Type and stability of critical point of a linear first order differential system Suppose we have the system
$$ \begin{cases} \frac{ dx}{dt} = \alpha x - y \\ \frac{dy }{dt} = x + \alpha y \end{cases} $$
where $\alpha \in \mathbb{R}$. Identify the type and stability of the critical point $(0,0)$.
Try
First, we have
$$ J(x,y) = \left( \begin{matrix} \alpha & -1 \\ 1 & \alpha \end{matrix} \right) $$
we find the eigenvalues: We have that $\det( J - \lambda I) = 0 $ iff
$$ (\alpha - \lambda)^2 + 1 = 0 $$
Thus,
$$ \lambda_{1,2} = \alpha \mp i $$
So, the type will depende according to $\alpha$. If $\alpha = 0$, then we have $\mathbf{centers}$. If $\alpha \neq 0$, then we have $\mathbf{spirals}$. But, Im still stuck on how do we decide the direction of the solutions?
 A: The general way is to look to the eigenvectors of $J$ as suggested by caverac. Anyway, there is a faster way.
Suppose that $\alpha <0$ and at time $t$ both $x(t)$ and $y(t)$ are positive (i.e the trajectory is passing through the first quadrant). Then:
$$\frac{ dx}{dt} = \alpha x - y < 0,$$
as long as $x(t) > 0$ and $y(t) > 0$.
This means that $x(t)$ will decrease and the trajectory will move from the $1$st to the $2$nd quadrant. Then, it seems that the trajectory is moving counterclockwise. Remember... the trajectory is a spiral around $0$!

More in general, you can depict the flow in the phase plane by looking to the "derivative vectors", defined as 
$$u(x(t), y(t)) = \left[ \begin{array}{c}
\displaystyle  \frac{ dx}{dt} \\
\displaystyle  \frac{ dy}{dt} 
\end{array}\right]_{x=x(t), y=y(t)}.$$
In particular, if you draw the vectors $u$ starting from $(x(t), y(t))$ (you need a few in general), then you will get an idea about the flow of the system and hence you can understand what is the direction of motion.
