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Suppose that $x_1,\ldots,x_n \in \mathbb{R}$ are $n$ distinct real numbers such that $x_i \neq 0, 1 \leq i \leq n.$ Let $V_n$ be the following Vandermonde matrix $$ V_n= \begin{bmatrix} x_1 & x_2 & \ldots & x_n \\ x_1^2 & x_2^2 & \ldots & x_n^2 \\ \vdots & \vdots& \ldots & \vdots \\ x_1^n & x_2^n & \ldots & x_n^n \\ \end{bmatrix}. $$ It is well known that $\mathrm{det}(V_n)= \prod_i x_i \prod_{i<j}(x_i-x_j)$ and hence $V_n$ is invertible. Suppose I construct the following modified matrix $$ V_{n-1}= \begin{bmatrix} x_1-x_n & x_2-x_n & \ldots & x_{n-1}-x_ n \\ x_1^2-x_n^2 & x_2^2-x_n^2 & \ldots & x_{n-1}^2-x_ n^2 \\ \vdots & \vdots& \ldots & \vdots \\ x_1^{n-1}-x_n^{n-1} & x_2^{n-1}-x_n^{n-1} & \ldots & x_{n-1}^{n-1}-x_n^{n-1} \\ \end{bmatrix}. $$ My goal is to prove that $\mathrm{rank}(V_{n-1})=n-1$, or equivalently, $V_{n-1}$ is invertible. One can also assume that $x_1, \ldots,x_n$ are generated from a continuous distribution if the above statement doesn't hold for all $x_1,\ldots,x_n$.

The following is my strategy: For $n=2$, $$ V_1=[x_1 - x_2] \Rightarrow \mathrm{det}(V_1)=x_1 -x_2 \neq 0, $$ and for $n=3,4$ respectively, $$ V_2=\begin{bmatrix} x_1 -x_3 & x_2 -x_3 \\ x_1^2 -x_3^2 & x_2^2 -x_3^2 \end{bmatrix} \Rightarrow \mathrm{det}(V_2)=-(x_1 -x_2)(x_1-x_3)(x_2-x_3) \neq 0, \\ V_3=\begin{bmatrix} x_1 -x_4 & x_2 -x_4 & x_3-x_4 \\ x_1^2 -x_4^2 & x_2^2 -x_4^2 & x_3^2 -x_4^2 \\ x_1^3 -x_4^3 & x_2^3 -x_4^3 & x_3^3 -x_4^3 \end{bmatrix} \\\Rightarrow \mathrm{det}(V_3)=-(x_1 -x_2)(x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4) \neq 0. $$

I think it should be possible to prove this through induction for general $n$.

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    $\begingroup$ The general statement isn't true. The determinant of $I_2 + \pmatrix{a\\b} \pmatrix{1&1}$ is $a + b + 1$. If we set $a = b = -\frac{1}{2}$, the resulting vector is neither zero nor any column of $I_2$, but the determinant is zero so the matrix is noninvertible. $\endgroup$ – Travis Jun 16 '17 at 20:08
  • $\begingroup$ @Travis:But is it possible for the rank of $V_{n-1}$ to be still $n-1$? given the Vandermonde structure? $\endgroup$ – pikachuchameleon Jun 16 '17 at 20:09
  • $\begingroup$ Have you tried a small scale example? Have you considered using induction? $\endgroup$ – Fimpellizieri Jun 16 '17 at 20:10
  • $\begingroup$ @Fimpellizieri: It seems to be true for at least $3 \times 3$ matrices. I don't have a formal proof yet. $\endgroup$ – pikachuchameleon Jun 16 '17 at 20:11
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    $\begingroup$ Yes, a priori, and I'd guess that the statement is true---I claim only that it can't be proved in the way proposed. There is a general formula for the determinant of a rank-$1$ update of a matrix: en.wikipedia.org/wiki/Matrix_determinant_lemma . There is a general formula for the inverse of a Vandermonde matrix: proofwiki.org/wiki/Inverse_of_Vandermonde%27s_Matrix $\endgroup$ – Travis Jun 16 '17 at 20:11

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