How can I prove that $ \begin{vmatrix} \frac{1}{(x_1-y_1)^2} & \frac{1}{(x_1-y_2)^2} & \cdots & \frac{1}{(x_1-y_n)^2} \\ \frac{1}{(x_2-y_1)^2} & \frac{1}{(x_2-y_2)^2} & \cdots & \frac{1}{(x_2-y_n)^2} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{(x_n-y_1)^2} & \frac{1}{(x_n-y_2)^2} & \cdots & \frac{1}{(x_n-y_n)^2} \end{vmatrix} \ne 0 $?

For $x_i$ and $y_i$ are all distinct.

  • 1
    $\begingroup$ Is there any unspecified assumptions about $x_i$ and $y_i$, such as $x_i$ are all distinct? $\endgroup$
    – Element118
    Commented Nov 14, 2015 at 14:52
  • $\begingroup$ If $x_1=x_2$ and $y_1=y_2$ then you fail for $n=2$. $\endgroup$
    – Jimmy R.
    Commented Nov 14, 2015 at 15:09
  • $\begingroup$ Usually, what one does is to fix the values for $y$ (all distinct because otherwise the matrix is singular), clear your fractions, find the degree of the corresponding polynomial in the $x_i$'s and try to see if there are enough obvious answers to to the polynomial to account for all the easy solutions. $\endgroup$ Commented Nov 14, 2015 at 15:11
  • $\begingroup$ We need all $x_i$ and all $y_j$ to be distinct. If $x_i =x_k$ for some $i,k$, then two of the rows would be identical and so the determinant would be zero. The same follows for the $y_j$'s except that two or more columns would be identical. $\endgroup$
    – Mnifldz
    Commented Nov 14, 2015 at 15:13

1 Answer 1


Even if all the $x_i$ and $y_j$ are distinct, this is not true.

For instance, if $x_1=2/3$, $x_2=2$, $y_1=1$, $y_2=0$, we have $$ \begin{vmatrix}\frac1{(x_1-y_1)^2}&\frac1{(x_1-y_2)^2}\\\frac1{(x_2-y_1)^2}&\frac1{(x_2-y_2)^2}\end{vmatrix} =\begin{vmatrix} \frac1{(2/3-1)^2}&\frac1{(2/3-0)^2}\\ \frac1{(2-1)^2}&\frac1{(2-0)^2} \end{vmatrix} =\begin{vmatrix} 9&9/4\\ 1&1/4 \end{vmatrix}=0 $$


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