Finding a Jordan canonical form $J$ and an invertible matrix $Q$ such that $J = Q^{-1}AQ$ $$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
The characteristic polynomial can be found to be:  $p(t)= -(t-1)(t-2)^2$.
I found $E_{\lambda_1} = N(A - I) = (1, 2, 1)$.
I'm having problems finding $E_{\lambda_2}$:
$$
(A - 2I) = \begin{bmatrix}
-2 & 1 & -1 \\
-4 & 2 & -2 \\
-2 & 1 & -1
\end{bmatrix}
$$
and
$$
(A - 2I)^2 = \begin{bmatrix}
2 & -1 & 1 \\
4 & -2 & 2 \\
2 & -1 & 1
\end{bmatrix}
$$
But there are numerous $\beta$'s that can be found for $(A - 2I)$, such as $\beta = (1, 1, -1)$ or $\beta = (1, 0, -2)$, etc., etc. 
Then for $(A - 2I)^2$, these same $\beta$'s can be used, so I already know that I have messed up somewhere. Can anyone offer some suggestions?
 A: We asked to find the Jordan Normal Form of the matrix:
$$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
We start out by solving $|A - \lambda I| = 0$ to find the eigenvalues, which gives:
$$-(\lambda - 2)^2(\lambda-1) = 0 \rightarrow \lambda_1 = 1,~ \lambda_2 = 2, ~\lambda_3 = 2$$
We have one eigenvalue that has algebraic multiplicity one and a second one that has multiplicity two.
Next, we want to find the three linearly independent eigenvectors $[A - \lambda_i I]v_i = 0$.
For $\lambda_1 = 1$, we get:
$$\begin{bmatrix} -1 & 1 & -1 \\ -4 & 3 & -2 \\ -2 & 1 & 0 \end{bmatrix}v_1 = 0$$
The RREF yields:
$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}v_1 = 0$$
This leads to the eigenvector $v_1 = (1,2,1)$.
For $\lambda_{2,3} = 2$, we get:
$$\begin{bmatrix} -2 & 1 & -1 \\ -4 & 2 & -2 \\ -2 & 1 & 1 \end{bmatrix}v_{2,3} = 0$$
The RREF yields:
$$\begin{bmatrix} 1 &-\dfrac{1}{2} & \dfrac{1}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}v_{2,3} = 0$$
This fortunately leads to two independent eigenvectors (not always the case):


*

*$v_2 = (-1,0,2)$

*$v_3 = (1,2,0)$


We are now able to find the diagonal matrix as:
$$J = Q^{-1} \cdot A \cdot Q$$
$Q$ is made up of the column eigenvectors as:
$$Q = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$$
This should yield:
$$J = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Now, you can also do this the way you were trying and should get an identical result (can be different as eigenvectors are not unique).
