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I proved that the determinant of a rotational matrix(which represents a rotation in 3D) must be $+1$ or $-1$, but I am not sure why it can't be $-1$.

I proved it using the fact that the Matrix must be orthogonal, therefore the determinant squared must be $1$.

projectilemotion
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ghc1997
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  • What does determinant mean to you, geometrically? – Arthur Aug 10 '17 at 08:24
  • What is your definition of rotation matrix? The determinant of $$\left( \begin{matrix} \cos \theta & \sin \theta \ - \sin \theta & \cos \theta \end{matrix} \right)$$ is clearly $+1$ – Crostul Aug 10 '17 at 08:24
  • @SiongThyeGoh I saw that article, but it only mention about in two dimension – ghc1997 Aug 10 '17 at 08:29
  • @Arthur the scalar of change in volume? – ghc1997 Aug 10 '17 at 08:31
  • @projectilemotion the Matrix need to be orthogonal, and the determinant squared must be 1 – ghc1997 Aug 10 '17 at 08:33
  • @Crostul what about the Matrix in 3D that is not of this form? – ghc1997 Aug 10 '17 at 08:34
  • It is a matter of direction. If +1 is counterclockwise, -1 clockwise. – farruhota Aug 10 '17 at 08:36
  • @farruhota is counterclockwise -1? – ghc1997 Aug 10 '17 at 08:38
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    in any dimension, rotations do not change orientation, and this implies the claim. – Francesco Polizzi Aug 10 '17 at 08:38
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    @farruhota No, direction of the rotation itself is not related to the determinant, or even well-defined since a counterclockwise rotation by $\theta$ is the same as a clockwise rotation by $-\theta$. – anon Aug 10 '17 at 09:01
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    @anon, you are right, sorry, it is $\begin{vmatrix} \cos{\theta} & -\sin{\theta} \ \sin{\theta} & \cos{\theta} \ \end{vmatrix}=\begin{vmatrix} \cos{(-\theta)} & -\sin{(-\theta)} \ \sin{(-\theta)} & \cos{(-\theta)} \ \end{vmatrix}=1.$ – farruhota Aug 10 '17 at 09:23

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You can perform a rotation "in time". Let $T$ be the rotation by some angle $\phi>0$ around the axis ${\bf a}$. Then there is a continuous map $$f:\quad[0,\phi]\to O(3),\qquad t\mapsto A(t)$$ with $A(0)={\rm id}$ and $A(\phi)=T$, namely $A(t)$ is the rotation by the angle $t$ around ${\bf a}$. Since ${\rm det}\bigl(A(0)\bigr)=1$ and the determinant is a continuous function on the space of linear maps it follows that ${\rm det}(T)={\rm det}\bigl(A(\phi)\bigr)=1$.

Christian Blatter
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