Let $A$ be a non-singular matrix and let $s(A)$ be the sum of its entries. Under which conditions can it be assured that $s(A) \neq 0$? 


if you like, you can assume that $A$ is symmetric.


Here is an example with $s(A)=0$:


$A=\begin{bmatrix}1 & 2 & 3 \\\\ 2 & -4 & -1\\\\ 3 & -1 & -5 \end{bmatrix}$



matrices whose entries sum to zero
It seems that, as quid suggested, very little can be said, at least if we want to say something invariant under rotations of coordinates. Specifically, the following are equivalent for a symmetric real matrix $M$: 


(1) There is an orthogonal matrix $T$ such that $T^{-1}MT$ has entries summing to 0.


(2) The eigenvalues of $M$ do not all have the same sign.


To see this, begin with F. Ladisch's comment that the sum of the entries of $M$ is $eMe^t$, where $e$ is the all-ones vector. It follows that (1) is equivalent to the existence of some non-zero vector $v$ with $vMv^t=0$, as we can use $T$ to rotate a scalar multiple of $e$ to $v$. Clearly no such $v$ can exist if the quadratic form defined by $M$ is strictly positive definite or strictly negative definite, i.e., if all the eigenvalues have the same sign. Conversely, if there is an eigenvector $x$ with positive eigenvalue $\lambda$ and there is another eigenvector $y$ with negative eigenvalue $-\mu$, and if we normalize $x$ and $y$ to be unit vectors, then, since $x$ and $y$ are orthogonal, $v=\sqrt\mu x+\sqrt\lambda y$ does the job.


