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Consider the Vandermond matrix $$ V (x_1, x_2, \ldots , x_n) = \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & x_1^n & x_1^{n+1} & \cdots \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & x_2^n & x_2^{n+1} & \cdots \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} & x_3^n & x_3^{n+1} & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \cdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & x_n^n & x_n^{n+1} & \cdots \end{pmatrix} $$ with the specific choice of $x_1 = e^{\lambda_1}, \dots x_n = e^{\lambda_n}$, where $\lambda_1, \dots, \lambda_n$ are distinct real numbers.

I was wondering, if one will always get by picking arbitrary (not necessary consecutive) $n$ columns in the above "infinite" Vandermonde matrix linearly independent vectors.

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closed as off-topic by Chris Godsil, Ricardo Andrade, José Figueroa-O'Farrill, S. Carnahan Sep 29 '14 at 2:30

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This works, as pointed out by Samuel in a comment. Here's an easy direct argument: We are claiming that no non-trivial polynomial $$ p(x) = \sum_{j=1}^N a_j x^{n_j} $$ with $N$ terms (let me call $N$ the pseudo-degree) has $N$ or more distinct positive zeros.

This is immediate from an induction on $N$. First of all, we can assume that $p$ starts with a constant term. Then, if $p$ had $N$ distinct positive zeros, its derivative would be a polynomial of pseudo-degree $N-1$ with at least $N-1$ positive zeros.

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    $\begingroup$ This is a weak form of Decartes’ rule of signs. $\endgroup$ – Emil Jeřábek Sep 28 '14 at 11:33