Infinitesimal Rotation under Orthogonal Similarity Transformation I’m reading charpter 4.8 of Goldstein’s classical mechanics 3rd edition that deals with infinitesimal rotations, and the following is the part I got stuck:

(p.166~167) If $d\boldsymbol{\Omega}$ is to be a vector in the same sense as $\mathbf{r}$, it must transform under $\mathbf{B}$ in the same way. As we shall see, $d\boldsymbol{\Omega}$ passes most of this test for a vector, although in one respect it fails to make the grade. One way of examining the transformation properties of $d\boldsymbol{\Omega}$ is to find how the matrix $\boldsymbol{\epsilon}$ transforms under a coordinate transformation. The transformed matrix $\boldsymbol{\epsilon}’$ is obtained by a similarity transformation:
  \begin{equation} \boldsymbol{\epsilon}’=B\boldsymbol{\epsilon}B^{-1}\end{equation}
  As the antisymmetry property of a matrix is preserved under an orthogonal similarity transformation, $\boldsymbol{\epsilon}’$ consists of nonvanishing elements $d\Omega’_i$ such that \begin{equation}d\Omega'_i=|B|b_{ij}d\Omega_j.\end{equation}

How can I derive the last formula about $d\Omega’_i$? Also, why does the transformation law for an axial vectors which do not change sign under inversion have to be of the form of this formula? I appreciate your help.
 A: How is the angular velocity vector $\vec{\omega}$ is transferred ?
Lets look at this equation:
$$\vec{v}=\vec{\omega}\times \vec{r}=\tilde{{\omega}}\,\vec{r}\tag 1$$
where
$$\tilde{{\omega}}=
 \begin{bmatrix}
  0 & -\omega_z & \omega_y \\
  \omega_z & 0 & -\omega_x \\
  -\omega_y & \omega_x & 0 \\
\end{bmatrix}$$
a antisymmetric skew matrix.
we transformed the vector $\vec{v}$ and  $\vec{r}$ with arbitrary orthogonal transformation matrix $S$ where $\det(S)\ne 0=\pm 1$
$\vec{v'}=S\,\vec{v}$ ,$\vec{r'}=S\,\vec{r}$
and get with  equation (1)
$$\vec{v'}=S\,\vec{v}=S\,\tilde{\omega}\,\vec{r}=S\,\tilde{\omega}\,S^{T}\,\vec{r'}
\overset{!}{=}\tilde{\omega'}\,\vec{r'}
\quad $$
$\Rightarrow$
$$\tilde{\omega'}=S\,\tilde{\omega}\,S^{T
}\tag 2$$
Example:
$$S=\begin{bmatrix}
   -1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
 \end{bmatrix}$$
$\Rightarrow$ equation (2)
$$\begin{bmatrix}
  0 & -\omega'_z & \omega'_y \\
  \omega'_z & 0 & -\omega'_x \\
  -\omega'_y & \omega'_x & 0 \\
\end{bmatrix}=\begin{bmatrix}
   -1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
 \end{bmatrix}\,\begin{bmatrix}
  0 & -\omega_z & \omega_y \\
  \omega_z & 0 & -\omega_x \\
  -\omega_y & \omega_x & 0 \\
\end{bmatrix}\,\begin{bmatrix}
   -1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
 \end{bmatrix}= \begin{bmatrix}
  0 & \omega_z & -\omega_y \\
  -\omega_z & 0 & -\omega_x \\
  \omega_y & \omega_x & 0 \\
\end{bmatrix}$$
$\Rightarrow$
$$\vec{\omega'}=\begin{bmatrix}
   \omega_x \\
   -\omega_y \\
    -\omega_z\\
 \end{bmatrix}=-S\,\vec{\omega}=\det(S)\,S\,\vec{\omega}$$
$$\boxed{\vec{\omega'}=\det(S)\,S\,\vec{\omega}}$$
obviously is the transformation of the angular velocity vector $\vec{\omega}$ , if the determinate of the transformation matrix $S$ not equal one ,different from the transformation of a "regular" vector, this is why the angular velocity vector is a pseudovector.
A: How is the angular velocity vector $~\mathbf{\omega}~$ is transferred ?
with
$$\mathbf \omega=\begin{bmatrix}
\omega_1 \\
  \omega_2\\
  \omega_3
\end{bmatrix}\quad ,\mathbf{\tilde{\omega}}= \left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}}
\\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right]
$$
hence
$$ \mathbf{\tilde{\omega}}_{ij}=-\epsilon_{ijk}\,\mathbf\omega^k$$
where $~\epsilon_{ijk}~$ is the Levi-Civita tensor and no summation over $~ijk~$ indices.
from here you obtain that
$$\epsilon^{ijk}\,\mathbf{\tilde{\omega}}_{ij}=-\underbrace{\epsilon^{ijk}\,\epsilon_{ijk}}_{=1}\,\mathbf\omega^k\\
\mathbf\omega^k=-\epsilon^{ijk}\,\mathbf{\tilde{\omega}}_{ij}$$
for example $~k=1~$ and only $~\epsilon^{231}\ne 0=1~,~\epsilon^{321}\ne 0=-1$
$$\omega^1=-\epsilon^{231}\,\mathbf{\tilde{\omega}}_{23}=\omega_1\\
\omega^1=-\epsilon^{321}\,\mathbf{\tilde{\omega}}_{32}=\omega_1$$
now if you transformed the vector $~\mathbf\omega~$ with the transformation matrix $~\mathbf S~$  you obtain
$$\mathbf \Omega= \pm\,S\,\mathbf\omega=\det(S)\,S\,\mathbf\omega\quad \Rightarrow\\
\widetilde\Omega=\det(S)\,\mathbf{\widetilde{S\,\omega}}=\det(S)\,\mathbf S\,\tilde \omega\,\mathbf{S}^{-1}\quad\Rightarrow$$
$$\Omega^k=
-\det(S)\,\epsilon^{ijk}\,\left(\mathbf{\widetilde{S\,\omega}}\right)_{ij}\quad\quad (1)$$
you can check that equation (1) is equal to
$$\mathbf\Omega=\det(S)\,\mathbf S\,\mathbf\omega\quad\quad (2)$$
Example
$$\mathbf S=\left[ \begin {array}{ccc} -\frac{1}{2}\,\sqrt {2}&0&\frac{1}{2}\,\sqrt {2}
\\  0&1&0\\  \frac{1}{2}\,\sqrt {2}&0&\frac{1}{2}\,
\sqrt {2}\end {array} \right]
\quad,\det(S)=-1
$$
$$\mathbf\omega=\left[ \begin {array}{c} \omega_{{1}}\\ \omega_{{2}
}\\ \omega_{{3}}\end {array} \right]
\quad,
\mathbf{\widetilde{S\,\omega}}=\left[ \begin {array}{ccc} 0&\frac{1}{2}\,\sqrt {2}\omega_{{1}}+\frac{1}{2}\,\sqrt {2
}\omega_{{3}}&-\omega_{{2}}\\ -\frac{1}{2}\,\sqrt {2}\omega_
{{1}}-\frac{1}{2}\,\sqrt {2}\omega_{{3}}&0&-\frac{1}{2}\,\sqrt {2}\omega_{{1}}+\frac{1}{2}\,
\sqrt {2}\omega_{{3}}\\ \omega_{{2}}&\frac{1}{2}\,\sqrt {2}
\omega_{{1}}-\frac{1}{2}\,\sqrt {2}\omega_{{3}}&0\end {array} \right]
$$
with equation (1)
$$\Omega^1=-(\det(S)\epsilon^{231}\,\left(\mathbf{\widetilde{S\,\omega}}\right)_{23}=
\frac{\sqrt{2}}{2}(\omega_1-\omega_3)$$
with equation (2)
$$\mathbf\Omega=\det(S)\,\mathbf S\,\mathbf\omega=
\left[ \begin {array}{c} \frac{1}{2}\,\sqrt {2}\omega_{{1}}-\frac{1}{2}\,\sqrt {2}
\omega_{{3}}\\  -\omega_{{2}}\\ -1/
2\,\sqrt {2}\omega_{{1}}-\frac{1}{2}\,\sqrt {2}\omega_{{3}}\end {array}
 \right]
$$
notice that the transformation matrix $~\mathbf S~$ is orthogonal
