# A possible converse to the Cayley-Hamilton theorem?

Happy new year MSE! During my holiday vacation I had an interesting idea! The Cayley-Hamilton theorem states that if $f:\mathbb C^n\to\mathbb C^n$ is a linear function, then it is a root of its own characteristic polynomial $\chi_f(f) = 0$.

So I wondered: what if we had a function $f:\mathbb C^n \to \mathbb C^n$ satisfying a polynomial functionial equation (PFE), i.e. there is some polynomial $p=\sum a_k x^k$ such that $p(f)=0$, where we interpret

$$f^k = \underbrace{f\circ\ldots\circ f}_{k \text{ -times}}$$

and $f^0 = \text{id}$. (replace product of variables with composition of functions) E.g. if $p=x^2+ax+b$ then $p(f)=0$ iff $f(f(x)) + af(x) +bx=0$ for all $x$. This is motivated by the fact that after all matrix multiplication is nothing but the composition of linear functions.

Quesiton: Under which conditions would we be able to conclude that $f$ must be a linear function?

Here a few things are important to keep in mind

• If $f$ solves the PFE $p(f)=0$, then so does $\phi^{-1} \circ f\circ \phi$ for any bijective function $\phi$.

• The way we write down $p$ matters, e.g. although $x(x-1) = x^2 -x$, the resulting PFE $f(f(x)-x) = 0$ and $f(f(x)) - f(x)= 0$ are different. (This raises an interesting side question about under which conditions their solutions must coincide)

• General solutions to functional equations can be messy if no additional regularity assumptions are made (cf. Cauchy's equation).

With these caveats in mind I would question the validity of the following

Conjecture: Let $f\colon\mathbb C^n\to\mathbb C^n$ be an entire function satisfying a PFE $p(f) = 0$. Then $f$ is conjugate linear, i.e. there exists a holomorphic bijection $\phi\in\text{Aut}(\mathbb C^n)$ such that $\phi^{-1}\circ f\circ\phi$ is linear.

I did some digging in the literature and found this wonderful paper by Ahern and Rudin. They consider holomorphic $f$ that are functional roots of unity $f^m =\text{id}$ (also known as Babbage's equation), which is equivalent to the PFE given by $p=x^m-1$. Among other things they prove:

• If $f^m = \text{id}$ and $f$ is affine, i.e. $f(z)= Lz+c$, then $f$ is conjugate linear.

• If $f^m = \text{id}$ and $f$ has a fixed point, then $f$ is conjugate linear locally around it.

• If $f^m = \text{id}$ and $f$ is $\mathbb C^2\to\mathbb C^2$ and a finite composition of overshears, then $f$ is conjugate linear.

Here and overshear is a map of the form

$$\begin{pmatrix}x\\ y\end{pmatrix} \longrightarrow \begin{pmatrix}g(y)x+h(y)\\ y\end{pmatrix}$$

with $g,h$ entire and $g(y)\neq 0$ for all $y$; or more generally $f(x_i) = x_i$ for $i\neq j$ and $f(x_j) = g(x) x_j + h(x)$ where $g,h$ are entire, independent of $x_j$ and $g(x)\neq 0$ for all $x$. It is known that the set of all finite compositions of overshears forms a dense subgroup of $\text{Aut}(\mathbb C^n)$.

There are also some known negative results of non-linearizable holomorphic automorphisms (e.g. Derksen 1997) but I don't understand enough of the advanced algebra to really fathom this paper and its possible implications on the question at hand.

There are some simpler sub-problems that might be easier to track:

Problem 1: Let $f(z)=Lz+c$ be affine and satisfy a PFE $p(f)=0$. Does $f$ admit a fixed point?

In this case $f$ is conjugate linear by choosing $\phi$ to be the translation onto the fixed point. If false, this might be the easiest route towards a counter example. If true the next logical step should be to try

Problem 2: If $f:\mathbb C^n \to \mathbb C^n$ is entire and solves the PFE $p(f)=0$, then $f$ admits a fixed point.

Finally, a neat little observation I made is the following: if $f$ solves the PFE $p(f)=0$, and there exists a non-zero vector $v$ and entire function $g$ such that $f(\lambda v) = g(\lambda) v$ for all $\lambda \in \mathbb C$, then $f^k(\lambda v) = g^k(\lambda) v$, hence $g$ is a scalar function solution to the PFE $p(g)=0$. Maybe this indicates that some sort of Eigenvalue theory is possible?

Anyway, it seems like some of this stuff is still untapped terrain so it might be worth some further investigation. Thanks for reading!

The way we write down $p$ matters, e.g. although $x(x−1)=x^2−x$, the resulting PFE $f(f(x)−x)=0$ and $f(f(x))−f(x)=0$ are different.
This example suggests that you would consider the following to be a counter-example: Take $f(z) = z^2$ and observe that $$f \circ (2f) = 4 f \circ f.$$ Thus $f$ obeys the "polynomial" $x(2x) - 4 x^2$. Since $f$ is not bijective, it is certainly not conjugate linear.
More generally, if $f(x)$ is any polynomial of degree $d$, then the infinite list of functions $f(kf(x))$ will all be polynomials of degree $d^2$. The space of polynomials of degree $d^2$ is a vector space of dimension $d^2+1$, so there will be a linear relations between these compositions.
I was not able to find a non conjugate linear example of a function obeying $\sum c_j f^j(z)=0$ for nonzero $c_j$.