Stability of dynamical system with purely imaginary eigenvalues I have a 4 dimensional dynamical system and I want to know the stability of one of its fixed points, the problem is that all the eigenvalues of the Jacobian there are purely imaginary:
$$
0 - 5.64799 i, 0 + 5.64799 i, 0 - 5.71808 i, 0 + 5.71808 i
$$
The eigenvectors of the Jacobian at the fix point are linearly independent and are the following:
$$
\left(-5 \sqrt{\frac{5}{192 \sqrt{5}-125}},1,0,0\right) \\
\left(5 \sqrt{\frac{5}{192 \sqrt{5}-125}},1,0,0\right) \\
\left(0,0,-5 i \sqrt{\frac{5}{125+64 \sqrt{5}}},1\right) \\
\left(0,0,5 i \sqrt{\frac{5}{125+64 \sqrt{5}}},1\right)
$$
What is the stability of this fix point and how can one figure it out?
Extra info:
My system is a continuous system of ODEs describing a physical system.
(I found a similar question here but I did not fully understand how it would apply to my case since he is talking about an eigenvalue with multiplicity 2.)
In $\mathbb{R}^2$ the modulus of the eigenvalues of the Jacobian gives you the criteria for stability:
\begin{gather}
0 &lt; |\lambda_1| \leq |\lambda_2|&lt;1 \; \text{ stable} \\
1 &lt; |\lambda_1|\leq  |\lambda_2| \; \text{ unstable} \\
0 &lt; |\lambda_1| &lt; 1 &lt; |\lambda_2| \; \text{ saddle point} 
\end{gather}
But since you're working in $\mathbb{R}^4$, I don't think it should be too different. Whichever eigenvalues have modulus less than one will correspond to a local stable manifold of the fixed point (given by the corresponding eigenvector), whereas the ones with modulus more than one are unstable.
So in summary I guess it would be sort of weird to just call the fixed point &quot;stable&quot; or &quot;unstable&quot; since there are many more possibilities, but if all the eigenvalues' norms are less than one the fixed point is stable, and if all of them are more than one then it is unstable.
Good Luck! Hope this helps!
In $\mathbb{R}^2$ the modulus of the eigenvalues of the Jacobian gives you the criteria for stability:
\begin{gather}
0 &lt; |\lambda_1| \leq |\lambda_2|&lt;1 \; \text{ stable} \\
1 &lt; |\lambda_1|\leq  |\lambda_2| \; \text{ unstable} \\
0 &lt; |\lambda_1| &lt; 1 &lt; |\lambda_2| \; \text{ saddle point} 
\end{gather}
But since you're working in $\mathbb{R}^4$, I don't think it should be too different. Whichever eigenvalues have modulus less than one will correspond to a local stable manifold of the fixed point (given by the corresponding eigenvector), whereas the ones with modulus more than one are unstable.
So in summary I guess it would be sort of weird to just call the fixed point &quot;stable&quot; or &quot;unstable&quot; since there are many more possibilities, but if all the eigenvalues' norms are less than one the fixed point is stable, and if all of them are more than one then it is unstable.
Good Luck! Hope this helps!