Prove that two linear map are same --- from “A invitation to 3-d vision"

Question

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.

Answer 1

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.

Answer 2

Thanks for the answer of "user1551". From his help I come up with a way. But I don't use the scalar triple product formula(may be I haven't totally understand his proof). My proof is:

for the equation $A^T\hat{\omega}A=\widehat{A^{-1}\omega}$, multiply vector u on both right, we need to prove: $$ A^T(\omega \times(Au)) = (A^{-1}\omega)\times u$$ Since $A^T$ is a rotate matrix, and it preserve the angle between vectors, we got $$ A^T(\omega \times(Au)) = (A^T\omega) \times (A^TAu) = (A^{-1}\omega)\times u $$ So it is proved.