Question about Stillwell Naive Lie theory exercise 4.6.3

Question

Given that for $a,b$ reals and $a\neq 1$ $$ \begin{pmatrix} a & b \\ 0 & 1 \\ \end{pmatrix}^n=\begin{pmatrix} a^n & b \frac {a^n-1}{a-1} \\ 0 & 1 \\ \end{pmatrix} $$

Stillwell claims that these matrices lie on a line in $\mathbb R ^4$. This looks obviously false. What am I missing?

Later I understood what I missed. Consider the affine line in $\mathbb R^4$ $(1,0,0,1)+t(a-1,b,0,0)$. When $t=0$ we get $(1,0,0,1)$. When $t=1$, we get $(a,b,0,1)$. When $t=\frac {a^n-1}{a-1}$ we get $(a^n,\frac{a^n-1}{a-1}b,0,1)$. These vectors in $\mathbb R^4$ correspond in the obvious way to $2\times 2$ matrices.

Answer 1

Hint: Let us first identify $2\times 2$ matrices with $\mathbb{R}^4$ by

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (a,b,c,d).$$

Under this identification the collection of (orientation-preserving) affine automorphisms of $\mathbb{R}^2$ is sent to one half of the $2$-dimensional affine subspace of $\mathbb{R}^4$ defined by

$$c=0,d=1.$$

We can further identify this affine subspace with $\mathbb{R}^2$ by

$$(a,b,0,1)\mapsto (a,b).$$

Thus the iterates of an anonymous affine automorphism will lie in a two dimensional affine halfspace. If $a=1$, we are in the translational part of the affine group, and the iterates lie on a vertical line. Similarly if $b=0$, then we are in the linear part of the affine group and the iterates lie on a horizontal line (if $a=1$ and $b=0$ simultaneously we see only one point).

If $a\neq1$, then one can use one of the many ways of writing down the expression of a line from two or more points and verifying that all points, as well as $(1,0)$, lie on the line (e.g. one can write down the equation of the line passing through $(1,0)$ and $(a,b)$, and then verify that the $n$th iterate also satisfies the equation). Here is a humble interactive graph: https://www.desmos.com/calculator/whii61qcp1 (Here negative iterates are also drawn.)

Finally it's clear that any affine line in the last $\mathbb{R}^2$ corresponds to an affine line in $\mathbb{R}^4$.

Answer 2

$\newcommand\Mat{\rm Mat}$ The set

$$ S:=\left\{\begin{bmatrix} \alpha & \beta\\0 & 1 \end{bmatrix}\;:\;\alpha,\beta\in\mathbb R\right\} $$

is an affine subspace of the affine space $\Mat(2;\mathbb R)\;$ (with itself, considered as vector space, as the space of translations).

Indeed

$$ \begin{bmatrix}\alpha & \beta\\0 & 1\end{bmatrix} -\begin{bmatrix}\alpha' & \beta'\\0 & 1\end{bmatrix} = \begin{bmatrix}\alpha-\alpha' & \beta-\beta'\\0 & 0\end{bmatrix} $$

and the matrix on the right-hand side describes the vector subspace

$$ \vec F = \left\{\begin{bmatrix}r & s\\0 & 0\end{bmatrix}\;:\;r,s\in\mathbb R \right\} $$

of $\Mat(2;\mathbb R)$. Therefore the set $S$ lies in the plane

$$ F = \begin{bmatrix}a & b\\0 & 1\end{bmatrix} + \vec F $$

of the affine space $\Mat(2;\mathbb R)$.

Now,

$$ \begin{bmatrix}a^n & b\frac{a^n-1}{a-1}\\0 & 1\end{bmatrix}-\begin{bmatrix} a & b\\0 & 1\end{bmatrix} = (a^n-a)\begin{bmatrix}1 & \frac b{a-1}\\0 & 0\end{bmatrix} \tag{*}$$

so

$$ \begin{bmatrix}a & b\\0 & 1\end{bmatrix}^n\in \begin{bmatrix}a &b\\0 & 1\end{bmatrix}+\pmb\langle\begin{bmatrix}1 & \frac{b}{a-1}\\0 & 0\end{bmatrix}\pmb\rangle, $$

a line $L$ contained in $F\subseteq\Mat(2;\mathbb R)$, and therefore, by affine isomorphism, contained in $\mathbb R^4.\quad$QED

Cartesian equations of $\,\mathbf L$$\quad$ Let $X_{11},\ldots,X_{22}$ be affine coordinates on $\Mat(2;\mathbb R)$ with origin $O=\begin{bmatrix}a & b\\0 & 1\end{bmatrix}$ and canonical vectors $E_{ij}$ ($i,j=1,2$). The $({}^*)$ immediately gives

$$ X_{11} = a^n-a\qquad X_{12} = (a^n-a)\frac{b}{a-1}\qquad X_{21} = 0\qquad X_{22} = 0$$

thus the system of equations of the line $L\,$ is

\begin{cases} X_{12}=\frac{b}{a-1}X_{11}\\[1.5ex] X_{21}=0\\[1.5ex] X_{22}=0. \end{cases}