I'm a bit rusty in linear algebra and I'm having trouble understanding why in one of the examples in my notes the unstable subspace of a continuous-time dynamical system is equal to $\mathbb{R}^2$.
I am given a system of two first order non-linear ODEs:
\begin{align*}\begin{cases}\dot x = y\\ \dot y = x - x^3 + x^2y.\end{cases}\end{align*} At the equilibria $(x,y)= (\pm 1,0)$, I have found the following Jacobian matrix $$Df(\pm 1, 0) = \begin{pmatrix}0&1\\-2&1\end{pmatrix}.$$Then, the eigenvalues are given by $$\det(Df(\pm 1, 0) -\lambda I) = 0 \implies -\lambda(1-\lambda)+2 = 0 \implies \lambda_{1,2} = \frac{1}{2}\pm i\frac{\sqrt{7}}{2}.$$
The unstable subspace $E_u$ is defined as the span of all generalised eigenvectors whose eigenvalues have positive real part. Then, for $v := (v_1,v_2)^T$, I find that \begin{align*} \left(Df(\pm 1, 0) - \lambda_1 I\right)\cdot v = 0 &\implies \begin{pmatrix}-\lambda_1 & 1\\ -2 & 1-\lambda_1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\\ &\implies \begin{pmatrix}-\frac{1}{2} - i\frac{\sqrt{7}}{2} & 1\\ -2 & \frac{1}{2} - i\frac{\sqrt{7}}{2}\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\\ &\implies \begin{pmatrix}4 & -1+i\sqrt{7}\\ 0 & 0\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}. \end{align*} Thus, $$\begin{cases}v_1 = \frac{1-i\sqrt{7}}{4}v_2\\v_2 = v_2\end{cases}.$$ $\lambda_2$ yields $v=\left(\frac{1+i\sqrt{7}}{4},1\right)^T$. So, I would say that the subspace is defined as $E_u :=$ Span$\left(\left(\frac{1-i\sqrt{7}}{4}, 1\right)^T,\left(\frac{1+i\sqrt{7}}{4},1\right)^T\right),$ but that can't be right because $E_u \subseteq \mathbb{R}^2$. So do I write the vectors as $$(v_1, v_2)^T = v_2\left(\frac{1\pm i\sqrt{7}}{4}, 1\right)^T = v_2\left(\frac{1}{4},1\right)^T \pm iv_2\left(\frac{\sqrt{7}}{4},0\right)^T?$$Thus, $$E_u := \left\{v: v = \kappa_1\left(\frac{1}{4},1\right)^T + \kappa_2\left(\frac{\sqrt{7}}{4},0\right)^T, \; \kappa_i \in \mathbb{C} \right\} \overset{(?)}{=} \text{Span}\left\{\left(\frac{1}{4},1\right)^T, \left(\frac{\sqrt{7}}{4},0\right)^T \right\}.\tag{1}$$In the example, the solution states that $E_u = \mathbb{R}^2$. How is it correct to assume that the subspace I defined is in $\mathbb{R}^2$, if the resulting vector $v$ can be complex? Or does the "realness" only pertain to the vectors that compose the resulting vector in $E_u$, because I guess it makes sense that $\mathbb{R}^2$ spans $\mathbb{C}$.
Also, assuming that the equality in $(1)$ makes sense, is $E_u = \mathbb{R}^2$ because $\left(\frac{1}{4},1\right)^T$and $\left(\frac{\sqrt{7}}{4},0\right)^T$ are linearly independent, so any linear combination will result in a vector in $\mathbb{R}^2$?
I apologise if it's a dumb question. Any help is well appreciated!
Edited: Corrected typo as pointed out by Alp Uzman. The rest remains unchanged.
Without calculating the eigenvectors, once you verify that all eigenvalues of the linearization at an equilibrium point have positive real part you are done; by definition all directions are expanded under the dynamics.
To justify this algebraically, since the eigenvalues have complex parts the eigenvectors have complex parts as well. (I believe there is a minor typo somewhere; as $\lambda_2=\overline{\lambda_1}$, the (complex) eigenvectors associated to $\lambda_2$ ought to be the complex conjugate of the eigenvectors associated to $\lambda_1$). Thus to accommodate for complex numbers in equations one first doubles the dimension (in your first expression for $E_u$ you are defining a $2$ complex dimensional space, i.e. a $4$ real dimensional space); but then by exactly what you did one can separate the real and imaginary parts of the eigenvectors and restrict to real scalars to obtain the real $E_u$. You may think of your first expression as the complex unstable $E_{u,\mathbb{C}}$ and the second expression as the real unstable $E_u$ (which is a two dimensional slice out of the four real dimensional $E_{u,\mathbb{C}}$).