$n$ linearly independent rows of Vandermonde matrix Consider the "infinite" Vandermonde matrix 
$$
V (x_1, x_2, \ldots , x_n) =
\begin{pmatrix}
  1      & x_1    & x_1^2  & \cdots & x_1^{n-1} & x_1^n & x_1^{n+1} & \cdots \\
  1      & x_2    & x_2^2  & \cdots & x_2^{n-1} & x_2^n & x_1^{n+1} & \cdots \\
  1      & x_3    & x_3^2  & \cdots & x_3^{n-1} & x_3^n & x_1^{n+1} & \cdots \\
  \vdots & \vdots & \vdots & \ddots & \vdots    & \vdots & \vdots & \cdots \\
  1      & x_n    & x_n^2  & \cdots & x_n^{n-1} & x_n^n & x_n^{n+1} & \cdots
\end{pmatrix}
$$
with distinct $x_1, \dots, x_n$. It is well-known that if we pick the first $n$ columns, then they span the whole space.
I was wondering, if the same holds true if one picks arbitrary (not necessary consecutive) $n$ columns in the above "infinite" Vandermonde matrix. I looked at simple examples and they suggest that this is true. Is there a nice way of proving this?
 A: I think you're talking about a Generalized Vanderrmonde Matrix, so the determinant will always be positive and the matrix is always invertible.
A: I just thought of a simple counterexample.  
$$
\left[\begin{matrix}
x &x^3\\
-x &(-x)^3\\
\end{matrix}\right]
$$
If all the columns are an even power or all are an odd power, then $x$ and $-x$ rows are linearly dependent.
Also, if any x = 0, any minor including that row will be singular.
A: Another counterexample would be for complex variables any real multiple of a n:th root of unity $$rx = r e^{\frac{2\pi i}{n}}$$
then $$\begin{bmatrix} (rx)^n& (rx)^{2n} \\(rx)^{n+k}&(rx)^{2n+k}\end{bmatrix} = \begin{bmatrix} r^n& r^{2n} \\r^{n+k}x^k&r^{2n+k}x^{k}\end{bmatrix}$$
and we see the second row is $r^kx^k$ times the first. 
For real matrices the only possibility is even and odd because they are the only roots of unity there but for complex we can have any integer.
A: It's wrong quite often. First of all let me reformulate the condition that $n$ columns of such matrix are linearly dependent. Suppose the columns numbers are $i_1, \dots, i_n$ and coefficients of linear dependence are $c_1,dots, c_n$. Construct polynomial
$$ p(x) = \sum_{k=1}^n c_k x^{i_k}.$$
Then existence of linear dependence for this columns with coefficients $c_i$ is equivalent to system: $p(x_i)=0$ for all $i$. Thus finding polynomial p(x) leads to linear dependence for columns of $V(x_1,\dots,x_n)$, where $x_i$ are roots of p(x).
Such polynomial $p(x)$ is of special form. It's degree should be greater or equal than $n$ and it should have at most $n$ nonzero coefficients. But of course there are a lot of such polynomials even with distinct real roots.
For example, take any positive distinct $x_1,\dots,x_{n-1}$. Then take $\lambda = -\sum x_i$ and construct
$$ p(x) = (x-\lambda)\prod_{i=1}^{n-1} (x-x_i).$$
This one has distinct real roots and is of the form $p(x)= x^n + c x^{n-2}+ \dots$ thus having at most $n$ nonzero coefficients. This leads to linear dependence for columns of $V(x_1\dots,x_{n-1},\lambda)$. Columns $1,\dots,n-1, n+1$.
