The formula that describes the swing of a pendulum is $\ddot{\theta}(t)+\gamma \dot{\theta}(t)+\omega ^2 \sin (\theta (t))=0$ where $\gamma = \frac{c}{mL}$ and $\omega = \frac{g}{L}$.
We want to represent this formula as a system of linear equations, find the stationary points, and find if the system is stable, asymptotically stable, or unstable at those points.
Firstly, let's assume that $\theta \approx 0$, so we can say that $\sin (\theta (t)) \approx \theta (t)$ and now our equation is $\ddot{\theta}(t)+\gamma \dot{\theta}(t)+\omega ^2 \theta (t)=0$.
I defined $x_1=\theta$, $x_2 = \dot{x_1}=\dot{\theta}$.
Then we have the following equations: $\dot{x_1}=x_2$, and $\dot{x_2} + \gamma x_2 +\omega ^2 x_1 =0$, or in other words, $\dot{x_2} = -\omega ^2 x_1 -\gamma x_2$.
So the linear system is:
$\begin{pmatrix} \dot{x_1} \\ \dot{x_2}\end{pmatrix}=\begin{pmatrix}0 & 1 \\ -\omega ^2 & -\gamma \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$
At the stationary points we have $\dot{x_1} = \dot{x_2}=0$, but $\dot{x_1}=x_2$, so we can infer that in any stationary point, we will have $x_2=0$.
Furthermore, we need $\dot{x_2}=-\omega^2 x_1-\gamma x_2 = -\omega ^2 x_1 =0$. Since $\omega$ is known and is not zero, we must have $x_1=0$. so the only stationary point of this system is $(x_1,x_2)=(0,0)$.
Is this correct? And how can we check the stability? I know it has something to do with the eigenvalues, but the eigenvalues of the system are independent of $x_1,x_2$...
The first assumption you applied assumes that $\theta\approx 0$, which allows you to linearize the differential equation and turns it into a linear autonomous and homogeneous differential equation, however such differential equation will always have an equilibrium solution at zero.
The correct way to solve this is to first find the equilibrium solution of the non-linear differential equation and then use linearisation in order to find the stability of the system at those equilibrium solutions.