# $A \in SO(3,\mathbb R)\setminus\{I\}$ , then there are exactly two points in $S^2:=\{(x,y,z)\in \mathbb R^3:x^2+y^2+z^2=1\}$ which are fixed by $A$?

Let $A \in SO(3,\mathbb R)\setminus\{I\}$ , then is it true that there exist exactly two points in $$S^2:=\{(x,y,z)\in \mathbb R^3:x^2+y^2+z^2=1\}$$ which are fixed by $A$?

Or equivalently we have to show that $1$ is an eigen-value of $A$ and there are exactly two distinct eigenvectors corresponding to it. Now I can show that $\det (A-I)=0$ and $Ax=x$ implies $A\left(\dfrac x{||x||}\right)=\dfrac x{||x||}$ and $A \left(-\dfrac{x}{||x||}\right)=-\dfrac {x}{||x||}$ also; but how can I show that these are only two fixed points of $A$ on the unit sphere? Please help. Thanks in advance

Hint Consider the eigenvalues of $A \in SO(3, \Bbb R)$. In particular, (1) their product is $\det A = 1$, (2) they all have modulus $1$, and (3) any nonreal eigenvalues come in complex conjugate pairs. Now, if $A \neq I$, what can you say about the $1$-eigenspace of $A$?
• Umm , why should all the eigen-values have modulus $1$ ?
• Multiplication by an orthogonal matrix preserves the length of a vector, so $|A {\bf v}| = |{\bf v}|$ for any vector ${\bf v}$. On the other hand, if ${\bf v}$ is an eigenvector of $A$ of eigenvalue $\lambda$, $|A {\bf v}| = \cdots$. Commented Feb 7, 2016 at 15:43