Stability of Dynamical System with Imaginary Eigenvalues I have this matrix $$\begin{bmatrix}0 &-3& 0& 0\\ 3 &0& 0& 1\\ 0 &0& 0 &-3\\ 0& 0& 3& 0\end{bmatrix}$$ and I found the eigenvalues: $0 + 3i$, $0 - 3i$,$ 0 + 3i$, and$ 0 - 3i$. But I'm not sure how to find the stability of this system. 
 A: First note that for a matrix $A$ there are two cases:


*

*$\lambda$ is a real eigenvalue with eigenvector $v$, then there is a solution $x(t)=ve^{\lambda t}$.

*$\lambda=a\pm ib$ is a complex conjugate pair with eigenvectors $v=u\pm iw$ then $x_1(t)=e^{at}(u\cos bt - w \sin bt)$ and $x_2(t)=e^{at}(u\sin bt+w\cos bt)$ are two linearly-independent solutions.


Let $\lambda$ be an eigenvalue of multiplicity 2 such that $\Re(\lambda)=0$. If there are 2 linearly independent eigenvectors (say, $v_1$ and $v_2$), then your solution will be either a linear combination of 2 constants (if $\lambda\in\mathbb R$) or a linear combination of sines and cosines (if $\lambda\in\mathbb C$). The solution is thus uniformly stable.
If, on the other hand, the eigenvectors corresponding to $\lambda$ are linearly dependent ($v_1=cv_2$), then there will be a generalized eigenvector $v_3$ such that $Av_3=v_1$. If the projection of your initial value $x_0$ on $v_3$ is non-zero, then there will be a component of the solution containing $t$ as a factor. The solution is unstable.
For your example, the generalized eigenvector $v_3$ corresponding to $\lambda=3i$ is $v_3=[0,\,-1/3,\,2,\,-2i]^T$. You may check that $Av_3=v_1$, where $v_1=[1,\,-i,\,0,\,0]$.
If you set $x_0=v_3$, the real part of the solution will be $x(t)=\begin{bmatrix}t\cos(3t)\\ t\sin(3t) - \cos(3t)/3\\2\cos(3t)\\2\sin(3t)\end{bmatrix}$. We see that $\|x(t)\|\rightarrow \infty$ as $t\rightarrow \infty$.
A: In general it can be pretty difficult to find the stability of non-hyperbolic equilibria (i.e. one eigenvalue has a vanishing real part). If your system is linear and the eigenvalues would have been semi-simple we'd have gotten stability. But like this I'd say it depends on the original system. Now, for example let's take the following linear system
\begin{align} 
\dot a &= -3b \\ \dot b&=3a+d \\ \dot c&=-3d \\ \dot d&=3c
\end{align}
and its equilibrium in the origin; so we get this matrix as you described. We can now try to construct a Lypaunov function to show stability.
$$V(a,b,c,d)=\frac12 (a^2+b^2+c^2+d^2)$$
$$\frac{d}{dt} V=\langle (a,b,c,d)^T, (-3b,3a+d,-3d,3c)^T\rangle=-3ab+3a^2+bd\overset{!}{\leq} 0$$
So for example if we impose the condition that $d \leq \frac{3ab-3a^2}{b}$ on the original dynamical system we'd get that the origin is stable. Maybe we could have done it with a less constricting condition if we'd have chosen another Lyapunov function but this can be very tough.
A: As you've already seen, because the eigenvalues have no real part, one can't simply apply the stability criterion $Re \lambda < 0$ or $Re \lambda > 0$.
The 1 in the 2nd row also makes thing more complicated. If it were not there, the dynamics would split in the two planes $x_1, x_2$ and $x_3, x_4$. There, trajectories would be circles and the whole trajectory in $\mathbb{R}^4$ contained in a torus. In this case, the origin would be (Lyapunov) stable, though not asymptotically.
But with that 1, the equations are no longer decoupled. If you write the exponential matrix and your solutions explicitly, the third and fourth components are seen to stay bounded but the first and second definitely not, while also depending on the former. So maybe you can say something about the invariance of cylinders around $x_3=0,\ x_4=0$ (for every initial condition, solutions will remain trapped in such a cylinder).
