Find Jordan canonical form and basis of a linear operator. Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear operator such that: $T(x,y,z)=(-y-2z,x+3y+z,x+3z)$, I need to find a Jordan canonical form and a basis. 
This is what i did:
In the first place, I found the associated matrix to this linear operator in the canonical basis which is this one: 
$$A=\begin{pmatrix}
    0 & -1 & -2 \\
    1 & 3 & 1 \\
    1 & 0 & 3 \\
    \end{pmatrix}$$
After that i found the characteristic polynomial which is: $(\lambda-2)^3=0$ so we have this polynomial that has only one root with multiplicity 3.
After finding the eigenvalue I found the eigenvector associated to 2, which is $V_3=(-1,0,1)$.
Now the Jordan canonical form should be this one(If I have done it correctly):
$$J=\begin{pmatrix}
    2 & 0 & 0 \\
    1 & 2 & 0 \\
    0 & 1 & 2 \\
    \end{pmatrix}$$
We know the a Jordan basis is formed with these vectors: $B=v_1,v_2,v_3$
First I found $v_2$ :
$A.v_2=2.v_2+1.v_3$, which is equal to : 
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
 x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$So $v_2=(-1,-1,1)$
The problem is here with the last vector: After doing the same operation I get to this point:
$A.v_1=2.v_1+1.v_2$, which is equal to:
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
 x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$ and clearly this system does not have solution... what am I doing wrong? 

Well, $(A - 2 I )^3 = 0,$ but $(A - 2 I )^2 \neq 0.$ So the minimal polynomial and the characteristic polynomial agree, meaning each eigenvalue occurs in just a single block. Indeed
$$
(A - 2 I )^2 =
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 0 & 0 \\
-1 & -1 & -1
\end{array}
\right)
$$
Now find some $u$ such that $(A - 2 I )^2 u \neq 0.$ Since you want the extra $1$s below the main diagonal, we will put $u$ as the left hand column of $P,$ where we are solving $P^{-1} A P = J$ is the Jordan form you want.
I will take 
$$
u =
\left(
\begin{array}{c}
1 \\
0  \\
0 
\end{array}
\right)
$$
The middle column will be $v = (A - 2 I) u,$ or
$$
v =
\left(
\begin{array}{c}
-2 \\
1  \\
1 
\end{array}
\right)
$$
and finally  $w = (A - 2 I) v =(A - 2 I)^2 u ,$ so that  $ (A - 2 I) w =(A - 2 I)^3 u  = 0 u = 0,$ so that $w$ is a genuine eigenvector
$$
w =
\left(
\begin{array}{c}
1 \\
0  \\
-1 
\end{array}
\right)
$$
The columns of $P$ will be $u,v,w$ so
$$
P =
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
$$
next
$$
P^{-1} =
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
$$
With
$$
A =
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
$$
we get to
$$
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right) = 
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)  
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right) =
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
$$
If you would like to get the $1$s above the main diagonal instead, start over with the columns of $P$ in order $w,v,u,$ then calculate the new $P^{-1}$ Indeed, the fact that a single Jordan block is similar to its transpose gives a cheap proof that any matrix is similar to its transpose.
