Property of SO(3) 
Suppose $A\in SO(3).$
  Show that there exists a vector $v\in \mathbb{R}^3$ such that $Av=v$.

$ SO(3)={{A\in O(3)|detA=1}} $
and
$ O(3)={A:\mathbb{R}^3\rightarrow \mathbb{R}^3}|<Au,Av>=<u,v> for\space all\space u,v\in \mathbb{R}^3 $
I'm thinking proof by contradiction:
Suppose $Av\neq v \space \forall v$.
But I have no clue how properties of $SO(3) $ can help from here.
Would anyone give me any suggestions for the first step?
 A: We know that $A^T=A^{-1}$ and $\det A=1$. Therefore
$$
\begin{aligned}
\det(A-I)&=\det(A-AA^T)=\det(A(I-A^T))\\
&=\det(A)\det(I-A^T)\\
&=\det(I-A^T)=\det(I-A^T)^T\\
&=\det(I-A)\\
&=(-1)^3\det(A-I)=-\det(A-I),
\end{aligned}
$$
where the last equality is because $A$ has three rows.
Therefore $\det(A-I)=0$ and $\lambda=1$ is thus an eigenvalue.
A: You are asking whether $A$ has an eigenvalues 1:


*

*From the property $\langle A u, A v\rangle$ you can follow that the eigenvalues are on the unit circle.

*As the matrix is real, the eigenvalues come in complex conjugate pairs.

*So the eigenvalues are either $1$ or $-1$ or pairs $\lambda,\lambda^*$ with $\lambda\in\mathbb{C},$ $|\lambda|=1$ and $\lambda\neq \lambda^*$.

*The determinant is $\det A=1$, so $\lambda_1 \lambda_2 \lambda_3 =1$ with $\lambda_i$ the eigenvalues of $A$.

*Assume all $\lambda_i \neq 1$. There are either no pairs of eigenvalues, but then $\det A = (-1)^3=-1$. Alternative there is one pair and one eigenvalue $-1$, but then again $\det A = (-1) \lambda \lambda^* =-1$. 
So your matrix has an eigenvalue $1$.
A: The notion of orthogonal reflection will be used, see http://en.wikipedia.org/wiki/Reflection_%28mathematics%29


*

*Let $v$, $w$ two vectors on the unit sphere. Then there exist a reflection wr to a plane passing through the origin that takes $v$ to $w$. ( in fact, if $v\ne w$  the reflection is unique). If $v=w$ take the reflection wr to any plane containing $v$. If $v \ne w$ take the reflection wr to the plane passing through the origin and perpendicular to $v-w$.

*Let $v$, $w$ and $u$ three vectors on the unit sphere so that $\angle (v,u) = \angle (w,u)$. Then there exists a reflection that fixes $u$ and takes $v$ to $w$.  Apply the previous construction. 

*Let $v_1$, $v_2$ and $w_1$, $w_2$ points on the sphere so that $\angle (v_1,v_2) = \angle (w_1,w_2)$. Then there exists two reflections $R'$, $R$ so that the composition $R'\circ R$ takes $v_i$ to $w_i$, $i=1,2$. Indeed, let $R$ taking $v_1$ to $w_1$. Now $R v_2= w_2'$. Therefore we have $\angle (v_1 ,v_2) = \angle ( w_1, w_2')$ since $R$ preserves the angles. However, we have by hypotheses $\angle (v_1 ,v_2) = \angle ( w_1, w_2)$. Therefore, we can find $R'$ so that $R'w_1 = w_1$ and $R' w'_2= w_2$. We conclude that $R'\circ R v_i = w_i$ for $i=1,2$.

*Up to now we have not used that the dimension of the space is $3$. We assume from now on that the dimension of the space is $3$. Let's first note that if $B$ is a transformation in $O(3)$ that fixes each point of a plane passing through the origin then $B$ is either the identity or the reflection with respect to that plane. If moreover $B$ is in $SO(3)$ then $B$ is the identity transformation. 

*Let $A$ a transformation in $SO(3)$. Then $A$ is the composition of two reflections. Indeed, let $v_1$, $v_2$ non collinear vectors on the sphere and $w_i = A v_i$. Then $\angle (v_1, w_1) = \angle (v_2, w_2)$. By 3. there exist reflections $R$, $R'$ so that $R'\circ R v_i = w_i$. We claim that $A = R' \circ R$. Indeed, the transformation in $SO(3)$ $(R' \circ R )^{-1} \circ A$ fixes both vectors $v_1$, $v_2$ and therefore is the identity, by 4.

*The composition of two reflections wr to two distinct planes through the origin is the identity if the planes are identical, or a rotation wr to the axis that is the intersection of the two planes otherwise. The angle of rotation is double the angle between the planes.
All the above reasoning uses the spherical (hyperspherical ) geometry. One can effectively compose two rotations by noticing that for a two rotations $\rho_1$, $\rho_2$ we can choose a writing $\rho_1 = R'' \circ R'$ and $\rho_2 = R' \circ R$ ( each product of two reflections) and therefore $\rho_1 \circ \rho_2 = R'' \circ R$. This is based on spherical geometry.
A: The question in essence asks to show that $1$ is an 
eigenvalue of any
$O \in SO(3); \tag 0$
for such $O$ we have
$\langle Ox, Ox \rangle = \langle x, x \rangle; \tag 1$
if $\mu$ is an eigenvalue of $O$ there is a vector
$0 \ne x \in \Bbb R^3 \tag{1.5}$
with
$Ox = \mu x; \tag 2$
then from (1),
$\langle \mu x, \mu x \rangle = \langle x, x \rangle; \tag 3$
now $O$ is a $3 \times 3$ real matrix; thus its characteristic polynomial is a real polynomial of degree $3$, and as such has at least one real root; so we may assume
$\mu \in \Bbb R; \tag 4$
then
$\mu^2 \langle x, x \rangle = \langle x, x \rangle, \tag 5$
and in light of $x \ne 0$ we infer
$\mu^2 = 1 \Longrightarrow \mu = \pm 1; \tag 6$
now if
$\mu = 1, \tag 7$
done!!! Otherwise we note that
$\det O = 1; \tag 8$
so with 
$\mu = -1, \tag 9$
the two other eigenvalues $\nu_1$, $\nu_2$ satisfy
$\nu_1 \nu_2 = -1, \tag{10}$
since
$\mu \nu_1 \nu_2 = \det O; \tag{10.5}$
if, say,
$\nu_1 \in \Bbb C \setminus \Bbb R, \tag{11}$
then, again since the characteristic polynomial of $O$ lies in $\Bbb R[x]$,
$\nu_2 = \bar \nu_1, \tag{12}$
and so
$\nu_1 \nu_2 = \nu_1 \bar \nu_1 > 0, \tag{13}$
contradicting (10); thus as in (4)-(6),
$\nu_1, \nu_2 \in \Bbb R, \tag{14}$
and so
$\nu_1 = \pm 1 = \nu_2; \tag{15}$
then in light of (10) one of $\nu_1$, $\nu_2$ is $1$ and we are done!!!
