<s>[INST] 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? 
 [/INST] Usually the Jordan normal form is as follow
$$J=\begin{pmatrix}
    2 & 1 & 0 \\
    0 & 2 & 1 \\
    0 & 0 & 2 \\
    \end{pmatrix}$$
By Jordan theorem we know that a matrix $P$ exists such that $$P^{-1}AP=J$$
let $$P=[v_1,v_2,v_3,v_4]$$
then P has to satisfy the following system: $$AP=PJ$$ that is in this case $$Av_1=2v_1\implies (A-2I)v_1=0$$ $$Av_2=v_1+2v_2\implies (A-2I)v_2=v_1$$ $$Av_3=v_2+2v_3\implies (A-2I)v_3=v_2$$
Once we have $v_1$ we can find $v_2$ and finally $v_3$, that is
$$(A-2I)v_1=0 \implies\begin{pmatrix}
    -2 & -1 & -2 \\
    1 & 1 & 1 \\
    1 & 0 & 1 \\
    \end{pmatrix}v_1=0 \implies v_1=(1,0,-1)$$
$$(A-2I)v_2=v_1 \implies\begin{pmatrix}
    -2 & -1 & -2 \\
    1 & 1 & 1 \\
    1 & 0 & 1 \\
    \end{pmatrix}v_2=v_1 \implies v_2=(-1,1,0)$$
$$(A-2I)v_3=v_2 \implies\begin{pmatrix}
    -2 & -1 & -2 \\
    1 & 1 & 1 \\
    1 & 0 & 1 \\
    \end{pmatrix}v_3=v_2 \implies v_3=(0,1,0)$$
</s>