The subgroup of $GL_{n}(\mathbb{R})$ inside $B(1, 1)$ is $\lbrace I \rbrace$ 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: 


*

*The only complex eigenvalue of $g$ is $1$

*$g = I+N$ where $N$ is nilpotent

*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)

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 $\theta=0$.
