I am taking a course in differential equations and while I was doing some exercises I came across with the following statement: Let $A=(a_{ij})$ be a matrix such that $(a_{ij})>0$, $i\neq j$, then all entries of $e^A$ are positive.
I've been trying to find some pattern in the exponents of the matrix $A$ but without success. I also have been thinking of $$.^T:\mathcal{M}_{n\times n}\to \mathbb{R}_{n\times n}$$ as an operator and try to prove that $e^A=e^{\frac{A}{2}}e^{\frac{A}{2}}$ and then proving that $x^TAx>0$ but I'm not sure if this is what I really want. Does $x^TAx>0$ implies that all entries of the matrix $A$ are positive? Or is there a more "correct" way of proving the statement?
Thank you for your attention!
Presumably all diagonal entries of $A$ are real. As all off-diagonal entries of $A$ are positive, there exists a real number $a$ such that $B=A-aI$ is entrywise positive. Therefore $e^B$ is positive and in turn, $e^A=e^{aI+B}=e^ae^B$ is positive too.