Proof. (i) By Theorem 5.2.1, every entry of \(e^{Jt}\) has modulus at most \(t^ke^{\sigma t}/k!\) for some \(k < n\), and \(t^ke^{\sigma t} \leq C_ke^{\mu t}\) for \(t \geq 0\) because \(t^ke^{-(\mu - \sigma )t}\) is bounded. Then \(\|e^{At}\| \leq \|P\|\|e^{Jt}\|\|P^{-1}\| \leq Ke^{\mu t}\). (ii) If \(\sigma < 0\), take \(\mu = \sigma /2 < 0\) in (i). Conversely if \(\lambda \) is an eigenvalue with \(\Real \lambda \geq 0\) and eigenvector \(v\), then \(e^{\lambda t}v\) (or its real part) is a solution that does not tend to \(0\). (iii) The same solution grows like \(e^{(\Real \lambda )t}\). (iv) If an eigenvalue \(ib\) has only blocks of size one, its terms in \(e^{Jt}\) are \(e^{ibt}\), bounded; the other eigenvalues have negative real parts and contribute decaying terms. If it has a block of size \(m \geq 2\), the solution with initial value the second vector of the block’s Jordan chain contains \(te^{ibt}\), which is unbounded. □