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Let $A=$$ \begin{bmatrix} 1&1&1&1&1&1 \\ 0 & 1 & 0 & 0&0&1 \\ 0 & 0 & 1 & 0&0&1 \\ 0 & 0 & 0 & 1&0&1 \\ 0 & 0 & 0 & 0&1&1 \\ 0 & 0 & 0 & 0&0&1 \\ \end{bmatrix} $

I want to calculate the characteristic polynomial $p_A$, minimal polynomial $ m_A$ and the Jordan normal form J.

As a hint is given that almost no calculations have to be done!

A is upper triangular thus we can see $p_A=(t-1)^6$.

EDIT Correct solution according to comments and the accepted answer

$J=$$ \begin{bmatrix} 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&1&0 \\ 0 & 0 & 0 & 0&0&1 \\ \end{bmatrix} $
This jordan normal form has one block of size 3x3 and three blocks of size 1x1. Since there is only one eigenvalue and the biggest block has size 3 we get $m_A=(t-1)^3$.

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  • $\begingroup$ I will add some Details soon! $\endgroup$ – user519338 Mar 30 '18 at 17:59
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Denote the $k\times k$ Jordan block for an eigenvalue $\lambda$ by $J_k(\lambda)$. Clearly, $A-I$ is nilpotent and it has rank $2$. Therefore the Jordan normal form of $A-I$ is either $J_2(0)\oplus J_2(0)\oplus0_{2\times2}$ or $J_3(0)\oplus0_{3\times3}$. Now you may rule out one of the these two possibilities by considering the rank of $(A-I)^2$.

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  • $\begingroup$ Yes thanks for your answer I just did that. We get $rank(A-I)^0=6, rank(A-I)^1=2, rank(A-I)^2=0$ Then we get a partition $r=(4,2)$ and its dual partition is given by $r*=(2,2,1,1)$. Thus our jordan normalf orm is of the form $diag(J_2(1),J_2(1),J_1(1),J_1(1))$ this is exactly what I guessed. Am I right here? $\endgroup$ – user519338 Mar 30 '18 at 18:08
  • $\begingroup$ You have made some arithmetic errors. Calculate the ranks again. $\endgroup$ – user1551 Mar 30 '18 at 18:11
  • $\begingroup$ I do not understand your answer. I guess you mean $J_k(1)$ instead of $J_k(0)$, but still.. $\endgroup$ – user519338 Mar 30 '18 at 18:11
  • $\begingroup$ @J.D We may find the Jordan form of $A-I$ first. Once that is found, we can add $I$ to it to get the Jordan form of $A$. $\endgroup$ – user1551 Mar 30 '18 at 18:14
  • $\begingroup$ Yes, I have had an error. I get get $rank(A-I)^2=1, rank(A-I)^3=0$. Therefore our partition is given by $r=(6-2,2-1,1-0)=4,1,1$ and its dual partition is given by $r*=(3,1,1,1)$. So the jordan normal form of A is given by $diag(J_3(1),J_1,J_1,J_1)$. This coincides with your answer! Thanks again! $\endgroup$ – user519338 Mar 30 '18 at 18:20

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