If A is a positive matrix . Then how do we show that $e^{At}\geq 0$.

Question

If A is an $n \times n$ matrix consisting of $a_{ij}$ with $a_{ij}\geq 0 , \hspace{2mm} \forall \hspace{1mm} 1\leq i,j\leq n$, $i\neq j$. Then how do we show that for small enough $t\geq0$ we would have $e^{At}\geq 0$?

I tried to argue that since $$ e^{At}=I + At+ \cdots +\frac{A^nt^n}{n!}+\cdots$$ and we are given that $a_{ij}\geq 0 , \hspace{2mm} \forall \hspace{1mm} 1\leq i,j\leq n$, $i\neq j$ hence $e^{At}$ would be non negative.

Is this argument enough AND can there be an alternative argument to prove the above. Any help would be greatly greatly appreciated.

Answer 1

This argument is enough since you can show recursively that the coefficients of $A^n$ are positive.