Jordan Canonical form of $T(f(x))=f'''(x)+2f(x)$ For each linear operator T, find a Jordan canonical form $J$ of $T$ and a canonical basis $\beta $ for $T$
$T$ is the linear operatonr acting on $P_2( \mathbb{R})$

I guess I could throw in random polynomials until one gets the generelized eigenvectors magically. I am sure some here can do that no problem. uhm, this is how I have learned it and still interested in other ways as long at it is methodical.
here we go, first find the matrix representation of $T$. Basis for $P_3(\mathbb{R})= \{ 1,x,x^2,x^3\}$
$$
\begin{aligned} 
    T(1)&=0+2       &= 2+0x+0x^2+0x^3
\\  T(x)&=0+2x       &= 0+2x+0x^2+0x^3
\\  T(x^2)&=2+2x^2   &= 2+0x+2x^2+0x^3
\\  T(x^3)&=6x+2x^3 &= 0+6x+0+2x^3
\end{aligned}$$
making $$[T]_\beta = \begin{bmatrix}
         2 & 0 & 2 & 0
     \\ 0 & 2& 0& 6 
      \\ 0 & 0& 2& 0 
       \\ 0 &0& 0& 2
  \end{bmatrix} $$
It is upper triangular ofcourse so the char poly is $(\lambda-2)^4$, with e-vals $\lambda=1$ with algebreic multiplicity of 4.
we need to find e-vectors 
      $$ (T-2I)=\begin{pmatrix} 0& 0& 2& 0 
                          \\ 0& 0& 0&6 
                          \\ 0& 0& 0 &0
                           \\ 0& 0& 0& 0  \end{pmatrix}$$
$\dim(E_{\lambda_1}=2)$. there are 2 missing e-vectors that are generilized to make into jordan form aka $\dim(K_{\lambda_1})$
We need dot diagram for row 1 $$ \begin{matrix} * & * \end{matrix} $$
row2 $$\begin{aligned}
r_2 &=rank((T-\lambda_1 I )^{2-1})-rank((T-\lambda I)^2)
\\  &=rank((T-\lambda I))-rank((T-\lambda I)^2)
\\  &=2-rank \begin{pmatrix}  0& 0& 2& 0 
                      \\ 0 & 0& 0& 6 
                      \\ 0 &0& 0& 0
                     \\ 0 &0& 0& 0 \end{pmatrix}
                  \begin{pmatrix}  0& 0& 2& 0 
                      \\ 0 & 0& 0& 6 
                      \\ 0 &0& 0& 0
                     \\ 0 &0& 0& 0 \end{pmatrix}
\\&=2-0
\end{aligned}$$
So the Dot diagram for $T_1$ is $$ \begin{matrix} 
  * (T-I\lambda)v_1\ &  * (T-I\lambda)v_2
  \\*v_1 &*v_2
\end{matrix}$$  
So, picked $(1,0,-1,0)=(T-I\lambda)v_1$ as end gen e-vector with $v_1=(0,0,1/2,1/6)$ it was found by straight inverse also $(T-I\lambda)v_2=(0,1,0,-1)$ but could not find $v_2$ I dont think it exists. there is some trickery to get our $v_1,v_2$ that works but I can't duplicate it at this time. 
Do know from the dot product that the jorad form is 
$$J=\begin{pmatrix} 2 & 1& 0 &0 
             \\ 0 &2& 0 &0 
          \\ 0 &0 &2 &1 
         \\ 0 &0& 0& 2  \end{pmatrix} $$
having trouble finding $Q$ s.t $ J=Q^{-1}AQ$
 A: Well, the question is quite intriguing. Firstly, some notation, according to this wikipedia article.
$n =4:$number of columns
$\mu = 4$: algebraic multiplicity of the eigenvalue $\lambda = 2$.
$\gamma = 2:$ geometric multiplicity of the eigenvalue $\lambda = 2$. 
$m = 2:$ the smallest positive integer such that $\text{rank } (T-\lambda I)^m = n - \mu$.
$\rho_k = \text{rank }(T-\lambda I)^{k-1} - \text {rank } (T-\lambda I)^k:$ number of generalized eigenvectors of rank $k$, with $k = 1,\ldots, m$. 

We have that $\rho_1 = \rho_2 = 2,$ which means that we will have $2$ ordinary eigenvectors (of rank $1$) and 2 generalized eigenvectors of rank $2$.
Starting from the generalized eigenvectors $\mathbf{x} = [x_1 \quad x_2 \quad x_3 \quad x_4]^T$, we have that:
$$(T-\lambda I)^2\mathbf{x} = \mathbf 0\quad \text{and} \quad (T-\lambda I)\mathbf x\neq\mathbf 0. $$
Since $\text{rank } (T-\lambda I)^2 = 0 \implies (T-\lambda I)^2 = \mathbf [0]$, the only restriction we have is that $x_3,x_4$ cannot be zero at the same time. Hence, our $2$ linearly indepedent vectors can be of the form  $\mathbf u_2 = [0 \quad 0 \quad 1 \quad 0]^T$ and $\mathbf v_2 =[0 \quad 0 \quad 0\quad 1]^T.$


*

*Starting from the generalized eigenvector $\mathbf{u}_2$ we can derive the ordinary eigenvector 
$$\mathbf{u}_1 = (T-\lambda I)\mathbf{u}_2= [2 \quad 0\quad 0 \quad 0]^T.$$
Thus, we have created our first chain $\{\mathbf{u}_1,\mathbf{u}_2\}$.

*Starting from the generalized eigenvector $\mathbf{v}_2$ we can derive the ordinary eigenvector $$\mathbf{v}_1 = (T-\lambda I)\mathbf{v}_2= [0 \quad 6\quad 0 \quad 0]^T.$$
Thus, we have created our second chain $\{\mathbf{v}_1,\mathbf{v}_2\}$.
Consider the matrix $Q =[\mathbf{u}_1\quad \mathbf{u}_2\quad \mathbf{v}_1 \quad\mathbf{v}_2]$
Making the substitutions, we have:
$$J = Q^{-1}\cdot T\cdot Q.$$ 
A: I know it is snotty to answer your own question.But, It has been an hour or 2 since the question was ask . Turns out the ordering is important that is why it is called an ordered basis.
Basis for $N((t-I2)^2)$ is 
$$\left \{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
          , \begin{pmatrix} 0 \\ 1 \\ 0  \\ 0 \end{pmatrix}
          , \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} 
          , \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}  \right \} $$
Select a non e-vector $(0,0,1,0)$ as $v_1$
$$\begin{aligned}
 (T-2I)(v_1)
        &= \begin{pmatrix} 0 & 0 & 2 & 0 
             \\ 0& 0& 0& 6
              \\ 0 &0& 0& 0
               \\ 0&0&0&0 
              \end{pmatrix}
 \begin{pmatrix}0 \\ 0\\1\\0  \end{pmatrix}
=\begin{pmatrix} 2 \\ 0\\0\\0 \end{pmatrix}
   \end{aligned} $$
We will pick another that $(0,0,0,1)$
$$\begin{aligned}
 (T-2I)(v_2)
        &= \begin{pmatrix} 0 & 0 & 2 & 0 
             \\ 0& 0& 0& 6
              \\ 0 &0& 0& 0
               \\ 0&0&0&0 
              \end{pmatrix}
 \begin{pmatrix}0 \\ 0\\0\\1  \end{pmatrix}
=\begin{pmatrix} 0 \\ 6\\0\\0 \end{pmatrix}
   \end{aligned} $$
dot diagram is lookin like 
       $$\begin{matrix}
            * \begin{pmatrix} 2 \\ 0\\0\\0 \end{pmatrix} 
          & * \begin{pmatrix} 0 \\ 6\\0\\0 \end{pmatrix}
          \\  *  \begin{pmatrix}0 \\ 0\\1\\0  \end{pmatrix} 
            & *  \begin{pmatrix}0 \\ 0\\0\\1  \end{pmatrix} \end{matrix} $$
was selecting the ordered basis to be $\{ 2,6x,x^2,x^3 \}$ and keeped getting a matrix representation that was not quite jordan form but following the dot diagram closely 
consider the ordere basis of $P_3(\mathbb{R})=\{ 2,x^2,6x,x^3\}$ and finding $[T]$ like in the question but with our new basis we do get the Jordan form
