Understanding the Jacobian Matrix Taking the example of a two dimensional system, desribred by the following ODE's:
\begin{align}
\frac{dx_1}{dt}&=f_1(x_1,x_2)\\
\frac{dx_2}{dt}&=f_2(x_1,x_2)
\end{align}
The Jacobian Matrix JM is then given by: 
$$JM=\left( \begin{array}{cc}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}  \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{array}\right)$$
Now quoting from scholarpedia:

The stability of typical equilibria of smooth ODEs is determined by
  the sign of real part of eigenvalues of the Jacobian matrix. These
  eigenvalues are often referred to as the 'eigenvalues of the
  equilibrium'. The Jacobian matrix of a system of smooth ODEs is the
  matrix of the partial derivatives of the right-hand side with respect
  to state variables where all derivatives are evaluated at the
  equilibrium point x=xe . Its eigenvalues determine linear stability
  properties of the equilibrium.
An equilibrium is asymptotically stable if all eigenvalues have
  negative real parts; it is unstable if at least one eigenvalue has
  positive real part.



*

*Is there an intuitive way of understanding why the sign of eigenvalues of JM imply the state of stability of the system?

 A: *

*$f=-kx$ is stable whereas $f=kx$ is unstable.

*You can usually rewrite a matrix $A$ as $A=PDP^{-1}$ where $P$ is a matrix of eigenvectors and $D$ is a diagonal matrix of eigenvalues.

*If $F=Ax$, then by the above, $(P^{-1}F)=D(P^{-1}x)$. Now you have $n$ independent equations exactly of the form $f=kx$ or $f=-kx$. If any one of them is like $f=kx$, the solution will blow up to infinity under a tiny perturbation.


For example:
$
    \left(
\begin{array}{cc}
 2 & 2 \\
 -1 & 1 \\
\end{array}
\right)\left(
\begin{array}{cc}
 1 & 0 \\
 0 & -1 \\
\end{array}
\right)\left(
\begin{array}{cc}
 2 & 2 \\
 -1 & 1 \\
\end{array}
\right)^{-1}=\left(
\begin{array}{cc}
 0 & -2 \\
 -\frac{1}{2} & 0 \\
\end{array}
\right)=A$
Just from looking at the $PDP^{-1}$ diagonalization, I can tell - if $F=Ax$,then any $x$ going in the direction $(2,-1)$ will tend to be pushed away from the origin, and any $x$ going in the diretion $(2,1)$ will tend to be pushed back towards the origin. You should understand $P^{-1}x$ as a change of basis into a coordinate system where the axes are some $n$ arbitrary directions which act as lines of force.
Here is a plot of that system $F=\left(
\begin{array}{cc}
 0 & -2 \\
 -\frac{1}{2} & 0 \\
\end{array}
\right)x$.

A: Let us look at a one-dimensional example:

Recall that $f(x) = \dot{x}$, so $f$ encodes the time evolution of $x$. If $f < 0$, then $x$ will move to the left. If $f > 0$, then $x$ will move to the right. If $f = 0$, $x$ will not move at all, this is why $f(x_0) = 0$ is the equilibrium condition.
Now, look what happens if you perturb the equilibria $x_i$ slightly to the right: If $f'(x_i) < 0$, then, in a small surrounding of $x_i$, $f(x) < 0$ for $x > x_0$ in that surrounding, i.e. to the right of a equilibrium with $f'(x_i) < 0$, $x$ will move to the left - returning to equilibrium! Conversely, if $f'(x_i) > 0$, then $f(x) > 0$ for all $x$ to the right in the small surrounding, meaning that, after a small nudge to the right, $x$ will move even further right, leaving equilibrium!
The same line of reasoning can be applied to perturbations to the left, altogether showing that, if $f'(x_i) < 0$, then a small perturbation around $x_i$ will always move back into $x_i$, and if $f'(x_i) > 0$, then a small perturbation will become bigger and bigger, moving away from $x_i$.
The multi-dimensional case is less graphic, but the intuition is the same - negative eigenvalues of the Jacobian mean that the time evolution points back into equilibrium, positive eigenvalues mean that it points away from equilibrium.
