When is there a vector $D$ with positive coordinates such that $e^{Ct}D$ has a negative coordinate? Let $C$ be a $2 \times 2$ asymmetric matrix with real entries. Assume that $C$ has strictly negative, real eigenvalues. Fix $D\in\mathbb{R}^2$, where $D > 0$ (i.e., both coordinates are strictly positive). Let $t \geq 0$ be a scalar and define the vector valued function using the matrix exponential,
$$
H(t) = e^{C t} D\in\mathbb{R}^2
$$
Denote $\min(H(t))<0$ if at least one of the elements is less than $0$.  

Fixing $D > 0$, under what conditions on $C$ and $D$ does a $t^{*}> 0$ exist such that $\min(H(t^*)) < 0$?  Or conversely, given a numerical $C$ and $D$, how can I test whether the $H(t) > 0$ for all $t$?

Ideal conditions are checking signs of eigenvectors or diagonal elements of $C$, etc.  If they can be done without a particular $D$ matrix, so much the better.
A few notes:


*

*If possible, you can use the assumption that $-C^{-1} D \cdot \begin{bmatrix} 1 &1\end{bmatrix} = 1$.  i.e. the sum of $-C^{-1}D = 1$  This comes out of necessary equilibrium condition to ensure that $H(t)$ is a valid PDF, and may pin down requirements to make the answer only requirements on $C$.

*$C$ negative definite should be enough to ensure that $\lim\limits_{t\to\infty}H(t) = 0$

*If you wish, assume that C is diagonalizable and denote $C \equiv Q S Q^{-1}$, where $S$ the matrix of eigenvalues (both $< 0$).  In that case, 
$$
H(t)= Q e^{S t} Q^{-1} D
$$


Solution Approach?: Clearly, having negative eigenvalues of $C$ ensures convergence of $H(t) \to 0$, but what do the eigenvectors of $C$ tell us? Can they tell us from which direction it approaches $0$ (given the sign of $D$)?
I notice that in examples which break, one of the eigenvectors has the same signs for both coordinates, while the other has different signs?  Could $Q \leq 0$ be the answer?
Example where it goes below 0 with negative definite C:
See the following matlab code:
C = [- 2.2959 -1.5; -.1 -1.6918];
min(eig(-C)) %Can check the eigenvalues to ensure negative definite
D = [2.0; 0.6918];
F_p = @(z) expm(C * z) * D;
%Evaluate at a few z's
F_p(1) %Both > 0
F_p(3) %One of them < 0!!!
F_p(100) %They both converge to 0
%Could diagonalize C
[Q,B] = eig(C)
% Then define F''(z)
F_pp = @(z) Q * diag(exp(diag(B) * z))* inv(Q) * C * D

Added from Idea Given in Solution:
The following may be a sketch of a proof that $D$ as the eigenvector of $C$ fulfills the requirement, Preliminaries: 


*

*$A$ and $A^{n}$ have the same eigenvectors and if $\lambda $ is a root
of $A,\ \lambda ^{n\text{ }}\ $is a root of $A^{n}.\ $And $-A$ has roots $
-\lambda $

*$If$ $A$ is non-negative (with some positive elements) it has a
non-negative eigenvector $v\geq 0$ associated to a dominant root $\hat{
\lambda}>0,$  which is real, simple and larger in modulus than its other
roots.

*$A$ and $-A\ $\ have the same eigenvectors.


Now consider $e^{C t}D=\left[ I+C t+\frac{1}{2!}\left( C t\right) ^{2}+\frac{1}{
3!}\left( C t\right) ^{3}+\frac{1}{4!}\left( C t\right) ^{4}+\frac{1}{5!}
\left( C t\right) ^{5}+\frac{1}{6!}\left( C t\right) ^{6}+..\right] D$ where $
C\leq 0.$ Take $v$ to be the dominant eigenvector of $-C.$ So if $\lambda $
is the dominant root of $-C,$ then the minimum eigenvalue of $-C$ is $
-\lambda .$ Take $D=v.$ Then
\begin{eqnarray*}
C t v &=&\left( -\lambda t\right) v \\
\left( C t\right) ^{2}v &=&\left( -\lambda t\right) ^{2}v \\
\left( C t\right) ^{3}v &=&\left( -\lambda t\right) ^{3}v
\end{eqnarray*}
\begin{eqnarray*}
e^{C t}v &=&\left[ I+C t+\frac{1}{2!}\left( C t\right) ^{2}+\frac{1}{3!}\left(
C t\right) ^{3}+\frac{1}{4!}\left( C t\right) ^{4}+\frac{1}{5!}\left(
C t\right) ^{5}+\frac{1}{6!}\left( C t\right) ^{6}+..\right] v \\
&=&\left[ 1+\left( -\lambda t\right) +\frac{1}{2!}\left( -\lambda t\right)
^{2}+\frac{1}{3!}\left( -\lambda t\right) ^{3}+\frac{1}{4!}\left( -\lambda
t\right) ^{4}+\frac{1}{5!}\left( -\lambda t\right) ^{5}+\frac{1}{6!}\left(
-\lambda t\right) ^{6}+..\right] v
\end{eqnarray*}
But by definition
$$
e^{-\lambda t}=\left[ 1+\left( -\lambda t\right) +\frac{1}{2!}\left(
-\lambda t\right) ^{2}+\frac{1}{3!}\left( -\lambda t\right) ^{3}+\frac{1}{4!}
\left( -\lambda t\right) ^{4}+\frac{1}{5!}\left( -\lambda t\right) ^{5}+
\frac{1}{6!}\left( -\lambda t\right) ^{6}+..\right] 
$$
So
$$
e^{C t}v=\left[ I+C t+\frac{1}{2!}\left( C t\right) ^{2}+\frac{1}{3!}\left(
C t\right) ^{3}+\frac{1}{4!}\left( C t\right) ^{4}+\frac{1}{5!}\left(
C t\right) ^{5}+\frac{1}{6!}\left( C t\right) ^{6}+..\right] v=e^{-\lambda
t}v\geq 0
$$
since $v$ is the non-negative eigenvector. Choose $D=H\left(
0\right) =v$ and $H\left( t\right) =e^{Ct}D\geq 0.$
 A: If $D>0$ then the necessary and sufficient condition for $e^{C t}D$ to have a negative coordinate is for $e^{Ct}$ to have a negative entry. This is because when the row where the negative entry is gets multiplied by $D$ we can make the corresponding coordinate of $D$ so large and the rest so small that the result is negative. For $e^{Ct}$ to have non-negative entries for all $t\geq0$ it is necessary and sufficient that $C$ be non-negative off of the diagonal as I explained in my previous answer. So if $C$ has a negative entry off of the diagonal $e^{Ct}$ will have a negative entry in it, and so will $e^{Ct}D$ for some $D>0$. In fact, $e^{Ct}$ starts having negative entries right away in this case, for any $t>0$, because $e^{Ct}\approx I+tC$ for small $t$, and the negative entry from $C$ is inherited by $e^{Ct}$. For a particular $D>0$ the critical time $t^*$ after which $e^{Ct}D$ first has a negative coordinate will depend on $D$ of course. 
You may be interested in the general theory of positive and non-negative matrices in this sense that goes back to Perron and Frobenius. In particular, the theorem of Perron-Frobenius implies that positive matrices have an eigenvector with positive coordinates corresponding to the eigenvalue with the largest absolute value. If $C$ is non-negative off diagonal then $e^{Ct}$ is non-negative and they share eigenvectors, so $C$ also has an eigenvector with positive coordinates corresponding to the eigenvalue with the largest real part. This explains why one of the eigenvectors has the same signs. If $C$ is also symmetric then the other eigenvector has to be orthogonal to it, hence have the opposite signs. 
We can now say something for $C$ negative off of the diagonal. Then $-C$ has the same eigenvectors as $C$, but is positive off of the diagonal, so by the Perron-Frobenius theorem both have an eigenvector $u$ with positive coordinates. It's exactly the eigenvector corresponding to the eigenvalue $\lambda$ of $C$ with the smallest real part. Since $e^{C t}u=e^{\lambda t}u$ when $D=u$ the entries of $e^{C t}D$ remain positive for all $t$. 
However, if $C$ is symmetric negative definite then this property is unstable. The other eigenvector $v$ is then orthogonal to $u$, and so has both positive and negative coordinates. Moreover, its eigenvalue $\mu$ is bigger, i.e. less negative than $\lambda$. So if $D=\alpha u+\beta v$ in the eigenbasis then $e^{C t}D=\alpha e^{\lambda t}u+\beta e^{\mu t}v$, and the second term dominates for large $t$. Regardless of the sign of $\beta$ the vector $\beta e^{\mu t}v$, and therefore $e^{C t}D$ for large $t$, will have a negative coordinate. In other words, as long as $D$ is not proportional to the positive eigenvector, $e^{C t}D$ will have a negative coordinate eventually, and the limit of $\min H(t)$ will be approached from the negative side.
On the other hand, if we only know that $C$ is diagonalizable with negative eigenvalues then the second eigenvector $v$ may also have positive coordinates, and then so will any positive linear combinations of $u$ and $v$. Therefore, $e^{C t}D=\alpha e^{\lambda t}u+\beta e^{\mu t}v$ remains positive for all $t\geq0$ if $\alpha,\beta>0$, and this property is stable, e.g. if say $D=u+v$ then all its small enough perturbations have the same property.
