$$\begin{pmatrix}1&0&-4&4\\ \:\:\:\:0&2&0&0\\ \:\:\:\:0&1&1&0\\ \:\:\:\:0&1&0&1\end{pmatrix}$$
I need to find a matrix $P$ where $PAP^{-1}=D$ where $D=\begin{pmatrix}X&0\\ 0&Y\end{pmatrix}$ is a block matrix of blocks $X,Y$ are blocks.
My solution:
I found the minimal polynomial which is $m\left(x\right)=\left(x-1\right)^2\left(x-2\right)$ and as I know, the matrix $A$ in base of the eugenvectors should give me the D I want and then the $P$ is the matrix of the eugenvectors, but I didn't got this result:
$$\begin{pmatrix}1&0&0&0\\ 0&-1&1&0\\ 0&-1&0&1\\ 0&1&0&0\end{pmatrix}\begin{pmatrix}1&0&-4&4\\ 0&2&0&0\\ 0&1&1&0\\ 0&1&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\ 0&0&0&1\\ 0&1&0&1\\ 0&0&1&1\end{pmatrix}=\begin{pmatrix}1&-4&4&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&2\end{pmatrix}$$
in this case $$P=\begin{pmatrix}1&0&0&0\\ 0&0&0&1\\ 0&1&0&1\\ 0&0&1&1\end{pmatrix}$$
because: $$Ker\left(A-2I\right)=sp\left\{\left(0,1,1,1\right)\right\}$$
and $$Ker\left(\left(A-I\right)^2\right)=sp\left\{\left(1,0,0,0\right),\:\left(0,0,1,0\right),\:0,0,0,1\right\}$$
but $D$ isn't as needed, and I don't see what did I wrong in this solution path.