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Let $f,g:U\subseteq \mathbb{R}^{m}\to \mathbb{R}^{m}$ differentiables in the open $U$ that contains $0$ and $f(0)=g(0)=0$. If $h\circ f=g\circ h$ and $h'(0): \mathbb{R}^{m}\to \mathbb{R}^{m}$ is an isomorphism. is it true that $f '(0)$ and $g' (0)$ have the same eigenvalues?

I've been trying to find a counterexample in $\mathbb{R}$ without success. Is there a counterexample?

David G. Stork
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User1997
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1 Answers1

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By the Chain Rule, $$h'(f(0)) \circ f'(0) = (h \circ f)'(0) = (g \circ h)'(0) = g'(h(0)) \circ h'(0),$$ which is the same as saying that $h'(0) \circ f'(0) \circ [h'(0)]^{-1} = g'(h(0))$. So, if $h(0) = 0$, then $f'(0)$ and $g'(0)$ has the same eigenvalues.

azif00
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  • Could you explain to me why $h(0) = 0 $ implies that they have the same eigenvalues? – User1997 Sep 19 '20 at 00:50
  • If $h(0) = 0$ and $T := h'(0)$, the equation $h'(0) \circ f'(0) \circ [h'(0)]^{-1} = g'(h(0))$ can be written as $T \circ f'(0) \circ T^{-1} = g'(0)$. So, if $h(0) = 0$, $f'(0)$ and $g'(0)$ are similar linear maps. Now, you can see this for example. – azif00 Sep 19 '20 at 01:37
  • Thanks for your answer, it is very helpful!. I see that the condition $h (0) = 0$ is necessary. – User1997 Sep 19 '20 at 01:43