I want to obtain the eigenvalues of the following nonlinear system
${\displaystyle {\dot {x}}_{1}(t)=x_{2}(t)}$
$ {\displaystyle{\dot {x}}_{2}(t)=-{\frac {g}{\ell }}\sin {x_{1}}(t)-{\frac {k}{m\ell }}{x_{2}}(t)}$
I have tried to convert to matrix form in order to find $A-I\lambda$
$A= [0, 1$
$-g/l\sin(x_1), -k/ml]$
But this doesn't seem right due to he sin term still being present, any advice?
Normally what you do is to linearize the system about an equilibrium point, and then calculate eigenvalues and eigenvectors. Here, our vector field is $F: \Bbb{R}^2 \to \Bbb{R}^2$, defined by \begin{align} F(x_1,x_2) = \left(x_2, \, -\frac{g}{l} \sin(x_1) - \frac{k}{ml}x_2 \right). \end{align} An equilibrium point is one where $F$ vanishes. It is easy to see that $F(0,0) = (0,0)$, so the origin is an equilibrium point. Now, let's compute the Jacobian matrix of $F$ at the origin:
\begin{align} JF_{(0,0)} = \begin{pmatrix} 0 & 1 \\ -\frac{g}{l} & -\frac{k}{ml} \end{pmatrix} \end{align} So, you can consider the linearised system \begin{align} x'(t) = JF_{(0,0)} \cdot x(t) \end{align} where for convenince I denote $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$. Now, you can definitely compute the eigenvalues and eigenvectors of $JF_{(0,0)}$.
You've been given a non-linear equation $x'(t) = F(x(t))$, so it doesn't make much sense to talk about eigenvalues of such a system. What you have done is to artificially (and incorrectly) "force" $F(x(t))$ into a product of a certain $2 \times 2$ matrix and the column vector $x(t)$, which conceptually isn't very meaningful.
On the other hand, what I suggested is (usually) meaningful near the the equilibrium point, because you're approximating the non-linearity of $F$ by its derivative, which gives you a simple linear equation to analyze. And in general, knowing the behavior of the linear system tells you the behavior of non-linear system near the equilibrium point.