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The linearized perturbation obeys $\dot{\eta }=D\mathbf{f}(\mathbf{X}(t))\eta$, so its fundamental matrix satisfiesThe 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 $(\det \Phi {)}^{\prime }=(\mathrm{tr}D\mathbf{f})\det \Phi$. 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 exactly this exponential.