# 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 $s(A)=0$:

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

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I don't think this is a well stated question. –  Alexandre Eremenko Aug 10 '12 at 13:03
Well, they are not necessarily singular!! If you look at my example, you'll that it's determinant is 63. As for why I study them - I cam up against an expression of the form $j^{T}R^{-1}j$ in my work, where $R$ is some particular matrix related to signless Laplacians and I need to eliminate the case that this quantity is zero. I thought originally, as you did, that nonsigularity precludes zero sum, I can analyze my own very special $R$ but I'm curious if there is something general to be said here - and this is what mathematics is about in the final analysis, isn't it? –  Felix Goldberg Aug 10 '12 at 13:15
Yes, I see that frowning is a common expression here. :( –  Felix Goldberg Aug 10 '12 at 13:31
Anyway, I edited the question. Is this form more acceptable? Thanks. –  Felix Goldberg Aug 10 '12 at 13:34
The entry sum is $e Ae^t$, where $e=(1,1,\dots, 1)$. Does this help? –  Frieder Ladisch Aug 10 '12 at 13:44

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.

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Very nice! <<>> –  Felix Goldberg Aug 10 '12 at 16:12

If we are allowing coordinate transformations then a necessary condition is that $0\not\in FOV(A)$.

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FOV?${}{}{}{}{}$ –  Gerry Myerson Dec 9 '12 at 4:27