Multiplication by the displayed <Vandermonde matrix> maps the coefficient vector of a <polynomial> of degree at most $n$ to its vector of values at the nodes. Part (a) proves that every vector of node values is attained uniquely, so this square linear map is bijective. Therefore \b[$A_n$ is invertible]. Equivalently, a <kernel> vector would define a <polynomial> of degree at most $n$ with $n+1$ distinct roots and would have to vanish.

Solution

Past exam of the options II course of the University of Oxford / 2009 / P / P3 / b / Solution

 Solution (source code)