# Cauchy's integral formula for Cayley-Hamilton Theorem

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.

The idea is to use holomorphic functional calculus and to show that for a matrix $A$ and a polynomial $p(z)$ we have for $r \gt \|A\|$ $$\tag{\ast} p(A) = \frac{1}{2\pi i} \int_{|z| = r} p(z) \cdot (z - A)^{-1}\ \,dz$$ 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).

• Ah, that's great. Thanks a lot, Theo!
– Sam
Feb 7, 2011 at 3:36
• What a beatiful proof. :)
– Sam
Feb 7, 2011 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, 2011 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. Jul 16, 2011 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, 2011 at 8:02