# Find Jordan form of powers of Jordan matrix

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?

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