# Generalized eigenvectors for Jordan canonical form (and theory)

I am learnnig how to find a Jordan normal form of a matrix, and I got stuck on four items. Can you please explain to me and check if I am right with my understanding? (I do not have a real example on hand, so it is pure theory.)

So here are the steps I usually take:

1. Find $\det|A-\lambda{I}|$ where $A$ is the given matrix. It gives me some eigenvalues $\lambda_i.$
2. For each eigenvalue I find eigenvectors.
3. As far as I understand, to find the Jordan normal form, I have to use the formula $B=P^{-1}\cdot A\cdot P$ where

$B$ is the matrix we are looking for,

$P$ is a matrix that consists of all the eigenvectors we've found,

$P^{-1}$ is an inverse matrix to $P,$ and

$A$ is our original matrix.

Now let's imagine this case:

A $4 \times 4$ matrix is given.

We have its characteristic polynomial $(\lambda -3)^3(\lambda-2).$

Therefore $\lambda_1=3$ and $\lambda_2 = 2.$

Let's now imagine that for $\lambda_1,$ we've found only two eigenvectors.

My questions are:

1. Do I understand correctly that the number of eigenvectors for an eigenvalue should match the power of its eigenvalue? (Like for $\lambda_1,$ we found two eigenvectors, and one more is missing because $\lambda_1$ has a power equal to $3$.)
2. If the number of eigenvectors for an eigenvalue(s) does not match its power, we have to find so-called "generalized eigenvectors" for these value(s)?
3. Do I understand correctly that to find one more generalized eigenvector for my case, I have to solve the system of equations $(A-3I)v_3 = v_2$ where $3$ is the eigenvalue we are finding eigenvectors for, $v_3$ is the third eigenvector because the two previous are known, and $v_2$ gives the coordinates of the second eigenvector for eigenvalue $\lambda_1.$
4. Am I right with the formula for finding the canonical form I gave at item 3?
• Take a look here en.wikipedia.org/wiki/Generalized_eigenvector#Example_2 and here en.wikipedia.org/wiki/Jordan_normal_form – Raffaele Sep 17 '17 at 13:59
• @Raffael, I did, that's why I created the post, to check if I am right or not – M.Mass Sep 17 '17 at 14:19
• @Raffaele M.Mass's case includes a two-dimensional eigenspace, so it is not as easy to find the generalized eigenvectors as in the Wikipedia Example 2, for which the eigenspaces are one-dimensional. The method in item 3 of my answer works for a two-dimensional eigenspace if only one additional generalized eigenvector is needed. If more are needed, see Example 3 in the same Wikipedia article. – Maurice P Aug 4 '18 at 20:07

3. A general method for your case is to solve (A – 3 I)v$_3$ = av$_1$ + bv$_2$ for v$_3$ and scalars a and b simultaneously. (Your method works if a = 0, and b = 1.) Then v$_3$ is a second-order generalized eigenvector, and av$_1$ + bv$_2$ is the (first-order generalized) eigenvector in the chain.