My goal is to prove that $\det(A) \neq 0$, I took this statement as to be be true because comes from a scientific paper (see below).
Define $A \in \mathbb{R}^{2k \times 2k} $ to be:
$$ A = \begin{bmatrix} x_1 & x_1^2 & \cdots & x_1^k & y_1 x_1 & y_1 x_1^2 & \cdots y_1 x_1^k \\ x_2 & x_2^2 & \cdots & x_2^k & y_2 x_2 & y_2 x_2^2 & \cdots y_2 x_2^k \\ \vdots \\ x_{2k} & x_{2k}^2 & \cdots & x_{2k}^k & y_{2k} x_{2k} & y_{2k} x_{2k}^2 & \cdots y_{2k} x_{2k}^k \\ \end{bmatrix} $$
where all the $x_i$'s are distinct and different from 0, $x_i \neq y_i$ and not all $y_i$ values are equals. Additionally, there is a polynomial $S$ of degree $2k$ such that $y_i = S(i)$ for every i.
I looked similar proofs based on Vandermonde matrix and Kronecker product, but I can't get to a demonstration.
More context about this demonstration
I am trying to prove Lemma 1 of Distributed oblivious transfer.
Given two inputs $m_0, m_1$, define $$ Q(x,y) = \sum_{j=0}^{d_x} \sum_{l=0}^{d_y} a_{j,l} x^j y^l $$ where $a_{0,0} = m_0$ and $\sum_{l=0}^{d_y} a_{0,l} = m_1$.
I want to show that retrieving $k=2 d_x + 1$ values doesn't allow to learn more than a single linear combination of $m_0$ and $m_1$.
The beginning of the proof is detailed in the proof of Lemma 1, I am stuck at proving that the matrix $B'$ is not singular.
This can be easily singular. Pick two monic degree-$k$ polynomials $p(x)=\sum_{r=1}^kc_rx^r$ and $q(x)=\sum_{r=1}^kd_rx^r$ with zero constant terms at random. More often than not, we have $q(x_i)\ne0$ for every $i$ and $\frac{p(x_1)}{q(x_1)},\ldots,\frac{p(x_{2k})}{q(x_{2k})}$ are all distinct. Now let $y_i=-\frac{p(x_i)}{q(x_i)}$. Then $A$ is singular. In fact, $Av=0$ when $v=(c_1,\ldots,c_k,d_1,\ldots,d_k)^T\ne0$.