Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
I assume that the index of $y$ is $j$. –  Phira May 26 '12 at 18:26
@Phira : You're right. Let me edit that. –  Student May 26 '12 at 18:27
If one of the sets are integers, this follows from the Vandermonde determinant and the definition of a Schur function. –  Phira May 26 '12 at 18:30
Positivity of this matrix is equivalent to positivity of another matrix $(e^(x_i-x_j)^2)$ (expand the square then one gets $x_iy_j$). But the latter is well known to be positive, for example in the area of gaussian processes (it is the covariance matrix). There is actually a very neat lower bound for the determinant of the matrix you give, proved by Drury-Marshall (1987). –  Syang Chen May 30 '12 at 16:18
This has been proved in Theorem 6.5.2, pp.297-299 of Kung et al., Combinatorics: The Rota Way, Cambridge University Press. –  user1551 May 9 '13 at 7:26

3 Answers 3

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.

share|improve this answer

It suffices to prove that given $x_1<\ldots <x_n$ and $y_1<\ldots <y_n$ the determinant $D$ of the matrix $M=\left(e^{x_iy_j}\right)_{1\leq i,j\leq n}$ is nonzero.

Indeed, $D$ is a continuous real function on the open connected subset $$ U=\{(\bar{x},\bar{y}):x_1<\ldots <x_n\ \wedge\ y_1<\ldots <y_n\} \subset \mathbb{R}^n\times\mathbb{R}^n $$ , hence, if it never vanishes, it must be always positive or always negative. Then you check positivity for a particular choice of the variables (for example, for $x_j=j-1$ you get a Vandermonde determinant and positivity is an easy check).

To prove the nonvanishing of the determinant, we use induction on $n$, imagine to fix the values of $x_1,\ldots, x_{n-1}$ and of $y_1,\ldots ,y_n$ and let vary $x_n=z$ over $\mathbb{R}$. Consider the determinant $D(z)$ of $M$ as a function of $z$. Clearly, we have $D(x_i)=0$ for $i=1,\ldots n-1$ and we expect (if the thesis is true) that $D(z)$ does not vanishes elsewhere.

Using Laplace expansion for the determinant we see that $D(z)$ has the following form $$ D(z)= A_1 (e^{y_1})^z+\ldots A_n (e^{y_n})^z $$ with coefficients $A_j\neq 0$ (by induction on $n$).

Now to conclude it is natural to recall a well-known fact about exponential polynomials.

Lemma 1 Let $\rho_1\ldots,\rho_n$ positive distinct real numbers and $c_1,\ldots, c_n\in\mathbb{R}$ non vanishing real numbers. Then the equation $\displaystyle\sum_{i=1}^n c_i\rho_i^z=0$ has at most $n-1$ solutions $z\in\mathbb{R}$

Lemma 1 it is a direct corollary of the more general.

Lemma 2 Let $\rho_1\ldots,\rho_n$ positive distinct real numbers and $c_1,\ldots, c_n\in\mathbb{R}[x]$ non trivial polynomials with real coefficients. Then the equation $\displaystyle\sum_{i=1}^n c_i(z)\rho_i^z=0$ has at most $\displaystyle\sum_{i=1}^n (deg(c_i)+1)-1$.

Lemma 2 is easily proven using only induction and Rolle's theorem.

share|improve this answer

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...)

share|improve this answer
Thanx to Davide, I already edited my answer and changed the notation. @Anon, I don't quite understand what you mean by X independent to A...? One is a vector, the other one is a matrix. –  DonAntonio May 26 '12 at 19:02
@Davide, I'm not sure. I think we get something like $$X^TAX=\sum_{k=1}^n v_k\sum_{i=1}^n v_ie^{x_ky_i}$$ –  DonAntonio May 26 '12 at 19:08
Re: your comment to me, the matrix $(\exp x_iy_j)_{ij}$ is a function of the vector $(x_i)$; they are not independent of each other and so saying that $X^TA(X)X>0$ would not necessarily mean $A(X)$ was pos-def for a specific $X$. Of course you wanted $X$ independent of $A$, and so changed this in light of Davide's comment. –  anon May 26 '12 at 23:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.