does not matter) that has zeros in all positions except $i_1,\ldots,i_r$ and has values $\alpha_1,\ldots, \alpha_r$ given by Claim \[cla-rtuples\] in those positions. Since the rows of $M$ are the points $u_1,\ldots,u_n$, the third item of Claim \[cla-rtuples\] guarantees that $A \cdot M = 0$ as we wanted. The next claim asserts that $A$ is a design matrix.

\[cla-Adesign\] The matrix $A$ constructed above is a $(r,k,2)$-design matrix.

Clearly, each row of $A$ contains at most $r$ non zero coordinates. Since each point $v \in V$ is in at least $k$ $r$-tuples from $R$ we have that each column of $A$ contains at least $k$ non-zero coordinates. The size of the intersection of the supports of two distinct columns in $A$ is at most two by item (2) of Claim \[cla-rtuples\].

We now use Eq. (\[eq-rank1\]) from Theorem \[thm-rankbound\] to get $$\rank(A)\ge n-\frac{2nr^2}{k}.$$ This implies (using $r \geq 4$) that $$\rank(M) \leq \frac{2nr^2}{k}  \leq \left(\frac{r-2}{d}\right)^d < \m(d,r-2),$$ if $$k \geq  2(2d)^d \cdot \frac{n}{r^{d-2}}.$$

If we have the additional assumption that each point is in at most $8k$ lines then, using the bound $m= |R| \leq 16nk/r $ in Eq. (\[eq-rank2\]) of Theorem \[thm-rankbound\]. We get

$$\rank(A) \geq n-\frac{2mr^2}{k^2} \geq n 