Jordan normal form - Real matrices Given the matrix $A = \begin{pmatrix} 7 &1 &2 &2 \\ 1 &4 &-1 &-1 \\ -2 &1 &5 &-1 \\ 1 &1 &2 &8 \end{pmatrix}$. I found its characteristic polynomium, which is $f(x) = (x-6)^4$, and I found this eigenvectors: $(-1,-1,1,0),(-1,-1,0,1)$. My questions are:


*

*Am I right?

*How can I do to find another matrix $C$ such that $\mathcal{J} = C^{-1}AC$, where $\mathcal{J} = \begin{pmatrix} 6 &1 &0 &0 \\ 0 &6 &0 &0 \\ 0 &0 &6 &1 \\ 0 &0 &0 &6 \end{pmatrix}$


Thanks for your help.
 A: Let $B = A - 6 I$ 
$$
B=\pmatrix{  1&
 1&
 2&
 2\\
 1&
-2&
-1&
-1\\
-2&
 1&
-1&
-1\\
 1&
 1&
 2&
 2
}$$
Verify that $B^3=0$ but $B^2 \ne 0$. So it must have a jordon block of size $3$. So you will have one block of size 3 and the other of size 1, i.e of the form
$$
\pmatrix{  
 6&
 1& 0 &
 0\\
0&
6&
1&0\\
0&0&
 6&
0\\
 0&
 0&
 0&
 6
}$$
How to find the transformation matrix?
First find the eigenvectors by solving 
$$
B e = 0
$$
and you have already done it to get $e_1$ and $e_2$.
Now try and solve
$$
B e_3 = e_1$$
If this is not possible, try
$$
B e_3 = e_2$$
Only one of them is possible.
Next solve
$$
B e_4 = e_3
$$
Now the four vectors are the columns of $C$.
Note in all this it goes without saying that $e_i$ should not be zero.
Sorry, I pulled a fast one on this
In general, if you start with $e_1$ and $e_2$ you may not be able to get $e_3$. However, there is one linear combination of $e_1$ and $e_2$ that is in the column space of $B$. That is the one you want to use to get $e_3$. I.e. a vector in the intersection of the space spanned by columns of $B$ and the space spanned by $e_1$ and $e_2$.
