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Is this Jordan form possible?

$$J=\begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$$

Motivation

I was trying to find how can I know the Jordan form of a 7x7 matrix $A$ with one eigenvalue of multiplicity 7.

Suppose that $\dim(\ker(A-\lambda\mathbb{I}))=3$, which means that there will be 3 Jordan blocks. And $\dim(\ker(A-\lambda\mathbb{I})^3)=\dim(\ker(A-\lambda\mathbb{I})^4)$, i.e, the biggest Jordan block is 3x3.

This yields to possibilities:

$J_1 =\begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$

and

$J_2 =\begin{pmatrix} \lambda & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$

What information do I need in order to distinguish both cases? I think that $J_1$ is not possible, so the only possibility is $J_2$, but I can't prove it.

Should I look the dimension of $\dim(\ker(A-\lambda\mathbb{I})^2)$?

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  • 4
    $\begingroup$ Why on earth would one have to be ruled out? Do Jordan matrices adhere to the laws of the Highlander universe? By writing out the matrix, you've shown it's possible. There's no reason to conclude that it suddenly can't actually be such a decomposition. (Unless there's some conditions you aren't mentioning.) $\endgroup$ – rschwieb Dec 13 '13 at 20:08
  • $\begingroup$ @rschwieb, I think the OP is trying to determine the JCF using the various data collected, and trying to distinguish between the two cases presented. $\endgroup$ – vadim123 Dec 13 '13 at 20:12
  • $\begingroup$ @vadim123 OK, so I guess the answer to my question is "more conditions." Look forward to any clarifications along those lines. $\endgroup$ – rschwieb Dec 13 '13 at 20:13
  • $\begingroup$ This may be helpful: wiki.math.toronto.edu/TorontoMathWiki/images/1/12/… $\endgroup$ – vadim123 Dec 13 '13 at 20:16
  • $\begingroup$ Should "Show" in the last line be "Should"? $\endgroup$ – Jakub Konieczny Dec 13 '13 at 20:23
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Both options are possible. Just notice that if both $A = J_1$ and $A = J_2$ satisfy all the assumptions you impose.

To distinguish the two cases, you would need to be able to say something more about the sizes of Jordan blocks. For easier notation, let $B := A - \lambda I$. It would help if you could say something about ranks (codimensions of kernels) of powers of $B$. You already know that $rank \ B = 4$, $rank \ B^3 = 0$. For case $J_1$ you have $ rank \ B^2 = 1$; for case $J_2$ you have $ rank \ B^2 = 2$. So, if you can figure out $rank B^2$, you have the solution.

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  • $\begingroup$ In that case, how can I distinguish both possibilities? $\endgroup$ – jinawee Dec 13 '13 at 20:12
  • $\begingroup$ Does $rank\ B^2 = 1$ mean that $\dim\ker \ B^2=6$ and $rank \ B^2 = 2$, $\dim\ker \ B^2=5$? Right? $\endgroup$ – jinawee Dec 13 '13 at 20:36
  • $\begingroup$ Yes. $rank = 7 - dim \ ker$. It's just that it's $4$ letters rather than $6$. $\endgroup$ – Jakub Konieczny Dec 13 '13 at 20:38

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