Proof that $\mathbf{R}[\omega]_\times\mathbf{R} = [\mathbf{R}\omega]_\times$

Question

I have to prove that

$$\mathbf{R}[\omega]_\times\mathbf{R}^\mathrm{T} = [\mathbf{R}\omega]_\times$$

Herein $\omega$ is a vector with elements. The notation $[\mathbf{a}]_\times$ is a conversion of the vector $\mathbf{a}$ to to a matrix to compute the cross-product e.g. $\mathbf{a} \times \mathbf{b} = [\mathbf{a}]_\times \mathbf{b} = [\mathbf{b}]_\times^\mathrm{T} \mathbf{a}$. With:

$$[\mathbf{a}]_\times = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}$$

Further $\mathbf{R}$ is an 3D rotation matrix.

Because I didn't really know how to proof this I tried just to write everything out in terms of $\mathbf{r}_{ij}$ and $\omega_i$

This gives for $[\mathbf{R}\omega]_\times$: $$\begin{bmatrix} 0 & -r_{31}w_1 - r_{32}w_2 - r_{33}w_3 &r_{21}w_1 + r_{22}w_2 + r_{23}w_3 \\ r_{31}w_1 + r_{32}w_2 + r_{33}w_3 & 0 & -r_{11}w_1 - r_{12}w_2 - r_{13}w_3 \\ -r_{21}w_1 - r_{22}w_2 - r_{23}w_3 & r_{11}w_1 + r_{12}w_2 + r_{13}w_3& 0 \end{bmatrix}$$

And $\mathbf{R}[\omega]_\times \mathbf{R}^\mathrm{T}$ becomes $$\begin{bmatrix} \begin{bmatrix} r_{13}(r_{11}w_2 - r_{12}w_1) - r_{12}(r_{11}w_3 - r_{13}w_1) + r_{11}(r_{12}w_3 - r_{13}w_2) \\ r_{23}(r_{11}w_2 - r_{12}w_1) - r_{22}(r_{11}w_3 - r_{13}w_1) + r_{21}(r_{12}w_3 - r_{13}w_2) \\ r_{33}(r_{11}w_2 - r_{12}w_1) - r_{32}(r_{11}w_3 - r_{13}w_1) + r_{31}(r_{12}w_3 - r_{13}w_2) \end{bmatrix}^T\\ \begin{bmatrix}r_{13}(r_{21}w_2 - r_{22}w_1) - r_{12}(r_{21}w_3 - r_{23}w_1) + r_{11}(r_{22}w_3 - r_{23}w_2) \\ r_{23}(r_{21}w_2 - r_{22}w_1) - r_{22}(r_{21}w_3 - r_{23}w_1) + r_{21}(r_{22}w_3 - r_{23}w_2)\\ r_{33}(r_{21}w_2 - r_{22}w_1) - r_{32}(r_{21}w_3 - r_{23}w_1) + r_{31}(r_{22}w_3 - r_{23}w_2)\end{bmatrix}^T \\ \begin{bmatrix}r_{13}(r_{31}w_2 - r_{32}w_1) - r_{12}(r_{31}w_3 - r_{33}w_1) + r_{11}(r_{32}w_3 - r_{33}w_2) \\ r_{23}(r_{31}w_2 - r_{32}w_1) - r_{22}(r_{31}w_3 - r_{33}w_1) + r_{21}(r_{32}w_3 - r_{33}w_2)\\ r_{33}(r_{31}w_2 - r_{32}w_1) - r_{32}(r_{31}w_3 - r_{33}w_1) + r_{31}(r_{32}w_3 - r_{33}w_2)\end{bmatrix}^T \end{bmatrix}$$

This however leads to nothing. So how do I proof this any tips and help is appriciated?

Quick matlab code

R = sym('r',3);
syms w1 w2 w3;
w = [w1; w2; w3];
W = [  0 -w3  w2; ...
      w3   0 -w1; ...
     -w2  w1   0];
R*w % note: not in [a]_x form
R*W*transpose(R)

Answer 1

For convenience, I write $w$ in place of $\omega$ and $R$ in place of $\mathbf{R}$.

\begin{align*} R[w]_\times R^T = [Rw]_\times \Leftrightarrow& R[w]_\times = [Rw]_\times R \\ \Leftrightarrow& z\cdot(R[w]_\times x) = z\cdot([Rw]_\times Rx)\quad \forall x,z\\ \Leftrightarrow& Rv\cdot(R[w]_\times x) = Rv\cdot([Rw]_\times Rx)\quad \forall v,x\\ \Leftrightarrow& Rv\cdot R(w\times x) = Rv\cdot(Rw\times Rx)\quad \forall v,x\\ \Leftrightarrow& v\cdot (w\times x) = Rv\cdot(Rw\times Rx)\quad \forall v,x\\ \Leftrightarrow& \det(v,w,x) = \det(Rv,Rw,Rx). \end{align*} Now the last line is true because $$\det(Rv,Rw,Rx)=\det\left(R(v,w,x)\right)=\det(R)\det(v,w,x)=\det(v,w,x).$$