Prove that two linear map are same --- from “A invitation to 3-d vision" I found a question from the book to prove the following equation:
$A^T\hat{\omega}A=\widehat{A^{-1}\omega}$.
where $\hat{}$ means turn a vector $(x_1, x_2, x_3)$ into a skew-symmetric matrix
$
  \left[\begin{matrix}
   0 & -x_3 & x_2 \\
   x_3 & 0 & -x_1 \\
   -x_2 & x_1 & 0
  \end{matrix}\right]
$
and A is a rotation matrix. How to prove it? One hint the book gives is, prove two linear maps, $A^T\hat{(.)}A$ and $\widehat{A^{-1}(.)}$ are the same.
Thank you for reading the question anyway.

Note that $\widehat{x}$ is the cross product matrix such that $\widehat{x}y=x\times y$ for every vector $y$. Therefore, the problem boils down to proving that
$$
u\cdot\left(A^T(\omega\times Av)\right)=u\cdot\left((A^{-1}\omega)\times v\right)
$$
for every pair of vectors $u$ and $v$. The scalar triple product formula should be useful here.
