The question I'm going through lists the eigenvector for $\lambda = \frac3 2 \pm \frac {3\sqrt{3}}{2}i$ as \begin{pmatrix}1-\lambda\\0\\-1 \end{pmatrix}
It then says that the 2d unstable eigenspace is
$$\begin{Bmatrix}\begin{pmatrix}1\\0\\0 \end{pmatrix},\begin{pmatrix}0\\0\\1 \end{pmatrix}\end{Bmatrix}$$
I understand that the real part of $\lambda$ is positive (so that's why it's the unstable eigenspace) but I can't seem to get the same answer as what's listed. Any help would be great thank you.
Edit: The transformation is $$\begin{pmatrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_3 \end{pmatrix}=\begin{pmatrix}2 &0&-7\\0&0&0\\1&0&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$
Let $A=\begin{bmatrix} 2 & 0 & -7\\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$, $\det{(A-\lambda I)}=-\lambda(\lambda^2-3\lambda +9)$.
The eigenvalues of $A$ are: $\lambda_1=0$, $\lambda_2=\frac{3+3\sqrt{3}i}{2}$, $\lambda_3 = \frac{3-3\sqrt{3}i}{2}$. Since $\text{Re}(\lambda_2)>0$ and $\text{Re}(\lambda_3)>0$, the solution of this differential equation is $\bf{}\text{unstable spiral}$.
Now, find the eigenvector w.r.t eigenvalue $\lambda_2$ as an example. Consider the equation $(A-\lambda_2I)v=0$, and solve $v$, that is,
$\begin{bmatrix} 2-\lambda_2 & 0 & -7 \\ 0 & -\lambda_2 & 0 \\ 1 & 0 & 1-\lambda_2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$.
We can easily find that $v_2 = 0$, and if $v_3 = k$, then $v_1=(\frac{1-3\sqrt{3}i}{2})k$. So, write $v=\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = k \left( \begin{bmatrix} \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} \frac{-3\sqrt{3}}{2} \\ 0 \\ 0 \end{bmatrix}i \right)$.
You can follow the same argument to find the eigenvector w.r.t eigenvalue $\lambda_3$.
For the real part of eigenvectors, the simplest component to construct eigenvectors in this case is $\{ [1,0,0]^T, [0,0,1]^T\}$. (They are linear combination of $\{ [1,0,0]^T, [0,0,1]^T\}$.)