While studying the positivity of mechanical systems, we land on the following conjecture but don't know how to prove this.

If a square matrix $A \in \mathbb{R}^{n,n}$ satisfies both the two following identities \begin{align} & \sum_{i=0}^{\infty} A^n \dfrac{t^{2n}}{(2n)!} \geq 0 \ \mbox{ for all } \ t\geq 0 \label{1} \\ & \sum_{i=0}^{\infty} A^n \dfrac{t^{2n+1}}{(2n+1)!} \geq 0 \mbox{ for all } t\geq 0, \end{align} then $A \geq 0$. Here $\geq 0$ means that all the elements of a matrix are non-negative, but not in the sense of positive definiteness.

It is quite clear, that by letting $t$ sufficiently small, from the 1st inequality, we see that matrix A is Metzler, i.e., $a_{ij} \geq 0$ for all $i \not= j$. However, I don't know how to handle diagonal elements $a_{ii}$.

Any suggestions or comments are very welcome. Thanks a lot.

While studying the positivity of mechanical systems, we land on the following conjecture but don't know how to prove this.

If a square matrix $A \in \mathbb{R}^{m\times m}$ satisfies both the two following inequalities: \begin{align} & \sum_{n=0}^{\infty} A^n \dfrac{t^{2n}}{(2n)!} \geq 0 \ \mbox{ for all } \ t\geq 0; \label{1} \\ & \sum_{n=0}^{\infty} A^n \dfrac{t^{2n+1}}{(2n+1)!} \geq 0 \mbox{ for all } t\geq 0, \end{align} then $A \geq 0$. Here $\geq 0$ means that all the elements of a matrix are non-negative, but not in the sense of positive definiteness.

It is quite clear, that by letting $t$ sufficiently small, from the 1st inequality, we see that matrix A is Metzler, i.e., $a_{ij} \geq 0$ for all $i \not= j$. However, I don't know how to handle diagonal elements $a_{ii}$.

Any suggestions or comments are very welcome. Thanks a lot.