I have the system (took very basic example on purpose, to understand the idea):
$$\begin{cases} \dot{x} = x \\ \dot{y} = 2x -y \end{cases}$$
so I have plot phase plane. what have been done so far:
$$A = \begin{pmatrix} 1 && 0 \\ 2 && -1 \end{pmatrix}$$
$$\det A = \begin{vmatrix}1 - \lambda && 0 \\ 2 && -1 -\lambda\end{vmatrix} = (1-\lambda)(-1-\lambda) = -1-\lambda + \lambda + \lambda^2 = \lambda^2 -1 = 0$$
therefore $\lambda_1 = 1$ and $\lambda_2 = -1$ (we have saddle here)
Corresponding eigenvectors are $v_1 = \begin{pmatrix}0 \\ 1\end{pmatrix}$ and $ v_2\begin{pmatrix} 1 \\1\end{pmatrix}$
what should I do next? I am confised from here.
A picture is worth a thousand words (taken from Wikipedia)
This is of course just an illustration, and in your case the actual plot is different. The general solution to the ODE $\dot{x}=Ax$ is given by
$$x\left(t\right)=C_{1}v_{1}e^{\lambda_{1}t}+C_{2}v_{2}e^{\lambda_{2}t}$$
where $\lambda_{1,2}$ are the eigenvalues of $A$ associated with the eigenvectors $v_{1,2}$. In the case of a saddle point, $\lambda_{2}<0<\lambda_{1}$ without loss of generality. Thus, if $C_{1}\neq 0$ and $C_{2}=0$ you have a diverging solution along $v_{1}$, while if $C_{1}=0$ and $C_{2}\neq 0$ you have a converging solution along $v_{2}$. So the phase space in your case is very similar to the picture above, but the blue/green lines are now $v_{1,2}$ respectively instead.