The nth power of a jordan block involving binomial coefficients

Question

For $0\le k \le n$, write $ \binom{n}{k} = \frac{n! }{ k!(n-k)! } .$ For $0\le n < k$, we define $ \binom{n}{k} = 0$. Let $\lambda$ $\in$ $\Bbb C$ and write a Jordan Block as $$ J_k(\lambda) = \begin{bmatrix} \lambda & 1 & 0 & \cdots & 0 \\ & \lambda & 1 & \cdots & 0 \\ & &\lambda & \cdots & 0\\ & & & \cdots & \cdots\\ & & & & \lambda \end{bmatrix}\in \Bbb C^{k,k}.$$

Assume $k, n$ $\ge$ $0$. Prove that $$ J_k(\lambda)^n = \begin{bmatrix} \binom{n}{0}\lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1} \\ & \binom{n}{0}\lambda^n & \binom{n}{1}\lambda^{n-1} & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & \binom{n}{0}\lambda^n & \binom{n}{1}\lambda^{n-1}\\ & & & & & \binom{n}{0}\lambda^n \end{bmatrix}.$$

Please can anyone help me out here?

Answer 1

I proposed the most natural solution as a comment, but actually you can think in terms of graphs. Consider the graph:$([n],\{(i,i+1,1):i\in [n-1]\}\cup \underbrace{\{(i,i,\lambda):i\in [n]\}}_{\text{loops}}),$ where the triples describe the edges as $(out, in, weight)$ in the following way:

Then, $(J_k(\lambda)^n)_{i,j}=$ number of ways to go from $i$ to $j.$
then there are two ways if you are in an intermediate node $k,$ to stay in loop weighted $\lambda$ or advance one step, of course if $i-j=r$ then you have to take $r$ steps ahead and then you will have to stay in the loop $n-r$ steps, so the number is $$\lambda ^{n-r}\binom{n}{r}.$$

Answer 2

Write $J = J_k(\lambda)$ as $J = \Lambda + K$ where $\Lambda$ is the diagonal matrix with $\lambda$ in the diagonal and $K$ the matrix with ones on the upper subdiagonal. Since $\Lambda$ and $K$ commute we can write: $$ J^n = (\Lambda+K)^n = \Lambda^n + {n \choose 1}\Lambda^{n-1}K + \ldots +{n \choose i}\Lambda^{n-i}K^i+\ldots+ K^n $$ and note that the ones in the subdiagonal in $K$ move to the right upper corner as the exponent of $K$ increases, and that from $i = k$ on we have $K^i = 0$.