Can matrix exponentials ever be negative? If so, under what conditions? Let $C$ be a $2 \times 2$ matrix with real entries, and $x\in\mathbb{R}^2$. We write  $x > 0$ if both coordinates are strictly positive.

Suppose $x>0$, under what conditions on $C$ and $x$ do we also have $e^Cx>0$? Note that the inequalities are strict. 

(In my example, $C$ is negative-definite, but don't use that if you don't need to.)
 A: The strict inequality is trickier but for $e^C x \geq0 $ to hold with any $x\geq0$ the matrix $e^C$ must have only non-negative entries. This is different from positive or negative definiteness, there is no relation in either direction. For $C$ it is sufficient to have non-negative entries off-diagonal. This is because $e^{tC}=I+tC+o(t)$, so for small positive $t$ the identity $I$ dominates on the diagonal, and the $C$ entries dominate off the diagonal. For $t=1$ we use the exponential property $e^{C}=(e^{\frac1nC})^n$ to see that positivity still holds. Indeed, $e^{\frac1nC}$ has non-negative entries for large $n$, and this is preserved under matrix multiplication. 
The non-negative off-diagonal is also necessary if we ask to have $e^{tC}x\geq0$ for all numbers $t\geq0$ (and all $x \geq0 $). If $C$ has negative off-diagonal entries then for small $t$ we have $e^{tC}\approx I+tC$ and off-diagonal entries of $A:=e^{tC}$ will also be negative. Let's say $a_{12}<0$. Multiplying it by $x=(\varepsilon, R)^T$ with small enough $\varepsilon>0$ and large enough $R>0$ we can ensure that the first coordinate of $Ax$ is negative.
To have $e^C x>0$ for all $x>0$ it is sufficient, by the same reasoning as above, that $e^C$ have strictly positive entries, and that $C$ correspondingly have strictly positive entries off-diagonal. But it can probably be somewhat relaxed.
