Properties of matrix exponential and Lyapunov equation 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 $A$?
 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.
A: A bit aside, but regarding the if and only if part it is not true. The Sylvester equation (see e.g. wiki) $AX + XB =C$  has a unique solution iff $A$ and $-B$ has no eigenvalues in common. In your case this reduces to the condition that for any two eigenvalues $\lambda_i,\lambda_j$ of $A$ we must require $\lambda_i+\lambda_j\neq 0$. If you have access to numerically inverting matrices it is in fact very simple to solve the Sylvester (or Lyapunov) equation. Just construct $X \mapsto f(X) = AX + XB$ as a linear map on an $n^2$ dimensional space and invert it. 
