4

Find the determinant of the $m \times m$ matrix $K$ where $$K_{ij} = {1 \over 1 - x_ix_j} $$ for any values of $x_1,x_2,\dotsc, x_m$.

My first thought is to make each component polynomial by scaling the rows of the matrix: $$K_{ij} \prod_{k=1}^m(1-x_ix_k)=\prod_{k\neq j}(1-x_ix_k)$$

Since the determinant of the scaled matrix is a polynomial you can determine its degree, and find its roots by using the fact that the determinant of a matrix with two columns the same is zero.

What happens is that the degree of the polynomial in $x_1$ is $2(m-1)$. And playing around with Wolfram Alpha shows that $m-1$ of them are double roots. Finding those $m-1$ roots is easy but showing they have multiplicity $2$ seems hard; it seems I have to differentiate.

Is there an easier way?

wlad
  • 8,185
  • 1
    Maybe a long shot, but writing $\frac{1}{1-x_i x_j} = \frac{x_i^{-1}}{x_i^{-1}-x_j} = \frac{z_i}{z_i - y_j}$ for $y_i=x_i$ and $z_i=x_i^{-1}$, can't the question be cast as a special case of the determinant of a Cauchy matrix? – Clement C. Nov 24 '15 at 15:40
  • 1
    That looks right. $z_i$ is a common factor in the $i$th row, so do a row-operation to get rid of it. – Empy2 Nov 24 '15 at 15:47

2 Answers2

0

If any two $x_i$ are the same, the determinant is zero, so the numerator of $\det(K)$ contains all the factors $(x_i-x_j)$.
To swap $x_i$ with $x_j$, you swap the two rows, and the two columns, so the determinant stays the same. So the numerator must contain all the factors $(x_i-x_j)^2$.
I expect $(1-x_ix_j)$ is a factor in the denominator, and it appears twice if $i\neq j$.
Is the answer $$\frac{\prod_{i<j}(x_i-x_j)^2}{\prod_{i,j}(1-x_ix_j)}$$

Empy2
  • 50,853
0

Indeed the answer is \begin{equation} det(K)=\frac{\underset{{\scriptscriptstyle i\neq j}}{\prod}\left(x_{i}-x_{j}\right)}{\underset{{\scriptscriptstyle i,j=1..n}}{\prod}(-1+x_{i}x_{j})} \end{equation} To prove it you just have to proceed with induction. For $n=1$ is trivially right and assuming it right for $n$ the rightness for $n+1$ it should follow easily since you just have to sum $n+1$ determinant of the same form multiplied by $\frac{1}{-1+x_{1}x_{j}}$. Might be a mess in notation but it should be pretty straightforward.

Dac0
  • 9,154
  • But the point is — how? ${}{}{}$ – Clement C. Nov 24 '15 at 15:42
  • In which sense "how"? In the same way determinants are computed... Once you've proofed the determinant is that one I'm not sure what you're asking... – Dac0 Nov 24 '15 at 15:57
  • Your answer originally gave an expression only. The main point of this website (as I understand it) and more broadly of mathematics (again, as I understand it) is how you get to the answer, more than the answer itself. Hence my comment. – Clement C. Nov 24 '15 at 15:58
  • Good for you NaN! But You didn't get the point: "use induction"... – Dac0 Nov 25 '15 at 15:01
  • Anyway since you're so good in Wolfram Alpha why didn't you give us directly the formula? – Dac0 Nov 25 '15 at 15:07
  • And so now you finally know I didn't use "Wolfram Alpha"... – Dac0 Nov 26 '15 at 22:08