Let $G \subset GL_{n}(\mathbb{R})$ be a subgroup such that $G \subset B$ where $$B = \lbrace M \in \mathcal{M}_{n}(\mathbb{R}), ||M-I|| < 1\rbrace $$

Let $g \in G$.

I'm asked to show:

  1. The only complex eigenvalue of $g$ is $1$
  2. $g = I+N$ where $N$ is nilpotent
  3. and finally that $g=I$.

I have shown that if $\lambda \in \mathbb{C}$ is an eigenvalue, then $|\lambda -1| < 1$. But I don't really know if it helps..

(note sure for the tags btw)


2 Answers 2


For 1: note that $g^n \in G \subset B$ for each $n$. It follows that if $\lambda$ is an eigenvalue of $g$, then we have $$ |\lambda^n - 1| < 1 $$ for every $n \in \Bbb Z$ (since $\lambda$ can't be zero, negative exponents are fine). If $|\lambda| \neq 1$, then either $|\lambda^n|$ or $|\lambda^{-n}|$ diverges to $\infty$ as $n \to \infty$. You'll need to handle the $|\lambda| = 1$ (with $\lambda \neq 1$) case separately, noting that $\lambda^n$ circles around the unit disk.

For 2: this is an immediate consequence of the Cayley-Hamilton theorem.

For 3: consider the binomial expansion $(I + N)^n$ as $n \to \infty$.

For the case $|\lambda| = 1$, we have $\lambda = e^{i \theta}$, and $$ |\lambda^n - 1|^2 = [\cos(n \theta) - 1]^2 + \sin^2(n \theta) = 2[1- \cos(n \theta)] $$ it suffices to state that for any $\theta \in (0,2\pi)$, there exists some $n$ for which $\cos(n \theta) < 0$.

The arguments for 2 and 3 work over real matrices, since we were considering their complex eigenvalues.

  • $\begingroup$ Thanks for your answer. It's ok for point 3. I still can't figure out the case $|\lambda | = 1$ though... And for point 2, it's pretty clear for me if $g \in \mathcal M_{n} (\mathbb C)$ but what if $g \in \mathcal M_{n} (\mathbb R)$ ? $\endgroup$
    – krirkrirk
    Commented Sep 13, 2015 at 0:20
  • $\begingroup$ It was so obvious.... I'm king of rusty with complex numbers. Thank you very much!! $\endgroup$
    – krirkrirk
    Commented Sep 13, 2015 at 11:39

This is an explicit instance of the No Small Subgroups theorem for real or complex Lie groups, namely, that there is a sufficiently small neighborhood of $I$ so that the only subgroup contained in it is $\{I\}$.

The specific estimate is perhaps not so crucial in the bigger scheme of things, but is tangible. Indeed, if there were an eigenvalue $\lambda$ other than $1$, with non-zero eigenvector $v$, then $|(M-I)v|=|(\lambda-1)v|<|v|$ implies $|\lambda-1|<1$, as you noted. But/and $M^2$ is also in the subgroup, and has $v$ as $\lambda^2$ eigenvector, and similarly for $M^3$, ... That is, $|\lambda^n-1|<1$ for all integers $n$. Then there are various choices for showing that $\lambda=1$. E.g., let $-\pi<\theta<\pi$ be the argument of $\lambda$, using $|\lambda-1|<1$ just slightly. For $|\theta|<\pi/2$, the argument of $\lambda^2$ is $2\theta$. If the argument of $\lambda^2$ is still less than $\pi/2$ in absolute valued, the argument of $\lambda$ is $4\theta$. By induction, either the argument of some $\lambda^{2^n}$ is in the range $[{\pi\over 2}\pi]\sqcup[-\pi,{-\pi\over 2}]$, or $\theta=0$.

  • $\begingroup$ Thanks for your answer. Yet I'm sorry but I don't understand your point, why can't the argument of some $\lambda^{2^{n}}$ be in the range you mentionned ? $\endgroup$
    – krirkrirk
    Commented Sep 13, 2015 at 0:17
  • $\begingroup$ $|\lambda-1|<1$ forces $\lambda$ to be in the right half-plane... but powers will get out of it ... Lotta ways to talk about this. $\endgroup$ Commented Sep 13, 2015 at 3:37
  • $\begingroup$ Thank you very much! The theorem you mentionned seems really interesting. I'm gonna check it out. $\endgroup$
    – krirkrirk
    Commented Sep 13, 2015 at 11:40

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