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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 ?

Shuzheng
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    This is a special case of the formula $$ (Ma)\times(Mb)=\det(M) (M^T)^{-1}(a\times b)\tag{1} $$ for a general $3\times3$ invertible matrix $M$, or a special case of the even more general formula $(Ma)\times(Mb)=\operatorname{cof}(M)(a\times b)$ for a possibly singular $M$. See the thread Cross product: matrix transformation identity for several proofs of $(1)$.

    When $M$ is orthogonal, $M^T=M^{-1}$ and $\det(M)=\pm1$. Hence the result.

     – user1551 Feb 04 '15 at 22:30

2 Answers2

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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.

Jason
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  • Hi Jason. I've tried to use your hints, but after a couple of hours I must give it. Maybe, this exercise is above my geometry-level at the moment. You are welcome to supply the proof, if you like. Or describe it to end. Thanks. – Shuzheng Feb 04 '15 at 14:29
  • I'll edit in a full solution. – Jason Feb 05 '15 at 00:26
  • You can shortcut the last part of the argument by using that the length of $Av\times Aw$ is the area of the parallelogram spanned by $Av$ and $Aw$, which coincides with that area of the parallelogram spanned by $v$ and $w$ since $A$ is orthogonal. So you must have $\alpha=\pm 1$ from the beginning, and then it is just a matter of orientation. – Andreas Cap Feb 09 '15 at 05:46
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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.

Raskolnikov
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