Next you need to find the minimal polynomial. It is one of $x^2,x^3,x^4$. Use direct verification- here clearly matrix multiplied by itself gives zero, hence the minimal polynomial is $x^2$.Therefore largest Jordan block $J(0,l)$ is of size 2 (that occurs in JNF ) and 4 is the sum of the sizes of these Jordan blocks.
Also the dimension of the eigenspace for an eigenvalue equals the number of Jordan blocks with that eigenvalue. Here there are therefore two Jordan blocks corresponding to the eigenvalue zero. So basically the matrix is already in Jordan Normal Form.
We have:
$$A = \begin{bmatrix}1&1&1\\0&2&0 \\ 0&0&2\end{bmatrix}$$
We find the eigenvalues of $|A - \lambda I| = 0$, hence:
$$\lambda_1 = 1, \lambda_{2,3} = 2$$
That is, we have a single root and a double root eigenvalue, algebraic multiplicity.
To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$.
For $\lambda_1 = 1$, we get the eigenvector:
$$v_1 = (1,0,0)$$
For $\lambda_{2,3} = 2$, we get the eigenvectors (normally, we do not get two linearly independent eigenvectors):
$$v_2 = (1,0,1), v_3 = (1,1,0)$$
We now can write $P$ using the eigenvectors as columns. We have,
$$P = [v_1 | v_2 | v_3 ] = \begin{bmatrix}1&1&1\\0&0&1 \\ 0&1&0\end{bmatrix}$$
We can write the Jordan Normal Form (notice that we do not have any Jordan blocks), $J$, using the corresponding eigenvalues:
$$J = P^{-1} A P$$
However, we can also write this straight off from the eigenvalues and knowing we do not need any Jordan blocks.
$$J = \begin{bmatrix}1&0&0\\0&2&0 \\ 0&0&2\end{bmatrix}$$
Lastly, we should verify:
$$A = P J P^{-1}$$
I purposely left things so you can fill in the details of the calculations.
Best Answer
Hints, to find the Jordan Normal Form, we can use the following approach (and others are possible too):
Updates
To find the eigenvalues/eigenvectors, we set up and solve the characteristic polynomial $|A - \lambda I| = 0$, so we get ($v_i's$ are the eigenvectors):
Jordan Form
The JF is given by:
$$A = PJP^{-1} = \begin{bmatrix}3 & -1/2\\2 & 0\end{bmatrix} \cdot \begin{bmatrix}-1 & 1\\0 & -1\end{bmatrix} \cdot \begin{bmatrix}0 & 1/2\\-2 & 3\end{bmatrix}$$
Note what makes up $J$ and the columns of $P$.