Logarithm of matrix with positive entries For matrices with positive entries (or more generally, irreducible matrices with non-negative entries), we have the Perron-Frobenius theorem, which tells us that there will be a unique eigenvector with positive entries and a positive eigenvalue, and that this will be the eigenvalue with the greatest magnitude.
One nice interpretation of this is in terms of a simple dynamical system, often described as a model of economic activity or of biological populations. Let $x_i$ be the amount of some "economic activity" at a given time. The model is
$$
\mathbf{x}(t+1) = \mathrm{A}\mathbf{x}(t). \tag{1}
$$
$\mathrm{A}$ has positive entries, meaning that in this model every economic activity has a positive effect on every other activity, or something like that. In this case the Perron-Frobenius theorem tells us that the the system will converge to a situation where the relative amounts of each activity are proportional to the values of the leading eigenvector, and the activity as a whole will be exponentially increasing (or decreasing) at a rate determined by the leading eigenvalue.
I'm interested in whether one can derive a continuous version of this model. I note that if we let
$$
\frac{d}{dt}\mathbf{x} = \mathrm{B}\mathbf{x},\tag{2}
$$
where $\mathrm{B} = \ln\mathrm{A}$ (the matrix logarithm of $\mathrm{A}$, assuming it exists) then for integer values of $t$, $(1)$ and $(2)$ will give the same values for $\mathbf{x}$.
However, the matrix $\mathrm{B}$ will not generally have positive entries. From doing a few examples numerically it seems $\mathrm{B}$'s elements are always real, but beyond that I'm not sure what effect $\mathrm{A}$ having positive entries will have on $\mathrm{B}$. Of course $B$ will have the property that the entries of $e^B$ are positive, but how can I interpret that intuitively in terms of the dynamical system $(2)$?
In summary, my question is, what can be said in general about a matrix $B$ such that $e^B$ has positive entries? (Or more generally, such that $e^B$ is irreducible with non-negative entries.)
 A: $e^{t B}\geq 0$ for all $t\geq 0$ is equivalent to $b_{ij}\geq0$ for $i\not=j$. In 
http://www.fa.uni-tuebingen.de/research/publications/one-parameter-semigroups-of-positive-semigroups/Nagel%20One-parameter%20Semigroups%20of%20Positive%20Operators.pdf
you can find a proof (pp. 123) in the context of Banach lattices. Your situation is mentioned in Example 1.4 (p. 124).
A: If $\lambda_i$ is an eigenvalue of $B$, then $e^{\lambda_i}$ is an eigenvalue of $A$ with the same eigenvector (this holds in general). Then, by using Perron-Frobenius theorem there exists a unique eigenvector of $B$ with positive entries (which is the same eigenvector of $A$) and corresponding real eigenvalue $\lambda$ such that $\text{Re} \lambda_i < \lambda$, provided that exponential map is bijective, i.e. logaritm of $A$ is unique.
Also, stability of the system is determined by this eigenvalue, since all eigenvalues of $A$ needs to be in the unit circle for stability, which is equivalent to all eigenvalues of $B$ being in the left half plane.
A: As pointed out in Uwe Stroinski's answer, $e^{tB}\ge0$ for all $t\ge0$ if and only if all off-diagonal elements of $B$ are nonnegative.
Suppose $e^{tB}\ge0$ for all $t\ge0$. Since $e^{tB}=I+tB+o(t)$, the off-diagonal part of $B$ must be nonnegative.
Conversely, suppose $B=D+F$, where $D$ and $F$ are respectively the diagonal and off-diagonal parts of $B$ with $F\ge0$. By Lie product theorem, $e^{tB}=e^{tD+tF}=\lim_{m\to\infty}(e^{tD/m}e^{tF/m})^m$. The result now follows because both $e^{tD/m}$ and $e^{tF/m}$ are nonnegative matrices when $t\ge0$.
