Finding the Jordan form for a $4\times4$ matrix $$A:= \begin{bmatrix}4 & -4 & -11 & 11\\3 & -12 & -42 & 42\\ -2 & 12 &  37 & -34 \\ -1 & 7 & 20 & -17  \end{bmatrix}$$
I'm struggling with this matrix: it has $p_A(x) = (x-3)^4 $, yet $\ker(A - 3I)^3 $ is already the whole space $ \mathbb C^4 $. I read on another post that (for the $ \mathbb C^3 $ case of a similar matrix) I should take this to be the size of the biggest Jordan block, thus leaving me with $1$ standard eigenvector and a Jordan chain (image of the image of $\ldots$ ) of a vector $v$ not in $\ker(A - 3I)^2$. But this doesn't work for any $v$. 
What steps should I follow to find the Jordan form of $A$?

It maybe of interest to have the explicit decomposition 
$A=PJP^{-1}$ with
$$J=\left(\begin{array}{rrr|r}3  &   1   &  0   &  0\\
     0   &  3  &   1  &   0\\
     0   &  0  &   3  &  0\\ \hline
     0   &  0  &   0  &   3\end{array}\right) \ \ \text{and} \ \ P=\begin{pmatrix} 
     \ \ 3   &  3  &  14  &   0\\
     \ \ 9   &  0   &  0   &  0\\
    -6   &  6  &   1  &   1\\
    -3   &  6  &   0   &  1\end{pmatrix}.$$
Let $B=A-3I_4$. We have $\ker(B) \subsetneqq \ker(B^2) \subsetneqq
\ker(B^3)$ with:
$\ker(B)=\langle \begin{pmatrix} 
     0\\0\\1\\1\end{pmatrix},\begin{pmatrix}
     2\\6\\-1\\1\end{pmatrix}\rangle, \ \ \ker(B^2)=\langle \begin{pmatrix} 
     0\\0\\1\\1\end{pmatrix},\begin{pmatrix}
     2\\6\\-1\\1\end{pmatrix},\begin{pmatrix}
     1\\0\\0\\0\end{pmatrix}\rangle, \ \ \ker(B^3)=\mathbb{R}^4.$
