Prove there exist a non-zero vector $v$ such that $Av=v$. I need to prove that if $A^t=A^{-1}$, $\det(A)>0$ for a matrix $A_{3\times3}$ then exist a vector $v$ non-zero such that $Av=v$.
I tried taking a general matrix A,
$
\left( \begin{array}{cccc}
 a_{11} & a_{12} & a_{13} \\ 
 a_{21} & a_{22} & a_{23} \\
  a_{31}& a_{32} & a_{33} 
\end{array} \right) $
but I don't get anything, and I tried using the definition of eigenvalue $\lambda$ and eigenvector $v$, $\{v \in V \mid Av=\lambda v\}$, and I did not get anything important, so, any hint is very valuable, thanks!
 A: If $v\in\mathbb R^3$, then\begin{align}\lVert Av\rVert^2&=\langle Av,Av\rangle\\&=\langle v,A^tAv\rangle\\&=\langle v,v\rangle\text{ (since $A^t=A^{-1}$)}\\&=\lVert v\rVert^2\end{align}and therefore $\lVert Av\rVert=\lVert v\rVert$. On the other hand, the characteristic polynomial of $A$ is a cubic polynomial and therefore it has a real root $\lambda$. So, $\lambda$ is an eigenvalue of $A$; let $v$ be a corresponding eigenvector. We have\begin{align}\lVert v\rVert&=\lVert Av\rVert\\&=\lVert\lambda v\rVert\\&=\lvert\lambda\rvert.\lVert v\rVert\end{align}and therefore $\lambda=\pm1$. If $\lambda=1$, we're done. If $\lambda=-1$, let $W=v^\perp=\{w\in\mathbb R^3\,|\,\langle v,w\rangle=0\}$. Then $A.W\subset W$. The determinant of $A|_W$ must be negative, since $0<\det(A)=(-1)\times\det(A|_W)$. But a $2\times2$ matrix with negative determinant always has eigenvalues (the discriminant of the characteristic polynomial is greater than $0$). Therefore one of the eigenvalue of $A|_W$ is positive and the other one is negative. Since we found a positive eigenvalue, and since we have already proved that the only positive eigenvalue that $A$ can have is $1$, we're done.
A: If $A^tA=I$, $\det(A)>0$ for a  matrix $A_{3\times3}$  then obviously $A$ is   some rotation matrix. 
Such matrix has many representations, one of them  is    Rodrigues formula.
$A=I+\sin\theta K(v)+(1-\cos\theta) K^2(v)$ 
where $K(v)$ is skew-symmetric matrix linked with some vector $v$. 
Image for this matrix is orthogonal to $v$, consequently by direct calculations we have from the formula $Av=v$.
A: Such matrices are called orthogonal.  
It is well known that orthogonal matrices act as isometries of Euclidean space,  such as rotations, reflections or rotoreflections. 
Read about them here.
In fact,  we have a rotation,  since $\operatorname{det}A\gt0$. (We have an element of $\operatorname{SO}(3)$, the special orthogonal group.)  The claim follows.
