Matrix with $i,j$ entry $a_i/(a_i + a_j)$ Suppose $a_1 > a_2 > \cdots > a_n > 0$, consider an $n\times n$ matrix $A$ with $i,j$ entry $\frac{a_i}{a_i + a_j}$. I am wondering if the matrix has a name/ any insight about the eigenvalue/eigenvectors? All I can observr is $A - 1/2$ is a skew-symmetric matrix, but nothing else.
Thanks a lot!

$A$ is the product of two positive definite matrices $\operatorname{diag}(\mathbf a)$ and $\left(\frac{1}{a_i+a_j}\right)_{i,j\in\{1,2,\ldots,n\}}=\int_0^\infty e^{-x\mathbf a}e^{-x\mathbf a^\top}dx$, where $\mathbf a=(a_1,\ldots,a_n)^\top$ and $e^{\mathbf v}$ denotes the entrywise exponential of a row/column vector $\mathbf v$. Therefore $A$ has a positive spectrum. 
It follows that each eigenvalues $\lambda$ of $A$ has a corresponding real unit eigenvector $\mathbf v$. However, as $A+A^\top=\mathbf e\mathbf e^\top$ (where $\mathbf e=(1,1,\ldots,1)^\top$),
$$
2\lambda
=\mathbf v^\top(A+A^\top)\mathbf v
=\mathbf v^\top\mathbf e\mathbf e^\top\mathbf v
=(\mathbf v^\top\mathbf e)^2
\le\|\mathbf v\|_2^2\|\mathbf e\|_2^2
=n.
$$
Therefore $\rho(A)\le\frac n2$. Since $A$ is entrywise positive, by Perron-Frobenius theorem, $\rho(A)$ is a simple eigenvalue of $A$ and it has a corresponding eigenvector that is entrywise positive.
