how to derive determinant of a rotation matrix is 1 from rotation preserving orientation

Question

According to the norm preservation of rotation matrix, we get: $$ (\mathbf R \mathbf u)^{\top}(\mathbf R \mathbf v) = \mathbf u^{\top} \mathbf v; \forall \mathbf u,\mathbf v\in \mathbb R^3 \text{and } \mathbf R \in \mathbb R^{3 \times 3} \\ \Rightarrow \mathbf R^{\top}\mathbf R = \mathbf I \\ \Rightarrow \det(\mathbf R) = \pm 1 $$ My question is how to derive $\det(\mathbf R) = 1$ further with rotation preserving orientation: $$ (\mathbf R \mathbf u) \times (\mathbf R \mathbf v) = \mathbf R(\mathbf u \times \mathbf v) $$

Answer 1

If $Ru\times Rv=R(u\times v)$, then $(Ru\times Rv)\cdot Rw=R(u\times v)\cdot Rw$ for every vector $w$. However, by the scalar triple product formula $(x\times y)\cdot z = \det(x,y,z)$, we have \begin{cases} (Ru\times Rv)\cdot Rw =\det(Ru,Rv,Rw) =\det(R)\det(u,v,w),\\ R(u\times v)\cdot Rw =(u\times v)\cdot w =\det(u,v,w). \end{cases} Therefore $\det(R)\det(u,v,w)=\det(u,v,w)$ for every $u,v,w\in\mathbb R^3$. In particular, if $u,v,w$ are linearly independent, we get $\det(R)=1$.

Answer 2

There’s probably a more elegant, coordinate-free way to do this, but here’s one way:

$$\det\mathbf R=\sum_{ijk}\epsilon_{ijk}R_{i1}R_{j2}R_{k3}\;,$$

where $\epsilon$ is the Levi–Civita symbol, and

$$ (\mathbf u\times\mathbf v)_i=\sum_{jk}\epsilon_{ijk}u_jv_k\;. $$

The condition for preserving orientation becomes

$$ \sum_{jk}\epsilon_{ijk}\left(\sum_lR_{jl}u_l\right)\left(\sum_mR_{km}v_m\right)=\sum_nR_{in}\left(\sum_{lm}\epsilon_{nlm}u_lv_m\right)\;. $$

As this holds for all $\mathbf u$ and $\mathbf v$, we can compare coefficients to obtain

$$ \sum_{jk}\epsilon_{ijk}R_{jl}R_{km}=\sum_nR_{in}\epsilon_{nlm}\;. $$

Now set $l=1$, $m=2$:

$$ \sum_{jk}\epsilon_{ijk}R_{j1}R_{k2}=\sum_nR_{in}\epsilon_{n12}\;. $$

The right-hand side is $R_{i3}$, so

$$ \sum_{jk}\epsilon_{ijk}R_{j1}R_{k2}=R_{i3}\;. $$

Multiply by $R_{i3}$ and sum over $i$ to obtain

$$ \det\mathbf R=\sum_{jki}\epsilon_{ijk}R_{j1}R_{k2}R_{i3}=\sum_iR_{i3}^2\ge0\;. $$

Answer 3

If $\mathbf R$is an orthogonal matrix, $$\mathbf R=\text {cof}(\mathbf R)/\det(\mathbf R). $$ $$\text { Thus if } \det(\mathbf R)=1, \text { then }\mathbf R=\text {cof}(\mathbf R).$$ $$\text { If } \det(\mathbf R)=-1, \text { then }\mathbf R=-\text {cof}(\mathbf R)$$ For any 3x3 matrix $M$ and any 3-element column-vectors $\mathbf {u,v},$ $$(M \mathbf u)\times(M\mathbf v)=\text {cof}(M)(\mathbf {u \times v})$$ Thus, for a 3x3 orthogonal matrix $\mathbf R,$ $$(\mathbf R \mathbf u)\times(\mathbf R\mathbf v)=\mathbf R(\mathbf {u \times v})\forall \mathbf {u,v} \text { if }\det(\mathbf R)=1 \text { and}$$ $$(\mathbf R \mathbf u)\times(\mathbf R\mathbf v)=-\mathbf R(\mathbf {u \times v})\forall \mathbf {u,v} \text { if }\det(\mathbf R)=-1.$$ Suppose $\mathbf R $ is a 3 x 3 orthogonal matrix,
$$(\mathbf R \mathbf u)\times(\mathbf R\mathbf v)=\mathbf R(\mathbf {u \times v})\forall \mathbf {u,v}$$ and $\det(\mathbf R)=-1.$ We shall derive a contradiction. Let $\mathbf {i,j,k}$ be the standard basis, written as column-vectors, in $\mathbb R^3$ and let $\mathbf {u=i,v=j}$. Then $$(\mathbf R \mathbf i)\times(\mathbf R\mathbf j)=\mathbf R(\mathbf {i \times j})=\mathbf {Rk}$$ and $$(\mathbf R \mathbf i)\times(\mathbf R\mathbf j)=-\mathbf R(\mathbf {i \times j})=-\mathbf {Rk}$$ So $\mathbf {Rk}=-\mathbf {Rk}$. Since $\mathbf R$ is non-singular, $\mathbf {k=-k},$ an absurdity. Thus, for a 3x3 orthogonal matrix $\mathbf R$ $$(\mathbf R \mathbf u)\times(\mathbf R\mathbf v)=\mathbf R(\mathbf {u \times v})\forall \mathbf {u,v} \text { iff }\det(\mathbf R)=1.$$