Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having a tough time understanding how to find the Jordan canonical form of a $5\times5$ matrix. This is a problem from my linear algebra book. Any help would be great or suggestions on how to get started.

Let $A$ be a $5\times5$ matrix with complex entries such that $A^{3} =0$. Find all the possible Jordan Canonical forms of $A$.

share|improve this question
2  
What does $A^3=0$ tell you about the eigenvalues? –  copper.hat Jul 23 '12 at 7:22
    
Try to compute $A^3$ for different matrices in Jordan normal form, then you can get a feeling what $A^3=0$ means. –  Julian Kuelshammer Jul 23 '12 at 7:30
1  
That the only eigenvalue is 0. –  Kelly Jul 23 '12 at 7:30
    
Thanks for the suggestions. I guess I don't really understand how you would compute $A^{3}$ for different matrices in Jordan normal form. –  Kelly Jul 23 '12 at 7:35
2  
Hint: if you have an $k \times k$ Jordan block for eigenvalue $0$, it means there are $k$ vectors $v_1, \ldots, v_k$ such that $A v_1 = v_2,\ A v_2 = v_3, \ \ldots, A v_{k-1} = v_k, \ A v_k = 0$. So what is $A^3 v_1$? –  Robert Israel Jul 23 '12 at 7:45

2 Answers 2

Here is a short list of the facts you need to solve this question:

  1. $A^3=0$ if and only if the cube of its Jordan form is zero.
  2. The $n^{\rm th}$ power of a Jordan form is upper triangular and the eigenvalues in the diagonal are the $n^{\rm th}$ powers of the original eigenvalues.
  3. A power of a Jordan form is calculated by taking that power for each of its Jordan blocks.
  4. If $J$ is a Jordan block of size $m$ and eigenvalue zero, then $J^n=0$ if and only if $n\geq m$.
share|improve this answer

If $A^3=0$, then the minimal polynomial divides $t^3$, and that means that the characteristic polynomial must be $t^5$ (alternatively, if $\lambda$ is an eigenvalue of $A$, then $\lambda^3$ is an eigenvalue of $A^3$, so $\lambda^3=0$, hence $\lambda=0$).

Now remember that the highest power of $t-\lambda$ that divides the minimal polynomial gives you the size of the largest Jordan block associated to $\lambda$ in the Jordan canonical form. So, for example, if the minimal polynomial were $t^4$, that means that there must be at least one block of size $4$, and no blocks of larger size. For a $5\times 5$ matrix, this would mean that the Jordan form must consist of a single $4\times 4$ Jordan block associated to $0$, and a $1\times 1$ block associated to $0$ (since that is all that is left).

Now consider the three possibilities for the minimal polynomial and what that tells you. Enumerating the possible Jordan forms from that information is fairly straightforward.

As a way to check your work, there should be five different possible Jordan canonical forms.

share|improve this answer

Your Answer

 
discard

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.