Find $c, M > 0$ such that $\lvert e^{tA}x_0\lvert \le Me^{ct}\lvert x_0\lvert$ In a system of differential equations $x'=Ax$, where $A$ is a constant matrix, and the equation is a sink (all eigenvalues of $A$ have negative real parts), I need to find constants $c,M>0$ such that $\lvert e^{tA}x_0\lvert \le Me^{ct}\lvert x_0\lvert$, $ \forall t\ge 0$, $\forall x_0\in \mathbb{R}^2$.
I would appreciate it very much if someone could please recommend me a source to read about how to do, why one does it, and what it all means. Unfortunately, it's not clear at all from the course notes what this is all about, and the reference textbook is unclear as well, and doesn't seem to contain any concrete information on this problem (I think it assumes that the student already knows what this means). But I'm an undergrad, and the textbook is a grad text. Completely lost with this problem.
 A: First of all: you can take $c<0$, although of course you can also take $c>0$.
What you ask simply means that in a sink of a linear equation $x'=Ax$ everything goes exponentially fast to the origin (it could go slower and still be a sink).
In order to show this, it is sufficient to observe that each entry of $e^{At}$ is bounded by $p(t)e^{-dt}$ for some polynomial $p$ and some constant $d>0$ (this follows from the computation of the exponential). Note also that for all these entries (which are finitely many), given $\varepsilon>0$ there exists $C>0$ such that $$p(t)e^{-dt}\le Ce^{(-d+\varepsilon)t}.$$By taking $\varepsilon$ sufficiently small you can ensure that $c=-d+\varepsilon<0$.
Now observe that since all entries of $e^{At}$ are bounded by $Ce^{(-d+\varepsilon)t}=Ce^{ct}$ (you can take the same constants $C$ and $N$ for all entries), you get $$\|e^{At}\|\le C'e^{ct}$$ for some other constant $C'>0$. This readily implies what you ask, in view of the definition of $\lVert\cdot\rVert$.
