# Some fun with a (very) simple matrix

Posted by: Gary Ernest Davis on: February 20, 2011

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The $2\times 2$ matrix $A=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}$ is very simple: it has entries consisting only of 0’s and 1’s.

As a $2\times 2$ matrix we can think of $A$ as transforming points in the euclidean plane. These are points with coordinates $(x,y)$ where $x \textrm{ and } y$ are real numbers.

The matrix $A$ transforms the point $(x,y)$ into the point $A\cdot (x,y) = \begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (x,y):= (y, x+y)$:

We can look at what the matrix $A$ does to geometric shapes in the plane.

For example, the square with corners $(0,0), (1,0), (0,1), (1,1)$ – show in yellow in the diagram below – gets transformed into the parallelogram with corners $(0,0), (0,1), (1,1), (1,2)$ – shown in purple dots in the diagram below:

The area of the parallelogram is 1 unit – the same as the area of the original square. Because area of any plane figure is based on counting squares – and taking a limit with vanishingly small squares if the figure is somewhat round, such as a circular region – the matrix $A$ is an area-preserving transformation of the plane.

However if we imagine traveling counter-clockwise from the point $(1,0)$ to the point $(0,1)$ we see that the transformation $A$ has us traveling clockwise from $A\cdot(1,0) = (0,1)$ to $A\cdot(0,1) = (1,1)$:

In other words, the matrix $A$ is an orientation-reversing transformation of the plane.

#### Appearance of the Fibonacci numbers

What happens when we apply the matrix over and over to a point?

For instance, if we apply the matrix $A$ to the point $(x,y)$Â  we get the point $A\cdot (x,y)=(y,x+y)$, and if we then apply $A$ to this new point we get the point $A\cdot (y,x+y) =(x+y,x+2y)$.

This is the same result as if we had applied the square $A^2=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix} =\begin{pmatrix} 1 & 1 \\1 & 2 \end{pmatrix}$ to the point $(x,y)$.

This is true more generally: to figure out what applying the matrix $A$ over and over does to a point $(x,y)$, it is the same as working out the appropriate power of the matrix $A$ and applying that matrix to $(x,y)$.

So can we characterize, or identify, the powers of the matrix $A=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}$ ?

Here are the first few powers:

$A^1=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}$

x

$A^2=A\cdot A=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}$

$=\begin{pmatrix} 1 & 1 \\1 & 2 \end{pmatrix}$

x

$A^3=A\cdot A^2=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1 \\1 & 2 \end{pmatrix}$

$= \begin{pmatrix} 1 & 2 \\2 & 3 \end{pmatrix}$

x

$A^4=A\cdot A^3=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 2 \\2 & 3 \end{pmatrix}$

$= \begin{pmatrix} 2 & 3 \\3 & 5 \end{pmatrix}$

x

$A^5=A\cdot A^4=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} 2 & 3 \\3 & 5 \end{pmatrix}$

$= \begin{pmatrix} 3 & 5 \\5 & 8 \end{pmatrix}$

These matrix entries are looking suspiciously familiar: they are looking like the Fibonacci numbers $1, 2, 3, 5, 8, \ldots$

Have we just been lucky so far, or is this something we can prove?

A good proof technique for this sort of problem is proof by induction, because we know inductively how to work out the powers ofÂ  the matrix $A$, namely $A^1=A \textrm{ and } A^{n+1}=A\cdot A^n$.

First, we need a precise statement of what it is we are trying to prove.

It’s looking like:

$A^n= \begin{pmatrix} F_{n-1} & F_n \\F_n & F_{n-1} \end{pmatrix}$ where $F_n$ are the Fibonacci numbers given by $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}\textrm{ for } n\geq 2$, namely $0, 1, 1, 2, 3, 5, 8, \ldots$

x

This is certainly true for $n=1$ because $A^1=A=\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}=\begin{pmatrix} F_0 & F_1 \\F_1 & F_2 \end{pmatrix}$

So let’s suppose that it’s true for some particular value of $n$, say $n=k$ (for example: $k=1$).

Then $A^{k+1}=A\cdot A^k = \begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot \begin{pmatrix} F_{k-1} & F_k \\F_k & F_{k+1} \end{pmatrix}$

$=\begin{pmatrix} F_k & F_{k+1} \\F_{k-1}+F_k & F_k+F_{k+1} \end{pmatrix}=\begin{pmatrix} F_k & F_{k+1} \\F_{k+1} & F_{k+2} \end{pmatrix}$

so our statement is true also for $n=k+1$. By the principle of mathematical induction, it is true for all $n\geq 1$.

#### Moving to a torus

The matrix $A$ has the property that it sends points in the plane with integer coordinates to similar points: $A\cdot (m,n)=(n,m+n)$ so if $m, n$ are integers then so are the coordinates of $A\cdot (m,n)$.

This means we can restrict $A$ as a transformation of the unit square with corners $(0,0), (1,0), (0,1), (1,1)$ so long as we keep only the fractional parts of the coordinates of points.

For example, the point $(\frac{3}{4},\frac{2}{3})$ lies in the unit square but it’s transformation by the matrix $A$ does not: $A\cdot (\frac{3}{4},\frac{2}{3}) = (\frac{2}{3},\frac{17}{12})$. So we reduce the result by keeping only the fractional parts of the coordinates: $(\frac{2}{3},\frac{5}{12})$.

This operation, of reducing coordinates by keeping only the fractional parts is called reduction modulo 1.

Wwe write the result of keeping only the fractional parts of the coordinates of the point $(x,y)$ as $(x,y) \textrm{ mod }1$

When we think about this for a moment we see that some apparently different points on the unit square are actually the same, modulo 1.

For example, any point of the form $(t,0) \textrm{ where } 0\leq t\leq 1$ is the same, modulo 1, as the point $(t,1)$.

Ditto, any point $(0,t) \textrm{ where } 0\leq t\leq 1$ is the same, modulo 1, as the point $(1,t)$.

This means the sides of the squares, are really the same line, modulo 1:

We can get a different picture of the unit square, modulo 1,Â  if we glue together points on the square that are the same modulo 1.

First we can glue corresponding points on the blue sides to get a cylinder:

Now, when we glue together the red ends of the cylinder, as indicated, we get the surface of a donut, otherwise known as a torus:

So now we can use the matrix $A$ to transform points on the torus, so long as we remember to reduce the coordinates of the points modulo 1.

What happens if, for example, we begin with the point $(\frac{1}{2},\frac{1}{3})$ and keep applying the matrix $A$ over to the result?

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{2},\frac{1}{3})=(\frac{1}{3},\frac{5}{6})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{3},\frac{5}{6})=(\frac{5}{6},\frac{1}{6})$ (remember we are reducing modulo 1)

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{5}{5},{1}{6})=(\frac{1}{6},0)$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{6},0)=(0,\frac{1}{6})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (0,\frac{1}{6})=(\frac{1}{6},\frac{1}{6})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{6},\frac{1}{6})=(\frac{1}{6},\frac{1}{3})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{6},\frac{1}{3})=(\frac{1}{3},\frac{1}{2})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{3},\frac{1}{2})=(\frac{1}{2},\frac{5}{6})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{2},\frac{5}{6})=(\frac{5}{6},\frac{1}{3})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{5}{6},\frac{1}{3})=(\frac{1}{3},\frac{1}{6})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{3},\frac{1}{6})=(\frac{1}{6},\frac{1}{2})$

x

$\begin{pmatrix} 0 & 1 \\1 & 1 \end{pmatrix}\cdot (\frac{1}{6},\frac{1}{2})=(\frac{1}{2},\frac{1}{3})$

x

and we are back to where we started, after 13 steps.

The collection 12 points we collected along the way: $(\frac{1}{3},\frac{5}{6}), (\frac{5}{6},\frac{1}{6}), \ldots$ is called the orbit of the initial point $(\frac{1}{3},\frac{5}{6})$.

Any of the points in the orbit of $(\frac{1}{3},\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\begin{pmatrix} -1 & 1 \\1 & 0 \end{pmatrix}$

Just as for the point we chose, any point on the torus with rational number coordinates will have a finite orbit.

Question 1: Can you see why this is so?

Question 2: Can you figure out the size of the orbit of the point $(\frac{a}{b},\frac{c}{d})$ from the numbers $a, b, c, d$ ?

Wherever we are on the torus there is a point nearby – as close as we wish – that has rational coordinates. Therefore, wherever we are on the torus there is a point nearby – as close as we wish – with a finite orbit.

#### Arnold’s cat map

The square of the matrix $A$, namely the matrix $\begin{pmatrix} 1 & 1 \\1 & 2 \end{pmatrix}$ is known as Arnold’s cat map.

This matrix gives us an area-preserving, orientation-preserving transformation of the torus, that like the matrix $A$ is technically chaotic.

What this means in practice is that when we take two points on the torus that are very close together, and examine points in their orbit we will soon find corresponding points that are quite far apart.

Then as we progress through the orbits we will find the points get back close together again.