Exponential convergence to equilibrium for finite continuous-time Markov chains Setting: Consider a continuous-time Markov chain on a finite state space $\mathcal Z$ with irreducible (constant) generator matrix $L\in\mathbb R^{|\mathcal Z| \times |\mathcal Z|}$. Let $\rho_t\in\mathcal P(\mathbb R^{|\mathcal Z|})$ be the probability law of $X_t$ (here subscript denotes dependence on time and $\mathcal P(\cdot)$ is the space of probability measures).
This time-$t$ distribution evolves according to
\begin{equation}
\partial_t\rho_t = L^T \rho_t,
\end{equation}
with some initial distribution. Since $L$ is irreducible there exists a stationary measure $\mu\in \mathcal P(\mathbb R^{|\mathcal Z|})$, i.e. $L^T\mu = 0$.
Question: Are there some results in this setting which state that the time-$t$ distribution $\rho_t$ converges exponentially fast to $\mu$, i.e.
\begin{equation}
\|\rho_t - \mu\|_{TV} \leq Ce^{-D t},
\end{equation}
for some positive constants $C,D$. Here $\|\cdot\|_{TV}$ is the total-variation norm.
Remarks: When I search online, typical results require that the Markov chain is reversible and then they provide estimates on convergence, however I do not want to make any such assumption. Also, there is often discussion about spectral gap, but I don't yet understand what I need to assume for $L$ so as to get a spectral gap.
 A: The solution $P(t)=P(0)e^{Lt}$ is valid for finite state space
(see for example, here). As explained for example here, $e^{Lt}$ is stochastic for each $t$. If $L$ is irreducible, then $e^{L}$ is primitive (actually positive, but I'm blanking out how to show this; in any case, the discrete-time Markov chain corresponding to $e^L$ is irreducible, and can not have any periodicity other than 1, so is aperiodic, hence primitive, which is enough).  Then by Perron-Frobenius there is a spectral gap - i.e. the maximal eigenvalue is 1, corresponding generalized eigenspace is simple and contains unique eigenvector $\mu$  with positive entries that sum to 1, and all other eigenvalues have norm strictly $<1$. Since there are finitely many of them they all have norm at most some fixed $\lambda<1$. Then it follows that $L$ has the same generalized eigenspaces, one of eigenvalue $0$ (spanned by $\mu$) and all others with real part of eigenvalues $<\ln \lambda<0$. The $e^{tL}$ again has same generalized eigenspaces, one of eigenvalue $1$ spanned by $\mu$ and all other with eigenvalues of norm $<\lambda^t$. Writing a vector $\rho$ with entries summing to 1 using this generalized eigenspace decomposition we have
$$\rho=\mu+\sum c_i \rho_i$$
and then
$$e^{Lt}\rho=\mu+\sum c_i e^{Lt}\rho_i$$
But on each generalized eigenspace $e^{Lt}$ is the product of the scalar  $\lambda_i^t$ and a matrix polynomial in $t$ (as in here), so each entry of $e^{Lt}\rho_i$ is of the form $\lambda_i^t P(t)$ for some fixed polynomial $P$, and in particular they are all bounded by $C\lambda^t$ for some $C$ (since $(\lambda_i/\lambda)^t<|C/P(t)|$).   Since this is true for all pieces $\rho_i$, we get that each entry of $|e^{Lt}\rho-\mu|$ is bounded by $C\lambda^t$. Taking $D=-\ln \lambda$ this bound becomes $Ce^{-Dt}$ (with $C$ but not $D$ dependent on $\rho$).
In this finite-dimensional setting, it follows from this that $|e^{Lt}\rho-\mu|_{TV}<\hat{C}e^{-Dt}$ for some $\hat{C}$.

Why is the coefficient of $\mu$ equal to $1$?
Lemma: If $x$ is left generalized eigenvector of $A$ with generalized eigenvalue $a$ and $y$ right eigenvector of $A$ with eigenvalue $b$ and $a\neq b$ then $x \cdot y=0$.
Proof: $0=y^T \vec{0}= y^T(A-aId)^nx= (b-a)^n(y^Tx)$, so $y^Tx=0$.
Now because all other left generalized eigenspaces of $P(t)$ are orthogonal to the right eigenvector $\mathbb{1}=(1,...,1)$, so the sum of entries of $\rho$, aka $\rho\cdot\mathbb{1}$, is the coefficient of the $\mu$:
$$1=\rho\cdot\mathbb{1}=(c\mu +\sum c_i \rho_i)\cdot \mathbb{1}=c (\mu\cdot \mathbb{1})+0=c$$
