Show that if $A ∈ O(2)$ has $|A| = 1$, then $A$ = the rotation matrix.

Question

Show that if $A ∈ O(2)$ has $|A| = 1$, then $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$.
I read in "Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix" That there is a proof using the unit circle, however I was curious if there is a way to do it without the unit circle.

Answer 1

I'm not sure exactly what you mean by "a proof [not] using the unit circle," so I hope this post is actually useful to you.

The goal will be to explain why an orthogonal matrix with determinant $1$ acts as a rotation about the origin, without talking about the structure of entries of the matrix at all. We only need to talk about circles insofar as we can't possibly talk about rotations without (at least implicitly) mentioning circles.

We can simply characterize rotation about the origin as follows:

1. it preserves norm
2. it's linear
3. it preserves orientation of the vector space

I leave it to you to check that these conditions are necessary (any rotation possesses these properties). As for sufficiency:

1. Condition $1$ says that a vector on the circle of radius $r$ is mapped by $A$ to somewhere else on the circle of radius $r$.

2. Condition $2$ will firstly imply continuity (linear maps between finite dimensional vector spaces are continuous), which we clearly need, but most importantly... Condition $2$ taken with condition $1$ implies that any two points $x,y$ on a circle of radius $r$ are actually rotated: the distance between them remains unchanged. To see this consider that if $R$ is linear and norm preserving, then $|Rx - Ry| = |R(x-y)| = |x-y|$. Then if $x$ is $\theta$ radians from $y$ on the circle with corresponding euclidean distance $d$, applying $R$ to $x$ and $y$ preserves the extrinsic distance $d$ between them and also by symmetry the intrinsic distance $\theta$. The only worry is that originally $x$ was $\theta$ radians from $y$ measured clockwise, but after applying $R$ it is $\theta$ radians from $y$ measured counterclockwise.

3. That's why we require condition $3$ to ensure we have a pure rotation and not a reflection.

So, we have characterized rotation about the origin as a linear map that preserves orientation and norm.

Now suppose $A$ is orthogonal with determinant $1$. We'll show that $A$ satisfies all three conditions. First off, condition $2$ is satisfied just because matrix multiplication is linear.
Since $A$ is orthogonal ($A^T A = I$), we have that $A$ preserves dot products: $\langle Ax, Ay \rangle = (Ax)^T (Ay) = x^T A^T A y = x^T I y = \langle x, y \rangle$. Condition $1$ follows since $|Ax|^2 = \langle Ax, Ax\rangle$. Thus $A$ satisfies condition $1$.
Finally $|A|=1$ implies that $A$ preserves orientation.

So the big thing about orthogonal matrices is that they preserve dot products and hence vector norms. That coupled with linearity is already enough to show that the matrix is either a pure rotation or a rotation with reflection. Then finally requiring the determinant to be $1$ excludes reflections, which reverse orientation.