# Why does the $n$-th power of a Jordan matrix involve the binomial coefficient?

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?

• Have you tried a few simple cases to see what happens? – Bruno Joyal Mar 10 '13 at 18:36

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: $$J_k(\lambda)^n=(\lambda I+N)^n=\sum_{r=0}^\color{red}{n} \binom{n}{r}\lambda^{n-r} N^r=\sum_{r=0}^\color{red}{\min(n,k-1)} \binom{n}{r}\lambda^{n-r} N^r.$$

• It's necessary to mention that $\lambda I$ and $N$ commute. – Git Gud Mar 10 '13 at 18:46
• Does $N$ becomes $0$ (in the sum) as soon as $r = k$, right? – user34295 Mar 10 '13 at 20:02
• Found it, $N$ should be the canonical nilpotent matrix, that is of order $n$ and degree $n$... correct? – user34295 Mar 10 '13 at 20:03
• @Gremo $N$ is actually identical to $J_k(0)$. It is $k\times k$, i.e. its size conforms to the size of $J_k(\lambda)$. The first exponent that makes $N^r=0$ is $r=k$. For $1\le r<k$, the power $N^r$ is the matrix that has all ones in the $r$-th superdiagonal and zero elsewhere. – user1551 Mar 10 '13 at 20:12

user1551 already gave the "simple explanation", so let me just mention the general result for evaluating matrix functions when the argument is a Jordan block. Given the $n\times n$ Jordan block with eigenvalue $\lambda$,

$$\mathbf J=\begin{pmatrix}\lambda&1&&\\&\lambda&\ddots&\\&&\ddots&1\\&&&\lambda\end{pmatrix}$$

the matrix $f(\mathbf J)$ looks like this:

$$f(\mathbf J)=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$

This general result is proven in a number of references; e.g. this.

For the power function $f(x)=x^k$, we have the general result (easily proven inductively):

$$\frac1{j!}\frac{\mathrm d^j}{\mathrm dx^j}x^k=\binom{k}{j}x^{k-j}$$

Making the necessary replacements gives the formula for the power of a Jordan block.