<s>[INST] Finding the Jordan form and basis for a matrix I have this matrix $A=\left[ {\begin{array}{ccc}
-3&3&-2\\
-7&6&-3\\
1&-1&2\\
\end{array} } \right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=\left[ {\begin{array}{ccc}
-5&3&-2\\
-7&4&-3\\
1&-1&0\\
\end{array} } \right]$ 
Using row reduction $\left[ {\begin{array}{ccc}
1&-1&0\\
0&1&1\\
0&0&0\\
\end{array} } \right]x=0$ gives me eigenvectors having the form $e=r\left[ {\begin{array}{c}
1\\
1\\
-1\\
\end{array} } \right]$
$v=\left[ {\begin{array}{c}
1\\
1\\
-1\\
\end{array} } \right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=\left[ {\begin{array}{ccc}
2&1&0\\
0&2&0\\
0&0&1\\
\end{array} } \right]$
Now to find the generalized eigenvector I can take an augmented matrix
$\left(\begin{array}{ccc|c}  
 5&3&-2&1\\  
 -7&4&1&1\\
1&-1&0&-1  
\end{array}\right)$
I row reduced this to the form
 $\left(\begin{array}{ccc|c}  
 1 & -1 & 0&-1\\  
 0 & 1 & 1&2\\
0&0&0&0  
\end{array}\right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=\left(\begin{array}{c}  
 0\\  
 1\\
1 
\end{array}\right)$ which seems to work, so I have a basis for the generalized eigenspace for $\lambda=2$
then I need to find a solution for $(A-I)x=0$ for $\lambda=1$.
I row reduce $A-I=\left[\begin{array}{ccc}  
 -4&3&-2\\  
 -7&5&1\\
1&-1&1  
\end{array}\right]$ to get $\left[\begin{array}{ccc}  
 1&0&-1\\  
 0&1&-2\\
0&0&0  
\end{array}\right]$ and using $(A-I)x=0$ found that all eigenvectors for $\lambda=1$ have the form $v=s\left(\begin{array}{c}  
1\\  
2\\
1 
\end{array}\right)$
So I can make a matrix $Q=\left[\begin{array}{ccc}  
 1&1&0\\  
 2&1&1\\
1&-1&1  
\end{array}\right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
 [/INST] Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$. 
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
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