Rotation of a matrix

Question

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!

Answer 1

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}}