$$ y = \frac{ax+b}{cx+d} \Longleftrightarrow x = \frac{dy-b}{-cy+a}, ad-bc \ne 0 $$ on the other hand $$ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{-1} = \frac{1}{ad-bc} \left( \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right), ad-bc \ne 0 $$ This looks amazing to me. Is there any meaning to this connection between rational functions and matrices? Can it be generalized for matrices of higher ranks?

  • $\begingroup$ $$\left( \begin{array}{cc}1&-y\end{array} \right) \left( \begin{array}{cc}a&b\\c&d\end{array} \right) \left( \begin{array}{c}x\\1\end{array} \right) = \left( \begin{array}{cc}1&-y\end{array} \right) \left( \begin{array}{c}ax+b\\cx+d\end{array} \right) =ax+b-cxy-dy. $$ By setting this equal to zero you get the "rational equation" your question starts with. But I don't see quite how to expand this to larger matrices or more variables, or how the inverse can apply. $\endgroup$ Feb 1, 2014 at 16:21

2 Answers 2


Let's say that when $A$ is a $2\times 2$ matrix $\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$ we will write $M_A(x)$ for the rational function $$ \frac{a_{11}x + a_{12}}{a_{21}x + a_{22}}.$$ Then notice that $$M_F(M_G(x)) = M_{FG}(x).$$

That is, to compose two functions of this type, you multiply their matrices.

(You can see this by just writing out the expression for $M_F(M_G(x))$ and simplifying it until it is in the form $\frac{ax+b}{cx+d}$, whereupon the coefficients $a,b,c,d$ will have the right values for $M_{FG}(x)$.)

This kind of correspondence:

$$\begin{array}{ccc} 2\times2\text{ matrices} & \implies & \text{rational functions} \\ \text{matrix multiplication} & \implies & \text{function composition} \end{array}$$

is called a homomorphism. Group theory tells us that if there is a homomorphism from the group of $2\times 2$ matrices to the group of functions as there is here, then the two groups' identity elements and inverses must also correspond. In this case that means that the identity function $f(x) = x$ will correspond to the identity matrix $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, and the inverse of a function $M_A(x)$ will be the function $M_{A^{-1}}(x)$, which is your observation.

I don't know if there is any corresponding homomorphism between rational functions and matrices of higher order; I suspect it would not be too hard to prove not. (For one thing, the set of functions $x\mapsto \frac{ax+b}{cx+d}$ is closed under composition, but this is not true of higher-degree rational functions.) It would suffice to rule out such a correspondence for $3\times 3$ matrices, since the $n\times n$ matrices for $n>3$ contain the $n=3$ matrices as a special case. So all one needs to do is find some algebraic property of $3\times 3$ matrices that can't be reflected in the group of functions.

(The next paragraph is to give you some keywords to search for if you are interested in pursuing this in more detail.)

The rational functions $x\mapsto \frac{ax+b}{cx+d}$ are called Möbius transformations. This relation between the Möbius transformations and the $2\times 2$ matrices is well known. The group of Möbius transformations has the curious name $PGL(2,K)$, where $K$ is $\Bbb R$ if you consider them to be functions of a real argument, and $\Bbb C$ if you consider them to be functions of a complex argument. (In general $K$ can be any system in which the expression $\frac{ax+b}{cx+d}$ makes sense; such a system is called a field.) The “$2$” here refers to the fact that the matrices are square matrices of rank 2. The $PGL(2, K)$ groups are well-studied and $PGL(2, \Bbb C)$ in particular is of great importance in complex analysis and mathematical physics. Your question about the matrices of higher rank, put into the language of group theory, would seem to be about $PGL(n, \Bbb R)$ for $n≥3$. I see that the Wikipedia article on the PGL groups mentions the correspondence between $PGL(2, K)$ and the rational linear functions, and that the page on Möbius transformations mentions it also, but neither place mentions any analogous correspondence with higher-rank $PGL(n,K)$ groups, which suggests either that none is known, or that it only appears in a very abstract way.


Let me use MJD's notation: a $2 \times 2$ matrix $A = \left( \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right)$ is associated with the Möbius transformation $$ M_A(x) = \frac{a_{11} x + a_{12}}{a_{21} x + a_{22}}. $$ The connection between $A$ and $M_A$ indeed has a meaning, and you can actually use this connection to prove that $M_A \circ M_B = M_{A B}$. Yes, it's not hard to check that with a simple computation anyway.

Let me stick to matrix with coefficients in $\mathbb{C}$, but you can use $\mathbb{R}$ if you like. The connection between $A$ and $M_A$ comes from looking at the projective line $\mathbb{C}P^1$, which is defined as the quotient $\mathbb{C}^2 \setminus \{0\}$ where you identify proportional points, i.e. $(z,w)$ is identified with $(\lambda z, \lambda w)$ for any $\lambda \neq 0$. It's customary to use the notation $(z : w)$ for a point in $\mathbb{C}P^1$ corresponding to $(z,w) \in \mathbb{C}^2 \setminus \{ 0 \}$.

Points $(z : w)$ are almost in correspondence with points in $\mathbb{C}$, just by taking $z/w \in \mathbb{C}$ (notice that it's well defined, as $(\lambda z : \lambda w)$ would give the same result). The only exception is $(1 : 0)$, which we associate with infinity (whatever that means, but you can make sense of it if you like). In result $\mathbb{C}P^1$ corresponds to the complex plane $\mathbb{C}$ with an added point at infinity.

Now consider an invertible linear map $A \colon \mathbb{C}^2 \to \mathbb{C}^2$. Because it's linear, we have $A \cdot (\lambda \cdot v) = \lambda \cdot (A \cdot v)$, and thus $A$ gives rise to a function $M_A \colon \mathbb{C}P^1 \to \mathbb{C}P^1$. To see this, first consider the composition $$ \mathbb{C}^2 \setminus \{ 0 \} \ni (z,w) \stackrel{A}{\longmapsto} (z',w') \longmapsto (z' : w') \in \mathbb{C}P^1 $$ and then notice that it always maps $(z,w)$ and $(\lambda z, \lambda w)$ to the same point in $\mathbb{C}P^1$, allowing us to define a function on $\mathbb{C}P^1$ instead of $\mathbb{C}^2$. Now the main point: the correspondence $A \mapsto M_A$ is a group homomorphism by the very definition.

OK, let me just connect this construction with the things you mentioned in your question. First, by choosing the standard basis, we can associate invertible linear maps $A \colon \mathbb{C}^2 \to \mathbb{C}^2$ with invertible $2 \times 2$ matrices. If we identify points in $\mathbb{C}P^1$ with points in $\mathbb{C}$ (with infinity added), then the composition above is $$ (z,w) \stackrel{A}{\longmapsto} (a_{11}z+a_{12}w, a_{21}z+a_{22}w) \longmapsto (a_{11}z+a_{12}w : a_{21}z+a_{22}w) \cong \frac{a_{11}z+a_{12}w}{a_{21}z+a_{22}w} = \frac{a_{11}\frac{z}{w}+a_{12}}{a_{21}\frac{z}{w}+a_{22}}. $$ If we remember that $(z:w) \in \mathbb{C}P^1$ corresponds to $z/w \in \mathbb{C}$, then the resulting function $M_A$ is just $$ u \stackrel{M_A}{\longmapsto} \frac{a_{11}u+a_{12}}{a_{21}u+a_{22}}, $$ the Möbius transformation we know and love.

PS. You can use the same construction in $\mathbb{C}^n \setminus \{ 0 \}$, which results in the projective space $\mathbb{C}P^{n-1}$, but its points no longer correspond to numbers. Here, $\mathbb{C}P^1$ is topologically a sphere (called the Riemann sphere), whereas $\mathbb{R}P^1$ is the real line $\mathbb{R}$ with infinity attached, so topologically a circle. The projective plane $\mathbb{R}P^2$ is also an example you may have heard of for other reasons.


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