= Solution
The linearized perturbation obeys $\dot\eta=D\mathbf f(\mathbf X(t))\eta$, so its <fundamental matrix> satisfies
$$
\boxed{\dot\Phi=D\mathbf f(\mathbf X(t))\Phi,\qquad\Phi(0)=I.}
$$
The <Floquet multipliers> are the <eigenvalues> of the <monodromy matrix> $\Phi(T)$. Differentiating the orbit equation shows that $\dot{\mathbf X}(t)=\mathbf f(\mathbf X(t))$ is a solution of the variational equation. For a nonconstant <periodic orbit> this tangent solution is nonzero and periodic, giving one multiplier equal to $1$.
The <Liouville formula for a fundamental matrix> gives $(\det\Phi)'=(\operatorname{tr}D\mathbf f)\det\Phi$. Since $\det\Phi(0)=1$, the product of the two multipliers is $\exp(\int_0^T\nabla\cdot\mathbf f(\mathbf X(t))\,dt)$. The other multiplier is therefore \b[exactly this exponential].
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