For a matrix $R \in \mathbb{R}^{3 \times 3}$ to be a proper rotation $\in$ SO($3$), it has to satisfy two constraints:
- Orthonormality: $R^TR = RR^T = I$
- Orientation preservation: det($R$) = $1$
Some sources write orientation preservation constraint as if $R = [u_1, u_2, u_3]$, then orientation preservation: $$u_i \times u_j = u_k $$ , where $\times$ denotes the vector cross product and $(i,j,k) = \text{cycle}(1,2,3)$.
I am trying to preve that these two (the unit determinant and the crossproduct) constraints are equivalent for $3$-dimensional case, but no progress yet. Can anyone help me
Since the conditions $RR^T=I$ and $\det(R)=1$ are preserved under a cyclic permutation of the columns of $R$, we may assume without loss of generality that $(i,j,k)=(1,2,3)$. For convenience, let us drop the subscripts and write $R=[u,v,w]$. The cross product of $u$ and $v$ is defined as the unique vector $u\times v$ such that $$ \det(u,v,r)=(u\times v)\cdot r\quad\forall r\in\mathbb R^3. $$ It follows that $$ R^T(u\times v)=\pmatrix{u^T\\ v^T\\ w^T}(u\times v) =\pmatrix{(u\times v)\cdot u\\ (u\times v)\cdot v\\ (u\times v)\cdot w} =\pmatrix{\det(u,v,u)\\ \det(u,v,v)\\ \det(u,v,w)} =\pmatrix{0\\ 0\\ \det(R)}=\det(R)\pmatrix{0\\ 0\\ 1} $$ and hence $$ u\times v=RR^T(u\times v)=\det(R)R\pmatrix{0\\ 0\\ 1}=\det(R)w.\tag{1} $$ As $R$ is non-singular, $w\ne0$. Therefore $(1)$ shows that $u\times v=w$ if and only if $\det(R)=1$.