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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.

Community
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2 Answers2

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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.

Seirios
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  • I am still stuck on your first point, why If $A$ and $BAB^{-1}$ commute then they are diagonalizable over C ? –  May 12 '14 at 14:32
  • I can juste prove that they have a common eigenvector –  May 12 '14 at 14:37
  • It is a classical result, see for example here: http://math.stackexchange.com/questions/56307/simultaneous-diagonalization – Seirios May 12 '14 at 15:09
  • You are right. But I have to prove before that $A$ and $BAB^{-1}$ are diagonalizable over $\mathbb{C}$, no ? For $A$ is ok using trigonalization but for $BAB^{-1}$ ? –  May 12 '14 at 15:13
  • Any element of $O_n(\mathbb{R})$ is diagonalizable over $\mathbb{C}$; in fact, it is sufficient to consider $2 \times 2$-rotation matrices. – Seirios May 12 '14 at 15:19
  • Yes I know that, so are you saying that $BAB^{-1}$ is an element of $O_n(\mathbb{R})$ ? –  May 12 '14 at 15:23
  • Of course: $O_n(\mathbb{R})$ is a group, so $A,B \in O_n(\mathbb{R})$ implies $BAB^{-1} \in O_n(\mathbb{R})$. – Seirios May 12 '14 at 17:15
  • Ahh yes! It slipped my mind. Thank you very much. –  May 12 '14 at 19:30
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    This is really slick... – Gabriel Romon May 12 '14 at 20:45
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Unfortunately I can't give you solid proofs for the following, so this is really just a series of (hopefully) educated guesses....but I suspect the following, after a bit of investigation:

  1. if $\|B-I_n\|<\sqrt{2}$ then $B \in $SO$(n)$, so in other words $B$ is a "rotation". On dimensions 2 and 3 this certainly seems to hold...you can test it on a rotation matrix, and you will also find if you exchange columns so that the transformation is not a rotation you will always get $\|B-I_n\|>\sqrt{2}$. but I am not sure how this generalizes to higher dimensions...
  2. the proof will be complete if you can show that $BAB^T=A$...
  3. I'm guessing the fact that $A$ commutes with $BAB^T$ indicates that it is also in the group SO$(n)$, since SO$(n)$ is also a lie algebra, but I am not sure how this will help you to derive the result.

again some of this may be completely wrong, but could maybe help you to start looking in the right places for answers...

Christiaan Hattingh
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  • How can we prove your first point ?(then $B\in SO(n)$) It is this inequality worries me.. –  May 12 '14 at 10:55
  • Okay, I had not see your edit. I will try to prove this fact tonight. (+1) For the effort. Thanks. –  May 12 '14 at 10:58
  • @Edwin thank you for the question - i am very interested in the answer myself...i certainly hope there are others here who can help contribute – Christiaan Hattingh May 12 '14 at 11:00