Jordan normal form in $n$-th root

Question

How to find $\bigl(Jk(λ)\bigr)^n$ ($n \in\mathbb N)$. I've found the solution for $\lambda=0$. I need the look of matrix and the steps

Answer 1

We have $Jk(\lambda)=\lambda I+(Jk(\lambda)-\lambda I)$ and define $M=Jk(\lambda)-\lambda I$, then $I$ and $M$ are commuting. You can use binomial theorem to get $$Jk(\lambda)^n=(\lambda I+M)^n=\sum_{j=0}^n\binom{n}{j}(\lambda I)^jM^{n-j}=\lambda^nI+\sum_{j=0}^{n-1}\binom{n}{j}(\lambda I)^jM^{n-j}$$ Maybe you should try to find out what happens for $k=3$ in the binomial theorem: $$\left(\begin{pmatrix}\lambda &0 &0\\ 0 &\lambda & 0\\ 0 & 0 & \lambda \end{pmatrix}+\begin{pmatrix}0 &1 &0\\ 0 &0 & 1\\ 0 & 0 & 0 \end{pmatrix}\right)^n=T$$ This is the sum of a diagonalizable ($\lambda I$) and a nilpotent ($M$) matrix.

The result is

\begin{align}T=\begin{pmatrix}\lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2}\\0 & \lambda^n & \binom{n}{1}\lambda^{n-1}\\ 0 & 0 & \lambda^n \end{pmatrix}\end{align}