# Jordan Form of a matrix

I'm trying to find a matrix $P$ such that $J=P^{-1}AP$, where $J$ is the Jordan Form of the matrix: $$A=\begin{pmatrix} -1&2&2\\ -3&4&3\\ 1&-1&0 \end{pmatrix}$$

The characteristic polynomial is: $p(\lambda)=(\lambda-1)^3$, and a eigenvector for $A-I$ is $\begin{pmatrix} 0 \\ 1 \\-1 \end{pmatrix}$. Now, how can I find other $2$ vectors?

-
See this example: en.wikipedia.org/wiki/Jordan_normal_form#Example –  wj32 Oct 20 '12 at 6:37

Maybe your fault is, that there is a second eigenvector, the Jordan normal form is $$\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}.$$ You can find a second eigenvector $w$ and a vector $z$ such that $(A-I)z=w$.

-

Call the eigenvector you found, $v$. Then you need $w$ such that $(A-I)w=v$, and $x$ such that $(A-I)x=w$.

EDIT: Sorry, I was taking it for granted that the eigenspace for $\lambda=1$ is 1-dimensional, but in fact it's 2-dimensional. That is, the matrix $A-I$ has nullity 2, so you can find a second linearly independent eigenvector, call it $u$. Then you have to find $w$ such that $(A-I)w$ is in the span of $u$ and $v$.

More edit: Julian noticed this 2 minutes before I did.

-
maybe I am mistaken, but my problem is that I can not find $w$ because the system has no solution. Obviously I have an error, but I can not see it. thanks –  Hiperion Oct 20 '12 at 6:52
@Hiperion: We can't tell you what your error is unless you show us your working. –  wj32 Oct 20 '12 at 7:31