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Consider a Vandermonde matrix $$V = \begin{bmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ & & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{bmatrix}$$ It is a simple exercise to prove that if the rank of the Vandermonde is $< n$, then the determinant is $0$ and not all of the $x_i$ can be distinct. But is the following more fine-grained statement true?

Question: If a Vandermonde matrix has rank $k$, then does the set $\{ x_1,\cdots,x_n \}$ contain exactly $k$ distinct elements?

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    $\begingroup$ A similar statement which can be used to answer your question is quoted in the Wikipedia entry for Vandermonde matrices; so presumably this result is well-known. $\endgroup$
    – Willie Wong
    Commented Feb 26, 2020 at 20:05
  • $\begingroup$ @WillieWong I see! Thanks for the reference. I was not aware of this result. $\endgroup$
    – Television
    Commented Feb 26, 2020 at 20:09
  • $\begingroup$ See also (for a more general, noncommutative setting): T. Y. Lam, A general theory of Vandermonde matrices, Expositiones Mathematicae 4 (1986), pp. 193--215. $\endgroup$
    – darij grinberg
    Commented Feb 26, 2020 at 20:24
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    $\begingroup$ If $\{x_1,\ldots,x_n\}$ has exactly $k$ distinct elements, then $V$ has a $k \times k$ invertible submatrix, and has only $k$ distinct lines, so the rank must be $k$. $\endgroup$
    – François Brunault
    Commented Feb 26, 2020 at 22:51
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    $\begingroup$ Well what I wrote says that the rank is equal to the cardinality of $\{x_1,\ldots,x_n\}$, which is what you want. $\endgroup$
    – François Brunault
    Commented Mar 3, 2020 at 22:04

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