Q:

Jordan Canonical form $3\times3$

Let $$A= \begin{pmatrix} 3& 1& -2 
                      \\ -1& 0& 5
                      \\ -1& -1& 4\end{pmatrix} $$
express it as Jordan Canonical form.

This is an example in book by Friedberg,Insel,Spence 2 on sec 7.1. Was able to follow that characteristic poly of A is 
 $$ f(t)=det(A-tI_{3x3})=-1(t-3)(t-2)^2$$
let $\lambda_1=3$ with mult 1 and $\lambda_2=2$ with mult 2. by thm $dim(K_{\lambda_1}(T))=1$ and $dim(K_{\lambda_2}(T))=2$ 
and by some proposition 
$$ \begin{aligned} 
 K_{\lambda_1}&=N(T-3I)=E_{\lambda_1}
\\K_{\lambda_2}&=N((T-2I)^2) &&[\text{the book has maybe 1 ed. older }N((T-3I)^2) ]
 \end{aligned}$$
I was able to find nullpace $N(T-3I)$ by rref it is $\{ (-1,2,1)\}=K_{\lambda_1}$.
By rref $N(T-2I)$ to be  $\{ -1,3,-1\}$
cannot easily verify the following 
 when it comes to finding $E_{\lambda_2}$

the generalized eigenspace has a basis consisting of a union of cycles,
  the basis is either union of 2 cycles each length 1 or a single cycle
  of length 2. The former case is impossible because the basis elements
  would be eigevectors- contradicting the fact that
  dim$(E_{\lambda_2})=1$ which is easily verified.

now it stated $$\beta_2=\{ (A-2I)v,v\}
=\left \{  \begin{pmatrix} 1\\-3\\1 \end{pmatrix}
                          , \begin{pmatrix}1\\2\\0   \end{pmatrix} \right \}$$
Not sure how $ (1,2,0)$ was derived. besides doing inverse  

A:

Let $B = A - 2I$.  What you want is a vector $v$ so that $v$ and $Bv$ are nonzero but $B^2v = 0$.  You know that the nullspace of $B^2$ is dimension $2$ and the nullspace of $B$ is dimension $1$, so pick $v$ to be any vector in $N(B^2) \setminus N(B)$.  The vector $(1, 2, 0)$ is one of many possible choices.

