Let $m_i\in \mathbb{N}, 1\leq i \leq n $ such that wlog if $m_i < m_j\in \mathbb{Z^+}$ 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 $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{m_1} & x_2^{m_1} & x_3^{m_1}\\ x_1^{m_2} & x_2^{m_2} & x_3^{m_2} \end{bmatrix}$$ $det(V_3) = x_2^{m_2} x_1^{m_1}-x_3^{m_2} x_1^{m_1}-x_2^{m_1} x_1^{m_2}+x_3^{m_1} x_1^{m_2}-x_2^{m_2} x_3^{m_1}+x_2^{m_1} x_3^{m_2}$.

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. One can guess that over $\mathbb{C^n}$, if the $|x_i|$ are distinct and have no-non trivial relations, then the schur function is non-zero. Which brings me back to my question:

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

Well yeah, a little bit.There are some results one could see from Mitchel's result about the positivity of the coefficients of the Schur function. If we are working over $\mathbb{R^n}$, then as a consequence of the positivity of the coefficients mentioned by Mitchel, one can deduce that if $x_i$ are distinct and positive for $1 \leq i \leq n$, then the schur function takes on positive values and therefore non-zero. This result is mentioned in this Paper.

Examples:

1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1 & x_2 & x_3\\ x_1^{3} & x_2^{3} & x_3^{3} \end{bmatrix}$$ where $m_1 = 1, m_2 = 3$, then $det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3) (x_1+x_2+x_3)$, so our schur function is the elementary symmetric function $x_1+x_2+x_3$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

2. 1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{4} & x_2^{4} & x_3^{4}\\ x_1^{6} & x_2^{6} & x_3^{6} \end{bmatrix}$$ where $m_1 = 4, m_2 = 6$, then

$det(V)=-x_2^4 x_1^6+x_3^4 x_1^6+x_2^6 x_1^4-x_3^6 x_1^4+x_2^4 x_3^6-x_2^6 x_3^4$

$det(V)=-(x_1-x_2) (x_1+x_2) (x_1-x_3) (x_2-x_3) (x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$, so our schur function is $ (x_1+x_2)(x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

3. 1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{4} & x_2^{4} & x_3^{4}\\ x_1^{7} & x_2^{7} & x_3^{7} \end{bmatrix}$$ where $m_1 = 4, m_2 = 7$, then

$det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3)$ *

$(x_2^3 x_1^5+x_3^3 x_1^5+x_2 x_3^2 x_1^5+x_2^2 x_3 x_1^5+x_2^4 x_1^4+x_3^4 x_1^4+$ $2 x_2 x_3^3 x_1^4+2 x_2^2 x_3^2 x_1^4+2 x_2^3 x_3 x_1^4+x_2^5 x_1^3+x_3^5 x_1^3+$ $2 x_2 x_3^4 x_1^3+3 x_2^2 x_3^3 x_1^3+3 x_2^3 x_3^2 x_1^3+2 x_2^4 x_3 x_1^3+x_2 x_3^5$ $x_1^2+2 x_2^2 x_3^4 x_1^2+3 x_2^3 x_3^3 x_1^2+2 x_2^4 x_3^2 x_1^2+x_2^5 x_3 x_1^2+$ $x_2^2 x_3^5 x_1+2 x_2^3 x_3^4 x_1+2 x_2^4 x_3^3 x_1+x_2^5 x_3^2 x_1+x_2^3 x_3^5+$ $x_2^4 x_3^4+x_2^5 x_3^3)$

So our schur function is everything that we have after dividing out by $V_p$. This function is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

My goal is to really answer the following interesting question, Let $S\subset\mathbb{N}$ be infinite and arbitrary. Then Let the set $Z_d={zeros of the schur function with degree d|d\in S}$. Now Is $\Cup_d\in S Z_d$ empty, assuming that the variables of the schur function generate a free abelian group of rank equal to the number of variables.