Almost a Vandermonde matrix

Question

I would like to simplify

$\det\begin{pmatrix}1&x_1^2&x_1^3&x_1^4\\1&x_2^2&x_2^3&x_2^4\\1&x_3^2&x_3^3&x_3^4\\1&x_4^2&x_4^3&x_4^4 \end{pmatrix}$

which is not exactly like the Vandemonde matrix by the exponents, but I was wondering if there is a way to compute it in $\mathbb{Q}[x_1,x_2,x_3,x_4]$

I tried computing it like a normal matrix but after a very long computation I don't find an "easy" formula to describe it, I also tried following the general proof of the Vandermone matrix but without success. I think there is more behind it, does anyone know about this?

Answer 1

$$\begin{split} &\det\begin{pmatrix}1&x_1^2&x_1^3&x_1^4\\1&x_2^2&x_2^3&x_2^4\\1&x_3^2&x_3^3&x_3^4\\1&x_4^2&x_4^3&x_4^4 \end{pmatrix} =\det\begin{pmatrix} x_2^2-x_1^2 &x_2^3-x_1^3 &x_2^4-x_1^4 \\ x_3^2-x_1^2 &x_3^3-x_1^3 &x_3^4-x_1^4\\ x_4^2-x_1^2 &x_4^3-x_1^3 &x_4^4-x_1^4 \end{pmatrix}\\ &= (x_2-x_1)(x_3-x_1)(x_4-x_1) \det\begin{pmatrix} x_2+x_1 &x_2^2+x_2x_1+x_1^2 &x_2^3+x_2^2x_1+x_2x_1^2+x_1^3 \\ x_3+x_1 &x_3^2+x_3x_1+x_1^2 &x_3^3+x_3^2x_1+x_3x_1^2+x_1^3\\ x_4+x_1 &x_4^2+x_4x_1+x_1^2 &x_4^3+x_4^2x_1+x_4x_1^2+x_1^3 \end{pmatrix}\\ &= (x_2-x_1)(x_3-x_1)(x_4-x_1) \det\begin{pmatrix} x_2+x_1 &x_2^2 &x_2^3+x_2^2x_1 \\ x_3+x_1 &x_3^2 &x_3^3+x_3^2x_1\\ x_4+x_1 &x_4^2 &x_4^3+x_4^2x_1 \end{pmatrix}\\ &= (x_2-x_1)(x_3-x_1)(x_4-x_1) \det\begin{pmatrix} x_2+x_1 &x_2^2 &x_2^3 \\ x_3+x_1 &x_3^2 &x_3^3\\ x_4+x_1 &x_4^2 &x_4^3 \end{pmatrix} \end{split} $$ where $$ \det\begin{pmatrix} x_2+x_1 &x_2^2 &x_2^3 \\ x_3+x_1 &x_3^2 &x_3^3\\ x_4+x_1 &x_4^2 &x_4^3 \end{pmatrix} = \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ x_3+x_1 &x_3^2 &x_3^2(x_3-x_2)\\ x_4+x_1 &x_4^2 &x_4^2(x_4-x_2) \end{pmatrix}\\ = \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ x_3-x_2 &x_3^2-x_2^2 &x_3^2(x_3-x_2)\\ x_4-x_2 &x_4^2-x_2^2 &x_4^2(x_4-x_2) \end{pmatrix}\\ =(x_3-x_2)(x_4-x_2) \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ 1 &x_3+x_2 &x_3^2\\ 1 &x_4+x_2 &x_4^2 \end{pmatrix}\\ =(x_3-x_2)(x_4-x_2) \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ 1 &x_3+x_2 &x_3^2\\ 0 &x_4-x_3 &x_4^2- x_3^2 \end{pmatrix}\\ =(x_3-x_2)(x_4-x_2)(x_4-x_3) \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ 1 &x_3+x_2 &x_3^2\\ 0 &1 &x_4+x_3 \end{pmatrix}\\ =(x_3-x_2)(x_4-x_2)(x_4-x_3) \det\begin{pmatrix} x_2+x_1 &x_2^2 &0\\ 1 &x_3+x_2 &x_3^2\\ 0 &1 &x_4+x_3 \end{pmatrix}\\ = (x_3-x_2)(x_4-x_2)(x_4-x_3)(x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4+x_2x_3x_4) $$

Answer 2

Hint:

The determinant is a polynomial of degree nine in the unknowns, and has the factors

$$(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4).$$