Given the Jordan form $J$, find matrix $P$ In a question set in my linear algebra course, I'm asked the following: 

Find $P$ such that $P^{-1}AP=J$, where $$A = \begin{pmatrix}6&5&-2&-3\\ -3&-1&3&3\\ 2&1&-2&-3\\-1&1&5&5\end{pmatrix}$$ and $$J = \begin{pmatrix}2&1&0&0\\ 0&2&0&0\\ 0&0&2&1\\0&0&0&2\end{pmatrix}$$ 

I can see that there's only one eigenvalue and two eigenvectors associated to it, but I don't have a clue as to how to proceed. While searching online, I didn't find any reasonable way to find $P$. Please give me a guideline as to how I should proceed. Thanks in advance.
 A: You want $P^{-1}AP=J$, which is equivalent to $P^{-1}(A-2I)P=J-2I$ or 
$$(A-2I)P=P(J-2I)$$
(assuming that $P$ is regular.)
We have
$$J-2I=
\begin{pmatrix}
  0 & 1 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 1 \\
  0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
Let us denote the columns of $P$ by $\vec p_1, \dots, \vec p_4$. Then you get
\begin{align*}
(A-2I)P&=PJ\\
(A-2I)\begin{pmatrix}\vec p_1&\vec p_2&\vec p_3&\vec p_4\end{pmatrix}&= \begin{pmatrix}\vec p_1&\vec p_2&\vec p_3&\vec p_4\end{pmatrix} J\\
\begin{pmatrix}(A-2I)\vec p_1&(A-2I)\vec p_2&(A-2I)\vec p_3&(A-2I)\vec p_4\end{pmatrix}&= \begin{pmatrix} \vec0 &\vec p_1&\vec 0&\vec p_3\end{pmatrix} 
\end{align*}
Let us have a look at $\vec p_1$ and $\vec p_2$. (The situation for $\vec p_3$ and $\vec p_4$ is similar.) We got
\begin{align*}
(A-2I)\vec p_1&=\vec p_2\\
(A-2I)\vec p_2&=\vec 0
\end{align*}
From this we see that $\vec p_2$ is an eigenvector. Moreover, we get
$$(A-2I)^2\vec p_1=\vec0.$$
Exactly in the same way we get $(A-2I)\vec p_3=\vec p_4$ and $(A-2I)^2\vec p_3=\vec0$.
You can notice that $(J-2I)^2=0$ and, consequently, $(A-2I)^2=0$. This means that $(A-2I)^2\vec x=\vec 0$ is true for any vector. However, for $\vec p_1$ and $\vec p_3$ we want to choose such vectors which are linearly independent and, moreover $(A-2I)\vec p_1$ and $(A-2I)\vec p_3$. (Since to get invertible matrix we need $\vec p_2$ and $\vec p_4$ to be linearly independent.)
It is certainly not that difficult to find two vectors such that $(A-2I)\vec p_1$ and $(A-2I)\vec p_3$ are linearly independent. We can then find the two remaining columns using $\vec p_2=(A-2I)\vec p_1$ and $\vec p_4=(A-2I)\vec p_3$.
