Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)

$A=\begin{pmatrix} 0 & -1 & 0\\  4 &  4 & 0\\  2 &  1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenvalue is $\lambda=2$.
Calculate the eigenvectors now:
$\begin{pmatrix}
-2 & -1 & 0\\  4  &  2 & 0\\  2  &  1 & 0 \end{pmatrix} \begin{pmatrix} x\\  y\\  z \end{pmatrix}= \begin{pmatrix} 0\\  0\\  0
\end{pmatrix}$
We get the equations:
$I: -2x-y=0 \Leftrightarrow y = -2x$
$II: 4x+2y=0$
$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$

We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.
$x\begin{pmatrix}
1\\ 
-2\\ 
z
\end{pmatrix}$ and this is the eigenspace...?
And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix}
1\\ 
-2\\ 
3
\end{pmatrix}$?
Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...

I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.

You must solve $(A - 2I)\vec{x} = \vec{0}$, i.e.
$$
\begin{pmatrix}
-2 & -1 & 0\\  4 &  2 & 0\\  2 &  1 & 0
\end{pmatrix}
\begin{pmatrix}x\\ y\\  z\end{pmatrix}
=\begin{pmatrix}0\\ 0\\  0\end{pmatrix}
$$
Note that all equations are multiple of each other, so let's leave the last one to use and eliminate all others. You get the constraint $2x+y = 0$. So Your solutions will have to have $y = -2x$, in other words,
$$
\begin{pmatrix}x\\ y\\  z\end{pmatrix}
 = \begin{pmatrix}x\\ -2x\\  z\end{pmatrix}
 = x \begin{pmatrix}1\\ -2\\  0\end{pmatrix}
 + z \begin{pmatrix}0\\ 0\\  1\end{pmatrix}
$$
and so you see that there are 2 eigenvectors that can be used as a basis.
As for the number of independent eigenvectors in the basis not equal the multiplicity of the root of the characteristic equation, this has to do with the fact that for your matrix, the geometric and algebraic multiplicities don't match for the eigenvalue $\lambda = 2$. You can read more about it here.
