Why does $A^TA=I, \det A=1$ mean $A$ is a rotation matrix? 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?


This depends on how we want to define "rotation" in the first place. Some people prefer a narrow definition where the only things that qualify as "rotations" are things that can be expressed as a $(2+n)\times(2+n)$ block matrix
$$\pmatrix{\pmatrix{\cos\theta&\sin\theta\\-\sin\theta&\cos\theta}&0\\0&I}$$ with respect to some orthogonal basis. Under this definition there are $4\times 4$ matrices that are orthogonal and have determinant 1 but are not rotations - for example,
$$\pmatrix{0.6&0.8&0&0\\-0.8&0.6&0&0\\0&0&0&1\\0&0&-1&0}$$
But one might also say that a "rotation" is any matrix $A$ such that there is a continuous family of matrices $(A_t)_{0\le t\le 1}$ such that $A_0=I$, $A_1=A$ and $A_t$ is always orthogonal. This captures the idea of a gradual isometric transformation away from a starting point. Such a definition immediately tells us that the determinant of a rotation must be 1 (because the determinant is continuous), but it is harder to see that we get all orthogonal matrices with determinant 1 this way.
I don't have a quick proof of the latter, but I imagine that it can be done fairly elementarily by induction on the dimension, first a series of rotations to make the first column fit, then sort out the remaining columns recursively working within the orthogonal complement of the first column.
