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How do I derive an n-dimensional rotation matrix from a geometric perspective? I have read on wikipedia that it preserves distance so that $Q^TQ = I$ but the explanation to be honest isn't very clear. I've had a thorough look on Google and can't find a decent explanation that starts from the geometry first. Also, on Wikipedia (see here: https://en.wikipedia.org/wiki/Rotation_matrix) it says that $det(Q) = 1$ but it isn't clear at all why! Thanks.

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  • $\begingroup$ You may find interesting youtu.be/0rEz9-6trHw $\endgroup$ – mfl Aug 20 '18 at 14:54
  • $\begingroup$ I'm not sure I understand what you mean by "deriving something from a geometric perspective", but you might find this question relevant: math.stackexchange.com/questions/2872052/…. $\endgroup$ – joriki Aug 20 '18 at 14:56
  • $\begingroup$ Beware: there is not any commonly accepted definition of "rotation matrix" when the size of the matrix is larger than 3 by 3. $\endgroup$ – user1551 Aug 20 '18 at 15:19
  • $\begingroup$ You can think about rotation matrices as members of the group of symmetries of a sphere centred at the origin. These are called in general "orthogonal matrices". Not all symmetries of the sphere are called 'rotations' - some are called 'reflections' - see also the comment above me - from user1551. $\endgroup$ – uniquesolution Aug 20 '18 at 16:29
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Suppose $Q$ is an $n\times n$ matrix that preserves distance. Heuristically, that would mean that $$ \Vert Qx-Qy \Vert = \Vert x - y \Vert $$ for all $n$-vectors $x$ and $y$. Now if you use the polarization identity $$ \left<x,y\right> = \frac{1}{4}\left(\Vert x+y \Vert^2 - \Vert x-y\Vert^2\right) $$ you can show that preserving distances is equivalent to preserving the inner product. So $Q$ preserves distances if and only if $$ \left<Qx,Qy\right> = \left<x,y\right> $$ for all $x$ and $y$. Using the defining property of the transpose, we can move it over: $$ \left<Qx,Qy\right> = \left<x,y\right> \implies \left<x,Q^TQy\right> = \left<x,y\right> $$ for all $x$ and $y$. From this follows that $Q^TQ = I$. A nice consequence is that if treat the columns of $Q$ as vectors, they form an othonormal set: each has unit length, and each pair are orthogonal (perpendicular).

Now using determinant properties, you have that $\det Q^2 = \det I = 1$. So $\det Q = \pm 1$.

But this all comes from the distance-preserving property of $Q$. If you consider $\mathbb{R}^n$ as an oriented vector space, you can determine if distance-preserving matrices preserve or reverse the orientation. The condition of preserving the orientation is equivalent to the determinant of $Q$ being $1$.

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  • $\begingroup$ This is a decent answer. But unfortunately I’m confused by the last paragraph. I don’t understand what you mean by an oriented vector space - so you lost me! I looked on google and saw a rough proof that a rotation matrix must have determinant 1 since determinant is a continuous function and thus rotation and identity matrices must have the same determinant (1). However, I’d still like to understand what you mean - so I will google ‘preserving orientation’ so I can understand your last paragraph. Thanks $\endgroup$ – Christian Aug 21 '18 at 2:05
  • $\begingroup$ Fair point. I had to be a bit vague in that last paragraph because, like user1551 says, there isn't really a notion of rotation matrix for dimensions higher than 3 that's independent of the determinant property. I'll try to put in something for $n=2$ that might help. $\endgroup$ – Matthew Leingang Aug 21 '18 at 13:15
  • $\begingroup$ To be honest, I’d prefer if any answer only relates to n by n matrices so that it’s generally applicable. I can mark yours as answer as it’s close enough $\endgroup$ – Christian Aug 21 '18 at 13:17

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