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Let $(x_1,y_1),...,(x_n,y_n)\in \mathbb{R}^2 $, where $x_i\neq x_j$ if $i\neq j$. Let $p$ be a polynomial such that $$\det\begin{pmatrix} p(x)& 1 & x & x^2 &\dots & x^n \\ y_0 & 1 & x_0 & x_0^2 &\dots &x_0^n \\ y_1 & 1 & x_1 & x_1^2 &\dots &x_1^n \\ \vdots & & & & & \vdots \\ y_n & 1 & x_n & x_n^2 &\dots & x_n^n \end{pmatrix}=0. $$ Then $p(x_k)=y_k ,\forall k=1,...,n$.

My ideas: We should compute the determinant using Laplace's formula, although I can't see a nice pattern to do a proof by induction or to conclude the proposition.

Thanks for the help

bests

bjn

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Hint. Put $x=x_0$ and subtract the second row from the first row. The determinant is then equal to $$ \det\pmatrix{p(x_0)-y_0&0_{1\times(n+1)}\\ \ast&V}=(p(x_0)-y_0)\det(V), $$ where $V$ is a Vandermonde matrix. In order that this is zero, ...

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