Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm just working through Conway's book on complex analysis and I stumbled across this lovely exercise:

Use Cauchy's Integral Formula to prove the Cayley-Hamilton Theorem: If $A$ is an $n \times n$ matrix over $\mathbb C$ and $f(z) = \det(z-A)$ is the characteristic polynomial of $A$ then $f(A) = 0$. (This exercise was taken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390-391)

Unfortunately, I was not able to find said paper. I'm completely lost with this exercise. I can't even start to imagine how one could possibly make use of Cauchy here...

Thanks for any hints.

Regards, S.L.

share|cite|improve this question
A brief google search gave this note. –  Jacopo Notarstefano Feb 6 '11 at 13:11

1 Answer 1

up vote 25 down vote accepted

The idea is to use holomorphic functional calculus and to show that for a matrix $A$ and a polynomial $p(z)$ we have for $r \gt \|A\|$ \begin{equation}\tag{$\ast$} p(A) = \frac{1}{2\pi i} \int_{|z| = r} p(z) \cdot (z - A)^{-1}\\,dz \end{equation} in complete analogy with the Cauchy formula for complex numbers. The integral of a matrix of holomorphic functions is defined by integrating each entry separately.

By Cramer's rule, the $(k,l)$-entry of $(z-A)^{-1}$ is $\displaystyle ((z-A)^{-1})_{k,l} = \frac{1}{\det(z-A)} c_{k,l}(z)$ where $c_{k,l}(z)$ is some polynomial in $z$. Let $p(z) = \det(z-A)$ be the characteristic polynomial of $A$. Conclude using $(\ast)$ by applying Cauchy's integral theorem to $c_{k,l}$.

To see that the identity $(\ast)$ holds, proceed as follows (this is a slight variant of McCarthy's argument):

  • The usual matrix norm induced by the Euclidean norm on $\mathbb{C}^{n}$ satisfies $\|A^{n}\| \leq \|A\|^{n}$.
  • Use this to show that $(z - A)^{-1} = \sum_{n = 0}^{\infty} \frac{A^{n}}{z^{n+1}}$, where the right hand side converges uniformly on $\{|z| \gt \|A\| + \varepsilon\}$.
  • It follows that we can interchange integration and summation. Conclude that $$ A^{k} = \int_{|z| = r} z^{k} (z - A)^{-1}\,dz$$ and $(\ast)$ follows by linearity.

Here's a link to McCarthy's article (you need a university subscription to download it, but the first page is almost the entire article).

share|cite|improve this answer
Ah, that's great. Thanks a lot, Theo! –  Sam Feb 7 '11 at 3:36
What a beatiful proof. :) –  Sam Feb 7 '11 at 4:24
@S. L.: I like it too, and I didn't know it before. Thanks for asking this question! –  t.b. Feb 7 '11 at 6:04
Beautiful indeed! Since I can't give one more upvote than I would've loved to give, let me mention instead what is (methinks) an apropos paper: The Equivalence of Definitions of a Matric Function by Rinehart (p. 399, in particular). This Cauchy construction has been used numerically as well. –  J. M. is back. Jul 16 '11 at 4:24
@J.M. Thanks a lot for these links! Good to see you're back in town, at least occasionally :) –  t.b. Jul 16 '11 at 8:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.