Is this matrix diagonalisable? Let $A$ be the following matrix:
$$A = \left(\begin{array}{rrr}
-1 & \hphantom{-}3 & \hphantom{-}0\\
0  & 2 & 0\\
-3 & 3 & 2
\end{array}\right).$$
I've found that the eigenvalues are -1 and 2 (multiplicity 2). However, when I try to find the eigenvectors for the eigenvalue 2, I can only find one, as the augmented matrix $A-2I$ reduces to 
$$\left(\begin{array}{rrr}
1 & -1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right).$$
So is there any other way to diagonalise this matrix, or have I made a mistake somewhere?
 A: First: not every matrix is diagonalizable; if the dimension of the eigenspace of a given eigenvalue is strictly smaller than the multiplicity of the eigenvalue, then the matrix is not diagonalizable (the condition is both necessary and sufficient for matrices whose characteristic polynomial splits: a matrix is diagonalizable if and only if the characteristic polynomial splits, and the geometric multiplicity of each eigenvalue equals the geometric multiplicity).
Second: this matrix is diagonalizable. You did everything right, but misinterpreted what $A-2I$ was telling you.
You are correct that your matrix has characteristic polynomial $\chi(t) = (2-t)^2(-1-t)$, so that $\lambda=2$ is an eigenvalue with multiplicity $2$, and $\lambda=-1$ is an eigenvalue with multiplicity $1$.
When you try to find the nullspace for $A-2I$ to find the eigenspace for $\lambda=2$, you get
$$\left(\begin{array}{rrr}
-3 & 3 & 0\\
0 & 0 & 0\\
-3 & 3 & 0
\end{array}\right) \rightarrow \left(\begin{array}{rrr}
1 & -1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right),$$
exactly what you described, so far so good.
But now you've misinterpreted what you found: this matrix has rank $1$, so the nullspace has dimension two, exactly what you need! The matrix corresponds to the system that has the unique equation $x-y = 0$. This  means that $z$ is completely free, and $x$ must equal $y$. So the solutions to this system are:
$$\begin{array}{rcl}
x & = &s\\
y & = &s\\
z & = &t
\end{array}\qquad\quad\mbox{$s$ and $t$ arbitrary.}$$
So you get two linearly independent vectors: one corresponding to $s=0$ and $t=1$, and one corresponding to $s=1$ and $t=0$. So you can find two linearly independent eigenvectors corresponding to $2$: $(0,0,1)$ and $(1,1,0)$. Multiply them by $A$ to verify that they are indeed eigenvectors of $\lambda=2$.
