Question about diagonal entries of inverse matrices? Assume $A$ is a symmetric positive semidefinite matrix with diagonal zero and all other entries are less than one. Also assume $D$ is a diagonal matrix with all entries in diagonal are positive and less than $1$. 
Can we say that the diagonal entries of $[D+A]^{-1}$ are all positive?
 A: Since $A$ is SPSD, $I+A$ is SPD since $x^T(I+A)x=x^Tx+x^TAx>0$ for all nonzero $x$. Inverse of an SPD matrix $B$ is SPD as well because
$$
x^TB^{-1}x=(B^{-1}x)^TB(B^{-1}x)=y^TBy>0
$$
for all nonzero $x$ (since $B$ is SPD and hence nonsingular, $y\neq 0$ iff $x\neq 0$ and for every $y\neq 0$ there's a nonzero $x$ such that $x=By$). Hence $(I+A)^{-1}$ is SPD too.
An SPD matrix $B$ always have positive diagonal entries since $0<e_i^TBe_i=b_{ii}$, where $e_i$ is the $i$th vector of the canonical basis and $b_{ii}$ is the $i$th diagonal entry of $B$.
P.S.: The assumption on the diagonal and off-diagonal entries of $A$ is not needed here. All what is required is that $A$ is SPSD. However, note that an SPSD matrix with zero diagonal entries is necessarily the zero matrix. To see this, note that the trace of $A$ (that is, the sum of the diagonal entries) is equal to the sum of the eigenvalues due to the invariance of the trace w.r.t. the change of basis. The eigenvalues of $A$ are non-negative because $A$ is SPSD and since they sum up to zero, they are necessarily all equal to zero.
