Exponential bound on norm of matrix exponential (of linear ODE) Consider a linear ODE: $\dot{x} = A x$ where $A$ is Hurwitz, i.e. all its eigenvalues have negative real parts. Thus the system is exponentially stable. We know that there exists positive numbers $\beta$ and $\alpha$ such that $\| e^{A t} \| \leq \beta e^{-\alpha t}$ for all $t$. I see this result being used in many analysis.
My question is how to (practically) compute these values? In particular, if I pick $\alpha$ so that $ -\alpha > \max_i \Re(\lambda_i)$, where $\lambda_i$ are the eigenvalues of $A$, then how to compute a tight value for $\beta$?
 A: Put $A$ in Jordan canonical form: $A = U J U^{-1}$.  Then $e^{At} = U e^{Jt} U^{-1}$ so $\|e^{At}\| \le \|U\| \|U^{-1}\| \|e^{Jt}\|$.  Of the eigenvalues with greatest real part (say $r = \max_i \Re(\lambda_i)$), take one with the largest Jordan block (say of size $m$).  Then for $t \ge 0$, 
$\|e^{Jt}\| \le e^{rt} \sum_{k=0}^m t^k/k!$.
EDIT: This was not quite right.  It is true for sufficiently large $t$.
But in general, you have to say 
$$ \|e^{Jt}\| \le \max_i e^{r_i t} \sum_{k=0}^{m_i} t^k/k!$$
where the $i$'th Jordan block has size $m_i$ and eigenvalue $\lambda_i$ with $r_i = \Re(\lambda_i)$.  
A: I have the same problem here. What I can do is to obtain $\alpha $ and $\beta$, but I am not sure whether their values, especially the one for $\beta$, is tight or not. 
At the moment my proof only concerns semi-simple matrices (but I think you can easily extend to the general case using the idea from this answer) and induced 2-norm of matrices (note the equivalence in norms). And it is easy to show that
$$ \begin{align*}
\left\Vert e^{At}\right\Vert _{2} & =\left\Vert e^{T^{-1}JtT}\right\Vert _{2}\\
 & =\left\Vert T^{-1}e^{Jt}T\right\Vert _{2}\\
 & \leq\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}\left\Vert e^{Jt}\right\Vert _{2}\\
 & =\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}e^{\max_{i}\mathfrak{R}\left(\lambda_{i}\left(A\right)\right)t}\\
 & \triangleq\beta e^{-\alpha t}
\end{align*} $$
where 
$$ A=T^{-1}JT $$
It is easy to see that under this scenario $ \alpha=-\max_{i}\mathfrak{R}\left(\lambda_{i}\left(A\right)\right) $ is tight, but as I have checked $\beta=\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}$ can be way off.
It's been years since this question is posted, not sure if you have found a better answer @Truong.
