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We are given $A$ is a 6 by 6 matrix with values from $\mathbb C$, we are also given $rank(A-3I_6)=4$, and the minimal polynomial of $A$ is $m_A=(x-1)^2(x-3)^2$

We are asked to find all possible jordan forms of $A$.

My solution

We know that $rank((A-\lambda I)^{i-1})-rank((A-\lambda I)^i) =$ number of blocks corresponding to the eigenvalue $\lambda$ of order that is at most i.

$rank((A-3I)^0) - rank(A-3I) = rank(I_6)-4=6-4=2$

So there are 2 blocks corresponding to the eigenvalue $3$ and we know that there is a block sized 2x2 corresponding to the eigenvalue $3$ because that is the exponent of $3$ in the minimal polynomial. And due to the same logic, we can say the same thing about the eigenvalue 1.

Summary: We must have a jordan block sized 2x2 corresponding the eigenvalue 1, we must have a jordan block sized 2x2 corresponding to the eigenvalue $3$, we must have overall 2 jordan blocks corresponding to eigenvalue $3$, and the matrix has to be a 6x6 matrix.

Solution:

$\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 &3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{pmatrix}$

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    $\begingroup$ Looks fine to me...! +1 $\endgroup$ – DonAntonio Dec 4 '13 at 21:29
  • $\begingroup$ Solution is written in the body of the question. Will close this question whenever I can. $\endgroup$ – Oria Gruber Dec 4 '13 at 21:36
  • $\begingroup$ Yes, seems correct. $\endgroup$ – Berci Dec 4 '13 at 21:41
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    $\begingroup$ @OriaGruber: Move it from the body of the question to an actual answer. $\endgroup$ – Mariano Suárez-Álvarez Dec 5 '13 at 1:09
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There is an error in your answer. All you can conclude are

a) Largest block corresponding to $\lambda=1$ is size 2

b) Largest block corresponding to $\lambda=3$ is size 2

c) Total size of all the blocks corresponding to $\lambda=3$ is 4

So the Jordon blocks for $\lambda=3$ can be two blocks of size 2 (your answer) or one block of size 2 and two of size 1.

Note that minimal polynomial only tells you the size of the largest block. So the second possibility is $$ \begin{pmatrix}1 & 1 & 0 & 0 & 0 & 0\cr 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 3 & 1 & 0 & 0\cr 0 & 0 & 0 & 3 & 0 & 0\cr 0 & 0 & 0 & 0 & 3 & 0\cr 0 & 0 & 0 & 0 & 0 & 3\end{pmatrix} $$

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