In my notes under "Fractional Linear Transformations" of Complex Numbers, it says the following:
Let $M$ be a $2\times 2$ complex matrix
$$M=\begin{pmatrix} a & b \\ c & d\end{pmatrix}.$$
We write $T_M$ for the associated fractional linear transformation:
$$T_M(z) = \frac{az+b}{cz+d}.$$

Observe now that $T_{\lambda M} = T_M$ if $\lambda \neq 0$, and recall that $\det (\lambda M) = \lambda ^2 \det (M)$. This means that when $\det (M) \neq 0,$ if we set $M'$ = $(\det M)^{-\frac{1}{2}}M$, then $\det M' = 1$ and $T_M = T_M'$. Thus, we may restrict our attention to fractional linear transformations associated to matrices with determinant $1$.

I don't see why this equality $T_{\lambda M} = T_M$ holds. I feel like the $\lambda$ shouldn't be a subscript. However, assuming it's true, when it says "we may restrict our attention...." is it saying that if we are ever given a transformation matrix $M$ with non-unit and non-zero determinant, then we can just "unify" it by slamming the constant $(\det M)^{-\frac{1}{2}}$ in front and calling the new transformation $M'$ (will this new transformation actually be the same as the old one??)?

I hope someone could help clear this up for me

  • $\begingroup$ maybe the subscript is the matrix $\lambda M$ and not two subscripts: $\lambda, M$ $\endgroup$ Commented Jul 16, 2017 at 8:39
  • $\begingroup$ Hmm, not sure what the difference is; the subscript in the original post is indeeed just $\lambda M$. $\endgroup$ Commented Jul 16, 2017 at 8:40
  • $\begingroup$ The difference I meant appears in the answer. The matrix $\lambda M$ compared to the matrix $M$, they will mean the same as in the expression the lambdas up and down in the fraction cancel. $\endgroup$ Commented Jul 16, 2017 at 8:42
  • $\begingroup$ Ah right... for some reason I didn't realise that they would cancel upon substitution $\endgroup$ Commented Jul 16, 2017 at 8:44

1 Answer 1


It's simple! $\;\lambda\begin{pmatrix}a&b\\c&d \end{pmatrix}=\begin{pmatrix}\lambda a&\lambda b\\\lambda c&\lambda d \end{pmatrix}$, so $$T_{\lambda M}(z)=\frac{\lambda az+\lambda b}{\lambda cz+\lambda d}=\frac{az+b}{cz+d}.$$ There indeed results from the considerations in the post that $$\det M'=\det\bigl((\det M)^{-1/2}M)=(\det M)^{-1}\det M=1.$$

  • $\begingroup$ Thank you!! Didn't realise it for some reason... would you know what the point would that matrix $M'$ (making the matrix $M$ have determinant $1$)? $\endgroup$ Commented Jul 16, 2017 at 8:45
  • 2
    $\begingroup$ Since we can multiply by any scalar without the interpretation of the matrix becomes altered we might choose a scalar so that the determinant becomes one. It reduces the degrees of freedom for the matrix. $\endgroup$ Commented Jul 16, 2017 at 8:50

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