For example, given Generating Matrix G for a 3 state Markov Process X(t), where
$G = \begin{bmatrix} q_{11}&q_{12}&q_{13}\\ q_{21}&q_{22}&q_{23}\\ q_{31}&q_{32}&q_{33}\\ \end{bmatrix}$
what can one say about the time the process stays in a particular state or the probability of the next state?
If $X(t)$ is the Markov Chain with transition function $p_t(i,j)$ associated with $G$ (and $X(t)$ has suitable properties), for $G$ holds: $$g_{ij} = \frac{d}{dt} p_t(i,j)\Big|_{t=0}$$ So let us interpret this relation: We know $p_0(i,i)=1$ and for $s>0$ we have $p_s(i,i) \leq 1$. The last transition probability is the probability to go from $i$ to $i$ in a step with length $s$. Due to that, for a small step we could say that this is the probabilty to not leave the state $i$. We see that $$g_{ii} = \frac{d}{ds} p_s(i,i)\Big|_{s=0} \leq 0$$ (Of course we have $g_{ii} = 0$, if $p_t(i,i) = 1$ for all $t$.) So the diagonal elements $g_{ii}$ of $G$ are giving us informations about how likely $X(t)$ wants to leave the state $i$. Indeed (in the right setting), if we define $\tau := \inf \{t>0 : X(t) \neq X(0)\}$ as first time $X(t)$ is leaving the initial state we get:
$$\Bbb{P}(\tau > t | X(0) = i) = \exp(g_{ii}t)$$ With this we see with a bigger value on $|g_{ii}|$ the chain $X(t)$ leaves $i$ more likely to a earlier time.
On the other hand, for $i\neq j$ it holds $p_0(i,j)=0$ and $p_t(i,j) \geq 0$. With analogue thoughts we derive an analogue interpretation for $$g_{ij} = \frac{d}{dt} p_t(i,j)\Big|_{t=0} \geq 0$$