What eigenvectors can be used to find the Jordan canonical form? I have a matrix $A$ and I want to put it into Jordan canonical form.
Let $A$ be a $3x3$ matrix and have one eigenvalue, $\lambda$ with algebraic multiplicity 3.
To create a matrix $P$ that is a basis for $\mathbb R^3$, I find the eigenvector $v_1$ for $\lambda$, and then I need to find two more generalized eigen vectors for $\lambda$ as well correct?
But from working problems it seems that I can not just use any two other generalized eigenvectors to form $P$, correct? They have to be two specific ones, that when I carry out:
$PAP^{-1}$ 
I will have $A$ in its Jordan form. 
I thought it could be any two generalized eigen vectors?
 A: Any two generalize eign vector  should work.
You can. Find them just by solving 
(A-λI)X1=X, (A-λI)X2=X1,   where X is the eigenvector .
And X1 X2 are corresponding generalize eign vectors..
A: You have to proceed backwards.
The general strategy is this:
 you have inclusions
$$0\varsubsetneq\ker(A-\lambda I)\varsubsetneq(A-\lambda I)^2\varsubsetneq\dots\varsubsetneq\ker(A-\lambda I)^k=\ker(A-\lambda I)^{k+1}=\cdots$$
(for a $3\times3$matrix, $k\le3$).
Take the maximal number of linearly independent vectors $u_i$ in $\ker(A-\lambda I)^k\smallsetminus \ker(A-\lambda I)^{k-1}$. They make up the beginning of a Jordan basis. The vectors $v_i=(A-\lambda I)u_i$ are linearly independent vectors in $\ker(A-\lambda I)^{k-1}\smallsetminus \ker(A-\lambda I)^{k-2}$. 
Complete this second set of vectors to a maximal number of linearly independent vectors in  $\ker(A-\lambda I)^{k-1}\smallsetminus \ker(A-\lambda I)^{k-2}$, then map them to a third set of vectors $w_i=(A-\lambda I)v_i$, and so on untile you arrive at the eigenspace.
