29

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?

Rodrigo de Azevedo
  • 18,977
  • 5
  • 36
  • 95
user34295
  • 675
  • 1
  • 8
  • 15

2 Answers2

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

Marc van Leeuwen
  • 107,679
  • 7
  • 148
  • 306
user1551
  • 122,076
  • 7
  • 103
  • 187
  • 16
    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
  • 3
    @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 – user1551 Mar 10 '13 at 20:12
28

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.

J. M. ain't a mathematician
  • 71,951
  • 6
  • 191
  • 335