Linear approximation of a stable manifold 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?
 A: 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:
$$
x = y
$$
A: Use the Jacobain matrix $A$
$$
A = \frac{\partial f}{\partial \mathbf{x}} = 
\begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\
\end{bmatrix}
=
\begin{bmatrix}
-1 & 2y \\
1 & -2+2y
\end{bmatrix}_{x=y=1}
=
\begin{bmatrix}
-1 & 2 \\
1 & 0
\end{bmatrix}
$$
Now the linear approximation system is 
$$
\begin{bmatrix}
\dot{x} \\ \dot{y}
\end{bmatrix}
=
\begin{bmatrix}
-1 & 2 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
$$
Or 
$$
\begin{align}
\dot{x} &= -x + 2y \\
\dot{y} &= x
\end{align}
$$
