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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 $x_2$?
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):
