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{align} \label{eq:ex1} \dot{x} = \begin{bmatrix} 0 & 1\\\\0 & 0 \end{bmatrix}x \end{align} The system in \eqref{eq:ex1} 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 \eqref{eq:ex1}. 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{align} A= \begin{bmatrix} 0 & 1 \\\
-1 & 0 \end{bmatrix}. \end{align} The Jordan normal form of \(A\) is \begin{align} \begin{bmatrix} i & 0\\\\0 & -i \end{bmatrix}. \end{align} It would be nice if we can find a real matrix whose Jordan normal form is \begin{align} \label{eq:jordan} \begin{bmatrix} i & 1 & 0 & 0\\\
0 & i & 0 & 0\\\
0 & 0 & -i & 1\\\
0 & 0 & 0 & -i \end{bmatrix}. \end{align} Actually, it is indeed possible. According to this document (Real Jordan Form), the matrix in \eqref{eq:ex2} has exactly the Jordan normal form in \eqref{eq:jordan}.

\begin{align} \label{eq:ex2} A = \begin{bmatrix} 0 & 1 & 1 & 0\\\
-1 & 0 & 0 & 1\\\
0 & 0 & 0 & 1\\\
0 & 0 & -1 & 0 \end{bmatrix}. \end{align} Therefore, the system \(\dot{x} = Ax\) with \(A\) in \eqref{eq:ex2} serves as a counter example for our revised question.