Positivity of a determinant

Question

I'm stuck to prove the following exercise : Given real numbers $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$, show that $$ \det(e^{\large{x_iy_j}})_{i,j=1}^n>0 $$ provided that $x_1<\cdots<x_n$ and $y_1<\cdots<y_n$.

Any idea ?

Answer 1

In the following I give a partial answer: I prove the inequality for rational $y$ which only implies the weak inequality for real $y$.

First suppose that $y_j$ are integers. Set $\lambda_j=y_j-j$ which is a partition.

Set $a_i=e^{x_i}$. Your determinant becomes $\det(a_i^{y_j})=\det(a_i^{j-1}) s_{\lambda}(a_1,\dots,a_n)$, a product of the Vandermonde determinant which is positive by the monotonicity of your variables (and exp respects this, of course) and a Schur function of positive variables (since exp is always positive) which is also positive because the Schur function is a sum of monomials in its variables.

Now, suppose that the $y_j$ are rational. Let their common denominator be $D$ and change $x_i$ and $y_j$ to $x_i/D$ and $y_iD$ which brings us back to the the integer case.

Finally, for real numbers, you can take the limit which unfortunately just gives you $\det \ge 0$, but maybe you have another way of knowing that your matrix is not singular.

Answer 2

Some ideas:

1) What about induction on n? For $\,n=1\,$ the claim is trivial. For $\,n=2\,$ we have to prove that $\,\displaystyle{e^{x_1y_1+x_2y_2}>e^{x_1y_2+x_2y_1}\Longleftrightarrow (x_1-x_2)(y_1-y_2)>0}\,$, check...etc. (this looks really awful)

2) For any $\,X:=(v_1,v_2,...,v_n)\in\mathbb{R}^n\,$ , show that $\,X^TAX>0\,$ , with $\,A:=\left(e^{x_iy_j}\right)\,$ , making this matrix positive definite and thus its determinant is positive (this looks slightly better...)