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I recently got the question whether it is true that if we have a linear system of ode's given by a real matrix with only pure imaginary eigenvalues, the solution of our system is bounded. I answered that if the $n \times n$ matrix $A$ that governs our system is diagonalisable over the complex numbers with only purely imaginary eigenvalues, the answer is yes.

But I'm not sure about the fact if it's true or not when $A$ fails to be diagonalisable over the complex numbers. By which I mean that the sum of the dimensions of the eigenspaces isn't $n$. But I'm also not even sure whether such a matrix exists, I tried constructing such a matrix by playing around with the real Jordan normal form but with no success. So my question is: do such real matrices exist?

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$$ A = \left( \begin{array}{rrrr} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array} \right) $$ First note $(A^2 + I)^2 = 0.$ Also when $$ P = \left( \begin{array}{rrrr} i & 0 & 1 & 0 \\ 1 & 0 & i & 0 \\ 0 & i & 0 & 1 \\ 0 & 1 & 0 & i \end{array} \right) $$ then $$ P^{-1} = \left( \begin{array}{rrrr} -i/2 & 1/2 & 0 & 0 \\ 0 & 0 & -i/2 & 1/2 \\ 1/2 & -i/2 & 0 & 0 \\ 0 & 0 & 1/2 & -i/2 \end{array} \right) $$ and $$ P^{-1} A P = \left( \begin{array}{rrrr} -i & 1 & 0 & 0 \\ 0 & -i & 0 & 0 \\ 0 & 0 & i & 1 \\ 0 & 0 & 0 & i \end{array} \right) $$

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