0
$\begingroup$

Let $$ A= \begin{bmatrix} x_1^{4} & x_2^{4} & x_3^{4} \\ x_1^{2} & x_2^{2} & x_3^{2} \\ 1 &1 &1 \end{bmatrix} .$$ I want to show that the $$\det A \times \frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)} = x_1^2x_2+x_1^2x_3+x_2^2x_3+x_1x_2^2+x_1x_3^2+x_2x_3^2+2x_1x_2x_3.$$

The determinant of $A$ is trivial but trying to manipulate $\frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}$ is where I am having some problems.

We can think of $$\frac{1}{(x_1-x_2)} = \frac{1}{x_1(1-\frac{x_2}{x_1})}=\frac{1}{x_1}(1+ \frac{x_2}{x_1}+\frac{x_2^2}{x_1^2}+\cdots)$$ but I still dont end up getting the answer after multplying the whole thing out. Is there a trick to this? any hint will be appreciated.

P.S (Combinatorics): This is an expression from the definition of Schur Functions.

$\endgroup$
  • $\begingroup$ By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem. $\endgroup$ – Jack D'Aurizio Nov 21 '18 at 22:33
  • $\begingroup$ Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator. $\endgroup$ – B. Goddard Nov 21 '18 at 22:38
  • $\begingroup$ @B.Goddard Thanks a lot. Done :) $\endgroup$ – Jaynot Nov 21 '18 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.