Square of Vandermonde Matrix

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 validity of the material on this page is questionable. In particular: 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. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Proof

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.