Is this matrix diagonalisable or not? We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).
We have matrix $A=\begin{pmatrix}
3  & -1 &  0\\ 
2  &  0 &  0\\ 
-2 &  2 & -1
\end{pmatrix}$
The eigenvalues are $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=-1$
We will only concentrate on $\lambda_{3}=-1$ because I have checked the others already and I'm very sure they are fine (they show that the matrix is diagonalisable).
We insert $\lambda_{3}=-1$ here $\begin{pmatrix}
3-\lambda  & -1        &  0\\ 
2          &  -\lambda &  0\\ 
-2         &  2        & -1-\lambda
\end{pmatrix} \begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}$
Then we get $\begin{pmatrix}
 4  & -1 & 0\\ 
 2  &  1 & 0\\ 
-2  &  2 & 0
\end{pmatrix} \begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}$
$I: 4x-y=0 \Leftrightarrow y=4x$
$II: 2x+y=0$
$III: -2x+2y=0$
We see that there is no value for $z$, so we can choose an arbitrary $z$. We get the eigenspace $E_{A}(-1)=\left\{ \left. \begin{pmatrix}
x\\ 
4x\\ 
z
\end{pmatrix} \right | x,z \in \mathbb{R} \right\}$
Two variables mean the basis is made up by $2$ vectors and this means the dimension of the eigenspace is $2$ and this means the matrix is not diagonalisable because $\lambda_{3}=-1$ is just a single eigenvalue?

Can you please tell me if this is correct? 

From I and II, we have $y = 4x$ and $y = -2x$.  The only way these can simultaneously hold is if $x = y = 0$.  However, $z$ is indeed a free variable.  We find that
$$
E_A(-1) = \{(0,0,z)^T : z \in \Bbb R\}
$$
