Let $ A \in M_{n\times n}\left(\mathbb{C}\right) $, such that $ A^{2008}=O $ and $ 0 \lt 3\rho\left(A^2\right)\lt2\rho\left(A\right)\lt5 $. I need to find its Jordan Normal Form. Did I do it correctly?
$ A^{2008} = O \Rightarrow A$ is a nilpotent matrix $ \Rightarrow P_{A}\left(\lambda\right)=\lambda^7$. $ M_{A}\left(\lambda\right) \vert P_{A}\left(\lambda\right) $ and every root of $ P_{A}\left(\lambda\right) $ is a root of $ M_{A}\left(\lambda\right) $. Therefore, there exists a $ 0\lt k \leq 7 $ such that $ M_{A}\left(\lambda\right) = \lambda^k $. $ k $ is the minimal number such that $ \rho(A^k)=0 $.
$ 0 \lt 2\rho(A) \lt 5 \Rightarrow 0 \lt \rho(A) \leq 2 \Rightarrow \rho(A) \in \{1, 2\} $.
Suppose $ \rho(A) = 1 $. $ 0<\rho(A^2)\Rightarrow \rho(A^2) < \rho(A) \Rightarrow \rho(A) = 0 $ - contradiction. Therefore, $ \rho(A) = 2, \rho(A^2) =1 $ and $ \rho(A^3) = 0 $, and so $ M_{A}(\lambda) = \lambda^3 $.
The number of Jordan Blocks of the eigenvalue $ 0 $ in the Jordan Normal Form is $ 7 - \rho(A) = 5 $, and the largest Jordan Block of $ 0 $ is of order $ 3 $, and therefore the Jordan Normal Form is:
$ \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} $