# Collision of eigenvalues of a linear ODE (Krein collisions)

I am trying to understand the so called Krein collisions in Hamiltonian mechanics but I shall formulate the question in a rather general way.

Suppose we have the following linear ODE:

$\dot{v}= Kv$

in a space of dimension $2n$ where $K$ is a constant coefficient matrix with real valued coefficients satisfying the following properties: the characteristic polynomial of $K$ is even and if $\lambda$ is an eigenvalue of multiplicity $k$, then $-\lambda$ is also an eigenvalue of multiplicity $k$ and if $0$ is an eigenvalue, then it has even multiplicity. Now suppose that the eigenvalues depend continuously on a parameter $\mu$ and suppose the system has an equilibrium point for which it is linearly stable (i.e, all the eigenvalues start off on the imaginary axis because the even nature of the characteristic polynomial prevents a scenario where the eigenvalues can only have a negative real part). Assume also that zero cannot be an eigenvalue associated with the equilibrium point.

It is then asserted that the eigenvalues cannot leave the imaginary axis unless they first collide. Why is this necessarily true?

I am trying to envision the following scenario. Consider a four dimensional system so that at $\mu=\mu_0$, there are two eigenvalues $\pm 2i$ of multiplicity $2$ which then move, as $\mu$ varies towards $\pm i$ and at $\mu=\mu_0 + T$, they are at $\pm i$. Now why cannot they split off so into a quartet of eigenvalues (the so called Krein quartet) $\pm i \pm \varepsilon$ at the next instant of time.

Thank you.

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Asserted by whom and where? Are you sure you did not pose this in a too general way? Symplectic matrices have determinant 1 (and reciprocal characteristic polynomial), so they would not be able to have eigenvalues $2i$ without also having $i/2$. –  user31373 Jul 29 '12 at 13:12
One might also say that in the situation 2i, 2i, -2i, -2i collision already took place since we have double eigenvalues. –  user31373 Jul 29 '12 at 13:27