<s>[INST] What's the Jordan form of $J^2$? $$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.  
So it could be 
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or } 
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?
 [/INST] Nice question! Here's another perspective:
You have an ordered basis $\{v_1,v_2,v_3,v_4,v_5\}$ for $\mathbb{R}^{5}$ such that $Jv_{5} = v_{4}$, $Jv_{4} = v_{3}$, $Jv_3 = v_{2}$, $Jv_2 = v_{1}, Jv_{1} = 0$. This is called a cycle of generalized eigenvectors.
If you consider $J^{2}$, you get two cycles:
$$J^{2}v_{5} = v_{3}; \quad J^{2}v_{3} = v_{1}; \quad J^{2}v_{1} = 0$$
$$J^{2}v_{4} = v_{2}; \quad J^{2}v_{2} = 0$$
One cycle has length 3 and the other has length 2, meaning the Jordan form has a block of size 3 and a block of size 2. Also, $\{v_1,v_3,v_5,v_2,v_4\}$ is a Jordan basis for $J^{2}$. 
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