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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!

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    $\begingroup$ Well, how do you define orientation and proper rotation? Start from there $\endgroup$ – Hagen von Eitzen Apr 27 '17 at 7:33
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    $\begingroup$ I think you need to explain what is meant by "a rotation matrix of general finite dimension" first. Some people (me included) refer to matrices in $SO(n;\mathbb R)$ as "rotation matrices" (note: this is a nonstandard usage; many people reject it when $n>3$), but in this case, $\det R=1$ by definition. $\endgroup$ – user1551 Apr 27 '17 at 7:48
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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.

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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.

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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)

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