TITLE: Let $A$ and $B$ in $O_n(\mathbb{R})$. Show that $A$ and $B$ commute.
QUESTION [9 upvotes]: Let $A$ and $B$ in $O_n(\mathbb{R})$ (orthogonal matrices) such that $|||B-I_n|||<\sqrt{2}$ (subordinate norm) and $A$ commute with $BAB^{-1}$.
Show that $A$ and $B$ commute.

My 'attempt':
I know that $B^{-1}=B^{T}.$
We have $$ABAB^{T}=BAB^{T}A.$$
Since $A,B \in On(\mathbb{R})$ then $AA^{T}=I_n$ and $BB^{T}=I_n$.
Unfortunately I do not see how can I use the fact that $|||B-I_n|||<\sqrt{2}$.
Thank you in advance for your help.

REPLY [4 votes]: If $A$ and $BAB^{-1}$ commute, then they are diagonalizable (over $\mathbb{C}$) in a same orthonormal basis; let $$\mathbb{R}^n = E_1 \overset{\bot}{\oplus} \cdots \overset{\bot}{\oplus} E_r$$ be the associated decomposition. 
If $e_i \in E_i$ then there exists $\lambda_i \in \mathbb{C}$ such that $BAB^{-1}e_i= \lambda_i e_i$, hence $A(B^{-1}e_i)= \lambda_i (B^{-1}e_i)$. Therefore, $B^{-1}$ (and a fortiori $B$) permutes the eigenspaces of $A$, that is the $E_i$'s. 
Suppose by contradiction that there exist $i \neq j$ such that $BE_i = E_j$. Then for any $e_i \in E_i$ satisfying $\|e_i \|=1$,
$$2= \| Be_i \|^2+ \| e_i \|^2 = \| Be_i-e_i \|^2 \leq \| B- \operatorname{Id} \|^2< 2,$$
a contradiction. Therefore, $BE_i=E_i$ for every $1 \leq i \leq r$. Now, it can be deduced that $A$ and $B$ commute.