Geometric intuition behind an n-dimensional rotation matrix

Question

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.

Answer 1

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 $1$.