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Let $A$ be a Jordan matrix with blocks $J_5(0),J_6(0)$ with $J_m(\lambda)$ having size $m\times m$. I am to find the Jordan form of $A^2$. Since $A$ is in Jordan form and powers of $A$ have the same transition matrices $P$ as $A$, I think this implies $A^2$ must be in Jordan form, and to find it I have no choice but to calculate the squares of $J_5(0),J_6(0)$. Is this correct, and is there an easier way?

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    $\begingroup$ Observe that in $\;J_5(0)\;$ you already have $\;1\;$ in the entry $\;1-3\;$ , so it can't be in Jordan normal form. $\endgroup$ – user351910 Jul 5 '16 at 12:23
  • $\begingroup$ Are you asking only about Jordan forms of powers, or also about the posibly revised transition matrices? $\endgroup$ – hardmath Jul 5 '16 at 12:24
  • $\begingroup$ @AntoineNemesioParras I don't understand - $J_5(0)$ is basically by definition in Jordan form - why are you saying it isn't? $\endgroup$ – linalg Jul 5 '16 at 12:29
  • $\begingroup$ @hardmath both! $\endgroup$ – linalg Jul 5 '16 at 12:30
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    $\begingroup$ @linalg Sorry about that. It should have been $\;J_5(0)^2\;$ . $\endgroup$ – user351910 Jul 5 '16 at 12:36

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