# Using the Vandermonde Determinant to calculate the Jacobian of a transformation

determinantjacobianlinear algebramatricesmultivariable-calculus

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.

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}