Linearization (Lyapunov's indirect method)

Since every eigenvalue of $A$ has negative real part ($A$ is Hurwitz), the matrix equation $A^{T}P+PA=-I$ has a unique symmetric positive-definite solution $P$, given explicitly by the convergent integral $P=\int_0^{\infty}e^{A^{T}t}e^{At}\,dt$.

Define the candidate Lyapunov function $V(\xi)=\xi^{T}P\xi\ge0$ for the deviation $\xi=x-x^{*}$. Writing the nonlinear system as $\dot\xi=A\xi+g(\xi)$ with $g(\xi)=o(\|\xi\|)$ as $\xi\to0$ (the Taylor remainder), compute the time-derivative of $V$ along nonlinear trajectories.

Using $\dot\xi=A\xi+g(\xi)$: $\dot V=\dot\xi^{T}P\xi+\xi^{T}P\dot\xi=\xi^{T}(A^{T}P+PA)\xi+2\xi^{T}Pg(\xi)=-\|\xi\|^{2}+2\xi^{T}Pg(\xi)$. Since $g(\xi)=o(\|\xi\|)$, there is a neighborhood of $x^{*}$ where $2\xi^{T}Pg(\xi)\le\frac{1}{2}\|\xi\|^{2}$, so $\dot V\le-\frac{1}{2}\|\xi\|^{2}<0$ for $\xi\ne0$ in that neighborhood.

Because $V$ is positive-definite and $\dot V$ is negative-definite near $x^{*}$, Lyapunov's direct method (proved next) concludes trajectories starting in that neighborhood converge to $x^{*}$: asymptotic stability. For the unstable case, if $A$ has an eigenvalue with positive real part, a symmetric construction (Chetaev's instability theorem) exhibits a direction along which $V$ increases, showing trajectories are pushed away from $x^{*}$ no matter how close they start.