Direction Field and Trajectories I am wondering how to draw a direction field and trajectories of a system of linear equations:
$$ x'= \left[
  \begin{array}{ c c }
     4 & -2 \\
     8 & -4
  \end{array} \right] x
.$$
I remember how to do them for a equation from the first part of ODE where we would have $$y'=2x-y$$ but I do not really understand how to begin with the system in this form. Would it just be:
$$  x'=4x_1 -2x_2 $$ $$ x'=8x_1-4x_2 $$ and we would just plot different values of $x_1$ and $x_2$?
 A: In $1-D$, it is called a direction field plot. 
In higher dimensions, it is called a phase portrait and these two sets of notes 1 and notes 2 provide a procedure for sketing it by hand.
We look at a handful of items (see the two sets of notes):


*

*Write the equation as $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dx}}{\dfrac{dx}{dt}} = 2$ 

*Sketch the isoclines of horizontal slope (where $y' = \dfrac{dy}{dt} = 0 \implies y = 2x$) - these are typically called nullclines

*Sketch the isoclines of vertical slope (where $x' = \dfrac{dx}{dt} = 0 \implies y = 2x$) - these are typically called nullclines

*Put together the horizontal and vertical arrows and sketch the resultant direction field (note direction and magnitude).

*Trace some sample trajectories.

*You can also find the critical points and evaluate the eigenvalues of the Jacobian at each critical point. For this system, we have eigenvalues $\lambda_{1,2} = 0, 0$, so these are just lines.
A phase portrait shows:
$~~~~~~~~~$ 
For this system, we can also explicitly solve it and arrive at:
$$x(t) = c_1(4t+1) - 2c_2 t, y(t) = 8c_1 t + c_2(1-4t)$$
You can parametrically plot these solutions and what do you notice? For example, let $x(0) = y(0) = 1$, yielding $x(t) = 2t+1, y(t) = 4t+1$. A parametric plot shows (compare to the phase portrait and add a bunch more of these for different initial conditions):

