# Fractional Linear Transformation Question

1. Jul 5, 2011

### joeblow

I am a graduate assistant and was asked a question about FLTs (Mobius Transformations). The student was asked to prove that any FLT can be written as an FLT with determinant 1.

However, I can't make sense of that. If I look at the possible Jordan Canonical forms of 2-by-2's, it would seem that the matrix
[x 0]
[0 y]
where x and y are distinct eigenvalues cannot be represented as an FLT with determinant 1 (since it would require finding a complex number that when multiplied with both x and y gives 1 which violates the uniqueness of multiplicative inverse).

2. Jul 5, 2011

### henry_m

To fix notation: a Mobius map is a map of the form
$$z\mapsto \frac{az+b}{cz+d}$$
for some complex constants $a,b,c,d$ satisfying $ad-bc\neq0$. It is this number $ad-bc$ that is meant by the 'determinant' in the problem (though that name is not justified quite yet, see later).

The constants are not, however, unique for any one map: if you multiply all of them by a nonzero number, $\lambda$, you get the same map. The 'determinant' is then multiplied by $\lambda^2$, so by choice of $\lambda$ we can pick it to be one.

The name determinant comes from the fact that there is an isomorphism between Mobius maps and the projective linear group $SL(2,\mathbb{C})/\{1,-1\}$ where a,b,c,d become the entries of the matrix (the freedom to still pick $\lambda=\pm1$ is why we have to take the quotient).