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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!

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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 $\alpha^2 I=L^TL$).

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  • $\begingroup$ Would a rotation of $-R$ not change the orientation of the angles though? $\endgroup$ Mar 23, 2020 at 15:26
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    $\begingroup$ @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. $\endgroup$
    – user762437
    Mar 23, 2020 at 15:33
  • $\begingroup$ 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? $\endgroup$ Mar 23, 2020 at 16:15
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    $\begingroup$ @MathsMatador Yes. You can can just assume that you always take $\alpha\geq0$. $\endgroup$
    – user762437
    Mar 23, 2020 at 16:17

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