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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 $A$.

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    What does $A^3=0$ tell you about the eigenvalues?2012-07-23
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    Try to compute $A^3$ for different matrices in Jordan normal form, then you can get a feeling what $A^3=0$ means.2012-07-23
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    That the only eigenvalue is 0.2012-07-23
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    Thanks for the suggestions. I guess I don't really understand how you would compute $A^{3}$ for different matrices in Jordan normal form.2012-07-23
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    Now that you have the eigenvalues what can you say about the size of the Jordan blocks?2012-07-23
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    For example compute $\begin{pmatrix}0&1\\&0&1\\&&0&1\\&&&0&1\\&&&&0\end{pmatrix}^3$2012-07-23
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    Ok. I'll go out on a limb here, the size is 0? If it is, then I don't really know what it's supposed to look like. Maybe I am blocked tonight or something.2012-07-23
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    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 $A^3 v_1$?2012-07-23
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    You can see how power of Jordan form looks like at [Wikipedia](http://en.wikipedia.org/wiki/Jordan_normal_form#Powers).2012-07-23

2 Answers 2

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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 $n\geq m$.
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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.