If a matrix $A$ is diagonalizable, then the solutions to the system of difference equations $${\bf x}(n) = A {\bf x}(n-1)$$ take the general form $$\sum_j c_j \lambda_j^n {\bf b_j},$$ where $\lambda_1, \ldots, \lambda_m$ are eigenvalues of $A$, ${\bf b_1}, \ldots, {\bf b_m}$ are eigenvectors of $A$, and $c_1, \ldots, c_m$ are arbitrary constants.

1. If $|\lambda_j|<1$, then $\lambda_j^n$ goes to zero, and that term disappears from the sum over time. We call such modes stable.
2. If $|\lambda_j|>1$, then $\lambda_j^n$ grows without bound. We call such modes unstable.
3. If $|\lambda_j|=1$, then $|\lambda_j^n|=1$ for all $n$. We call such modes neutral or borderline.
4. The term with the largest $|\lambda_j|$ will eventually grow larger than all the rest. We call this the dominant mode. The long-time behavior of the whole system is described by this mode. If this mode is stable, the whole system shrinks away to 0. If this mode is unstable, then ${\bf x}$ grows without bound. Either way, the direction of ${\bf x}(n)$ approaches the direction of the dominant eigenvector.

The solutions to the system $\frac{d{\bf x}}{dt} = A{\bf x}$ of differential equations behave similarly to the solutions to difference equations, with the only change being that we have $e^{\lambda t}$ instead of $\lambda^n$.

1. The general solution is $\sum_j c_j e^{\lambda_j t} {\bf b_j}$.
2. If $\lambda > 0$ (or if $\lambda = a+bi$ with $a>0$), then $e^{\lambda t}$ blows up and we have an unstable mode.
3. If $\lambda<0$, or if $\lambda = a+bi$ with $a<0$, then $e^{\lambda t}$ goes to 0 and we have a stable mode.
4. If $\lambda=0$, or is pure imaginary, then $e^{\lambda t}$ neither grows nor shrinks, and we have a neutral or borderline mode.
5. The dominant mode corresponds to the eigenvalue with the greatest real part, not to the one with the greatest magnitude. (That is $3$ beats $2 + 75i$, which beats $-10000$.)

Stability for second-order systems $\frac{d^2{\bf x}}{dt^2} = A {\bf x}$ is different in two ways. First, we don't have to worry about complex eigenvalues, since these problems almost always have real eigenvalues. Second, there are no truly stable modes. When $\lambda <0$, our modes oscillate with frequency $\omega = \sqrt{-\lambda}$. When $\lambda >0$, we get exponential growth like $e^{\kappa t}$, where $\kappa = \sqrt{\lambda}$. So the modes with $\lambda < 0$ are neutral and the modes with $\lambda > 0$ are unstable, as explained in the third video.