How to draw the trajectories of the solutions in the canonical basis? I'm trying to understand the steps to solve this exercise:

Consider the following differential system $$\begin{cases}{x'=-2 x-y} \\ {y'=3 x+y}\end{cases}$$
Draw the trajectories of the solutions in the canonical basis.

Here is the solution given by my professor:

In the basis $\left \{\begin{pmatrix}2 \\ -3 \\ \end{pmatrix}, \begin{pmatrix}0 \\ \sqrt 3 \\ \end{pmatrix} \right \}$, we have to solve $$\begin{bmatrix}f' \\ g' \\ \end{bmatrix} = \begin{bmatrix} -\dfrac{1}{2} & \dfrac{\sqrt 3}{2} \\ -\dfrac{\sqrt 3}{2} & -\dfrac{1}{2} \\ \end{bmatrix} \begin{bmatrix}f \\ g \\ \end{bmatrix}$$
Setting $z = f + i g$, we have to solve $$z' = \left  (-\frac{1}{2} - i \frac{\sqrt 3}{2} \right)z$$


We have $$\begin{bmatrix}x' \\ y' \\ \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 3 & 1 \\ \end{bmatrix} \begin{bmatrix}x \\ y \\ \end{bmatrix}$$
The characteristic polynomial has two complex roots $$\lambda = -\frac{1}{2} \pm i \frac{\sqrt 3}{2}$$
Could you please explain how to come up with the basis $\left \{\begin{pmatrix}2 \\ -3 \\ \end{pmatrix}, \begin{pmatrix}0 \\ \sqrt 3 \\ \end{pmatrix} \right \}$?
Thank you so much for your help!

Update: I include the screenshot of solution.


 A: These are the real and imaginary part of one of the complex eigenvectors. 
The problem that is tried to solve with this construction is that in the diagonalization you get complex numbers on the diagonal. If you wanted to draw this you would have to use 4 real dimensions. Fortunately, a system over the reals also has purely real solutions, so that you can leave out all solutions not inside the real plane. 
$\newcommand{\myvec}[1]{\mathbf{#1}}\newcommand{\va}{\myvec{a}}\newcommand{\vb}{\myvec{b}}\newcommand{\mA}{\mathfrak{A}}$
It is a question of taste if you take one basis solution $(\va+i\vb)e^{(\alpha+i\beta)t}$, $(\va+i\vb)$ an eigenvector to the eigenvalue $(α+iβ)$ and take real and imaginary part of it
$$
e^{αt}(\va\cos(βt)-\vb\sin(βt)),e^{αt}(\va\sin(βt)+\vb\cos(βt))
$$
or take the long way around in transforming the system to the adapted basis $(\va,\vb)$ where then the system matrix is $\pmatrix{α&β\\-β&α}$ and then compute its exponential $e^{αt}\pmatrix{\cos(βt)&\sin(βt)\\-\sin(βt)&\cos(βt)}$ via any means.

You get the same real basis solutions from the complex conjugate pair of eigenvalue $(α-iβ)$ and its eigenvector $(\va-i\vb)$.

Comparing the parts in $\mA(\va+i\vb)=(α+iβ)(\va+i\vb)$ gives $\mA\va=α\va-β\vb$ and $\mA\vb=β\va+α\vb$. Then any solution $y(t)=u(t)\va+v(t)\vb$ of $\myvec y'=\mA \myvec y$ leads to component equations
\begin{align}
u'(t)&=αu(t)+βv(t)\\
v'(t)&=-βu(t)+αv(t)\\
\hline
(u+iv)'&=(α-iβ)(u+iv)
\end{align}
so that indeed in the computations of your professor there appears some sign confusion.
