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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?
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  • $\begingroup$ Take a look here en.wikipedia.org/wiki/Generalized_eigenvector#Example_2 and here en.wikipedia.org/wiki/Jordan_normal_form $\endgroup$ – Raffaele Sep 17 '17 at 13:59
  • $\begingroup$ @Raffael, I did, that's why I created the post, to check if I am right or not $\endgroup$ – M.Mass Sep 17 '17 at 14:19
  • $\begingroup$ @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. $\endgroup$ – Maurice P Aug 4 '18 at 20:07
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  1. No, the number of eigenvectors associated with an eigenvalue does not necessarily equal the associated exponent of the characteristic polynomial in the form you give (as in the case you mention), but the number of generalized eigenvectors does.
  2. Yes.
  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.
  4. No, you do not have to find P. You can write down B as soon as you know the length of each generalized eigenvector chain. You can use your method, but if you do so, be careful how you order the generalized eigenvectors that make up the columns of P.
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