# Why $f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb C$, is a linear transformation?

Now I'm confused with what "a linear transformation" means.

In linear algebra textbook, I learned that a linear transformation is $T:V \to W$, where V,W are vector spaces, which satisfies additivity and homogeneity, in other words, $T(u+v)=Tu+Tv, T(av)=aTv$ for all $u,v \in V$ and $a \in F$

But in my complex analysis textbook, $\displaystyle f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb C$, is introduced as an example of a linear transformation.

However, this function $f$ doesn't seem to follow the definition from linear algebra. Indeed, $f(0) \neq 0$.

Is it like there are two kinds of linear transformations in mathematics, or they are actually the same thing but I don't get it well?

• In an intuitive sort of way, the way I visualize linear transformations is that they are some sort of stretching, moving, or rotation on the plane. (Speaking of $\mathbb{R}^2$ and $\mathbb{C}$, of course.) With this in mind, the transformation $f(z) = \frac{az + b}{cz + d}$ is an example of a linear transformation in the complex plane in that multiplication/division can be thought of as stretching followed by rotations of a complex vector, whereas addition can be thought of as a translation. So, perhaps in that sense, $f$ may be thought of as a linear transformation. Dec 2, 2012 at 20:15
• Yes, those are not linear transformations in the sense of linear algebra. In fact, their name is more usually mentioned to be fractional linear transformations or, simpler, Moebius transformations. The "linear" part is most probably due to the fact the set of all these things is a (projective) linear group Dec 2, 2012 at 20:17
• @DonAntonio Thank you very much for your answer! Dec 2, 2012 at 20:23
• @RaymondCheng Thank you very much for your explanation! Dec 2, 2012 at 20:24
• My complex analysis textbook called it " bilinear transformation" which is much more confusing! Haha!
– Ivy
Jun 16, 2018 at 1:57

$z \to \dfrac{az+b}{cz+d}$ is more properly called a fractional linear transformation (or linear fractional transformation, or Möbius transformation). It is not the same as a linear transformation, although abuse of the language sometimes does take place.

EDIT: For example, Ford's "Automorphic Functions", first published 1929, defines $z' = \dfrac{az+b}{cz+d}$ as a "linear transformation": in a footnote he says 'This is more properly called a "linear fractional transformation"; but we shall use the briefer designation.' http://books.google.ca/books?id=aqPvo173YIIC&pg=PA1

• I got it now! Thank you very much. Dec 2, 2012 at 20:19
• Thank you for additional information. I thought this happened because my textbook is kinda old, but it doesn't seem to be the case because this is not as old as the one you listed. Dec 2, 2012 at 20:45

Let $T=f$, then we have $$f(u+v)=\frac{a(u+v)+b}{c(u+v)+d}.$$

Likewise, $f(u)+f(v)=\frac{au+b}{cu+d}+\frac{av+b}{cv+d}.$ It is clear that $f(u+v)\ne f(u)+f(v)$, hence this is not a linear transform. However, does it satisfy the second property? Let's see:

$$f(\alpha u)=\frac{a(\alpha u)+b}{c(\alpha u)+d},$$

and $$\alpha f(u)=\alpha \cdot \frac{au+b}{cu+d}=\frac{a\alpha u+\alpha b}{cu+d}.$$

Since $f(\alpha u)\ne \alpha f(u)$, $f$ does not satisfy the second property either.

As Robert Israel first pointed out, this is the very definition of a linear fractional transform.

• Thank you for your explanation! Dec 2, 2012 at 20:23
• @Tengu, you're welcome!
– 000
Dec 2, 2012 at 20:24
• @anonymousdownvoter, do you care to explain your reason for the downvote?
– 000
Dec 2, 2012 at 23:27

The linear transformation you described in terms of linear algebra is a distinct notion in this case. If you are familiar with category theory, a linear transformation is a morphism in the category of vector spaces (if not disregard this sentence).

In the context of complex analysis, the type of transformation you have described is a called a Möbius transformation, $$f(z) = \frac{az +b}{cz + d}$$ with $a,b,c,d \in \mathbb{C}$, and traditionally, $ad - bc \neq 0$, since we don't want to consider the constant function. A Möbius transformation is a projective linear transformation (also called a homography) of the complex projective line. That is to say, it is a non-linear transformation in terms of Cartesian coordinates and generates a different group-theoretic structure (compare the projective linear group $PGL(V)$ to the general linear group $GL(V)$).

• I got it now. Thank you! Dec 2, 2012 at 20:21
• $ad-bc\neq 0$ is assumed because we want the transformation to be invertible. Dec 2, 2012 at 20:23
• @Alex Thank you for your answer. I was forgetting about it. Dec 2, 2012 at 20:28

I think perhaps one of the reasons it is called this is that composition of Möbius transformations acts like matrix multiplication on the coefficients. That is, if we have $$f(z) = \frac{a z + b}{c z + d},$$ we associate it with the matrix $$M_f=\begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$ Then, if $f(z)$ and $g(z)$ are Möbius transformations, then $$M_{f\circ g}=M_fM_g.$$ That is, the coefficients of $f(g(z))$ are exactly the elements of the product of the coefficients of $f(z)$ and $g(z)$ as matrices.

• Actually I tried to understand it that way. We can associate$\displaystyle w=f(z)=\frac {az+b}{cz+d}$ with the matrix $\displaystyle w= \begin {pmatrix} w_1 \\ w_2 \end {pmatrix}$ = $\begin {pmatrix} a & b\\ c & d \end {pmatrix}$ $\begin{pmatrix} z_1 \\ z_2 \end {pmatrix}$, where $z_1,z_2$ are chosen to be $\displaystyle z=frac{z_1}{z_2}$. Then $(z_1,z_2)=(0,0)$ would be the zero. Then it fulfills the requirement to be a linear transformation, but this didn't work well when I was working on exercises though Dec 2, 2012 at 22:12