Is the matrix $A_{nm} = 1/(2-a_n-a_m)$ with $0 \le a_n < 1$ for $n = 1,2,....,N$ positive? Is the $N \times N$ matrix $A_{nm} = 1/(2-a_n-a_m)$ where $0 \le a_n < 1$ for
$n=1,2,....,N$ non negative?    For the $2 \times 2$ case the answer is yes.
The diagonal entries are positive and the determinant is non negative.
I have played around with the $3 \times 3$ case and have found no counter examples.  I think the result may be true in general.  If so this must be well known.  This came up in the theory of E-0-semigroups.
Thanks for listening.
 A: The answer is Yes for the following reason. Use the fact that
$$\frac1K=\int_0^1t^{K-1}dt.$$
Then
$$A=\int_0^1S(t)dt$$
where
$$S(t)={\rm Mat}(t^{1-a_n-a_m})=tV(t)\otimes V(t),\qquad V(t)=\begin{pmatrix} t^{-a_1} \\ \vdots \\ t^{-a_N} \end{pmatrix}.$$
Since $S(t)$ is symmetric, positive semi-definite, $A$ is so. Actually, the vectors $V(t)$ span ${\mathbb R}^N$ and therefore $A$ is positive definite.
A: Notation: I will use the symbols $A>0$ and $A\geq 0$ to denote that $A$ is positive definite / semidefinite respecrtively (Löwner order).
Let $D > 0$ be the diagonal matrix with $D_{nn} = 1-a_n$. Then, direct verification shows that
$$DA+AD = E,$$ where $E$ is the matrix of all ones. This is a Lyapunov equation for $A$, and it is known that its solution is positive semidefinite when $D>0$ and $E \geq 0$. This result can be proved also directly in this case: let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$ (real because $A$ is symmetric); multiply the equation on the left and on the right by $v^*$ and $v$ to get
$$
2\lambda(v^*Dv) = v^*Ev.
$$
Hence
$$
\lambda = (v^*Ev) / 2(v^*Dv) \geq 0.
$$
Thus the eigenvalues of $A$ are non-negative. $A>0$ is false, there are counterexamples (e.g., all the $a_i$ are zero).
