Let $m_i\in \mathbb{N}, 1\leq i \leq n $ such that wlog if $m_i < m_j$ then $i < j$. Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 \\ x_1^{m_1} & x_2^{m_1} & x_3^{m_1} & \ldots & x_{n-1}^{m_1} \\ x_1^{m_2} & x_2^{m_2} & x_3^{m_2} & \ldots & x_{n-1}^{m_2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{m_n} & x_2^{m_n} & x_3^{m_n} & \ldots & x_{n-1}^{m_n} \\ \end{bmatrix}$$ Under What conditions is the generalized Vandermonde Matrix Invertible? It is clear that if one lets $x_i=x_j$,then the determinant is Zero, and therefore the principle Vandermonde Determinant $V_p=\prod_{1 \leq i < j\leq n}^n(x_i-x_j)$ can be factored from the $det(V)$. What remains is the schur function, which is a homogeneous polynomial of degree d which depends on $n$, which symmetric over its variables $x_i,1\leq i\leq n$. There is a proof of the fact that the coefficients of the schur function are positive integers which was given by Mitchel long ago.

When one looks at the simplest case where $x_i \in \mathbb{C}, 1\leq i \leq n$ and $n=2$, $$V= \begin{bmatrix} 1 & 1\\ x_1^n & x_2^n\\ \end{bmatrix}$$ one sees that the Vandermonde determinant is invertible when $\frac{x_1}{x_2}$ is not an $n^{th}$ root of unity.

For $n=3$ things become difficult already as $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{m_1} & x_2^{m_1} & x_3^{m_1}\\ x_1^{m_2} & x_2^{m_2} & x_3^{m_3} \end{bmatrix}$$ $det(V_3) = x_2^{m_2} x_1^{m_1}-x_3^{m_2} x_1^{m_1}-x_2^{m_1} x_1^{m_2}+x_3^{m_1} x_1^{m_2}-x_2^{m_2} x_3^{m_1}+x_2^{m_1} x_3^{m_2}$.

As far as I see (which is not very far), characterizing the zero set of the this polynomial is not at all simple in either $\mathbb{C^3}$ or $\mathbb{R^3}$ with exception of the trivial case we already mentioned where $x_i=x_j$ for some $i\neq j$ and perhaps some roots of unity. Are there any zeros other than those. Which brings me back to my question:

What are the conditions under which the generalized vandermonde matrix is invertible over $\mathbb{R^n}$ or $\mathbb{C^n}$?