Very soft question and I may be completely wrong about this, but does it make any sense to think about the Möbius transformation matrix as a change of basis for $\mathbb C$?
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$\begingroup$ C being the complex plane here $\endgroup$– JuliusL33tFeb 22, 2014 at 19:59
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$\begingroup$ Hint: Is that matrix a non-sigular one...? $\endgroup$– DonAntonioFeb 22, 2014 at 20:00
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$\begingroup$ yeah, since det(A) will always be nonzero, in order for it to be a mobious transf. $\endgroup$– JuliusL33tFeb 22, 2014 at 20:00
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$\begingroup$ And a square matrix determines a change of basis (taking it as a linear map) iff it is non-sigular, thus... $\endgroup$– DonAntonioFeb 22, 2014 at 20:05
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1 Answer
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No, the Möbius transformation is not a linear transformation of $\Bbb{C}$ even if the associated matrix might have other action on the space. Hence, the Möbius transformation matrix should not be thought of as a basis change for $\Bbb{C}$.
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$\begingroup$ OK, thanks, told you I might be waay off $\endgroup$ Feb 22, 2014 at 20:06
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$\begingroup$ Humans can always think of something as something else. That's imagination! Otherwise you could say that no, no, a matrix is an $n\times n$ array of numbers, it has nothing to do with a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$... $\endgroup$ May 5, 2015 at 8:19