TITLE: Giving bounds to the principal solution matrix
QUESTION [5 upvotes]: I've been trying to solve a problem from ODE and Dynamical System of G.Teschl and I got stuck in some linear algebra problem(linear algbra more than ODE I guess). 
Problem
For any matrix A(so it can be constant or depending on $t$), the matrix $Re(A) = \frac{1}{2}(A + A^{*})$ is symmetric and hence has only real eigenvalues. Let $\alpha_{0}$ be its largest eigenvalue.
Let $A(t)$ be given and and define $\alpha_{0}(t)$ as above, prove that $||\Pi(t,t_{0}) || \leq e^{\int_{t_{0}}^{t} \alpha_{0}(s)dx}$ for $t>t_{0}.$
My Problem
I know I can consider $\frac{d}{dt}|x(t)^{2}| = 2 <x(t),\dot x(t)> = 2<x(t),Ax(t)>$. After this, I cant find the way to relate $Re(A)$ with $A$ in order to bound my inner product(this is why I think my problem is with linear algebra more than ODE). After this I know I can use the fact that $<x,Re(A)x> \leq \alpha_{0}|x^{2}(t)|$ and then apply some gronewall's inequality.
Thanks so much for your answers!

REPLY [4 votes]: To do the complex case with a hermitean scalar product correctly, the first step in the calculation needs to be corrected to
$$
\frac{d}{dt}\|x(t)\|^2=⟨x,\dot x⟩+⟨\dot x,x⟩=2Re(⟨x,\dot x⟩)=2Re(⟨x,Ax⟩).
$$
Then you get
$$
2Re(⟨x,Ax⟩)=Re(⟨x,Ax⟩+⟨A^*x,x⟩)=Re(⟨x,Ax⟩+⟨x,A^*x⟩)=2⟨x,Re(A)x⟩
$$
Now you can use the spectral property of symmetric matrices that $⟨x,Re(A)x⟩\le α_0\|x\|^2$.

Alternatively, one could have used that $⟨x,Ax⟩\le \|A\|\,\|x\|^2$ where $\|A\|$ is the largest singular value. This however completely obscures the case where $α_0$ is negative everywhere, where the given method proves directly that the propagator is a contraction.