Affine Maps, Matricies, Invertibility, and Equivalence Relations

Question

A mapping $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is affine if there is an invertible $n$ x $n$ matrix $M$ and a vector $s \in \mathbb{R}$ such that $f(x)=Mx+s$ for every $x \in \mathbb{R}^n$.

1. Show that every affine map $f(x) = Mx+s$ is invertible and its inverse is also affine?

2. Show that the composition of affine maps $\mathbb{R}^n \rightarrow \mathbb{R}^n$ is affine.

3. For two subsets, $S$, $T$ of $\mathbb{R}^n$, write $S \sim T$ if there exists an affine map $f$ such that $f(S) =T$. We then say that $S,T$ are affine-related. Prove that $\sim$ is an equivalence relation on the set of subsets of $\mathbb{R}^n$.

Answer 1

1.) Let

$f(x) = Mx + s \tag{1}$

be an affine map $\Bbb R^n \to \Bbb R^n$. Suppose

$f(x_0) = y_0 \tag{2}$

for some $x_0 \in \Bbb R^n$. Then we have

$Mx_0 + s = y_0, \tag{3}$

or

$Mx_0 = y_0 - s, \tag{4}$

or, since $M$ is invertible,

$x_0= M^{-1}(y_0 - s) = M^{-1}y_0 - M^{-1}s. \tag{5}$

(4) suggests we set

$t = -M^{-1}s \tag{6}$

and consider the affine map

$g(x) = M^{-1}x + t; \tag{7}$

we have, for any $x \in \Bbb R^n$,

$g(f(x)) = M^{-1}(f(x)) + t = M^{-1}(Mx + s) + t$ $ = M^{-1}Mx + M^{-1}s + t = Ix + M^{-1}s - M^{-1}s = x, \tag{8}$

and

$f(g(x)) = f(M^{-1}x + t) + s = M(M^{-1}x + t) + s$ $ = MM^{-1}x + Mt + s = Ix - s + s = x, \tag{9}$

using (6). This shows that any affine map $f(x)$ has an affine inverse

$g(x) = f^{-1}(x). \tag{10}$

2.) Taking

$g(x) = Nx + r, \tag{11}$

where $N$ is invertible, we see that

$g(f(x)) = Nf(x) + r = N(Mx + s) + r = NMx + Ns + r = NMx + (Ns + r); \tag{12}$

since $M$ and $N$ are both invertible, so is $NM$; thus $g(f(x))$ is affine, showing that the composition of two affine maps is also affine.

3.) This is actually a special case of a much more general observation about collections of invertible maps on arbitrary sets:

Generality: Let $\Sigma$ be any set, and let $\mathcal C$ be any non-empty collection of invertible mappings $f:\Sigma \to \Sigma$ which is closed under both composition and the taking of inverses, that is

$f, g \in \mathcal C \implies f \circ g:\Sigma \to \Sigma \in \mathcal C, \tag{13}$

and

$f \in \mathcal C \implies f^{-1}:\Sigma \to \Sigma \in \mathcal C. \tag{14}$

If, for $\Delta_1, \Delta_2 \subset \Sigma$ we write

$\Delta_1 \sim \Delta_2 \Longleftrightarrow \exists f \in \mathcal C, f(\Delta_1) = \Delta_2, \tag{15}$

then $\sim$ is an equivalence relation on $2^\Sigma$, the power set (set of all subsets) of $\Sigma$.

Proof of Generality: Note $\mathcal C \ne \emptyset$ by hypothesis; denoting the identity map from $\Sigma \to \Sigma$ by $Id$, that is, $Id(x) = x$ for $x \in \Sigma$, we have $Id(\Delta_1) = \Delta_1$ for any $\Delta_1 \subset \Sigma$; furthermore, for $f \in \mathcal C$ and $x \in \Sigma$, since $f^{-1} \in \mathcal C$ by (14) and $f \circ f^{-1}, f^{-1} \circ f \in \mathcal C$ by (13), and

$Id(x) = f^{-1} \circ f(x) = f \circ f^{-1}(x), \tag{16}$

we see that $Id \in \mathcal C$. Thus $\Delta_1 \sim \Delta_1$ showing $\sim$ is reflexive. Now if $\Delta_1 \sim \Delta_2$, there is an $f \in \mathcal C$ with

$f(\Delta_1) = \Delta_2, \tag{17}$

whence

$f^{-1}(\Delta_2) = \Delta_1 \tag{18}$

with $f^{-1} \in \mathcal C$ by (13). Thus

$\Delta_2 \sim \Delta_1 \tag{19}$

and so $\sim$ is reflexive. Finally, suppose further that $\Delta_3 \subset \Sigma$ and

$\Delta_1 \sim \Delta_2, \Delta_2 \sim \Delta_3; \tag{20}$

then there are $f, g \in \mathcal C$ with

$f(\Delta_1) = \Delta_2, g(\Delta_2) = \Delta_3, \tag{21}$

whence

$g \circ f(\Delta_1) = g(\Delta_2) = \Delta_3, \tag{22}$

that is, since $g \circ f \in \mathcal C$ by (13),

$\Delta_1 \sim \Delta_3, \tag{23}$ and we see that $\sim$ is transitive. Thus $\sim$ is an equivalence relation on $2^\Sigma$. End of Proof of Generality.

Applying this Generality to the case at hand, taking $\Sigma = \Bbb R^n$ and $\mathcal C$ to be the set of affine mappings $\Bbb R^n \to \Bbb R^n$, we immediately see that the resulting relation $\sim$ is an equivalence relation on $2^{\Bbb R^n}$, by virtue of items (1.) and (2.) above.

Answer 2

Part 1: It suffices to find a formula for the inverse $$ y = Mx + s \implies Mx = y - s \implies x = M^{-1}(y-s) = M^{-1}y - M^{-1}s $$ So, define $g(x) = M^{-1}x - M^{-1}s$. Verify that $g$ is indeed an inverse to $f$, which is to say that $f(g(x)) = g(f(x)) = x$. $g$ is an affine transformation since it takes the input, multiplies it by $M^{-1}$, then adds the vector $-M^{-1}s$.

Part 2: Suppose that $f_1(x) = M_1x + s_1$ and $f_2(x) = M_2x + s_2$. Note that $$ f_2(f_1(x)) = M_2(M_1x + s_1) + s_2 = [M_2M_1]x + [M_2s_1 + s_2] $$ so, $f_2 \circ f_1$ is affine.

Part 3: For transitivity, suppose that $S \sim T$ and $T \sim U$. That is, there are affine transformations $f,g$ such that $f(S) = T$ and $g(T) = U$. It follows that $$ g(T) = g(f(S)) = (g \circ f)(S) = U $$ Since $g \circ f$ is affine, this tells us that $S \sim U$, as desired.