Say we have a non linear system $\dot{x}=f(x)$ and we linearise this around an equilibrium point $x_0$, to obtain the linear sytem: $\dot{x}=Df(x_0)\cdot x$.
Where $Df(x_0)$ is the jacobian (and in this case a $2\times 2$ matrix).
We find the eigenvalues, and for example say we have $\lambda_{1,2}=2\pm i$, so the equilibrium point is unstable ($Re(\lambda_{1,2})=2\gt 0$).
When we wish to sketch the dynamics, I realise that the "sketch" will be a spiral spinning out of $x_0$, but what determines the direction of the spin?
i.e. anticlockwise or clockwise.
Thanks!
If the eigenvalues are conjugate nonreal complex numbers, the off-diagonal coefficients $b$ and $c$ of the Jacobian matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ have opposite signs. If $c\gt0$ (and $b\lt0$), the solutions close to $x_0$ are going anticlockwise. If $b\gt0$ (and $c\lt0$), the solutions close to $x_0$ are going clockwise.