How to prove that $\dot{x}=ax,\space x(0)=1$ has a unique solution if $a,x$ are infinite dimensional matrices?

More specifically, let $Q$ be a bounded infinitesimal generator, i.e. $Q=(q_{i,j})_{i,j\in\mathbb{N}_0}$, its entries comprise a bounded set of real numbers, all of its entries not on the main diagonal are non-negative and each of its rows sums to $0$.

A family $(M_t)_{t\in[0,\infty)}$, $M_t=(m_{i,j}^{(t)})_{i,j\in\mathbb{N}_0}$ of real-valued, infinite-dimensional matrices shall be called a $Q$-system iff it is a Markov semigroup [i.e. $M_0=I$ (the identity matrix), for each $s\in[0,\infty)$, $M_s$ is stochastic and $\forall s,t\in[0,\infty),\space M_{s+t}=M_sM_t$)], it is entry-wise differentiable on $[0,\infty)$ (right differentiable at $0$) and $M_t'=QM_t$ for all $t\in[0,\infty)$.

How can it be shown that if $(P_t), (\tilde{P}_t)$ are $Q$-systems, $P_t=\tilde{P}_t$ for all $0\leq t$?

Here's what i've done so far in an effort to solve this problem.

  1. I have actually found a solution in a probability textbook, but could not make sense of the proof.

  2. I have investigated some properties of infinite dimensional matrices with an emphasis on infinite generators and Markov semigroups. While some useful results were obtained in the process, it hasn't brought me any closer to solving the present problem.

Edit I have found the following article that answers my question. Unfortunately, it is not very clear to me, since it relies on external results.


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