Finite no-jump propagator and survival identity

The matrix exponential supplier gives $V_t^{\mathrm{\prime }}=B{V}_t$, $V_t^{-1}=V_{-t}$ and $V_0=I$. For every positive matrix $X$ and every ancilla dimension, conjugation by $V_t$ tensored with the ancilla identity preserves positivity, so $S_t$ is completely positive. In particular $S_t(\sigma )$ is positive and nonzero, since $V_t$ is invertible; its trace is positive.

Differentiate the two finite matrix factors and use cyclicity of trace: $s_{\sigma }^{\mathrm{\prime }}(t)=\mathrm{Tr}((B+B^{†})S_t(\sigma ))=-\mathrm{Tr}(Q{S}_t(\sigma ))\le 0$. Thus $s_{\sigma }(t)\le s_{\sigma }(0)=1$. If $Q\le rI$, positivity gives $s_{\sigma }^{\mathrm{\prime }}(t)\ge -r{s}_{\sigma }(t)$, so the derivative of $e^{rt}{s}_{\sigma }(t)$ is nonnegative. Integrating this derivative gives $s_{\sigma }(t)\ge e^{-rt}$. This includes $r=0$ and the empty jump family.