I am trying to find a matrix $P$ such that $P^{-1}AP$ is the Jordan canonical form. But I am getting confused. I found the characteristic polynomial to be $(x-1)(x-2)^5$ and minimum polynomial to be $(x-1)(x-2)^3$. The eigenvectors that are related to 2 are $v_1=(-4,0,2,2,2,0),v_2=(9,1,-5,-4,0,2)$ and that is related to 1 is $v_3=(36,0,-18,-18,-9,4)$.
Now I believe Jordan form should have five 2s and one 1 on the diagonal and two 1s on the lower diagonal.
I am kind of confused how to get the columns of $P$. I think the last column should be the eigenvector of 1. I thought I had to solve $(A-2I)x_1=v_1$ then find a vector x_1. Then solve $(A-2I)x_2=x_1$ and find vector x_2 until I get a nonlinear vector and then do the same thing for $v_2$. But It doesn't seem to work. What am I doing wrong?
Done carefully, your method should work. The denominators come out a little large. I prefer this: columns I wil cal $u,v,w,x,y,z.$ $z$ is the far right column, solves $(A-2I)^3 z = 0$ but $(A-2I)^2 z \neq 0.$ I picked $z = (0,1,0,0,0,0)^T.$ Next $y = (A-2I)z$ and $x = (A-2I)y$ is automatically an eigenvector. Next, $w $ independent from $y$ with $(A-2I)^2w = 0$ but $(A-2I)w \neq 0.$ I chose $w = (0,0,0,1,0,0)^T.$ Then $x = (A-2I)w$ is another eigenvector. Finally, the far left column is $u,$ your eigenvector for $1.$ Put columns $u,v,w,x,y,z$ as $P,$ the determinant comes out $8,$ so $P^{-1}$ is $1/8$ times an integer matrix.