# 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?

## 2 Answers

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..

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.