For equation $A\cdot x=0$, what conditions on matrix $A$ will ensure that $x$ is not a positive or negative vector? Let $A$ be a $m\times n$ matrix and $x$ be a $n\times 1$ vector. Consider equation $A\times x=0$.  
Under what conditions, $x$ is not a positive or negative vector? That is, we cannot have all entries of $x$ are positive or all are negative.
I guess it's about the eigenvalues of $A^TA$. Perhaps, they can't be all positive or all negative? This is just my guess and I have been alway from linear algebra for a long time. Please correct me if I am wrong. Thank you.
 A: The absence of a positive (and negaitve) solutions to $Ax=0$ can be approached via convex duality where it is equivalent to the existing of a solution to the dual system of (in this case linear) constraints. Namely, one can use Farkas lemma.
We want the following system
$$
Ax=0,x>0
$$
to have no solution. Introducing the notation $e$ for the vector of all ones
$$
e=\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}
$$
the system may equivalently be rewritten as the second alternative in the link above
$$
\begin{cases}
Ax\ge 0,\\
-Ax\ge 0,\\
x\ge t\cdot e,\\
t>0
\end{cases}
\qquad\Leftrightarrow\qquad
\begin{cases}
\begin{bmatrix}
A & 0\\-A & 0\\I & -e
\end{bmatrix}
\begin{bmatrix}x\\t\end{bmatrix}\ge 0,\\
\begin{bmatrix}0 & -1\end{bmatrix}\begin{bmatrix}x\\t\end{bmatrix}<0.
\end{cases}
$$
Here zero blocks are columns/rows of corresponding dimensions to fit the block matrix.
Farkas lemma gives then the necessary and sufficient condition as that there exists a solution to the dual (first) alternative in the link
$$
\begin{cases}
\begin{bmatrix}
A^T & -A^T & I\\0 & 0 & -e^T
\end{bmatrix}
\begin{bmatrix}u\\v\\w\end{bmatrix}=
\begin{bmatrix}0\\-1\end{bmatrix},\\u,v,w\ge 0
\end{cases}\qquad\Leftrightarrow\qquad
\begin{cases}
A^T(v-u)=w,\\
e^Tw=1,\\
u,v,w\ge 0.
\end{cases}
$$
Introducing the notation $d=v-u$ we get rid of the positivity condition (the standard procedure in the linear programming), and the system becomes
$$
\begin{cases}
A^Td=w\ge 0,\\
e^Tw=1.
\end{cases}
$$
The last condition is the normalization which is equivalent to saying that $w\ne 0$ (because then we can always scale the solution by $e^Tw$). Therefore we arrive to the equivalent condition that 

there must exist a solution to the following system
  $$
A^Td\ge 0,\quad A^Td\ne 0.
$$

Interpretaion: it is the same as $d^TA\ge 0,\ne 0$, that is, there exists a linear combination of the rows of $A$ that is nonnegative and nonzero. It is easy to see that it is necessary (see the comment from Петя Нарышкин): $Ax=0$ $\Rightarrow$ $\underbrace{d^TA}_{\ge 0,\ne 0}x=0$ $\Rightarrow$ $x\not> 0$.
A test for that can be implemented e.g. as LP
$$
\gamma=\max_{d,w} e^Tw\quad\text{ subject to }A^Td=w,\ w\ge 0,\ -1\le d_i\le 1.
$$
If $\gamma>0$ then it is ok.
