In the following theorem A,P,Q in R^{n times n}, and P and Q are symmetric. The notation P>0 means that the matrix P is positive definite. Given any Q>0, there exists a unique P>0 satisfying A^TP+PA+Q=0 if and only if the linear system dotx=Ax is globally asymptotically stable. The quadratic function V(z)=z^T Pz is a Lyapunov function that can be used to verify stability.

Step 1
The problem of finding eigenvalues for non-hermitian (and non-symmetric) matrices is very unstable; it is quite hard to solve numerically (close to impossible) and can't be solved in radicals for matrices of size at least 5.
Finding eigenvalues for hermitian (symmetric) matrices is a much easier problem - and can be reliably solved numerically.
For asymptotic stability, we don't need the exact values of eigenvalues, we only need that their real part are strictly negative. The existence of $P=P^{\ast <!--\; \ast \; -->}>0$ satisfying the Lyapunov equation is a sufficient criterion.
We have explicit formulas for solutions of the Lyapunov equations.
The Lyapunov equation is linear in P (if we are talking about real matrices), hence, we obtain a large system of linear equations, so we can solve it numerically.
We are not looking for eigenvalues of P: we can use any other criterion of positive definness: for example, via main minors.
The most important one: we have estimations on the norm ∥x(t)∥ is terms of the matrices P and Q. A mere study of eigenvalues of A (given that we can find them...) does not give such information (we could use the eigenvectors to obtain that info, but hte problem would become even harder).
Step 2
The last result is quite standard for Lyapunov stability. Let x(t) be a solution of the differential equation. Then we can write
$(x^{\ast <!--\; \ast \; -->}(t)Px(t){)}^{\prime }=(x^{\ast <!--\; \ast \; -->}(t){)}^{\prime }Px(t)+x^{\ast <!--\; \ast \; -->}(t)P{x}^{\prime }(t)=x^{\ast <!--\; \ast \; -->}(t)(A^{\ast <!--\; \ast \; -->}P+PA)x(t)=-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}(t)QX(t)\le <!--\; \le \; -->q\Vert <!--\; \Vert \; -->x(t){\Vert <!--\; \Vert \; -->}^2\le <!--\; \le \; -->-<!--\; -\; -->\frac{q}{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->}{x}^{\ast <!--\; \ast \; -->}(t)Px(t),$
where $q>0$ is the minimal eigenvalue of Q.
The Gronwall's lemma gives us the inequality
$x^{\ast <!--\; \ast \; -->}(t)Px(t)\le <!--\; \le \; -->\exp <!--\; \; -->(-<!--\; -\; -->\frac{qt}{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->})x^{\ast <!--\; \ast \; -->}(0)Px(0)\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->\exp <!--\; \; -->(-<!--\; -\; -->\frac{qt}{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->})\Vert <!--\; \Vert \; -->x(0){\Vert <!--\; \Vert \; -->}^2.$
One the other hand, if $p>0$ is the minimal eigenvalue of P, then
$p\Vert <!--\; \Vert \; -->x(t){\Vert <!--\; \Vert \; -->}^2\le <!--\; \le \; -->x^{\ast <!--\; \ast \; -->}(t)Px(t)\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->\exp <!--\; \; -->(-<!--\; -\; -->\frac{qt}{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->})\Vert <!--\; \Vert \; -->x(0){\Vert <!--\; \Vert \; -->}^2,$
hence
$\Vert <!--\; \Vert \; -->x(t){\Vert <!--\; \Vert \; -->}^2\le <!--\; \le \; -->\frac{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->}{p}\exp <!--\; \; -->(-<!--\; -\; -->\frac{qt}{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->})\Vert <!--\; \Vert \; -->x(0){\Vert <!--\; \Vert \; -->}^2.$
The term $\frac{\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->}{p}$ is nothing but the conditionment number of the matrix P:
$cond(P)=\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->\Vert <!--\; \Vert \; -->P^{-<!--\; -\; -->1}\Vert <!--\; \Vert \; -->=\Vert <!--\; \Vert \; -->P\Vert <!--\; \Vert \; -->/p$