If all the real parts of the eigenvalues are negative, then indeed $e^{tA}→0$? Is there a short proof to show if all the real parts of the eigenvalues are negative, then indeed $e^{tA}→0$?  
Also I am wondering if $\lim_{t \rightarrow \infty} ||e^{At}||=0 \iff $ the real parts of all eigenvalues are LESS than 0 or can we have equal to 0 if the algabric and geometric multiplicities are the same?
 A: This is the shortest I could find:
Let $ A = P J P^{-1} $ be the Jordan decomposition of A. Then $ e^{tA} = e^{t P J P^{-1}} = P e^{tJ} P^{-1} $.
Since $J$ is a block diagonal matrix, $e^J$ is also a block diagonal matrix and each of its blocks is the exponential of a block of $J$.
Hence $ \lim_\limits{t \to \infty} e^{tA} = 0 \iff $ for each Jordan block $B$, $ \lim_\limits{t \to \infty} e^tB = 0 $.
A Jordan block $B$ has the form
$$ B(m,n)=
\begin{cases} 
\lambda &, n = m \\
1 &, n = m + 1  \\
0
\end{cases} $$
where $ \lambda = a + ib $ is an eigenvalue of $A$.
So
$$ e^B(m,n)=
\begin{cases} 
\frac{t^{n-m}}{(n-m)!} e^{\lambda t} &, m \le n \\
0
\end{cases} $$
Therefore $ \lim_\limits{t \to \infty} e^{tB} = 0 \iff \lim_\limits{t \to \infty} e^{\lambda t} = 0 $.
As $e^{i b t}$ is continuous and periodic, and consequently bounded
$$ \lim_\limits{t \to \infty} e^{\lambda t} = \lim_\limits{t \to \infty} e^{a t} e^{i b t} = 0 \iff a < 0 $$
Thus we can conlude that
$$ \lim_\limits{t \to\infty} ||e^{At}||=0 \iff \eta(A) < 0 $$
where $ \eta(A) $ is the spectral abscissa of $A$.
A: If $Av = \lambda v$ then $e^{tA}v = e^{t\lambda}v$.
Let $\lambda = a + ib$. Then $|e^{t\lambda}| = e^{t a}$.
