# Working with the Mobius transformatios and linear algebra.

Let $M=\left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right) \in GL_{2}(\mathbb{C})$ and we recall that the Möbius transformation attached to $M$ is the map:

$z \to \frac{az+b}{cz+d}$

then I have to show that the fix points correspond to the eigenspace of $M$.

My attempt is:

We must show that there is a one-one correspondence between the eigenvectors of $M$ and the fix points, so we have the following:

Let $f(z) = \frac{az+b}{cz+d}$. The we observe that:

$$\left(\begin{array}{ccc}a & b \\ c & d\end{array}\right) \left(\begin{array}{c}z \\ 1\end{array}\right) = \left(\begin{array}{c}az + b \\ cz + d\end{array}\right) = (cz+d)\left(\begin{array}{c}\frac{az + b}{cz + d} \\ 1\end{array}\right) = (cz+d)\left(\begin{array}{c}f(z) \\ 1\end{array}\right)$$

Is the above the big displayed equation?

let $z$ such that $f(z)=z$, then we have that:

$$(cz+d)\left(\begin{array}{c}f(z) \\ 1\end{array}\right)=(cz+d)\left(\begin{array}{c}z \\ 1\end{array}\right)$$

therefore if $z$ is a fix point we asociate it the eigenvector $(z,1)$ with eigenvalue $(cz+d)$.

Now we suppose that $\vec z = \left(\begin{array}{c}z_{1} \\ z_{2}\end{array}\right)$ is a eigenvector, then we scale it, so $\vec z = \left(\begin{array}{c} \frac{z_{1}}{z_{2}} \\ 1\end{array}\right)$

then we need to find:

$(cz+d) \left(\begin{array}{c}f(\frac{z_{1}}{z_{2}}) \\ 1\end{array}\right)= (cz+d)\left(\begin{array}{c}\frac{a\frac{z_{1}}{z_{2}}+b}{c\frac{z_{1}}{z_{2}}+d} \\ 1\end{array}\right)=(cz+d)\left(\begin{array}{c}\frac{\frac{az_{1}+bz_{2}}{z_{2}}}{\frac{cz_{1}+dz_{2}}{z_{2}}} \\ 1\end{array}\right)=(cz+d)\left(\begin{array}{c}\frac{az_{1}+bz_{2}}{cz_{1}+dz_{2}} \\ 1\end{array}\right) = \left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right) \left(\begin{array}{c} \frac{z_{1}}{z_{2}} \\ 1\end{array}\right)=\lambda \left(\begin{array}{c} \frac{z_{1}}{z_{2}} \\ 1\end{array}\right)$

MY question is, first of all Am I in the good way or what do I have to fix?,and the other one is what can be conclude from here (this is think that my eigenvector is not rigth, how can I fix that part,I dont know how to find the other side of the correspondence.) Can you help me to complete this problem please, thanks in advance.

• Don't compute the eigenvalues! Suppose something is an eigenvalue, and then scale it so its second coordinate is 1. Then it should fit into the big displayed equation with almost no more work (i.e. use the big equation, but in the reverse of how you were using it). – aes Nov 23 '14 at 22:07
• thanks for the help again :), but which is meaning of scale ? – user162343 Nov 23 '14 at 22:09
• and I compute the eigenvectors not values :) – user162343 Nov 23 '14 at 22:10
• You haven't shown it's a fixed point yet. Use $\left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right) \left(\begin{array}{c} \frac{z_{1}}{z_{2}} \\ 1\end{array}\right) = \lambda \left(\begin{array}{c} \frac{z_{1}}{z_{2}} \\ 1\end{array}\right)$. (Also, there's a minor problem with the way you've written it, since some terms are vectors and some are not.) – aes Nov 23 '14 at 23:59
• Replace the $f(z_1/z_2)$ at the left end of that equation with $(c z_1/z_2 + d) \left(\begin{array}{c} f(z_1/z_2) \\ 1 \end{array}\right)$. – aes Nov 24 '14 at 0:05