 Solution

Multiplying the Vandermonde matrix $A_n$ by a coefficient vector $(c_0,\dots ,c_n{)}^T$ evaluates the polynomial ${\sum }_j{c}_j{x}^j$ at all nodes. If the result is zero, part (a)'s uniqueness for the zero interpolation data forces every coefficient to be zero. Thus the square matrix has trivial kernel and is invertible. Equivalently the interpolation map is a linear bijection between two spaces of dimension $n+1$.