Suppose $J$ is an $n \times n$ Jordan block with eigenvalue $0$, what is the Jordan form of $J^2$?
My solution:
I squared the matrix, it follows that the eigenvalues are $0$ again, and $\dim(N(J^2)) = 2$, with eigenvectors $e_{n-1}$, $e_{n}$.
Therefore, the Jordan form of $J^2$ is an $(n-2) \times (n-2)$ block and two $1\times1$ blocks all with eigenvalue zero.
However, there is a hint with the question which says to treat odd and even values of $n$ separately but I fail to see why.
On
Let me illustrate this with the $n = 6$ case. Here, we have a basis $\{ e_1, e_2, e_3, e_4, e_5, e_6 \}$, and the action of $J$ on the basis vectors is given by $$ Je_1 = e_2, \ \ Je_2 = e_3, \ \ Je_3 = e_4, \ \ Je_4 = e_5, \ \ Je_5 = e_6, \ \ Je_6 = 0. $$ So the action of $J^2$ on this basis is given by
$$ J^2e_1 = e_3, \ \ J^2e_2 = e_4, \ \ J^2e_3 = e_5, \ \ J^2e_4 = e_6, \ \ J^2e_5 = 0, \ \ J^2e_6 = 0.$$
[You absolutely right that the kernel of $J^2$ has dimension two, and is spanned by $e_5$ and $e_6$.]
But here is the really key point. Notice that, under the action of $J^2$, the vector space decomposes into two invariant subspaces:
So $J^2$ has two Jordan blocks (corresponding to these two subspaces), and for each of these two blocks, the Jordan matrix is $$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$
That was the $n = 6$ case. Can you generalise this approach for $n = 7$? And from there, can you generalise for arbitrary $n$?
You're wrong. For example, $$ \pmatrix{0 & 1 & 0 \cr 0 & 0 & 1\cr 0 & 0 & 0\cr}^2 = \pmatrix{0 & 0 & 1\cr 0 & 0 & 0\cr 0 & 0 & 0\cr} \ \text{has Jordan form}\ \pmatrix{0 & 1 & 0\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$$ (a block of size $2$ and a block of size $1$) but $$ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr}^2 = \pmatrix{0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}\ \text{has Jordan form}\ \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr}$$ (two blocks of size $2$).