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How can we find the determinant of the following matrix $A$:

$\left( \begin{array}{cccccc} 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{array} \right) $

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.

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    $\begingroup$ This is a rank one matrice, and you should have a look here: math.stackexchange.com/questions/55165/… $\endgroup$ – Surb Sep 30 '14 at 10:48
  • $\begingroup$ Ok, my bad this is not a rank one matrix, take for example: $x = (1,2,3)$ and $y = (a,b,c)$. Then $$A= \begin{pmatrix} 1 \cdot a & 1\cdot b & 1\cdot c \\ 1\cdot b & 2 \cdot b & 2 \cdot c \\ 1\cdot c & 2 \cdot c & 3 \cdot c \end{pmatrix}$$ Setting $a = b = c$ leads to a rank $n$ matrix. $\endgroup$ – Surb Sep 30 '14 at 13:36
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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.

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  • $\begingroup$ That works! One can WLOG assume that $y_n \neq 0$ (we are proving a polynomial identity), and then reduce $A_{n+1}$ to $A_n$ by dividing column $n+1$ by $y_{n+1}$, dividing column $n$ by $y_n$, and subtracting column $n$ from column $n+1$. $\endgroup$ – darij grinberg Sep 30 '14 at 21:17
  • $\begingroup$ For the eigenvalue variant I get the following complicated expression $\text{det}(A_3-\lambda I) = 2x_1^2x_2y_2y_3^2-2x_1^2x_2^2y_3^2+\lambda x_2^2y_3^2+\lambda x_1^2y_3^2-x_1^2x_3^2y_2^2+\lambda x_1^2y_2^2+x_1^2x_2^2x_3^2-\lambda x_2^2x_3^2-\lambda x_1^2x_3^2+\lambda^2 x_3^2-\lambda x_1^2x_2^2+\lambda^2 x_2^2+\lambda^2 x_1^2-\lambda^3$ $\endgroup$ – andre Sep 30 '14 at 21:31
  • $\begingroup$ I wouldn't bother looking for the eigenvalues; they already have square roots for $n=2$. $\endgroup$ – darij grinberg Sep 30 '14 at 21:33
  • $\begingroup$ I think the case $n=3$ shows, that it could be very frustrating and unrewarding. $\endgroup$ – andre Sep 30 '14 at 21:37
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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 $x_1 y_n \prod_{i = 1}^{n-1} (x_{i + 1} y_i - x_i y_{i + 1})$.

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