate a degree $r$ polynomial in $x_1$ with the required properties using the invertibility of the Vandermonde matrix. Now, let $w \in \C^{\m(d,r)}$ be the point such that $f = f_w$. We know that ${\langle w,u_i \rangle}=0$ for $i=1\ldots r$ and thus, since $u_{r+1}$ is in the span of $u_1,\ldots,u_r$, we get that $f(v_{r+1}) = {\langle w,u_{r+1} \rangle}=0$ in contradiction. This completes the proof.

Polynomials vanishing on grids
------------------------------

We recall the Schwartz-Zippel lemma.

\[lem-SZ\] Let $S \subset \F$ be a finite subset of an arbitrary field $\F$ and let $f \in \F[x_1,\ldots,x_d]$ be a non-zero polynomial of degree at most $r$. Then $$\lvert \{ (a_1,\ldots,a_d) \in S^d \subset \F^d \,\,|\,\, f(a_1,\ldots,a_d)=0 \}\rvert \leq r \cdot |S|^{d-1}.$$

An easy corollary is the following claim about homogeneous polynomials.

\[lem-SZ-hom\] 