Rotation of a matrix Sorry for boring you my friends before the holiday. I am haunted by a question of rotation of a matrix. Suppose that we have a special matrix $\Omega$ takes the form of:
$\Omega =  \left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}}
\\ \omega_{{3}}&0&-\omega_{{1}}\\ 
-\omega_{{2}}&\omega_{{1}}&0\end {array} \right] 
$
and there is a abstract rotation matrix $R$ which verifies the property of $R^T=R^{-1}$. 
I would like to find out the geometry meaning behind the following equation: $R^T\cdot \Omega \cdot R$.
I tried to give an answer. This is the trail:
Firstly, I extracted a vector $\omega$ from $\Omega$ and it is given:
$\omega =  \left[ \begin {array}{c} \omega_{{1}}\\\omega_{{2}
}\\\omega_{{3}}\end {array} \right] 
$
, 
Then, I multiplied $\omega$ by $R^{T} $ on the left side and got $\bar{\omega}=R^{T}\cdot\omega$,  the $\bar{\omega}$ takes the form of:
$\bar{\omega} =  \left[ \begin {array}{c} \bar{\omega}_{{1}}\\\bar{\omega}_{{2}
}\\\bar{\omega}_{{3}}\end {array} \right] 
$
Finally, I returned the $\bar{\omega}$ back into the matrix form as $\bar{\Omega}$:
$\bar {\Omega} =  \left[ \begin {array}{ccc} 0&-\bar{\omega}_{{3}}&\bar{\omega}_{{2}}
\\ \bar{\omega}_{{3}}&0&-\bar{\omega}_{{1}}\\ 
-\bar{\omega}_{{2}}&\bar{\omega}_{{1}}&0\end {array} \right] 
$
Several simple numerical applications have been conducted and verified the above mathematical conjecture which is $\bar {\Omega} = R^T\cdot \Omega \cdot R$.
But I failed to demonstrate this conjecture mathematically, or probably the conjecture is wrong.
Thank you in advance for taking a look. Nice holiday!
 A: The matrix $\Omega$ has matrix elements of the following form:
$$\Omega_{i,j} = \sum_{k=1}^3 \epsilon_{i,j,k} \omega_k $$
where $\epsilon$ is the absolutely anti-symmetric Levi-Civita tensor.
For column vectors $x$, $y$ and $z$, the determinant of the matrix with these columns is
$$\det[x \mid y \mid z] = \sum_{i,j,k} \epsilon_{i,j,k} x_i y_j z_k$$
As such, the Levi-Civita tensor remain invariant under rotations:
$$
   \left(R \Omega R^t\right)_{i,j} = \det \left[ R_{i,\cdot}, R_{j,\cdot}, \omega \right] = \det \left[ R_{i,\cdot}, R_{j,\cdot}, R^t \left(R \omega\right) \right] = \det \left[ e_{i}, e_{j}, R \omega \right] 
$$
That is the vector $\omega$ under conjugation of $\Omega$ by rotation matrices simply gets the same rotation.
Confirming in Mathematica:
In[45]:= With[{axisVec = {1, 0, 0}}, 
 RotationMatrix[\[Theta], 
     axisVec].(LeviCivitaTensor[3].{Subscript[w, 1], Subscript[w, 2], 
       Subscript[w, 3]}).RotationMatrix[-\[Theta], axisVec] -
   (LeviCivitaTensor[3].RotationMatrix[\[Theta], axisVec].{Subscript[
      w, 1], Subscript[w, 2], Subscript[w, 3]}) // Simplify]

Out[45]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}

