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Suppose you have a symmetric real 3x3 Matrix $S$ and an orthogonal matrix $O$ such that $O$ commutes with $S$, i.e. $OS = SO$. Suppose that $O$ is a nontrivial rotation about an axis in direction of $n \in \mathbb{R}^3$, i.e. $On = n$ and $O \neq \mathrm{id}$.

Is it true that $n$ must be an eigenvector of $S$? If so, how to prove it?

If it is not true, are there slight modifications of the suppositions which make it true?

If it is true, is there a simple way to generalize this statement?

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Ok, so the point is that the only eigenspace of $O$ is $\mathbb{R}n$? So I can drop the symmetry requirement for $S$? –  student Jun 7 '12 at 9:01
    
Remark that $O$ has just one real eigenvalue: $1;$ and its eigenspace is $\mathbb{R}n.$ From $On=n$ and $SO=OS$ we derive $OSn=SOn=Sn.$ Therefore $Sn$ is in the eigenspace $\mathbb{R}n$ of $O$ i.e.: there exists a unique $\lambda\in\mathbb{R}$ s.t. $Sn=\lambda n.$ This means: $n$ is eigenvalue of $S.$ –  Giuseppe Tortorella Jun 7 '12 at 9:01
    
Could you turn your comment into an answer, then I can accept it. –  student Jun 7 '12 at 9:06

1 Answer 1

up vote 2 down vote accepted

Remark that $O$ has just one real eigenvalue: $1;$ and its eigenspace is $\mathbb{R}n.$

From $On=n$ and $SO=OS$ we derive $OSn=SOn=Sn.$
Therefore $Sn$ is in the eigenspace $\mathbb{R}n$ of $O$ i.e.: there exists a unique $\lambda\in\mathbb{R}$ s.t. $Sn=\lambda n.$ This means: $n$ is eigenvalue of $S.$

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