TITLE: Why does $A^TA=I, \det A=1$ mean $A$ is a rotation matrix?
QUESTION [18 upvotes]: I know
if $A^TA=I$, $A$ is an orthogonal matrix. Orthogonal matrices also contain two different types: if $\det A=1$, $A$ is a rotation matrix; if $\det A=-1$, $A$ is a reflection matrix.
My question is: what is the relationship between the determinant of $A$ and rotation/reflection. Can you explain why $\det A=\pm 1$ means $A$ is a rotation/reflection from the geometric perspective?
EDIT:
I think the following questions need be figured out first.

Do rotation matrices only exist in 2D and 3D space? That is: for any dimensional matrix, as long as it is orthogonal and with determinant as 1, the matrix represents a rotation transformation, right? Note the orthogonality and determinant are applicable to arbitrary dimensional matrices.
What is the most fundamental definition of a rotation transformation? 
Since an orthogonal matrix preserves length and angle, can we say an orthogonal matrix represents a "rigid body" transformation? "Rigid body" transformation contains two basic types: rotation and reflection?

REPLY [2 votes]: You could just write out the components to confirm that this is so -- a much more interesting way to understand things, however, is to write down the condition as:
$$A^TIA=I$$
The idea is that the matrix $A$ preserves the identity quadratic form -- note that $I$ is a quadratic form here and not a linear transformation, as this is the transformation law for quadratic forms ($A^TMA$ instead of $A^{-1}MA$).
The hyperconic section corresponding to the identity quadratic form is the unit sphere -- thus the orthogonal transformations are all those that preserve the unit sphere. Another way of putting this is that $(Ax)^TI(Ay)=x^TA^TIAy=x^TIy$, i.e. the Euclidean dot product $I$ is preserved by $A$. This is equivalent to preserving the unit sphere, because the unit sphere is determined by the dot product on the given space.
What sort of transformations preserve the unit sphere? 

The reason this is a good way of understanding things is that there are plenty of other "dot products" you can define. One elementary one from physics is the Minkowski dot product in special relativity, $\mathrm{diag}(-1,1,1,1)$ -- the corresponding quadric surface is a hyperboloid, and the transformations that preserve it, forming the Lorentz group, are boosts (skews between time and a spatial dimension), spatial rotations and reflections.

As for discriminating between rotations and reflections, suppose we define rotations in a completely geometric way -- for a matrix to be a rotation, all its eigenvalues are either 1 or in pairs of unit complex conjugates. 
What do the eigenvalues of orthogonal matrices look like? For each eigenvalue, you need $\overline{\lambda}\lambda=1$, i.e. all the eigenvalues are unit complex numbers. If a complex eigenvalue isn't paired with a corresponding conjugate, you will not get a real-valued transformation on $\mathbb{R}^n$. Meanwhile if an eigenvalue of -1 isn't paired with another -1 -- i.e. if there are an odd number of reflections -- you get a reflection. The orthogonal (or rather unitary) transformations that do not behave this way are precisely the rotations.
The similarity between unpaired unit complex eigenvalues and unpaired -1's is interesting, by the way -- when thinking about reflections, you might have gotten the idea that reflections are $\pi$-angle rotations in a higher-dimensional space -- like the vector was rotated through a higher-dimensional space and then landed on its reflection -- like it was a discrete snapshot of a process as smooth as any rotation. 
Well, now you know what this higher-dimensional space is -- precisely $\mathbb{C}^n$. And the determinant of a unitary matrix also takes a continuous spectrum -- the entire unit circle. In this sense (among other senses) complex linear algebra is more "complete" than real linear algebra.