Suppose $A\in {\mathbf R}^{n \times n}$ has real eigenvalue $\lambda$.
$A$ has a right eigenvector $v$ and a left eigenvector $w$ associated with $\lambda$.
$\chi(\lambda) = 0$, where $\chi(s) = \det (sI-A)$.
$w^Tv = 0$.
$w^Tv = 1$.
Suppose $A \in {\mathbf R}^{n \times n}$ is diagonalizable.
The eigenvalues of $A$ are distinct.
$A^T$ is diagonalizable.
$A^2$ is diagonalizable.
Suppose $A\in {\mathbf R}^{n \times n}$ is defective.
$A^2$ is defective.
$A$ does not have an independent set of $n$ eigenvectors.
$A$ has repeated eigenvalues.
$A$ is singular.
If the continuous-time system $\dot x(t) = Ax(t)$ is stable then the discrete-time system $y(t+1) = e^Ay(t)$ is stable. (In both cases, stable means all trajectories converge to zero as $t\to\infty$.)
If the discrete-time system $y(t+1) = e^Ay(t)$ is stable, then the continuous-time system $\dot x(t) = Ax(t)$ is stable.