Determine the Jordan Normal form of the Jordan block J^3 for n ≥ 9

Question

I have to determine the Jordan Normal form of the Jordan block J^3. Now I know that J^3 is (0,0,0,e_1,e_2,...,e_(n-3)). I also calculated that the Jordan Normal form of this block for any given n is always a matrix consisting of 3 Jordan blocks of size k/k+1 in the diagonal namely: [J(k),J(k),J(k)] when n=3k ; [J(k+1),J(k),J(k)] when n=3k +1 and [J(k+1),J(k+1),J(k)] when n=3*k+2. So if we write the dimension of the Jordan block n = 3k+r where k∈N, k≥3 und r∈{0,1,2}. I can determine the Jordan Normal form but I am having trouble proving why the 3 Jordan blocks are always of size k/k+1 when instead (example) for n=9 it could be J(4),J(3) and J(2) instead of all being J(3). Does anyone have any idea how to help me. It would be appreciated. Thank you!

Answer 1

Hint: note that if $n = 3k + r$, then we have $(J^3)^k = 0$ if $r=0$ and $(J^3)^{k+1} = 0$ if $r = 1,2$. Consider what this tells you about the sizes of the Jordan blocks in the normal form of $J^3$.