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Given the characteristic polynomial of a matrix $A \in \mathbb{C}^{6x6}$ with $p(A)=(\lambda-2)^2(\lambda-1)^4$, we were supposed to determine all Jordan normal forms that have this characteristic polynomial.

I determined 10 (is this correct?) and was wondering whether this is a general way to compute the number of them for an arbitrary characteristic polynomial.

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  • $\begingroup$ If $p(A)(\lambda)=\prod_{i=1}^k(\lambda-\lambda_i)^{m_i}$, where $\lambda_i$'s are pairwise distinct, then for each $\lambda_i$, the number $n_i$ of possibilities of Jordan blocks corresponding to $\lambda_i$ is the number of solutions of $\sum_{j=1}^{m_i}j\cdot a_j=m_i$, where $a_j$'s are nonegative integers. The meaning of $a_j$ is the number of Jordan blocks corresponding to $\lambda_i$ of size $j\times j$. Then the total number of Jordan normal forms is $\prod_{i=1}^kn_i$. In your case, it is $2\times 5=10$. $\endgroup$ – 23rd Jun 7 '13 at 22:49
  • $\begingroup$ The partition function, $p(n)$, is defined to be the number of ways of writing $n$ as a sum of positive integers, order not counting. E.g., $p(4)=5$ because you have $4,3+1,2+2,2+1+1,1+1+1+1$. Much information on it is available on the web and in Number Theory texts. In the notation of @Landscape, the answer to your question is the product of the numbers $p(m_i)$. In the specific example, that's $p(2)p(4)=2\times5=10$. $\endgroup$ – Gerry Myerson Jun 8 '13 at 4:38
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Let's see: up to order of the Jordan Blocks and the eigenvalues, we have, with $\,m_A(x)=$ the matrix's minimal polynomial:

$$m_A(x)=(x-1)(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2\end{pmatrix}$$

$$m_A(x)=(x-1)(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&1\\ 0&0&0&0&0&2\end{pmatrix}$$

$$(1)\;\;\;m_A(x)=(x-1)^2(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2\end{pmatrix}$$

$$(2)\;\;\;m_A(x)=(x-1)^2(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&1&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2\end{pmatrix}$$

$$m_A(x)=(x-1)^3(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2\end{pmatrix}$$

$$(1)\;\;\;m_A(x)=(x-1)^2(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix} 1&1&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&1&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&1\\ 0&0&0&0&0&2\end{pmatrix}$$

$$(2)\;\;\;m_A(x)=(x-1)^2(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix} 1&1&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&1\\ 0&0&0&0&0&2\end{pmatrix}$$

$$m_A(x)=(x-1)^3(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix} 1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&1\\ 0&0&0&0&0&2\end{pmatrix}$$

$$m_A(x)=(x-1)^4(x-2)\;\;\;:\;\;\;\;\begin{pmatrix} 1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&1&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&0\\ 0&0&0&0&0&2\end{pmatrix}$$

$$m_A(x)=(x-1)^4(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix} 1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&1&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&2&1\\ 0&0&0&0&0&2\end{pmatrix}$$

Yes, I also get $\,10\,$ different (i.e., non-similar) JCF's for that matrix, check we agree (I, you or both could be wrong), and I don't think there's a general method to come up with the different JCF's . I, for example, try to do it by checking the different possibilities for the minimal polynomial, according to the its different possbile degrees...

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  • $\begingroup$ I don't think minimal polynomial is much helpful in this question. Please see my comment under the question for reference. $\endgroup$ – 23rd Jun 7 '13 at 22:52
  • $\begingroup$ I beg to differ: the power to which each irreducible factor of the characteristic polynomial appears in the minimal pol. tells you what's the maximal size of a Jordan Block in the JCF of the matrix. That's the reason why there are two options in my answer correspondign to the same minimal polynomial...Perhaps there's a formula to tell how many possible JCF's are there, yet where you wrote in your comment the summatory seems to be a typo there, as you have $\,m_i\,$ both a upper limit of the sum and as result...It also isn't clear what the $\,n_i$'s are $\endgroup$ – DonAntonio Jun 7 '13 at 22:57
  • $\begingroup$ As you said, "two options in my answer corresponding to the same minimal polynomial", this suggests that minimal polynomial is not so helpful in this question. There is no typo in the summation in my comment. In fact I am concerned about the number of solutions of $\sum_{j=1}^\infty j\cdot a_j=m_i$, but since $a_j$'s are nonegative integers, so $a_j=0$ when $j>m_i$. $\endgroup$ – 23rd Jun 7 '13 at 23:08
  • $\begingroup$ I think it is, though a closed form for all the different possible JCF's is, I agree, better. Yet, Istill can't understand what is that $\,n_i$'s in your comment, and/or why don't you better write down a compelte answer instead of a simple comment. $\endgroup$ – DonAntonio Jun 7 '13 at 23:12
  • $\begingroup$ $n_i$ denotes the number of solutions of $\sum_{j=1}^{m_i} j\cdot a_j=m_i$, where $m_i$ is the multiplicity of $\lambda_i$. Here $a_j$ means the number of the $j\times j$ Jordan blocks with eigenvalue $\lambda_i$, so it is not hard to see that $n_i$ is precisely the number of possible Jordan blocks corresponding to $\lambda_i$. $\endgroup$ – 23rd Jun 7 '13 at 23:19

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