I am trying to work out a relation between the largest singular value $\sigma_1(Q)$ of a normal matrix $Q$, and the largest singular value of $e^{Q}$. In an ideal world for the wider context of my proof (looking at epidemic spread on a network), it would be that $$\sigma_1(e^Q) \leq e^{\sigma_1(Q)}$$
So far, I have got this far in working: $$ ||Q\mathbf{u}|| = \sqrt(\mathbf{u}^TQ^TQ\mathbf{u}) \leq \sigma_1(Q)||\mathbf{u}|| $$ $$\therefore ||e^{Q}\mathbf{u}|| = ||I\mathbf{u} + Q\mathbf{u} + \frac{Q^2}{2!}\mathbf{u} + ... ||$$ $$ \leq ||I\mathbf{u}|| + ||Q\mathbf{u}|| + \frac{1}{2}||Q^2\mathbf{u}|| + ...$$ $$ \leq ||\mathbf{u}|| + \sigma_1||\mathbf{u}|| + ... $$
but I cannot see a way to simplify any higher-order terms of the sequence to get to a factorisation to an exponential. Any help, signposting to other resources or just telling me it is impossible would be greatly appreciated!
Like this: $$\sigma_1(e^Q) = \|e^Q\|=\left\|\sum_{k=0}^\infty \frac{Q^k}{k!} \right\|\leq\sum_{k=0}^\infty\frac{\|Q\|^k}{k!}=e^{\|Q\|}=e^{\sigma_1(Q)},$$ where $\|\cdot\|$ is the spectral norm.