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Could you please explain why every conformal linear map is a scalar times a rotation matrix? I can prove that every scalar-rotation matrix is a conformal map but not the opposite.

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Let $C$ the matrix that represents the linear transformation, such transformation is conformal iff preserves the angles, so we must have, for allo vectors $x,y$: $$ \frac{\langle Cx,Cy \rangle}{|Cx||Cy|}=\frac{\langle x,y \rangle}{|x||y|} $$ now consider the standard orthonormal basis $\{e_i\}$, we have, for $i\ne j$:

$$ \langle e_i,e_j \rangle=0 \Rightarrow \langle Ce_i,Ce_j \rangle=0 \Rightarrow e_j^T(C^TC)e_i=0 $$ and this means that all the non diagonal elements of $C$ are null: $(C^TC)_{ij}=0$.

We have also: $$ \langle e_i-e_j,e_i+e_j \rangle=0 \Rightarrow \langle C(e_i-e_j),C(e_i+e_j) \rangle=0 $$

and, in the same way, this means that the diagonal elements of $C^TC$ are all identical: $(C^TC)_{ii}=(C^TC)_{jj}$, so we have that $$C^TC=\lambda I $$ Now let $C=\sqrt{\lambda}R$, we have: $$ C^TC=\lambda I \Rightarrow R^TR=I $$ and $R$ is a matrix that represents an orthogonal transformation and, if by preserving the angles we means also the orientation, $R$ is a rotation and $C$ is the product of scalar transformation and a rotation.

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  • $\begingroup$ Thank you. One question. Why is $R$ a rotation and not a reflection? $\endgroup$ Feb 21, 2016 at 11:51
  • $\begingroup$ Nevermind, I got it! Thanks again! $\endgroup$ Feb 21, 2016 at 12:36

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