Linear Algebra – Why the n-th Power of a Jordan Matrix Involves the Binomial Coefficient

Question

I've searched a lot for a simple explanation of this. Given a Jordan block $J_k(\lambda)$, its $n$-th power is:

$$
J_k(\lambda)^n = \begin{bmatrix}
\lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1} \\
& \lambda^n & \binom{n}{1}\lambda^{n-1} & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2} \\
& & \ddots & \ddots & \vdots & \vdots\\
& & & \ddots & \ddots & \vdots\\
& & & & \lambda^n & \binom{n}{1}\lambda^{n-1}\\
& & & & & \lambda^n
\end{bmatrix}$$

Why does the $n$th power involve the binomial coefficient?

Answer 1

Let $N$ denote the nilpotent matrix whose superdiagonal contains ones and all other entries are zero. Then $N^k=0$. Hence, by the binomial theorem: $$J_k(\lambda)^n=(\lambda I+N)^n=\sum_{r=0}^\color{red}{n} \binom{n}{r}\lambda^{n-r} N^r=\sum_{r=0}^\color{red}{\min(n,k-1)} \binom{n}{r}\lambda^{n-r} N^r.$$