Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Compute the Jordan Canonical form of $\begin{bmatrix}0 & 1 & 0 & 0\\0 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$

I don't know what to do after computing the characteristic polynomial, which i got $x^4$=0. This would give us an eigenvalue of x=0 of multiplicity 4.

Any thoughts/comments on where to go after this? Thanks so much!

share|cite|improve this question
Huh? Isn't the matrix already a Jordan form? – user1551 Nov 13 '13 at 9:07
up vote 1 down vote accepted

Next you need to find the minimal polynomial. It is one of $x^2,x^3,x^4$. Use direct verification- here clearly matrix multiplied by itself gives zero, hence the minimal polynomial is $x^2$.Therefore largest Jordan block $J(0,l)$ is of size 2 (that occurs in JNF ) and 4 is the sum of the sizes of these Jordan blocks.

Also the dimension of the eigenspace for an eigenvalue equals the number of Jordan blocks with that eigenvalue. Here there are therefore two Jordan blocks corresponding to the eigenvalue zero. So basically the matrix is already in Jordan Normal Form.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.