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:
- Find $\det|A-\lambda{I}|$ where $A$ is the given matrix. It gives me some eigenvalues $\lambda_i.$
- For each eigenvalue I find eigenvectors.
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:
- 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 $3$.)
- 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)?
- 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 $\lambda_1.$
- Am I right with the formula for finding the canonical form I gave at item 3?