matrices whose entries sum to zero 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 $s(A)=0$:
$A=\begin{bmatrix}1 & 2 & 3  \\\\ 2 & -4 & -1\\\\  3 & -1 & -5 \end{bmatrix}$
 A: 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.
A: If we are allowing coordinate transformations then a necessary condition is that $0\not\in FOV(A)$.
