Consider the system of ODEs
$$\frac{dx}{dt} = -2x -y +1$$ $$\frac{dy}{dt} = -y $$
The eigenvalues of the corresponding derivative are $-2$ and $-1$ with eigenvectors $e_1$ and $e_2$ respectively. There are two stable manifolds corresponding to the fixed point $(0.5,0)$, the strong one $U_{SS}$ orthogonal to $e_1$ and the weak one $U_{WS}$ orthogonal to $e_2$. I also know that an initial condition picked on $U_{SS}$ will stay on $U_{SS}$ with the fastest contraction rate, by the stable manifold theorem.
But how exactly do I get the graph of $U_{SS}$? In the figure of the linear ode simulated below, we see the density along where the stable manifold should be. Is this curve the global stable manifold?
In linear ODEs, the stable manifold is the stable eigenspace. The eigenspaces related to $e_1$ and $e_2$ are the eigenvectors related to $e_1$ and $e_2$, respectively. The whole space $R^2$ is also $E^s=W^s$ (the stable eigenspace or the stable manifold).
The eigenvector related to $e_1$ is $[1 \ 0]^T$ because $\begin{bmatrix} -2 & -1\\ 0 & -1 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=-2\begin{bmatrix} v_1\\ v_2 \end{bmatrix}$ implies that $v_2=0$.
Now the fact that $[1 \ 0]^T$ is the eignevector of $e_1$ implies that $U_{ss}$ is the x-axis or $y=0$