Square of Vandermonde Matrix
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The validity of the material on this page is questionable.
The case $n = 2$ left me clueless to what could possibly be intended here; only $\mathbf {V V}^T$ is trivially seen symmetric in the $x_n$, but this can hardly be called a square
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This theorem requires a proof..
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
Theorem
The square of the Vandermonde matrix of order $n$:
- $\mathbf V = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^n & x_2^n & \cdots & x_n^n \end{bmatrix}$
is symmetrical in $x_1, \ldots, x_n$.
The case $n = 2$ left me clueless to what could possibly be intended here; only $\mathbf {V V}^T$ is trivially seen symmetric in the $x_n$, but this can hardly be called a square
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
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Proof
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