# Deriving the matrices for a conformal linear transformation

I have been given the following definition for a conformal linear transformation (in $$\mathbb{R}^2$$):

A non-singular linear transformation $$L:\mathbb{R}^2 \rightarrow \mathbb{R}^2$$ is said to be conformal if the counter-clockwise angle from $$Lu$$ to $$Lv$$ is the same as the counter-clockwise angle from $$u$$ to $$v$$, for any pair of non-zero vectors $$u, v \in \mathbb{R}^2$$.

In the process of deriving the $$2 \times 2$$ matrices that correspond to these transformations (a product of a scaling and a rotation), I have shown explicitly that the non-diagonal entries of the $$L^\top L$$ are zero, and that its diagonal entries are equal. Moreover, if $$L = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies L^\top L = \begin{bmatrix} a^2 + c^2 & 0 \\ 0 & b^2 + d^2 \end{bmatrix},$$ then $$a^2 + c^2 = b^2 + d^2$$. But then this says that $$L^\top L = \alpha^2I$$ where $$\alpha \in \mathbb{R}, \alpha \neq 0$$. If I continue to let $$L = \alpha R$$, then I can show that $$R$$ is the rotation matrix. My problem is that if $$\alpha < 0$$, then does this not change the direction of vectors in $$\mathbb{R}^2$$ and if so does this contradict the given definition of a conformal linear transformation? I am asking this because all other information I can find on this is that the scaling is assumed to be positive, of which I am not sure why that is the case anyway.

Thanks for the help!

You can always assume that $$\alpha\geq0$$. If you happen to write $$L=\alpha R$$, with $$\alpha<0$$, then you can also write $$L=(-\alpha)(-R)$$. Here $$-\alpha\geq0$$ and $$-R$$ will be a rotation (if $$L$$ is conformal and $$\alpha^2 I=L^TL$$).
• Would a rotation of $-R$ not change the orientation of the angles though? Mar 23, 2020 at 15:26
• @MathsMatador No, when $L$ is conformal, and if you take $\alpha<0$, then and $R$ satisfying $L=\alpha R$ is the one that will change orientation. Then $-R$ will just be a rotation. In other words, you can just assume that $\alpha>0$, by changing both the signs of $\alpha$ and $R$ simultaneously.
• So if I have got this right, $R$ can be written as your regular rotation matrix with trig functions, but the "negative-ness" of $\alpha < 0$ incorporated into $R$ changes the orientation of angles. So $-R$ changes $R$ back into a rotation with the correct orientation? Mar 23, 2020 at 16:15
• @MathsMatador Yes. You can can just assume that you always take $\alpha\geq0$.