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More specifically, let $f：ℂ^n\to ℂ^n, f(z)=Az$ be linear and $ϕ：ℂ^n → ℂ^n$ entire and bijective, i.e. $ϕ∈\text{Aut}(ℂ^n)$. Is it possible that $ϕ^{-1}∘f∘ϕ$ is linear again when $ϕ$ is non-linear?

For example if $ϕ(z) = Lz+c$ is affine, then

$$ ϕ^{-1}∘f∘ϕ (z) = L^{-1}ALz + L^{-1}(A-I)c $$

is linear if and only if $c \in \ker(A-I)$. Are there more sophisticated examples where $\phi$ is truly non-linear, in the sense of having non-constant derivative?

Question: Does there exist a triple $(A, B, ϕ)$, such that

  1. $A≠B$ are complex $n×n$ matrices.
  2. $ϕ∈\text{Aut}(ℂ^n)$ with $ϕ'$ non-constant
  3. $ ϕ^{-1} ∘ A ∘ ϕ=B $
  4. $A∘ϕ$ is nonlinear (this avoids certain trivial examples)

In particular, is there an example where $A$ and $B$ are non-similar?

EDIT 2022:

  • Non-similarity is impossible as per Dap's comment ✔
  • I added the 4th condition to rule out certain trivial examples (see comments in Qiaochu Yuan's answer)
  • Dap provided an example in the comments with $A=B$
  • Note that if $(A, B, ϕ)$ is a valid triple, with $B=S^{-1}AS$, then we can actually non-linearly conjugate $A$ to any matrix $C=T^{-1}AT$ similar to $A$ via $ψ = ϕ∘S^{-1}∘T$. So we immediately get an example with $A≠B$ out of Dap's contraction.
  • An interesting follow-up question would be: Can we construct examples for generic $A$? So far we only have an example where $A$ is similar to a diagonal matrix and all eigenvalues have magnitude $1$.
Hyperplane
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  • What happens if $A=Id$? – Tsemo Aristide Jan 06 '18 at 22:33
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    @TsemoAristide this is a trivial case and not really interesting. I am interested in whether or not there exist any non-trivial examples. – Hyperplane Jan 06 '18 at 22:42
  • I think this fact might come handy, that, if $\phi$ is not a first degree polynomial as it is in your example, then $\phi^{-1}$ can never be entire analytic. See the link below: https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://math.ucr.edu/~res/math138A/inverse-series&ved=0ahUKEwjpj_udu8TYAhUfSo8KHZiOBX8QFggoMAE&usg=AOvVaw1GQZbFL59qys81eqQhoD7c – QED Jan 06 '18 at 23:23
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    @Abishanka: this is false in more than one variable. For example, the map $(z_1, z_2 \mapsto (z_1, z_2 + f(z_1))$ is an analytic bijection for any analytic function $f : \mathbb{C} \to \mathbb{C}$, with analytic inverse $(z_1, z_2) \mapsto (z_1, z_2 - f(z_1))$. – Qiaochu Yuan Jan 06 '18 at 23:42
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    $A$ and $B$ must be similar because the chain rule applied to $A\circ \phi=\phi\circ B$ gives $A\circ \phi'(0)=\phi'(0)\circ B,$ and $\phi'(0)$ must be non-singular. – Dap Jan 27 '18 at 10:10

1 Answers1

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Yes. For starters, let's take $n = 2$ and

$$\phi : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} z \\ w + f(z) \end{array} \right]$$

where $f$ is an analytic function $f : \mathbb{C} \to \mathbb{C}$ to be chosen later. $\phi$ is analytic with inverse

$$\phi^{-1} : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} z \\ w - f(z) \end{array} \right].$$

Now let $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ and let's write out what $\phi^{-1} \circ A \circ \phi$ is. After a little computation we get

$$\phi^{-1} \circ A \circ \phi : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} az + bw + b f(z) \\ cz + dw + d f(z) - f(az + bw + b f(z)) \end{array} \right].$$

To get rid of the nonlinear part of the first component we need to set $b = 0$. From here, to get rid of the nonlinear part of the second component the easiest option is to set $d = a = f(0) = 0$. So, explicitly, we can set $A = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$ and $f(z) = z^2$, giving

$$\phi^{-1} \circ A \circ \phi = A.$$

Qiaochu Yuan
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  • Ah yes, I actually considered a similar example. The map you used is known as a shear. Unfortunately this example is kind of uninteresting too, since essentially the non-linearities are avoided using the kernel of $A$.

    What I really want to know is if there is some example where the non-linearities sort of cancel out, or if this is impossible. I should have been more clear when I posed the question.

    Thanks for your post anyway!

     – Hyperplane Jan 07 '18 at 09:44
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    @Hyperplane: you can modify the example to get $\ker A=0,$ for example $f(z)=z^2$ and $A=\begin{pmatrix}-1&0\0&1\end{pmatrix}.$ – Dap Jan 27 '18 at 10:22