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Let $0<x_1<x_2<...<x_n<1$. Let us consider the matrix M defined such that: $$M_{i,j} = \min(x_i, x_j)^2 \left( 3 \max(x_i, x_j) - \min(x_i, x_j) \right)$$

I believe the determinant of M is different from zero. Any idea on how to prove it (if true)?

Of course, any reason why M would be invertible would suit me as well, no real need to get the formula of the determinant.

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Yuval's solution dropped an exponent; this is easily seen since his first expression is 6th degree and the later one is only 5th. It should be $$x_i^3(4x_2 - x_1) (x_2 - x_1)^2\;.$$ But his conclusion still holds; the claim is clearly true for $n = 2$.

For the general case, it may help to remember that this is a symmetric matrix, and all symmetric matrices can be diagonalized. Furthermore, since you restricted the $x_i$ to be in strictly increasing order, all the $\min$ and $\max$ stuff devolves to $M_{i,j} = x_i^2(3x_j - x_i) = 3x_i^2x_j - x_i^3$ whenever $i \leq j$. This is immediately recognizable as $(D_{x_i} - D_{x_j})(x_i^3x_j)$, where $D$ is the differential operator. I don't know if either of these facts are helpful, but it sure seems suspicious. You might try showing that all the eigenvalues are always positive, which would mean the matrix is positive definite (implying nonsingular).

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  • $\begingroup$ You mean "implying non-singular"? $\endgroup$ – joriki Mar 16 '11 at 1:40
  • $\begingroup$ er... yes, sorry, I just got confused toward the end. $\endgroup$ – Paul Z Mar 16 '11 at 1:49
  • $\begingroup$ Really interesting remark about the differential operator. $\endgroup$ – Wok Mar 16 '11 at 8:07
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I've verified for n <= 8 that the matrices are all positive definite (and in particular nonsingular).

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  • $\begingroup$ Perhaps the PSD square root has a nice form? $\endgroup$ – Yuval Filmus Mar 16 '11 at 1:56
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This checks out for $n=2$. The determinant is then [thanks, Paul Z] $$4x_1^3x_2^3 - x_1^4 (3x_2-x_1)^2 = -x_1^3(x_1-4x_2)(x_1-x_2)^2.$$ If $0 < x_1 = 4x_2$ then clearly $x_2 < x_1$.

Note that the order or the $x_i$ doesn't matter (as long as they're all different), and the upper bound $1$ makes no difference since everything is homogeneous (if you multiply everything by some constant then regularity/singularity of your matrix remains the same).

The next step would be to repeat the explicit calculation for $n=3$, and try to either factor it or numerically find a counterexample.

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