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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
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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
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Yes, I see that frowning is a common expression here. :( –  Felix Goldberg Aug 10 '12 at 13:31
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Anyway, I edited the question. Is this form more acceptable? Thanks. –  Felix Goldberg Aug 10 '12 at 13:34
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The entry sum is $e Ae^t$, where $e=(1,1,\dots, 1)$. Does this help? –  Frieder Ladisch Aug 10 '12 at 13:44
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2 Answers

up vote 15 down vote accepted

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
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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
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