Proof. By Theorem 5.1.2(iv) it suffices to treat a Jordan block \(C = \lambda I + N = \lambda (I + N/\lambda )\) with \(\lambda \neq 0\) and \(N^m = 0\). Choose a complex logarithm \(\log \lambda \) and set \[ B = (\log \lambda )I + L, \qquad L = \sum _{k=1}^{m-1}\frac {(-1)^{k+1}}{k}\Bigl (\frac N\lambda \Bigr )^k , \] the series for \(\log (1 + x)\) with \(x = N/\lambda \), which terminates because \(N\) is nilpotent. The formal power series identity \(\exp (\log (1 + x)) = 1 + x\) holds as an identity of polynomials in \(x\) modulo \(x^m\), and substituting the nilpotent matrix \(N/\lambda \) gives \(e^L = I + N/\lambda \). Since \((\log \lambda )I\) commutes with \(L\), \(e^B = \lambda e^L = C\). □

Proof. By Lemma 7.2.1 choose \(B\) with \(e^{BT} = M\), and define \(P(t) = \Phi (t)e^{-Bt}\), which is invertible. Then \(P(t + T) = \Phi (t + T)e^{-Bt}e^{-BT} = \Phi (t)Me^{-BT}e^{-Bt} = \Phi (t)e^{-Bt} = P(t)\), using \(e^{-BT}e^{-Bt} = e^{-Bt}e^{-BT}\). Every solution is \(x = \Phi (t)c = P(t)e^{Bt}c\), so \(y = P(t)^{-1}x = e^{Bt}c\) solves \(y' = By\), and conversely every solution \(e^{Bt}c\) of \(y' = By\) gives a solution \(x = P(t)y\). For the real form, \(M^2\) is real and has a real logarithm \(\tilde B\) with \(e^{2T\tilde B} = M^2\); then \(P(t) = \Phi (t)e^{-\tilde Bt}\) is real and \(2T\)-periodic by the same calculation. □