# Determining if a system of equations has an all negative solution

Consider the equation $f(x) = cx$. It can be seen that there is some $x$ such that $f(x)<0$ iff $c\not=0$.

If we write this system of one equation as a matrix, it's just the matrix $M=(c)$. So in this case, there is a negative assignment to $f$ iff $det(M)\not=0$.

I read something which seemed to indicate that this is true more generally: If you have $n$ equations over $n$ variables, there will be an assignment such that all $f$s are $<0$ iff $det(M)\not=0$.

I'm having trouble proving this even in the 2x2 case, much less in general.

1. Is there a name for this theorem? (If it's true)
2. Any hints on how to [dis]prove it?

EDIT: The paper I'm looking at is Quantum probabilities as Bayesian probabilities. On page 2, equation (2), they say "The bookie can choose values xA , xB , and xC that lead to R < 0 in all three cases unless [$\det(M)=0$]". Maybe I have misinterpreted their argument, but it seems like they're saying this is a general property of the determinant.

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Do you mean all the components of the vector $Mx$ negative? – matgaio May 1 '12 at 0:37
@matgaio: yes, exactly – Xodarap May 1 '12 at 0:45
Just to argue geometrically: we know that a non-singular matrix take $S^{n-1}\subset\mathbb{R}^{n}$ to an $n-1$-dimensional ellipsoid. centered at the origin (possibly with axes non-aligned to the canonical axes). Once this ellipsoid intersect the orthant with all coordinates negative, then there exist some $x$ with $Mx<0$. – matgaio May 1 '12 at 1:01

I take it you have $n$ linear forms in $n$ variables, $$f_i=a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n,\qquad1\le i\le n$$ and you want a condition to ensure that there exist $x_1,\dots,x_n$ such that $f_i(x_1,\dots,x_n)\lt0$ for all $i$. Letting $M$ be the matrix whose entries are the $a_{ij}$, it is certainly true that if $\det M\ne0$ then such $x_i$ exist, for if $\det M\ne0$ then the column space of $M$ is all of ${\bf R}^n$, including that part of ${\bf R}^n$ where all components are negative. However, even if $\det M=0$, such $x_i$ might exist, e.g., consider $f_1(x,y)=f_2(x,y)=x$. I don't know if there's any simple condition on $M$ which causes exclusion of the negative orthant.

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This certainly seems to be in line with the edit of OP's question and the quote from the paper - it's an if statement, not an iff. – process91 May 1 '12 at 2:02
@process91: good point, it's if not iff. This makes it a lot simpler, thanks. – Xodarap May 1 '12 at 12:56