$$ y = \frac{ax+b}{cx+d} \Longleftrightarrow x = \frac{dy-b}{-cy+a}, ad-bc \ne 0 $$ 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. $\endgroup$ – Jeff Snider Feb 1 '14 at 16:21

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.


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