Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 14E / a / Solution

Linearizing about the orbit gives $\dot{\eta }=A(t)\eta$, where $A(t)=Df(X(t))$. The fundamental matrix therefore satisfiesThe Floquet multipliers are the eigenvalues of the monodromy matrix $\Phi (T)$. Differentiating the orbit equation shows that its tangent $\dot{X}(t)$ solves the variational equation. Since $\dot{X}(T)=\dot{X}(0)\ne 0$, one multiplier is one.

For an invertible fundamental matrix, differentiating the determinant gives $(\det \Phi {)}^{\prime }=\mathrm{tr}({\Phi }^{-1}\dot{\Phi })\det \Phi =\mathrm{tr}A\det \Phi$. Integration from the identity yields $\det \Phi (T)=\exp ({\int }_0^T\mathrm{tr}A(t)dt)$. The determinant is the product of the two multipliers, so the other is