relation between conformal and orthogonal matrices in 2D I want to show that if a matrix $T \in \text{GL}(2, \mathbb{R})$ is conformal, i.e.
$$
\text{arccos} \left( \frac{\langle Tv,Tw \rangle}{|Tv||Tw|} \right) = \text{arccos} \left( \frac{\langle v,w \rangle}{|v||w|} \right),
$$
then there exists a $\lambda > 0$ with $\lambda T \in O(2)$. Starting with an orthogonal matrix and proving the other direction is no problem, but I am a bit lost how to use the equation to show that the matrix is orthogonal.
 A: This proof makes use of some geometry: Similarity.
Let $v,w \in \mathbb{R}^2$ be linearly independent vectors. The condition
$$
\text{arccos} \left( \frac{\langle Tv,Tw \rangle}{|Tv||Tw|} \right) 
= 
\text{arccos} \left( \frac{\langle v,w \rangle}{|v||w|} \right) \tag{$1$}
$$
states that the angle $\angle(v,w)$ will be preserved under the linear mapping $T$, i.e.
$$
\angle(v,w) = \angle(Tv,Tw) \tag{$2$}
$$
Now consider the triangle in $\mathbb{R}^2$ formed by the vertex triplet $\Delta := (0_{\mathbb{R}2}, v,w)$. Note that $T \in \text{GL}(2, \mathbb{R})$, i.e. $T$ has full rank, and as a result the images of the two linearly independent vectors $v,w$ (these are the two of three sides of the triangle) under the invertible linear map $T$ will be linearly independent again.
This guarantees that the image of our triangle will be a non-degenerate triangle again: the triangle consisting of vertices $\Delta' := (0_{\mathbb{R}^2},Tv,Tw)$.
Making use of the property $(2)$ gives us the similarity of the two triangles $\Delta$ and $\Delta'$. But another property of similar triangles is that the ratios of the lengths of the triangle sides coincide, i.e.
$$
\frac{\lvert v \rvert}{\lvert Tv \rvert} 
=
\frac{ \lvert w \rvert }{\lvert Tw \rvert} 
\quad\left(= \frac{ \lvert (v-w) \rvert}{\lvert T(v-w) \rvert}\right).
$$
This shows that $\lambda := \frac{\lvert v \rvert}{\lvert Tv \rvert} > 0$ is well defined and does not depend on the involved vectors (provided none of them is the zero vector).
Only thing left to do is to show that the mapping $\lambda  T$ is indeed orthogonal, i.e. length preserving. This follows from the bi-linearity of the scalar product and $(1)$:
$$
\langle \lambda Tv, \lambda Tw  \rangle
= 
\lambda^2 \langle Tv, Tw \rangle 
= 
\lambda^2 \langle v,w  \rangle\frac{\lvert Tv \rvert \lvert Tw \rvert }{\lvert v\rvert \lvert w \rvert}
=
\langle v,w \rangle.
$$
