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Consider the matrix $A$ whose coefficients are $A_{ij} = \frac{1}{1+x_i+x_j} $ where we have $ x_i \geq 0$ and $ x_j \geq 0$ for $ i,j=1,2,\dots,n$. How can I prove that this matrix is positive semidefinite for arbitrary $n$? Using first principal minors, I proved that this is positive semi-definite for $n=2$ but I could not generalize it.

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  • $\begingroup$ Have you tried induction? $\endgroup$ – Manos Feb 24 '13 at 19:35
  • $\begingroup$ I could not write the function which defines the relationship between $n$th and $n-1$th principal minor $\endgroup$ – neticin Feb 24 '13 at 19:38
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    $\begingroup$ I am not sure it will be easy to do this by calculating the minors. $\endgroup$ – Julien Feb 24 '13 at 19:48
  • $\begingroup$ The matrix being considered is a particular case of the Cauchy matrix. $\endgroup$ – J. M. is a poor mathematician Apr 4 '13 at 12:57
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Edit: I first foolishly proved that $A$ is positive definite in general. But that's obviously wrong, as $x_1=\ldots=x_n=0$ yields a rank $1$ matrix. So I removed the wrong argument.

Consider the matrix $$ A_t=\left(t^{x_i+x_j}\right)_{i,j} $$ and show this is a continuous path of symmetric positive semidefinite for $t\in[0,1]$.

To see this, introduce the matrix $$ B_t=\left(\frac{t^{x_j}}{\sqrt{n}}\right)_{i,j} $$ and observe that $$ A_t=B_t^*B_t. $$

Then integrate between $0$ and $1$, you find $$ \int_0^1A_tdt=A $$ where integration is performed componentwise.

Note that $A$ is obviously symmetric.

Now fix a vector $X$. Since $(X,A_tX)\geq 0$ for all $t\in[0,1]$, the integration of this continuous function yields $$\int_0^1(X,A_tX)dt=(X,AX)\geq 0$$ for all $X$.

So $A$ is positive semidefinite.

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  • $\begingroup$ $X$ is a vector and $A(t)A$ is a matrix. So how can you compute the dot product between a matrix and a vector $(X,A(t)A)$ ? $\endgroup$ – neticin Feb 24 '13 at 19:56
  • $\begingroup$ @neticin That was a typo, thanks. $\endgroup$ – Julien Feb 24 '13 at 19:57
  • $\begingroup$ @neticin Note that I found an error in my previous argument. It was not true that the $A_t$ are definite in general. See my edit. $\endgroup$ – Julien Feb 25 '13 at 1:30
  • $\begingroup$ @1015 $\int_0^1(X,A_tX)dt=(X,AX)$, why is this true? $\endgroup$ – Babai Aug 17 '16 at 19:45
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Let $X= \left( \begin{array}{c} a_1 \\ \vdots \\ a_n \end{array} \right)$. Then $$\begin{array}{ll} ^tXAX & = \sum\limits_{1 \leq i,j \leq n} \frac{a_ia_j}{1+x_i+x_j} = \sum\limits_{1 \leq i,j \leq n} a_ia_j \int_0^1 t^{x_i+x_j}dt \\ & = \sum\limits_{1 \leq i,j \leq n} a_ia_j \int_0^1 t^{x_i}dt \int_0^1 t^{x_j}dt = \left( \sum\limits_{i=1}^n a_i \int_0^1t^{x_i}dt \right)^2 \geq 0 \end{array}$$

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