Let $0<a_1<\ldots<a_n$ be real numbers, and let $\alpha>0$ be given. Consider the matrix $A=\begin{pmatrix}\frac1{(a_i+a_j)^\alpha}\end{pmatrix}_{1\leq i,j\leq n}$. Then is $A$ positive-definite?
I have come across this question, which considers the matrix with entries $\frac1{(a_i+a_j)}$ without $\alpha$.
I can show that, when $a_i\ne a_j,\,\forall i\ne j$, then the matrix (without $\alpha$) is positive-definite.
But I am unable to modify the answer in order to show the result for $A$.
I have tried to compare the matrix with $\alpha$ and the matrix without by the following equality:
$$
x^\top Mx=\sum_i\sum_j\frac{x_ix_j}{(a_i+a_j)^\alpha}=\sum_i\sum_j\left(\frac{(x_ix_j)^{\frac1\alpha}}{a_i+a_j}\right)^\alpha.
$$
But this does not make sense: $x_ix_j$ could be positive or negative and taking the exponent $\frac1\alpha$ results in a complex number, and then one cannot apply Jensen's inequality. Also the exponent $\alpha$ might be $\geq1$ or $<1$.
It seems the hypothesis that $0<a_1<\ldots<a_n$ should be used in some way. And I don't know how.
I have run out of ideas. Any help is sincerely appreciated. Thanks in advance.
Best Answer
We can directly adapt the proof from the original question: let $C_\alpha = \int_0^\infty \exp(-t^{1/\alpha}) \,dt$, so for any $b > 0$, we have $$\int_0^\infty e^{-bt^{1/\alpha}} \,dt = \int_0^\infty e^{-(b^\alpha t)^{1/\alpha}} \,dt = \frac{C_\alpha}{b^\alpha}$$ hence for $x \neq 0$ \begin{align*} x^TAx &= \sum_{i, j} x_i x_j \frac{1}{(a_i + a_j)^\alpha} \\ &= \frac{1}{C_\alpha} \sum_{i, j} x_i x_j \int_0^\infty e^{-(a_i + a_j)t^{1/\alpha}} \,dt \\ &= \frac{1}{C_\alpha} \int_0^\infty \left( \sum_i x_i e^{-a_i t^{1/\alpha}} \right)^2 \,dt \end{align*} which is positive, since all $a_i$ are distinct (meaning the integrand is somewhere nonzero).