Invertibility of generalized Vandermonde matrix

Question

I want to show that the following statement is false. Consider $n+1$ linearly independent functions on $\mathbf{C}$

$$ \phi_0, \cdots, \phi_n, $$

and the generalized Vandermonde matrix given by

$$ V_n = \phi_k(z_j) $$

for $n+1$ points in $\mathbf{C}$. Then $V_n$ is invertible IFF if $z_0, \cdots, z_n$ are pairwise distinct.

The only thing that came up in my mind would be to show that it is possible to have $det(V_n) = 0$ even if they are distinct, and that it is possible to have $det(V_n) \neq 0$ even if they are not pairwise distinct.

Nevertheless, I do not know if such a matrix is used for not pairwise distinct points, so maybe the second point is useless to prove.

Answer 1

In fact, given $n+1$ points $z_j$, there is a plenty of linearly independent functions which have equal values in these points (for example, polynomials of degree $d> n+1$). Taking these functions for the given points we get singular matrix $V_n = \phi_k(z_j)$.