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

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: $$(\mathbf R \mathbf u) \times (\mathbf R \mathbf v) = \mathbf R(\mathbf u \times \mathbf v)$$

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 $$\det(R)=1$$.

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 $$\mathbf R$$ $$(\mathbf R \mathbf u)\times(\mathbf R\mathbf v)=\mathbf R(\mathbf {u \times v})\forall \mathbf {u,v} \text { iff }\det(\mathbf R)=1.$$

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

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