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Let $A\in M_7(\mathbb{C})$ be a matrix in with the characteristic polynomial $p(A)=(\lambda+4)^5(\lambda-2)^2$.

I need to find all Jordan normal forms for this.

I think that i can use that the Jordan normal form of a given matrix $A$ is unique up to the order of the Jordan blocks, so because of that I think that there are 14 but i only find 10 so i'm almost sure that something is wrong.

I think that there are $14$ because if i undertood the above theorem there are $12$ for the $(\lambda+4)^5$ and $2$ for $(\lambda-2)^2$.

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    $\begingroup$ Where are you getting 14 from? $\endgroup$ – Git Gud Nov 24 '13 at 22:28
  • $\begingroup$ I believe 14 is correct. $\endgroup$ – Ted Shifrin Nov 24 '13 at 22:30
  • $\begingroup$ if i undertood the theorem right there are 12 for the $(\lambda+4)^5$ and 2 for $(\lambda-2)^2$, the problem is that i only find 10 :( $\endgroup$ – user111034 Nov 24 '13 at 22:31
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    $\begingroup$ For each eigenvalue I'm counting as many blocks as its algebraic multiplicty's partition. For instance, for $4$ I'm getting the possible combinations of jordan blocks: one $5\times 5$ block, one $4\times 4$ block plus one $1\times 1$ block,..., five $1\times 1$ blocks. See WA link. $\endgroup$ – Git Gud Nov 24 '13 at 22:38
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    $\begingroup$ I think $\;14\;$ is right, too: No. of partitions of $\;5\;\;(7)\;$ times the No. of partitions of $\;2\;\;(2)\;$ is fourteen... $\endgroup$ – DonAntonio Nov 24 '13 at 22:44
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The 14 forms are

$m(A)=(\lambda+4)^5(\lambda-2)^2$

$m(A)=(\lambda+4)^4(\lambda-2)^2$

$m(A)=(\lambda+4)^3(\lambda-2)^2$ with decomposition 3,2

$m(A)=(\lambda+4)^3(\lambda-2)^2$ with decomposition 3,1,1

$m(A)=(\lambda+4)^2(\lambda-2)^2$ with decomposition 2,2,1

$m(A)=(\lambda+4)^2(\lambda-2)^2$ with decomposition 2,1,1,1

$m(A)=(\lambda+4)(\lambda-2)^2$

$m(A)=(\lambda+4)^5(\lambda-2)$

$m(A)=(\lambda+4)^4(\lambda-2)$

$m(A)=(\lambda+4)^3(\lambda-2)$ with decomposition 3,2

$m(A)=(\lambda+4)^3(\lambda-2)$ with decomposition 3,1,1

$m(A)=(\lambda+4)^2(\lambda-2)$ with decomposition 2,2,1

$m(A)=(\lambda+4)^2(\lambda-2)$ with decomposition 2,1,1,1

$m(A)=(\lambda+4)(\lambda-2)$

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