Let $A = J_n (\lambda)$ be a $n \times n$ Jordan block. Determine the Jordan form of $A^k$

Question

Let $A = J_n(\lambda)$ be a $n \times n$ Jordan Block. Determine the Jordan form of $A^k$.

How to approach the problem? Need some hints

$A = J_n(\lambda) = J_n(0) + \lambda I$. Now $A^k = {(J_n(0) + \lambda I)}^k $ and the expansion follows the binomial expansion. Will this be any need?

Answer 1

Hint Continuing with your suggestion, your expression with the Binomial Theorem gives $$J_n(\lambda)^k = (J_n(0) + \lambda I_n)^k = \sum_{i = 0}^k {k \choose i} \lambda^{k - i} J_n(0)^i .$$ Now, since $\lambda$ is the only eigenvalue of $J_n(\lambda)^k$, the only eigenvalue of $J_n(\lambda)^k$ is $\lambda^k$. Thus, the Jordan form of $J_n(\lambda)^k$ is some direct sum of Jordan blocks $J_{n_a}(\lambda^k)$, and it remains just to determine what the block sizes $n_a$ are.

As usual, we can determine these sizes by analyzing the dimensions of the kernels of the powers of $$J_n(\lambda)^k - \lambda^k I_n .$$ Glancing back at our expansion of the first term we see that $\lambda^k I_n$ is precisely the first term in the sum, so this difference is just $$\sum_{i = 1}^k {k \choose i} \lambda^{k - i} J_n(0)^i$$ (NB this sum begins at index $i = 1$, and that it is empty in the special case $k = 0$, which is anyway trivial.)

Additional hint Notice that since $J_n(0)$ is strictly upper triangular, the last entry of $$\sum_{i = 1}^k {k \choose i} \lambda^{k - i} J_n(0)^i {\bf x}$$ is $0$, and, for example, the second-to-last entry is $k \lambda^{k - 1} x_n$. This suggests, among other things, splitting into the cases $\lambda = 0$ and $\lambda \neq 0$.