When is a vector subspace equal to $\mathbb{R}^2$? 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.
 A: 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 $E_{u,\mathbb{C}}$).
