It is a well-known result that for any positive-definite matrix $Q$, there exists a unique solution $P$ to the Lyapunov equation
$$ A^T P + PA = Q $$
if and only if all eigenvalues of $A$ have negative real parts.
A constructive proof suggests to choose
$$P:=\int \limits_{0}^{\infty} e^{A^Tt}Qe^{At} dt$$
But, in order for this to work, the integral needs to converge. In particular,
$$ \lim_{t \rightarrow \infty} e^{A t} = 0 $$
Can this result (in particular, convergence of the integral) be proven without resorting to a diagonal/Jordan normal form of $A$?
We know that $A$ is asymptotically stable, which is to say that there exists a $\mu \in \Bbb R$ with $\operatorname{Re}\lambda_i \leq \mu < 0$ for all $i$. We know that $A$ can be brought to normal form (since every matrix can be brought to Jordan form), which is to say that there exists an invertible $S$ such that $A = SJS^{-1}$, where $J$ is in Jordan form.
Let $\|\cdot\|$ denote the spectral norm (any matrix norm works, though) It can be shown that there exists a constant $C$ such that for all $t$, $\|e^{tJ}\| \leq C e^{\mu t}$. Moreover, we have $e^{tA} = Se^{tJ}S^{-1}$, so that $$ \left\|e^{tA} \right\| \leq \|S\|\cdot \left\|e^{tJ}\right\| \cdot \|S^{-1}\| = \kappa(S) \|e^{tJ}\| $$ where $\kappa(S)$ is the condition number. All together, we have $\|e^{tA}\| \leq C \kappa(S) e^{\mu t}$ (which is enough to state that $e^{tA} \to 0$). Let $D_1 = C\kappa(S)$.
Similarly, there is a $D_2$ such that $\|e^{tA^T}\| \leq D_2 e^{\mu t}$ From there, we have $$ \left\| \int_0^\infty e^{A^Tt}Qe^{At} dt \right\| \leq \int_0^\infty \left\| e^{A^Tt}Qe^{At} \right\|dt\\ \leq \int_0^\infty \left\| e^{A^Tt}\right\| \cdot \|Q\| \cdot \left\|e^{At} \right\|dt \\ \leq D_1D_2 \|Q\| \int_0^\infty e^{2 \mu t}\,dt $$ which converges.