Orthogonal Matrices And Rigid Body Dynamics Why can't orthogonal matrices whose determinant equal to $-1$ correspond to physical rotation or displacement of a Rigid Body?
 A: Orthogonal matrices have a determinant either equal to 1 or to -1. We need to be able to start from the identical matrix because it encodes "no motion". Then the motion of the rigid body involves a continuous path from the identity matrix to the orthogonal matrix describing the final state of rotation, a continuous path through the set of orthogonal matrices I mean. Since the determinant is a continuous function, it is impossible to end up on an orthogonal matrix of determinant -1 because that would involve a discontinuous jump from 1 to -1 at some point.
But actually, this is only half the truth: we could decide to start from a mirror image of some definite position X, i.e. start not from the identity matrix but from the matrix of a mirror. Then the motion would consist in a continuous path through the set of orthogonal matrix of determinant -1. Necessarily so, since again a discontinuous jump would be needed to reach a matrix of determinant 1. The final state would then be also a matrix of determinant -1, corresponding to the mirror image of another definite position Y. But then, the whole motion would just be the mirror image of a motion starting from X and ending in Y, i.e. a motion corresponding to a path through orthogonal matrices of determinant 1, starting from the identity matrix.
So eventually, the second scenario is usually not considered because the added complexity brings nothing to the table. But the lesson is that the only requirement is that either we stay all the time with orthogonal matrices of determinant 1, or we stay all the time with orthogonal matrices of determinant -1, but that we cannot jump between those two sets.
A: An orthogonal matrix $S$ with determinant $-1$ can always be written as
$$S=-IR,$$
where $I$ is the identity matrix and $R$ is a rotation matrix, i.e., $R\in\mathrm{SO}(3)=\{R\in\mathrm O(3);\mathrm{det}R=1\}$.
The matrix $-I$ actually corresponds to a reflection. In the passive point of view it reflects the coordinate axes $x_i\rightarrow -x_i$. There is no way this can be obtained by a rotation. Thus, the matrix $S$ corresponds to a rotation followed by a reflection and therefore it cannot represent a possible motion of the rigid body.
