Let $n\geq 2$ be a natural number. Consider the transformation $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ defined by
$$f(x_{1},…,x_{n})=\left ( \sum_{i=1}^{n}x_{i},\sum_{i=1}^{n}x_{i}^{2},…,\sum_{i=1}^{n}x_{i}^{n} \right ).$$
Find the Jacobian for every $(x_{1},…,x_{n})\in \mathbb{R}^{n}$.
Attempt:
Calculating the Jacobian Matrix yields
$$
\left[\begin{array}{ccccc}
1 & 1 & 1 & \ldots & 1 \\
2x_{1} & 2x_{2} & 2x_{3} & \ldots & 2x_{n} \\
3x_{1}^{2} & 3x_{2}^{2} & 3x_{3}^{2} & \ldots & 3x_{n}^{2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
nx_{1}^{n-1} & nx_{2}^{n-1} & nx_{3}^{n-1} & \ldots & nx_{n}^{n-1}
\end{array}\right].
$$
Since the determinant of the transpose of a square matrix is equal to the determinant of the matrix, the Jacobian will be equal to the determinant of the transpose of the Jacobian Matrix. This tranpose is
$$
\left[\begin{array}{ccccc}
1 & 2x_{1} & 3x_{1}^{2} & \ldots & nx_{1}^{n-1} \\
1 & 2x_{2} & 3x_{2}^{2} & \ldots & nx_{2}^{n-1} \\
1 & 2x_{3} & 3x_{3}^{2} & \ldots & nx_{3}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 2x_{n} & 3x_{n}^{2} & \ldots & nx_{n}^{n-1}
\end{array}\right],
$$
which is very similar to the Vandermonde Matrix.
Indeed if I compute the case for $n=3$, I get that the Jacobian is $-6 (x – y) (x – z) (y – z)$. If I compute more cases, I get that the Jacobian will always be the $n\times n$ Vandermonde determinant times $n!$. That is, the closed form of the Jacobian of the transformation is
$$n!\det(V),$$
where $V$ is the $n\times n$ Vandemonde Matrix.
However, I've not been successful in formulating a formal proof for this.
Best Answer
What you've done is fine. You have\begin{align}\begin{vmatrix}1 & 2x_{1} & 3x_{1}^{2} & \ldots & nx_{1}^{n-1} \\ 1 & 2x_{2} & 3x_{2}^{2} & \ldots & nx_{2}^{n-1} \\ 1 & 2x_{3} & 3x_{3}^{2} & \ldots & nx_{3}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \ldots & nx_{n}^{n-1}\end{vmatrix}&=2\begin{vmatrix}1 & x_{1} & 3x_{1}^{2} & \ldots & nx_{1}^{n-1} \\ 1 & x_{2} & 3x_{2}^{2} & \ldots & nx_{2}^{n-1} \\ 1 & x_{3} & 3x_{3}^{2} & \ldots & nx_{3}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \ldots & nx_{n}^{n-1}\end{vmatrix}\\&=(2\times3)\begin{vmatrix}1 & x_{1} & x_{1}^{2} & \ldots & nx_{1}^{n-1} \\ 1 & x_{2} & x_{2}^{2} & \ldots & nx_{2}^{n-1} \\ 1 & x_{3} & x_{3}^{2} & \ldots & nx_{3}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \ldots & nx_{n}^{n-1}\end{vmatrix}\\&=\vdots\\&=n!\begin{vmatrix}1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n-1} \\ 1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n-1} \\ 1 & x_{3} & x_{3}^{2} & \ldots & x_{3}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \ldots & x_{n}^{n-1}\end{vmatrix}.\end{align}