Why is the determinant of a proper rotation matrix 1?

Question

For any rotation matrix, R, I know that $RR^T = I$ and thus $det(R)=±1$. And I understand that proper rotations preserve orientation and that is why a proper matrix, R, is one such that $det(R)= 1$ However, I am unsure how to actually prove this for a rotation matrix of general finite dimension.

Thank you!

Answer 1

If you are talking about 2D or 3D rotations, the reason is simple: every proper rotation $R$ about an axis for some angle is the square of the proper rotation $Q$ about the same axis for the half angle; therefore $\det R=(\det Q)^2$ cannot be negative.

Answer 2

If it preserves orientation, so you begin from identity and get again identity after a full rotation. This means that the proper rotation must contain identity matrix for some special values. The determinant of identity matrix is $+1$. You can check that some sort of transformations like reflection about one axis has determinant $-1$ as it changes orientation.

Answer 3

For a 2D/3D rotation matrix one of the eigenvalues must be $1$ the other $2$ are complex conjugates of unit modulus $e^{\pm i \theta}$. Since the determinant is the product of the eigenvalues then $\det(R)=+1$

(The $2 \times 2$ $2$D rotation matrix is a shorthand for the $3\times3$ matrix rotating about and preserving the z axis)

Answer 4

The determinant of a 2D rotation is clearly one:

$$ det\begin{pmatrix} cos(\theta) & -\sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix} = cos^2(\theta) + sin^2(\theta) = 1$$

A general rotation in N dimensions can be obtained by composing (multiplying) $\begin{pmatrix} N \\ 2 \end{pmatrix}$ (N takes 2) 2-dimensional rotations, because we have a perpendicular plane on which it is possible to rotate for each pair of axes. (For example in 3D we have $\begin{pmatrix} 3 \\ 2 \end{pmatrix} = 3$ perpendicular planes on which to rotate).

Given that a 2D rotation in N dimensions (with identity for all the perpendicular axes) still has determinant 1 (in fact for example):

$$ det\begin{pmatrix} cos(\theta) & -\sin(\theta) & 0\\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix} = cos^2(\theta) + sin^2(\theta) = 1$$

and:

$$ det (A B ) = det (A) det(B) $$

We can see that a rotation in N dimensions still has determinant 1.