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Let $x_1, \dots, x_n > 0$ be positive real numbers.

From numerical experiments, it appears that the $n \times n$ matrix

$$A_{ij} = \frac{1}{x_i + x_j} $$

is always positive semidefinite.

Is this known or obvious??

In the $2 \times 2$ case, the minimum eigenvalue of $A$ is given by

$$\lambda_{min}=\frac{(x_1+x_2)^2 - \sqrt{(x_1+x_2)^4-4 x_1 x_2 (x_1 - x_2)^2}}{4 x_1 x_2 (x_1+x_2)}>0,$$

but I don't see how to prove positivity of $A$ for arbitrary $n$.

Jon Tyson
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    Note that if you had for $A_{ij}=1/(x_i-y_j)$ you'd have what's known as a Cauchy matrix. The sequence $x_i$ and $y_j$ need to be made up of distinct elements and obey $x_i \neq y_j$ of course. This matrix has a lot of nice properties which may help you. – kodlu Aug 30 '15 at 04:18
  • Thanks!! I think that helps. – Jon Tyson Aug 30 '15 at 04:55
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    See Achilles Hui's answer here http://math.stackexchange.com/questions/577197/how-prove-this-matrix-det-a-left-frac1-lna-ia-j-right-n-tim – Arin Chaudhuri Aug 30 '15 at 04:55
  • Thanks! Setting s=1 in Achilles Hui's answer solves it for x_i >1, but by homogeneity at is all that is needed. – Jon Tyson Aug 30 '15 at 05:12

1 Answers1

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The answer was provided by ACARCHAU's comment, above, since this is explicitly proven in Achilles Hui's answer by setting s=1 over there.

Equivalently, this is just a Grahamian matrix, since

$$\frac{1}{\lambda_i+\lambda_j}= (\exp(-\lambda_i t),\exp(-\lambda_j t))_{\ L^\infty(0,\infty)}$$

Community
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Jon Tyson
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  • Any idea what happened to the log in the Achilles answer? The question and answer do not seem to match, despite the many upvotes for both. – Michael Aug 30 '15 at 12:22
  • The answer here depends only on the very beginning of Achilles's proof, with no need to get into logs. – Jon Tyson Aug 31 '15 at 00:53
  • I know. My point is that the Achilles answer "accidentally" answers your question, but seems to have nothing to do with its own question, which is strange since it got a lot of attention. It is almost as if all those upvoters knew something else, like there was some "obvious" way of making his answer relevant to logs. – Michael Aug 31 '15 at 04:06