Conditions for $A$ that ensure the boundedness of the solutions of $x'=Ax$ I have to find conditions under which the solutions of $x'=Ax$ are bounded for $t\rightarrow \infty$ and $t\rightarrow \pm \infty$. We proved in our course that $\lim_{t \rightarrow \ \infty}  x(t)=0$ and $\lim_{t \rightarrow - \infty}  |x(t)|=\infty$, if all the eigenvalues of $A$ have negative real part (and for the last limit $x$ is not the zero solution); if $A$ has an eigenvalue greater that zero, then $\lim_{t \rightarrow  \infty}  |x(t)|=\infty$. 
So I think that the conditions are: $x$ is bounded for $t\rightarrow \infty$ , if the real part of all  the eigenvalues are negative (trivial, since it follows from my course) and bounded for $t\rightarrow \pm \infty$ if the real parts of all the eigenvalues vanishes. Do you think this is sufficient - or that my professor could want some other additional (finer) conditions ?
Side question: Since we didn't specify what $|x(t)|$ means, I took it means the norm of $x(t)$ and not the vector of the absolute values of the components of $x(t)$, because in the latter case $\lim_{t \rightarrow  \infty}  |x(t)|=\infty$ doesn't makes sense to me, because only a sequence of numbers can have the limit $\infty$. Am I right that it means the norm ?
 A: Yes, you are right that $|x(t)|$ means norm, although it is more common to denote it $\|x(t)\|$. Notice that all norms on $\mathbb{R}^n$ are equivalent so boundedness of $x(t)$ in any norm just means that the entries of $x(t)$ stay bounded. You can get boundedness of $\|x(t)\|$ on $[0, \infty)$ in essentially two different ways, you can read both of them off the formula for the exponential of $A$:
$$x(t) = \exp(tA) = P\exp(tD+tN)P^{-1} =  P\exp(tD)\exp(tN)P^{-1}$$
where
$$A = P(D+N)P^{-1}$$
is the Jordan decomposition of $A$. This means that $D$ is diagonal whose entries are generalized eigenvalues of $A$ and the entries of $N$ consists of $0$s, except for the entries on the super-diagonal which may (or may not) be $1$. Notice that $N$ and $D$ always commute so it is valid to say $\exp(tD+tN) =  \exp(tD)\exp(tN)$.
$N$ is nilpotent so you can calculate its exponential to be a matrix whose entries are polynomial in $t$. Therefore, your typical entry of $x(t)$ is of the form
$$\sum e^{\lambda t}p(t)$$
where $\lambda$ is an eigenvalue of $A$ and $p$ is a polynomial. If $\lambda$ has negative real part then this expression stays bounded. But it also stays bounded if $\lambda$ has real part equal to $0$ and $p$ is constant. This happens for example when $A$ is diagonalizable and the real parts of the eigenvalues of $A$ are less than or equal to $0$.
To get boundedness on $(-\infty, \infty)$ you would need that $A$ is diagonalizable and all of its eigenvalues have real part $0$. So you were right in what you said for the diagonalizable case, but in general you have to look at the impact of the polynomials caused by any generalized eigenvectors of $A$.
A: Yes you are right in both remarks: 1) for boundedness of $x$ you need all the eigenvalues of $A$ having negative real part. In that case, $A$ is called "stable" or "Hurwitz". 2) $\left|x(t)\right|$ is a vector norm.
