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.