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.


1 Answer 1


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).

  • $\begingroup$ Ah, that's great. Thanks a lot, Theo! $\endgroup$
    – Sam
    Feb 7, 2011 at 3:36
  • $\begingroup$ What a beatiful proof. :) $\endgroup$
    – Sam
    Feb 7, 2011 at 4:24
  • 2
    $\begingroup$ @S. L.: I like it too, and I didn't know it before. Thanks for asking this question! $\endgroup$
    – t.b.
    Feb 7, 2011 at 6:04
  • 4
    $\begingroup$ 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. $\endgroup$ Jul 16, 2011 at 4:24
  • 2
    $\begingroup$ @J.M. Thanks a lot for these links! Good to see you're back in town, at least occasionally :) $\endgroup$
    – t.b.
    Jul 16, 2011 at 8:02

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