On marginally stable systems

This post aims to answer a simple question in linear system theory: Is a system with all its poles on the imaginary axis always marginally stable?

The answer is NO. A well-known counter example is $$\begin{array}{}\text{(1)}& \dot{x}=\left[\begin{array}{cc}0& 1\\ \\ 0& 0\end{array}\right]x\end{array}$$ The system in $\text{(1)}$ has two poles at the origin, but it is not marginally stable since there is a single Jordan block corresponding to the eigenvalue $0$.

Now, let us make this question a bit more challenging: Is any system with all its poles satisfying (a) non-zero (b) on the imaginary axis always marginally stable?

This revised question excludes the counter example in $\text{(1)}$. According to Wikipedia,

a continuous-time state-space model is marginally stable if and only if the Jordan blocks corresponding to poles with zero real part are one-by-one matrices.

By Wiki (which might be wrong), the answer to the revised question seems negative, since the Jordan blocks corresponding to the non-zero poles on the imaginary axis may not be one-by-one. But I am not convinced by this argument unless someone can show me a counter example.

So, our goal now is to find a real matrix whose the Jordan block corresponding to an imaginary eigenvalue is not one-by-one. Have you seen such an example in your linear system theory course or linear algebra 101? I cannot recall anything like that. One famous example for a matrix with purely imaginary poles is $$\begin{array}{}\text{(2)}& A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right].\end{array}$$ The Jordan normal form of $A$ is $$\begin{array}{}\text{(3)}& \left[\begin{array}{cc}i& 0\\ \\ 0& -i\end{array}\right].\end{array}$$ It would be nice if we can find a real matrix whose Jordan normal form is $$\begin{array}{}\text{(4)}& \left[\begin{array}{cccc}i& 1& 0& 0\\ 0& i& 0& 0\\ 0& 0& -i& 1\\ 0& 0& 0& -i\end{array}\right].\end{array}$$ Actually, it is indeed possible. According to this document (Real Jordan Form), the matrix in $\text{(5)}$ has exactly the Jordan normal form in $\text{(4)}$.

$$\begin{array}{}\text{(5)}& A=\left[\begin{array}{cccc}0& 1& 1& 0\\ -1& 0& 0& 1\\ 0& 0& 0& 1\\ 0& 0& -1& 0\end{array}\right].\end{array}$$ Therefore, the system $\dot{x}=Ax$ with $A$ in $\text{(5)}$ serves as a counter example for our revised question.