Matrix inequalities in series form

Question

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.

Answer 1

A counterexample is given by $$A=\begin{bmatrix} -1 & a \\ a & -1 \\ \end{bmatrix}$$ with (say) $a=2$.

Indeed, then the sum of your first series is $$c(t):=\frac12\begin{bmatrix} c_+(t) & c_-t) \\ c_-(t) & c_+(t) \\ \end{bmatrix}$$ with $$c_\pm(t):=\cosh t\pm\cos \left(\sqrt{3} t\right)\ge1-1=0$$ for real $t$, and the sum of your second series is $$s(t):=\frac12\begin{bmatrix} s_+(t) & s_-t) \\ s_-(t) & s_+(t) \\ \end{bmatrix}$$ with $$s_\pm(t):=\sinh t\pm\frac{\sin \left(\sqrt{3} t\right)}{\sqrt{3}}\ge0$$ for real $t\ge0$, because $s_\pm(0)=0$ and $s'_\pm=c_\pm\ge0$.