Find the determinant of a symmetric matrix

Question

How can we find the determinant of the following matrix?

$$A = \begin{pmatrix} x_1y_1 & x_1y_2 & x_1y_3 & \cdots & x_1y_{n-1} & x_1y_n \\ x_1y_2 & x_2y_2 & x_2y_3 & \cdots & x_2y_{n-1} & x_2y_n \\ x_1y_3 & x_2y_3 & x_3y_3 & \cdots & x_3y_{n-1} & x_3y_n \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ x_1y_{n-1} & x_2y_{n-1} & x_3y_{n-1} & \cdots & x_{n-1}y_{n-1} & x_{n-1}y_n \\ x_1y_n & x_2y_n & x_3y_n & \cdots & x_{n-1}y_n & x_ny_n \end{pmatrix}$$

That is, the entry of $A$ is $a_{ij}=x_iy_j$ for $i\leq j$; and $a_{ij}=a_{ji}$ for $i>j$.

I do not have anything new. And I find also it is difficult to find its eigenvalues.

Answer 1

A hint, a quick computer algebra calculation gives for the determinant of the matrix $A_n$ \begin{align} \text{det}A_1 &= +x_1y_1\\ \text{det}A_2 &= -x_1y_2(x_1y_2-x_2y_1)\\ \text{det}A_3 &= +x_1y_3(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)\\ \text{det}A_4 &= -x_1y_4(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)(x_3y_4-x_4y_3)\\ \text{det}A_5 &= +x_1y_5(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)(x_3y_4-x_4y_3)(x_4y_5-x_5y_4) \end{align} So this suggests to show first the formula for $\text{det}A_2$, and then to somehow show the recursion \begin{equation} y_n \text{det}A_{n+1} = -y_{n+1}(x_ny_{n+1}-x_{n+1}y_n) \text{det}A_n, \quad\text{with}\quad n > 2. \end{equation} Then one has to see what kind of coefficients $x_k$ and $y_k$ one has.

Answer 2

I don't have a solution method, but the determinant appears as problem *222 (the star/asterisk apparently indicating a difficult problem) in Problems in Higher Algebra by D. K. Faddeev and I. S. Sominskii [translated by J. L. Brenner] (W. H. Freeman, 1965). The answer is given as $x_1 y_n \prod_{i = 1}^{n-1} (x_{i + 1} y_i - x_i y_{i + 1})$.