eigenspaces determining if diagonalisable 
I'm on (iii),
We defined Eigenspaces as follows; $$E_\lambda = \{ \underline{v} \in \mathbb{F^n} | A\underline{v} = \lambda \underline{v} \}$$
For the matrix I found the eigen values to be $0,0,2$ and solving $A\underline{v} = \lambda \underline{v}$ for $\lambda = 0$ I get 3 equations which all say the same thing:
$2x_1 + 3x_2 + x_3 = 0$, I don't really know what to do for this, but I gave it a go: 
$x_1 = -3/2 x_2 -1/2 x_3$, so $x_2,x_3$ are free to be anything, let them be $t_1,t_2$ respectively, then I stated that $E_0 =  \left \{ \underline{v} \in \mathbb{F^n} | \underline{v} = t_1 \begin{pmatrix} -\frac{3}{2} \\ 1 \\ 0 \end{pmatrix} + t_2 \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{pmatrix}, t_1t_2 \in \mathbb{R} \right \}$ and concluding with the dimension being 2. 
Is the above correct?
My second question is how can we determine from this if the matrix is diagonalisable? What am I missing?
Thank you
 A: A matrix is diagonalizable if, and only if, there is a basis of $\mathbb F^n$ in which all the elements are eigenvectors of the matrix at hand.
On the whole exercise you have $\mathbb F=\mathbb R$ and $n=3$.
Thus, each matrix will be diagonalizable if, and only if, you can find three linearly indepedent eigenvectors.
In $(iii)$ you have already found two linearly independent eigenvectors, (yes, what you did is correct). Since $n=3$, it is possible to guarantee that the matrix is diagonalizable because if you go find an eigenvector for the eigenvalue $2$ it will necessarily be linearly indepedent from $E_0$ because of this fact: eigenvectors corresponding to different eigenvalues are necessarily linearly independent.
A nitpick: $E_0$, as you wrote it, it's senseless you should choose one of this ways of writing it:
$$E_0 =  \left \{ \underline{v} \in \mathbb{R^n} \mid \exists  t_1t_2 \in \mathbb{R} \left(\underline{v} = t_1 \begin{pmatrix} -\frac{3}{2} \\ 1 \\ 0 \end{pmatrix} + t_2 \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{pmatrix}\right)\right \},$$
$$E_0 =  \left \{ t_1\begin{pmatrix} -\frac{3}{2} \\ 1 \\ 0 \end{pmatrix} + t_2 \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{pmatrix} \mid  t_1t_2 \in \mathbb{R} \right \},$$
$$\left\langle \begin{pmatrix} -\frac{3}{2} \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{pmatrix}\right\rangle\color{grey}{=\left\langle \begin{pmatrix} -3 \\ 2 \\ 0 \end{pmatrix} , \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}\right\rangle}.$$
A: An n×n matrix A over the field F is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n, which is the case if and only if there exists a basis of F^n consisting of eigenvectors of A. F here is the set of real numbers.
You have one eigenspace of dimension two and another of dimension one (check this fact for yourself). 2+1=3=n. So, your matrix is diagonalizable. The entries along the diagonal are the eigenvalues, in the order of your eigenbasis.
