# Is the _unit determinant_ constraint of a rotation matrix equivalent to handedness preservation constraint?

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

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

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 $$\det(R)=1$$.
• So, det$(R) = 1$ is equivalent to $u \times v = w$ for any $R \in \mathbb{R}^{3 \times 3}$ matrix? – zeeshan khan Jun 14 '19 at 11:06
• Under the assumption that $RR^T=I$, yes. In other cases, no in general. – user1551 Jun 14 '19 at 11:07
• Can you also include the proof of det$(u,v,w) = (u \times v) \cdot w$ for completeness? – zeeshan khan Jun 14 '19 at 11:09