Given $$ \begin{cases} \dot{x} = -x + y^2\\ \dot{y} = x - 2y +y^2 \end{cases} $$
Find a linear approximation of the STABLE manifold for the equilibrium $(1,1)$.
My attempt: By using the Principle of Linearized Stability, since $(1,1)$ is a hyperbolic equilibrium, $Df((1,1))$ has $2$ eigenvalues: $-2$ and $1$. This means $(1,1)$ is an unstable saddle, so only the $y$-axis is a stable manifold at $(1,1)$, based on the phase portrait of Andronov–Hopf bifurcation at unstable saddle equilibrium $(1,1)$.
My question: I find my answer kinda weird, so I just wonder if that's correct? If not, how are we supposed to find a stable manifold at an unstable equilibrium?
When you linearize, you are approximating your system by the first order terms around the equilibrium or, in matricial form: $$ \dot x = \begin{pmatrix} -1 & 2 \\ 1 & 0 \end{pmatrix} \, (x-x_0) $$
Now recall that the real meaning of finding eigenvalues is performing a change of basis. The matrix of eigenvectors is: $$B=\begin{pmatrix} -2 & 1 \\ 1 & 1 \end{pmatrix}$$ so that means that your diagonal form is expressed in a new coordinates $z$ related to the older by the change of basis $x=B z$, or $z=B^{-1} x$, or: $$ \begin{aligned} z_1 &= -x + y \\ z_2 &= x+2y \end{aligned} $$
(NOTE: if you substitute this change on the linearized equation, you obtain the uncoupled system $$ \begin{aligned} \dot z_1 &= -2 z_1 \\ \dot z_2 &= z_2 \end{aligned} $$ which can be solved as $z_1=k e^{-2 t}$, again showing that if you could restrict your space to the stable manifold, the system would be stable)
The equilibrium $(x,y)=(1,1)$ is moved to $z=(0,3)$, so you have the two linear subspaces: $$ \begin{aligned} 0 &= -x + y \\ 3 &= x+2y \end{aligned} $$ Since the first one corresponds to the negative eigenvalue, the stable subspace is: $$ x = y $$