How to prove $A( v \times w)$ = $\det A\cdot(Av \times Aw)$ , where $A\in \text {Mat}_{3 \times 3}(\mathbb R)$ is an orthogonal matrix. How to prove $A( v \times w)$ = $\det A\cdot(Av \times Aw)$, where $A\in \text {Mat}_{3 \times 3}(\mathbb R)$ is an orthogonal matrix ($\det A = \pm 1$) and $v,w\in \mathbb R^3$.
I've tried writing out the left hand side by using the definition of the cross-product and matrix-multiplication. However, I can't collect these terms into the right hand side.
Can anyone describe or give the proof ?
 A: You could just battle with all the nasty algebra, but I would recommend a different approach. The result is trivial if $v$ and $w$ are scalar multiples of one another, so assume they are linearly independent. Then $Av\cdot A(v\times w)=v\cdot(v\times w)=0$ and similarly $Aw\times A(v\times w)=0$, so $A(v\times w)$ is normal to the plane spanned by $Av$ and $Aw$. But so is $Av\times Aw$, so there exists a constant $\alpha\in\mathbb{R}$ such that
$$A(v\times w)=\alpha(Av\times Aw).$$
Now if $\theta$ is the angel between $v$ and $w$, then
$$\cos\theta=\frac{v\cdot w}{|v||w|}=\frac{Av\cdot Aw}{|Av||Aw|}$$
where we used $|Av|=\sqrt{Av\cdot Av}=\sqrt{v\cdot v}=|v|$. Hence $\theta$ is also the angle between $Av$ and $Aw$. This implies
$|Av\times Aw|=|Av||Aw|\sin\theta=|v||w|\sin\theta=|v\times w|=|A(v\times w)|$, so $\alpha=\pm1$. We have $\alpha=1$ if and only if $(Av,Aw,A(v\times w)$ is right-handed, which is true if and only if
$$0<\det[\begin{array}{ccc}Av&Aw&A(v\times w)\end{array}]=\det(A)\det[\begin{array}{ccc}v&w&v\times w\end{array}],$$
that is, if and only if $\det(A)>0$. Otherwise, $\alpha=-1$. Since $\det(A)=\pm1$ we get the required result.
A: It's relatively easy if you use the Levi-Civita symbol $\epsilon_{ijk}$ and the following identities
$$\sum_{ijk}\epsilon_{ijk}A_{ai}A_{bj}A_{ck}=\text{det}A \; \epsilon_{abc}$$
and
$$(v\times w)_i=\sum_{jk} \epsilon_{ijk} v_j w_k \; .$$
Then 
$$(A(v\times w))_a = \sum_{ijk} A_{ai}\epsilon_{ijk} v_j w_k = \sum_{ijk} A_{ai}\epsilon_{ijk} \sum_{t}\delta_{jt}v_t \sum_{s}\delta_{ks}w_s \; .$$
At the end, I introduced Kronecker's delta. Since $A$ is an orthogonal matrix, $A^TA=\mathbb{I}$ or in index notation
$$\sum_{b}A_{bj}A_{bt}=\delta_{jt}$$
which brings us to
$$(A(v\times w))_a=\sum_{ijkbcts}A_{ai}\epsilon_{ijk}A_{bj}A_{bt}v_tA_{ck}A_{cs}w_s$$
or after rearranging factors
$$(A(v\times w))_a=\sum_{ijkbcts}\epsilon_{ijk}A_{ai}A_{bj}A_{ck}A_{bt}v_t A_{cs}w_s=\sum_{ijkbc}\epsilon_{ijk}A_{ai}A_{bj}A_{ck}\sum_t A_{bt}v_t \sum_s A_{cs}w_s$$
The last sums are $(Av)_b$ and $(Aw)_c$, thus
$$(A(v\times w))_a=\sum_{bc}\left(\sum_{ijk}\epsilon_{ijk}A_{ai}A_{bj}A_{ck}\right)(Av)_b (Aw)_c = \text{det}A \sum_{bc}\epsilon_{abc}(Av)_b (Aw)_c$$
in which we used our very first identity. Finally, using the second identity
$$(A(v\times w))_a = \text{det}A \; (Av \times Aw)_a \; .$$
which implies the sought after identity.
