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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

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    $\begingroup$ Yes, this is true. A good way to prove it is by first showing that $\det R = (u_i\times u_j)\cdot u_k$. $\endgroup$ – user856 Jun 14 '19 at 7:39
  • $\begingroup$ Do you already know that orthonormality implies $u_i\times u_j = \pm u_k$? $\endgroup$ – Arthur Jun 14 '19 at 7:40
  • $\begingroup$ @Arthur, Oh yeah. That's true. Thank you $\endgroup$ – zeeshan khan Jun 14 '19 at 7:43
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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 $\det(R)=1$.

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  • $\begingroup$ So, det$(R) = 1$ is equivalent to $u \times v = w$ for any $R \in \mathbb{R}^{3 \times 3}$ matrix? $\endgroup$ – zeeshan khan Jun 14 '19 at 11:06
  • $\begingroup$ Under the assumption that $RR^T=I$, yes. In other cases, no in general. $\endgroup$ – user1551 Jun 14 '19 at 11:07
  • $\begingroup$ Can you also include the proof of det$(u,v,w) = (u \times v) \cdot w$ for completeness? $\endgroup$ – zeeshan khan Jun 14 '19 at 11:09
  • $\begingroup$ @zeeshankhan There is no proof. It is the definition of cross product. Some high school or university-level engineering textbooks will call this the scalar triple product formula, but in linear or multilinear algebra this is really the definition of cross product. $\endgroup$ – user1551 Jun 14 '19 at 11:09

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