This is the general concept:
Let $A$ be a Matrix with Eigenvalues $\lambda_1\dots,\lambda_n$ for an Eigenvalue $\lambda := \lambda_j$ define $G = A-\lambda \mathbb{I}$ Then $KerG\subset KerG^{2}\subset \dots \subset KerG^{d-1} = KerG^d$ (For some smallest $d$ the kernal doesn't change)
(1) Calculate the basis $B^{i}$ of $G^{i}$ starting from $i= 1$ upto $i =d$ such that $B^{i+1}$ always extends $B^{i}$ (The superscripts are just notation, the only place where I mean an actual power are for the $G$'s)
(1.1) To expand on the extending: Calculate $B^{i+1}$ and if necessary rewrite $B^{i+1}$ sucht that all the basisvectors in $B^{i+1}$ are exactly those in $B^i$ plus some new ones. (making the vectors match exactly is sometimes left out, as it is obvious which vectors extend but it is still done if only in the back of the head)
(2) Call the vectors that extend $B^{d}$: $v^d_{1},\dots,v^d_{s_d}$. And set $W_d = span\{v^d_{1},\dots,v^d_{s_d}\}$ (Those vectors are the $v_1$ you where referring to, i think, and no it does not matter which ones you choose they only have to expand $B^d$ in the correct way)
Observe that $G (W_d) := span\{G v^d_{1},\dots,G v^d_{s_d}\}$ are linear independent and $\not\in Ker G^{d-2}$
(4) In $Ker G^{d-1}$ choose all basisvectors $v^{d-1}_{1},\dots,v^{d-1}_{s_{d-1}}$ that are not in $Ker G^{d-2} \oplus G (W_d)$ and call the span of them $W_{d-1}$
Do this until you get to $Ker G$:
(5) Choose basisvectors of $Ker G$ that are not in $G(W_2)$: $W_1 := span\{v^1_{1},\dots,v^1_{s_1}\}$
(6) Now calculate the column vectors of the transformation matrix B: For each $W_i$, $i\in \{1,\dots,d\}$: you get $s_i * i$ vectors: $v^{i}_{1},\dots ,v^{i}_{s_i}, G v^{i}_{1},\dots ,G v^{i}_{s_i}, \dots \dots, G^{i-1} v^{i}_{1},\dots ,G^{i-1} v^{i}_{s_i}$
(7) Repeat (1-6) for all Eigenvalues and finally $B$ is the matrix of all the column vectors calculated in (6's)
Here is the way to go: consider the sequence of kernels:
$$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$
The sequence stops after step $2$ since
$$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix}
0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.
$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say
$$e_3=(0,1,0). $$
Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say
$$e'_1=(1,2,0).$$
The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the linear map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form:
$$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$
Best Answer
You need a generalized eigenvector for the third eigenvalue and it looks like that went wrong somehow, so lets fix it.
We have $\left(A - \lambda_2 I\right)v_3 = v_2$
From this, we get:
$\begin{pmatrix}3&2&1&1\\-1&1&1&-4\\4&1&0&5\end{pmatrix}$
The RREF yields:
$\displaystyle \begin{pmatrix}1&0&-\frac{1}{5}&\frac{9}{5}\\0&1&\frac{4}{5}&-\frac{11}{5}\\0&0&0&0\end{pmatrix}$
This gives us a generalized eigenvector of: $\displaystyle \left(\frac{9}{5}, -\frac{11}{5}, 0 \right)$.
To write the Jordan Normal Form, we form:
$\displaystyle A = S\cdot J\cdot S^{-1} = \begin{pmatrix} 1 & \frac{9}{5} & 1\\ -4 & -\frac{11}{5} & 0 \\ 5 & 0 & 1\end{pmatrix} \cdot \begin{pmatrix} -1 & 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} -\frac{11}{80} & -\frac{9}{80} & \frac{11}{80} \\ \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{11}{16} & \frac{9}{16} & \frac{5}{16}\end{pmatrix}$.
Notice the structure of the Jordan block. Also, notice what the columns of $S$ and $J$ are made of? Clear?