Let $m$ be an integer, let $x_j = \frac{j−1}{m},1 \leq j \leq m.$ Let $y_1, y_2 , . . . , y_m$ be real numbers, and let $p$ denote the polynomial of degree $\leq m − 1$ with $p(x_j ) = y_j$ for $1 \leq j \leq m$. Let $\xi_i = \frac{i-\frac{1}{2}}{m}, 1 \leq i \leq m$.Let $\eta_i = p(\xi_i )$.
Is there a matrix $A \in \mathbb R^{m×m}$ so that $$\begin{bmatrix} \eta_1\\ ...\\ \eta_m \end{bmatrix}=A \begin{bmatrix} y_1\\ ...\\ y_m \end{bmatrix}$$
(Does $\eta_i$ depend linearly on the $y_j$?)
My attempt: My intuition tells me that the polynomial $p(x_i)=y_i$ is uniquely determined and $p$ is a bijection, so there is a linear dependence. Can anyone give a rigorous proof of that? And how do we find matrix A then (maybe computing on MATLAB)?