Find a basis for the eigenspace of a complex eigenvalue $$(\lambda I - A)\vec x=\vec 0$$
This is the $(\lambda I - A)$:
$$\begin{bmatrix}\lambda+3 & -1 &1\\7 & \lambda-5 & 1\\-6 & -6 & \lambda+2 \end{bmatrix}$$
The characteristic polynomial is:
$$\lambda^3-64$$
My eigenvalues are: $$\lambda_1=4,\\ \lambda_2=-2+\sqrt 3 i,\\\lambda_3=-2-\sqrt 3 i$$ I solved for the real eigenvalue by finding its eigenvector $(0,1,1)^t$,but when I try to do the same with the complex eigenvalues I get that the eigenvector has to be $(0,0,0)^t$ (using a  matrix calculator ) which is false according to wolfram because the correct eigenvectors are:
$$\begin{pmatrix}\pm\frac{i}{2\sqrt3} \\ \pm\frac{i}{2\sqrt3} \\ 1\end{pmatrix}$$
 A: First note that the correct eigenvalues are:
$$
\lambda_1=4 \qquad \lambda_2=-2-2i\sqrt{3}\qquad \lambda_3=-2+2i\sqrt{3}
$$
Now , to find the  corresponding eigenvector, substitute the eigenvalue in $(\lambda I -A)x=0$ and, solving the homogeneous linear system, find the corresponding eigenspace; any vector in this eigenspace is an eigenvector.
E.g., for $\lambda_3$ we have the system:
$$
\begin{cases}
(1+2i\sqrt{3})x-y+z=0\\
7x+(2i\sqrt{3}-7)y+z=0\\
-6x-6y+2i\sqrt{3}z=0
\end{cases}
$$ 
adding the first two equation we find $x=y$ and substituting:
$$
-12x+2i\sqrt{3}z=0 \Rightarrow x=\frac{2i\sqrt{3}z}{12}=\frac{iz}{2\sqrt{3}}
$$
so the eigenspace is formed by vectors of the form 
$$
\begin{bmatrix}
\frac{iz}{2\sqrt{3}}\\\frac{iz}{2\sqrt{3}}\\z
\end{bmatrix}
$$
and one possible eigenvector is
$$
\begin{bmatrix}
\frac{i}{2\sqrt{3}}\\\frac{i}{2\sqrt{3}}\\1
\end{bmatrix}
$$
You can do the same for the other eigenvalue.
A: Substitute for example$\;\lambda=-2+2\sqrt3\,i\;$ in $\;\det(\lambda I-A)\;$ to obtain the homogeneous system (remember: the system is at most of rank two!):
$$\begin{cases}&(1+2\sqrt3\,i)x-&y&+&z=0\\{}\\&7x-&(7-2\sqrt3\,i)y&+&z=0&\end{cases}\implies(-6+2\sqrt3\,i)x+(6-2\sqrt3\,i)y=0$$$${}$$
$$\implies x=y\implies(1+2\sqrt3\,i)x-x+z=0\implies z=-2\sqrt3\,i$$
and an eigenvector for this eigenvalue is for example
$$\begin{pmatrix}1\\1\\-2\sqrt3\,i\end{pmatrix}$$
