Floquet's Theorem

Theorem
If: Then:
 * $$A(t)$$ is a continuous matrix function with period $$\omega$$, and
 * $$\Phi(t)$$ is a fundamental matrix of the Floquet system $$x' = A(t)x$$,
 * $$\Phi(t+\omega)$$ is also a fundamental matrix; moreover
 * There exists:
 * a nonsingular, continuously differentiable matrix function $$P(t)$$ with period $$\omega$$ and
 * a constant (possibly complex) matrix $$B$$ such that
 * $$\Phi(t) = P(t) e^{Bt} $$.

Proof
Allow the two hypotheses of the theorem by assumption.

Since
 * $$\frac{d}{dt} (\Phi(t+\omega)) = \Phi'(t+\omega) = A(t+\omega) \Phi(t+\omega) = A(t) \Phi(t+\omega)$$,

the first implication of the theorem obtains.

Because $$\Phi(t)$$ and $$\Phi(t+\omega)$$ are both fundamental matrices, there must exist some $$C$$ such that
 * $$\Phi(t+\omega) = \Phi(t)C$$,

hence by the existence of the matrix logarithm, there exists a $$B$$ such that
 * $$C = e^{B\omega}$$.

Defining $$P(t) = \Phi(t) e^{-Bt}$$, it follows that
 * $$P(t+\omega) = \Phi(t+\omega) e^{-Bt-B\omega}$$
 * $$=\Phi(t)C e^{-B\omega} e^{-Bt} $$
 * $$=\Phi(t) e^{-Bt}$$
 * $$=P(t)$$

and hence $$P(t)$$ is periodic with period $$\omega$$.

As $$\Phi(t) = P(t) e^{Bt}$$, the second implication also obtains.