4.7 Affine geometry · Matrix Theory

The solution set of a linear system $𝐴 𝐱 = 𝐲$ with $𝐲 \neq 𝟎$ is never a subspace, but by Proposition 3.1.19 it is the next best thing: the subspace $ker 𝑇_{𝐴}$ , slid over so that it passes through one solution. Lines and planes that miss the origin are sets of this kind, and so is the screen of a computer, which has no preferred origin either. On such a set the sum of two points means nothing (the sum of two solutions of $𝐴 𝐱 = 𝐲$ solves $𝐴 𝐱 = 2 𝐲$ ), yet some combinations of points, such as midpoints, make perfect sense. This section finds out which combinations these are, and which maps respect them. Its last part, homogeneous coordinates, turns this affine geometry back into linear algebra one dimension up, which is how graphics software moves and projects three-dimensional scenes.

4.7.1Affine subspaces

Recall the translate $𝐯 +$ $𝑈 =$ ${𝐯 + 𝐮 | 𝐮 \in 𝑈}$ of a subset $𝑈 \subseteq 𝑉$ (Notation 3.1.17).

Definition 4.7.1 (Affine subspace). Let

𝑉

be a vector space over

𝐹

. A subset

𝐸 \subseteq 𝑉

is an affine subspace of

𝑉

𝐸 = 𝐯 + 𝑈

for some

𝐯 \in 𝑉

and some subspace

𝑈

𝑉

An affine subspace contains $𝐯 = 𝐯 + 𝟎$ , so it is never empty. Some authors count $\emptyset$ as an affine subspace as well, which makes statements about intersections shorter; we will say “empty or an affine subspace” where it matters. We use the letters $𝐸$ , $𝐸^{'}$ , $𝐸_{𝑖}$ for affine subspaces and keep $𝐴$ and $𝐵$ for matrices.

Example 4.7.2 (Affine subspaces).

(a)Every subspace $𝑈$ is an affine subspace, since $𝑈 = 𝟎 + 𝑈$ . Every single point is one, since ${𝐯} = 𝐯 + {𝟎}$ .
(b)If the system $𝐴 𝐱 = 𝐲$ has a solution $𝐱_{0}$ , its solution set $𝐱_{0} + ker 𝑇_{𝐴}$ is an affine subspace of $𝐹^{𝑛}$ (Proposition 3.1.19). More generally, every nonempty fiber $𝑇^{- 1} (𝐰)$ of a linear map $𝑇 : 𝑉 \to 𝑊$ is one (Corollary 2.6.31).
(c)In $ℝ^{2}$ , the line $𝑥_{1} + 𝑥_{2} = 2$ is $(1, 1) +$ $span ((1, - 1))$ .
(d)In $ℝ [𝑥]$ , the polynomials $𝑝$ with $𝑝 (1) = 1$ form the fiber over $1$ of the linear map $𝑝 \mapsto 𝑝 (1)$ , so they form the affine subspace $1 +$ ${𝑝 \in ℝ [𝑥] | 𝑝 (1) = 0}$ . Here $1$ is the constant polynomial.

The vector $𝐯$ in $𝐸 = 𝐯 + 𝑈$ is not determined by $𝐸$ , but the subspace $𝑈$ is, and every point of $𝐸$ can play the role of $𝐯$ .

Proposition 4.7.3 (Direction and base points). Let $𝐸 = 𝐯 + 𝑈$ be an affine subspace of $𝑉$ .

(a) $𝑈 =$ ${𝐚 - 𝐛 | 𝐚, 𝐛 \in 𝐸}$ . In particular, if also $𝐸 = 𝐯^{'} + 𝑈^{'}$ for some $𝐯^{'} \in 𝑉$ and some subspace $𝑈^{'}$ , then $𝑈^{'} = 𝑈$ .
(b) $𝐸 = 𝐚 + 𝑈$ for every $𝐚 \in 𝐸$ .
(c)For $𝐰 \in 𝑉$ , the conditions $𝐰 + 𝑈 = 𝐸$ , $𝐰 \in 𝐸$ , and $𝐰 - 𝐯 \in 𝑈$ are equivalent.

Proof.routine

Uses. Definition 4.7.1 (Affine subspace), Notation 3.1.17 (Translates), Definition 1.4.1 (Subspace)

Techniques. Double containment (Technique 1.3.14), Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. For (a), write two points of

𝐸

𝐯 + 𝐮

and

𝐯 + 𝐮^{'}

. For the reverse containment, write

𝐮 \in 𝑈

as a difference of two points of

𝐸

Hint 2. For (b), write

𝐚 = 𝐯 + 𝐮_{0}

and prove both containments, moving

𝐮_{0}

from one side to the other.

Hint 3. For (c), go around the cycle of the three conditions; (a) and (b) do the work.

Sketch. (a)

(𝐯 + 𝐮) -

(𝐯 + 𝐮^{'}) =

𝐮 -

𝐮^{'} \in

𝑈

, and

𝐮 =

(𝐯 + 𝐮) -

(𝐯 + 𝟎)

. The set of differences depends only on

𝐸

, so it is

𝑈

however

𝐸

is written. (b) If

𝐚 = 𝐯 + 𝐮_{0}

, then

𝐚 + 𝐮 = 𝐯 + (𝐮_{0} + 𝐮)

and

𝐯 + 𝐮 = 𝐚 + (𝐮 - 𝐮_{0})

. (c)

𝐰 + 𝑈 = 𝐸

gives

𝐰 = 𝐰 + 𝟎 \in 𝐸

;

𝐰 \in 𝐸

gives

𝐰 - 𝐯 \in 𝑈

by (a); and

𝐰 - 𝐯 \in 𝑈

gives

𝐰 \in 𝐸

, so

𝐰 + 𝑈 = 𝐸

by (b).

(a) The points of $𝐸$ are the vectors $𝐯 + 𝐮$ with $𝐮 \in 𝑈$ . For two of them, $(𝐯 + 𝐮) -$ $(𝐯 + 𝐮^{'}) =$ $𝐮 -$ $𝐮^{'}$ , which lies in $𝑈$ because $𝑈$ is a subspace. Conversely, every $𝐮 \in 𝑈$ is the difference $(𝐯 + 𝐮) - (𝐯 + 𝟎)$ of two points of $𝐸$ . So $𝑈$ equals the set $𝐷$ of differences of points of $𝐸$ . The set $𝐷$ is defined in terms of $𝐸$ alone, so if also $𝐸 = 𝐯^{'} + 𝑈^{'}$ , the same argument gives $𝑈^{'} = 𝐷 = 𝑈$ .

(b) Let $𝐚 \in 𝐸$ , say $𝐚 = 𝐯 + 𝐮_{0}$ with $𝐮_{0} \in 𝑈$ . For $𝐮 \in 𝑈$ ,

𝐚 + 𝐮 = 𝐯 + (𝐮_{0} + 𝐮) \in 𝐯 + 𝑈 and 𝐯 + 𝐮 = 𝐚 + (𝐮 - 𝐮_{0}) \in 𝐚 + 𝑈,

because $𝐮_{0} + 𝐮$ and $𝐮 - 𝐮_{0}$ lie in $𝑈$ . So $𝐚 + 𝑈 = 𝐯 + 𝑈 = 𝐸$ .

(c) If $𝐰 + 𝑈 = 𝐸$ , then $𝐰 = 𝐰 + 𝟎 \in 𝐸$ . If $𝐰 \in 𝐸$ , then $𝐰 - 𝐯$ is a difference of two points of $𝐸$ (as $𝐯 = 𝐯 + 𝟎 \in 𝐸$ ), so it lies in $𝑈$ by (a). If $𝐰 - 𝐯 \in 𝑈$ , then $𝐰 = 𝐯 + (𝐰 - 𝐯) \in 𝐸$ , and $𝐰 + 𝑈 = 𝐸$ by (b). □

Technique 4.7.4 (Choose a base point). An affine subspace has no origin of its own, but by Proposition 4.7.3(b) any of its points can serve as one. Once a point $𝐚 \in 𝐸$ is fixed, $𝐱 \mapsto 𝐱 - 𝐚$ carries $𝐸$ onto the subspace $𝑈 = 𝐸 - 𝐚$ , and statements about $𝐸$ become statements about $𝑈$ , where linear algebra applies. Every point of $𝐸$ works equally well, so choose the most convenient one, such as a point that two affine subspaces have in common, or a fixed point of a map.

Reach for it whenever an affine question has a linear counterpart that you already know how to answer. The canonical example is Proposition 4.7.10. The same move drives Theorem 4.7.14, Proposition 4.7.17, and Theorem 4.7.37.

Used in 10 places and 2 practice exercises

Corollary 4.7.6, Proposition 4.7.10, Theorem 4.7.14, Proposition 4.7.17, Proposition 4.7.20, Theorem 4.7.37, Proposition 4.7.38, Theorem 4.7.39, Theorem 8.4.31, Exercise 8.4.46; 2 practice exercises.

Definition 4.7.5 (Direction, dimension, parallel). Let

𝐸

be an affine subspace of

𝑉

. The subspace

𝑈

with

𝐸 = 𝐯 + 𝑈

, which is unique by Proposition 4.7.3(a), is the direction of

𝐸

, written

dir 𝐸

. If

dir 𝐸

is finite-dimensional, the dimension of

𝐸

dim 𝐸 = dim dir 𝐸

. Affine subspaces of dimension

0

1

, and

2

are called points, lines, and planes; if

dim 𝑉 = 𝑛

, those of dimension

𝑛 - 1

are called hyperplanes. Two affine subspaces

𝐸

and

𝐸^{'}

𝑉

are parallel if

dir 𝐸 = dir 𝐸^{'}

A subspace $𝑈$ has direction $𝑈$ , so its dimension is the same whether we regard it as a subspace or as an affine subspace. A line is a set $𝐯 +$ $span (𝐮) =$ ${𝐯 + 𝑡 𝐮 | 𝑡 \in 𝐹}$ with $𝐮 \neq 𝟎$ .

Corollary 4.7.6 (Parallel affine subspaces are equal or disjoint). If

𝐸

and

𝐸^{'}

are parallel affine subspaces of

𝑉

, then

𝐸 = 𝐸^{'}

𝐸 \cap 𝐸^{'} = \emptyset

. Given a subspace

𝑈

𝑉

, every

𝐯 \in 𝑉

lies in exactly one translate of

𝑈

, namely

𝐯 + 𝑈

Proof.routine

Uses. Proposition 4.7.3 (Direction and base points)

Techniques. Choose a base point (Technique 4.7.4)

Hint. Suppose some

𝐜

lies in both. Use

𝐜

as the base point of each.

Sketch. If

𝐜 \in 𝐸 \cap 𝐸^{'}

, then

𝐸 = 𝐜 + 𝑈 = 𝐸^{'}

by Proposition 4.7.3(b), where

𝑈

is the common direction. For the second claim,

𝐯 \in 𝐯 + 𝑈

, and a translate

𝐰 + 𝑈

containing

𝐯

equals

𝐯 + 𝑈

by the same result.

Let

𝑈 = dir 𝐸 = dir 𝐸^{'}

. If

𝐜 \in 𝐸 \cap 𝐸^{'}

, then

𝐸 = 𝐜 + 𝑈

and

𝐸^{'} = 𝐜 + 𝑈

by Proposition 4.7.3(b), so

𝐸 = 𝐸^{'}

. For the second statement,

𝐯 = 𝐯 + 𝟎 \in 𝐯 + 𝑈

, and if

𝐯 \in 𝐰 + 𝑈

, then Proposition 4.7.3(b), applied to the affine subspace

𝐰 + 𝑈

and its point

𝐯

, gives

𝐰 + 𝑈 = 𝐯 + 𝑈

. □

So the translates of a subspace $𝑈$ cut $𝑉$ into disjoint parallel pieces, as the solution sets in Figure 3.1.4 do. They are the elements of the quotient space $𝑉 / 𝑈$ (Definition 4.6.20), which treats each translate as a single vector.

Non-example 4.7.7 (Sets that are not affine subspaces).

(a)The unit circle $𝐶 =$ ${𝐱 \in ℝ^{2} | 𝑥_{1}^{2} + 𝑥_{2}^{2} = 1}$ . If $𝐶$ were an affine subspace, then $(1, 0) -$ $(- 1, 0) =$ $(2, 0)$ would lie in $dir 𝐶$ by Proposition 4.7.3(a), and $𝐶 = (1, 0) + dir 𝐶$ by (b). Then $(3, 0) =$ $(1, 0) +$ $(2, 0) \in$ $𝐶$ , which is false.
(b)The union $𝐸$ of the lines $𝑥_{2} = 0$ and $𝑥_{2} = 1$ in $ℝ^{2}$ . If $𝐸$ were an affine subspace, then $(0, 1) - (0, 0) \in dir 𝐸$ , so $(0, 0) +$ $2 (0, 1) =$ $(0, 2)$ would lie in $𝐸$ .
(c)The three-point set ${(0, 0), (1, 0), (0, 1)} \subseteq$ $𝔽_{2}^{2}$ (Exercise 3.1.48).

Solution sets, revisited

Proposition 3.1.19 says that consistent linear systems have affine solution sets. With dimension theory we can now say how large they are, and show that every affine subspace of $𝐹^{𝑛}$ arises this way.

Theorem 4.7.8 (Affine subspaces of

𝐹^{𝑛}

are solution sets).

(a)Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐲 \in 𝐹^{𝑚}$ . If the system $𝐴 𝐱 = 𝐲$ is consistent, its solution set is an affine subspace of $𝐹^{𝑛}$ with direction $ker 𝑇_{𝐴}$ and dimension $𝑛 - rank 𝐴$ .
(b)Conversely, let $𝐸$ be an affine subspace of $𝐹^{𝑛}$ of dimension $𝑘$ . Then $𝐸$ is the solution set of a system $𝐴 𝐱 = 𝐲$ with $𝐴 \in 𝐹^{(𝑛 - 𝑘) \times 𝑛}$ , that is, of $𝑛 - 𝑘$ equations. No system of fewer than $𝑛 - 𝑘$ equations has solution set $𝐸$ .

Proof.standard

Uses. Proposition 3.1.19 (Solution sets are translates of the kernel), Corollary 4.2.23 (Dimension of solution sets), Proposition 4.4.16 (Complements exist), Theorem 4.4.19 (Dimension of a sum), Proposition 4.4.18 (Every subspace is a kernel and an image), Theorem 4.5.6 (Classification by dimension), Proposition 2.3.17 (Every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is some $𝑇_{𝐴}$ ), Theorem 4.2.7 (The image of $𝑇_{𝐴}$ ), Proposition 4.1.21 (Dimensions of subspaces)

Techniques. Particular plus homogeneous (Technique 3.1.18), Dimension counting (Technique 4.2.24)

Hint 1. (a) is Proposition 3.1.19 plus a dimension count.

Hint 2. For (b), write

𝐸 = 𝐯 + 𝑈

. It suffices to find

𝐴 \in 𝐹^{(𝑛 - 𝑘) \times 𝑛}

with

ker 𝑇_{𝐴} = 𝑈

, and then to take

𝐲 = 𝐴 𝐯

Hint 3. Every subspace is a kernel (Proposition 4.4.18), but of an operator

𝐹^{𝑛} \to 𝐹^{𝑛}

, whose image is a complement

𝑊

𝑈

. Follow it by an isomorphism

𝑊 \to 𝐹^{𝑛 - 𝑘}

Sketch. (a) The solution set is

𝐱_{0} + ker 𝑇_{𝐴}

, and

dim ker 𝑇_{𝐴} =

𝑛 -

rank 𝐴

. (b) Let

𝐸 = 𝐯 + 𝑈

and let

𝑊

be a complement of

𝑈

, so

dim 𝑊 = 𝑛 - 𝑘

. Take

𝑇 \in L (𝐹^{𝑛})

with

ker 𝑇 = 𝑈

and

im 𝑇 = 𝑊

, and an isomorphism

𝜑 : 𝑊 \to 𝐹^{𝑛 - 𝑘}

. Then

𝜑 \circ 𝑇

has kernel

𝑈

and equals

𝑇_{𝐴}

with

𝐴 \in 𝐹^{(𝑛 - 𝑘) \times 𝑛}

, and

𝐴 𝐱 = 𝐴 𝐯

has solution set

𝐯 + 𝑈 = 𝐸

. With

𝑚

equations, (a) gives

𝑘 =

𝑛 -

rank 𝐴 \geq

𝑛 -

𝑚

(a) By Proposition 3.1.19, the solution set is $𝐱_{0} + ker 𝑇_{𝐴}$ for any solution $𝐱_{0}$ , an affine subspace with direction $ker 𝑇_{𝐴}$ . By Corollary 4.2.23(a), $dim ker 𝑇_{𝐴} =$ $𝑛 -$ $rank 𝐴$ .

(b) Write $𝐸 = 𝐯 + 𝑈$ with $𝑈 = dir 𝐸$ , so $dim 𝑈 = 𝑘$ . By Proposition 4.4.16, $𝑈$ has a complement $𝑊$ in $𝐹^{𝑛}$ . Since $𝑈 \cap 𝑊 = {𝟎}$ and $𝑈 + 𝑊 = 𝐹^{𝑛}$ , Theorem 4.4.19 gives $dim 𝑊 = 𝑛 - 𝑘$ . By Proposition 4.4.18 there is $𝑇 \in L (𝐹^{𝑛})$ with $ker 𝑇 = 𝑈$ and $im 𝑇 = 𝑊$ , and by Theorem 4.5.6 there is an isomorphism $𝜑 : 𝑊 \to 𝐹^{𝑛 - 𝑘}$ . The map $𝑆 : 𝐹^{𝑛} \to 𝐹^{𝑛 - 𝑘}$ , $𝑆 (𝐱) = 𝜑 (𝑇 (𝐱))$ , is linear, and since $𝜑$ is injective, $𝑆 (𝐱) = 𝟎$ if and only if $𝑇 (𝐱) = 𝟎$ . So $ker 𝑆 = 𝑈$ . By Proposition 2.3.17, $𝑆 = 𝑇_{𝐴}$ for some $𝐴 \in 𝐹^{(𝑛 - 𝑘) \times 𝑛}$ . Put $𝐲 = 𝐴 𝐯$ . Then $𝐯$ solves $𝐴 𝐱 = 𝐲$ , and by Proposition 3.1.19 the solution set is $𝐯 +$ $ker 𝑇_{𝐴} =$ $𝐯 +$ $𝑈 =$ $𝐸$ .

Finally, suppose $𝐸$ is the solution set of a system $𝐴^{'} 𝐱 = 𝐲^{'}$ with $𝐴^{'} \in 𝐹^{𝑚 \times 𝑛}$ . By (a), $𝑘 = 𝑛 - rank 𝐴^{'}$ . Since $im 𝑇_{𝐴^{'}}$ is a subspace of $𝐹^{𝑚}$ , Theorem 4.2.7 and Proposition 4.1.21 give $rank 𝐴^{'} =$ $dim im 𝑇_{𝐴^{'}} \leq$ $𝑚$ . Hence $𝑘 \geq 𝑛 - 𝑚$ , that is, $𝑚 \geq 𝑛 - 𝑘$ . □

For $𝑘 = 𝑛 - 1$ this says that the hyperplanes of $𝐹^{𝑛}$ are exactly the solution sets of single equations $𝑎_{1} 𝑥_{1} +$ $\dots +$ $𝑎_{𝑛} 𝑥_{𝑛} =$ $𝑐$ with $(𝑎_{1}, \dots, 𝑎_{𝑛}) \neq 𝟎$ . (Such an equation is consistent, since $𝑥_{𝑗} = 𝑐 / 𝑎_{𝑗}$ and $𝑥_{𝑖} = 0$ for $𝑖 \neq 𝑗$ solves it when $𝑎_{𝑗} \neq 0$ , and its coefficient matrix has rank $1$ .)

Example 4.7.9 (A line in $ℝ^{3}$ as a solution set). Let $𝐸$ be the line through $(1, 2, 0)$ with direction $span ((1, 1, 1))$ . By Theorem 4.7.8(b), two equations cut it out. We need $𝐴 \in ℝ^{2 \times 3}$ of rank $2$ with $(1, 1, 1) \in ker 𝑇_{𝐴}$ . The rows $(𝑎_{1}, 𝑎_{2}, 𝑎_{3})$ must satisfy $𝑎_{1} + 𝑎_{2} + 𝑎_{3} = 0$ and be independent, so take $(1, - 1, 0)$ and $(0, 1, - 1)$ . Then $𝐴 (1, 2, 0) = (- 1, 2)$ , and $𝐸$ is the solution set of

𝑥_{1} - 𝑥_{2} = - 1, 𝑥_{2} - 𝑥_{3} = 2 .

Indeed, the solution set contains $(1, 2, 0)$ , and its direction $ker 𝑇_{𝐴}$ contains $(1, 1, 1)$ and has dimension $3 - 2 = 1$ , so it equals $span ((1, 1, 1))$ by Proposition 4.1.21(b).

Intersections

Proposition 4.7.10 (Intersections of affine subspaces). Let

{(𝐸_{𝑖})}_{𝑖 \in 𝐼}

be a family of affine subspaces of

𝑉

, indexed by a nonempty set

𝐼

. Then

⋂_{𝑖 \in 𝐼} 𝐸_{𝑖}

is either empty or an affine subspace with direction

⋂_{𝑖 \in 𝐼} dir 𝐸_{𝑖}

Proof.routine

Uses. Proposition 4.7.3 (Direction and base points), Proposition 1.4.16 (Intersections of subspaces)

Techniques. Choose a base point (Technique 4.7.4), Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. If the intersection is nonempty, pick a point

𝐜

in it. How does each

𝐸_{𝑖}

look when

𝐜

is used as its base point?

Hint 2.

𝐸_{𝑖} = 𝐜 + dir 𝐸_{𝑖}

, so

𝐱 \in 𝐸_{𝑖}

if and only if

𝐱 - 𝐜 \in dir 𝐸_{𝑖}

Sketch. With

𝐜

in the intersection,

𝐸_{𝑖} = 𝐜 + 𝑈_{𝑖}

for every

𝑖

. So

𝐱

lies in every

𝐸_{𝑖}

if and only if

𝐱 - 𝐜

lies in every

𝑈_{𝑖}

, and the intersection is

𝐜 + ⋂ 𝑈_{𝑖}

, a translate of a subspace.

Suppose $𝐜 \in ⋂_{𝑖 \in 𝐼} 𝐸_{𝑖}$ , and let $𝑈_{𝑖} = dir 𝐸_{𝑖}$ . By Proposition 4.7.3(b), $𝐸_{𝑖} = 𝐜 + 𝑈_{𝑖}$ for every $𝑖$ . For $𝐱 \in 𝑉$ we have $𝐱 \in 𝐜 + 𝑈_{𝑖}$ if and only if $𝐱 - 𝐜 \in 𝑈_{𝑖}$ (if $𝐱 = 𝐜 + 𝐮$ , then $𝐱 - 𝐜 = 𝐮$ ; conversely $𝐱 = 𝐜 + (𝐱 - 𝐜)$ ). Hence $𝐱$ lies in every $𝐸_{𝑖}$ if and only if $𝐱 - 𝐜 \in ⋂_{𝑖} 𝑈_{𝑖}$ , that is,

⋂_{𝑖 \in 𝐼} 𝐸_{𝑖} = 𝐜 + ⋂_{𝑖 \in 𝐼} 𝑈_{𝑖} .

By Proposition 1.4.16, $⋂_{𝑖} 𝑈_{𝑖}$ is a subspace, so the intersection is an affine subspace, and its direction is $⋂_{𝑖} 𝑈_{𝑖}$ by Proposition 4.7.3(a). □

The choice of base point was the whole proof: a point common to all the $𝐸_{𝑖}$ lets us write all of them as translates from one origin. Without such a point the intersection can be empty even when the directions have a large intersection, as for two distinct parallel lines. In $ℝ^{3}$ two lines can also miss each other without being parallel, for example the $𝑥_{1}$ -axis and $(0, 0, 1) + span (𝐞_{2})$ ; such lines are called skew. Exercise 4.7.53 characterizes skew lines, and Exercise 4.7.52 explains why a plane has none.

4.7.2Affine combinations

Intuition (Which combinations make sense without an origin). Suppose we only know positions relative to some reference point $𝐩$ , as on a screen or in a scene: the point $𝐯$ has position $𝐯 - 𝐩$ . Combining positions with coefficients $𝑐_{𝑖}$ and converting the result back into a point gives

𝐩 + \sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩) = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} + (1 - \sum_{𝑖} 𝑐_{𝑖}) 𝐩 .

For two reference points $𝐩$ and $𝐩^{'}$ the results differ by $(1 - \sum_{𝑖} 𝑐_{𝑖}) (𝐩 - 𝐩^{'})$ , so the result does not depend on the reference point exactly when the coefficients add up to $1$ . These are the weighted averages of points. Combinations whose coefficients add up to $0$ are meaningful in a different way: $\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ does not depend on $𝐩$ , and it is a displacement rather than a point, like the difference $𝐯 - 𝐮$ .

Definition 4.7.11 (Affine combination). An affine combination of

𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉

is a linear combination

𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}

whose coefficients satisfy

𝑐_{1} + \dots + 𝑐_{𝑘} = 1

Example 4.7.12 (Affine combinations).

(a)For $𝐮, 𝐯 \in 𝑉$ and $𝑡 \in 𝐹$ , $(1 - 𝑡) 𝐮 +$ $𝑡 𝐯 =$ $𝐮 +$ $𝑡 (𝐯 - 𝐮)$ . If $𝐮 \neq 𝐯$ , these points form the line through $𝐮$ and $𝐯$ . When $1 + 1 \neq 0$ in $𝐹$ , the case $𝑡 = 1 / 2$ is the midpoint $\frac{1}{2} 𝐮 + \frac{1}{2} 𝐯$ .
(b)If $𝑘 \cdot 1 \neq 0$ in $𝐹$ , the centroid $\frac{1}{𝑘} (𝐯_{1} + \dots + 𝐯_{𝑘})$ is an affine combination. Over $𝔽_{3}$ , three points have no centroid.
(c) $𝐮 - 𝐯 + 𝐰$ , with coefficients $1, - 1, 1$ , is the fourth vertex of the parallelogram with consecutive vertices $𝐮, 𝐯, 𝐰$ : its differences with $𝐰$ and with $𝐮$ are $𝐮 - 𝐯$ and $𝐰 - 𝐯$ .
(d)If $𝐴 𝐱_{𝑖} = 𝐲$ for $𝑖 = 1, \dots, 𝑘$ and $\sum_{𝑖} 𝑐_{𝑖} = 1$ , then $𝐴 (\sum_{𝑖} 𝑐_{𝑖} 𝐱_{𝑖}) =$ $(\sum_{𝑖} 𝑐_{𝑖}) 𝐲 =$ $𝐲$ . So affine combinations of solutions are solutions, while the sum of two solutions is not.

Lemma 4.7.13 (Affine combinations do not depend on the origin). Let $𝐯_{1}, \dots, 𝐯_{𝑘}, 𝐩 \in 𝑉$ and $𝑐_{1}, \dots, 𝑐_{𝑘} \in 𝐹$ .

(a)If $𝑐_{1} + \dots + 𝑐_{𝑘} = 1$ , then $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} =$ $𝐩 +$ $\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩)$ .
(b)If $𝑐_{1} + \dots + 𝑐_{𝑘} = 0$ , then $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} =$ $\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩)$ .

Proof.routine

Uses. Lemma 1.3.9 (Arithmetic of linear combinations)

Hint. Expand

\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩)

Sketch.

\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩) =

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} -

(\sum_{𝑖} 𝑐_{𝑖}) 𝐩

; substitute

\sum_{𝑖} 𝑐_{𝑖} = 1

0

By distributivity and Lemma 1.3.9,

\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐩) =

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} -

(\sum_{𝑖} 𝑐_{𝑖}) 𝐩

. If

\sum_{𝑖} 𝑐_{𝑖} = 1

, this is

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} - 𝐩

, and adding

𝐩

gives (a). If

\sum_{𝑖} 𝑐_{𝑖} = 0

, it is

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}

, which is (b). □

Subspaces are the nonempty sets closed under linear combinations. Affine subspaces have the analogous description.

Theorem 4.7.14 (Affine subspaces are the sets closed under affine combinations). A nonempty subset

𝐸 \subseteq 𝑉

is an affine subspace if and only if every affine combination of elements of

𝐸

lies in

𝐸

Intuition. The forward direction is a base point computation with Lemma 4.7.13(a). For the converse, the base point tells us what the direction must be: if

𝐸 = 𝐩 + 𝑈

with

𝐩 \in 𝐸

, then

𝑈 = {𝐞 - 𝐩 | 𝐞 \in 𝐸}

. So define

𝑈

this way and check the subspace criterion. Each closure property of

𝑈

should come from one affine combination in

𝐸

Proof.standard

Uses. Lemma 4.7.13 (Affine combinations do not depend on the origin), Proposition 4.7.3 (Direction and base points), Theorem 1.4.3 (Subspace criterion)

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. For (

\Rightarrow

), write

𝐸 = 𝐩 + 𝑈

and each

𝐞_{𝑖} = 𝐩 + 𝐮_{𝑖}

, and apply Lemma 4.7.13(a).

Hint 2. For (

\Leftarrow

), fix

𝐩 \in 𝐸

and put

𝑈 = {𝐞 - 𝐩 | 𝐞 \in 𝐸}

. Then

𝐸 = 𝐩 + 𝑈

, and it remains to show that

𝑈

is a subspace.

Hint 3.

𝑐 (𝐞 - 𝐩) =

(𝑐 𝐞 + (1 - 𝑐) 𝐩) -

𝐩

and

(𝐞 - 𝐩) +

(𝐞^{'} - 𝐩) =

(𝐞 + 𝐞^{'} - 𝐩) -

𝐩

. The vectors in parentheses are affine combinations of points of

𝐸

Sketch. (

\Rightarrow

)

\sum_{𝑖} 𝑐_{𝑖} (𝐩 + 𝐮_{𝑖}) =

𝐩 +

\sum_{𝑖} 𝑐_{𝑖} 𝐮_{𝑖} \in

𝐩 +

𝑈

. (

\Leftarrow

) With

𝑈 = 𝐸 - 𝐩

𝟎 = 𝐩 - 𝐩 \in 𝑈

𝑐 (𝐞 - 𝐩) =

(𝑐 𝐞 + (1 - 𝑐) 𝐩) -

𝐩 \in

𝑈

, and

(𝐞 - 𝐩) +

(𝐞^{'} - 𝐩) =

(𝐞 + 𝐞^{'} - 𝐩) -

𝐩 \in

𝑈

, because

𝑐 𝐞 + (1 - 𝑐) 𝐩

and

𝐞 + 𝐞^{'} - 𝐩

are affine combinations of elements of

𝐸

( $\Rightarrow$ ) Let $𝐸 = 𝐩 + 𝑈$ be an affine subspace, let $𝐞_{1}, \dots, 𝐞_{𝑘} \in 𝐸$ , and let $\sum_{𝑖} 𝑐_{𝑖} = 1$ . Since $𝐩 = 𝐩 + 𝟎 \in 𝐸$ , each $𝐮_{𝑖} = 𝐞_{𝑖} - 𝐩$ lies in $𝑈$ by Proposition 4.7.3(a). By Lemma 4.7.13(a),

\sum_{𝑖} 𝑐_{𝑖} 𝐞_{𝑖} = 𝐩 + \sum_{𝑖} 𝑐_{𝑖} 𝐮_{𝑖} \in 𝐩 + 𝑈 = 𝐸 .

( $\Leftarrow$ ) Suppose $𝐸$ is nonempty and closed under affine combinations. Fix $𝐩 \in 𝐸$ and put $𝑈 = {𝐞 - 𝐩 | 𝐞 \in 𝐸}$ . Then $𝐸 = 𝐩 + 𝑈$ , since each $𝐞 \in 𝐸$ equals $𝐩 + (𝐞 - 𝐩)$ . We check the conditions of Theorem 1.4.3 for $𝑈$ .

$𝟎 \in 𝑈$ , since $𝟎 = 𝐩 - 𝐩$ .

Scalar multiples. Let $𝐞 \in 𝐸$ and $𝑐 \in 𝐹$ . The vector $𝑐 𝐞 + (1 - 𝑐) 𝐩$ is an affine combination of $𝐞, 𝐩 \in 𝐸$ , so it lies in $𝐸$ , and

𝑐 (𝐞 - 𝐩) = (𝑐 𝐞 + (1 - 𝑐) 𝐩) - 𝐩 \in 𝑈 .

Sums. Let $𝐞, 𝐞^{'} \in 𝐸$ . The vector $𝐞 + 𝐞^{'} - 𝐩$ is an affine combination of $𝐞, 𝐞^{'}, 𝐩$ , with coefficients $1 + 1 - 1 = 1$ , so it lies in $𝐸$ , and

(𝐞 - 𝐩) + (𝐞^{'} - 𝐩) = (𝐞 + 𝐞^{'} - 𝐩) - 𝐩 \in 𝑈 .

So $𝑈$ is a subspace, and $𝐸 = 𝐩 + 𝑈$ is an affine subspace. □

Remark 4.7.15 (Two points or three). The proof used affine combinations of two points, for scalar multiples, and of three points, for sums. The three-point combination cannot be dropped in general. When

1 + 1 \neq 0

, Exercise 3.1.48 showed that closure under combinations of two points (the lines through pairs of points) already forces a nonempty set to be affine, by way of midpoints. Over

𝔽_{2}

, the only affine combinations of two points

𝐮, 𝐯

are

𝐮

and

𝐯

themselves. The set

{(0, 0), (1, 0), (0, 1)}

is closed under them, but

(1, 0) +

(0, 1) -

(0, 0) =

(1, 1)

is missing. Exercise 4.7.61 shows that over

𝔽_{2}

three points always suffice.

Definition 4.7.16 (Affine hull). The affine hull

aff 𝑆

of a subset

𝑆 \subseteq 𝑉

is the set of all affine combinations of finitely many elements of

𝑆

. For a list,

aff (𝐯_{0}, \dots, 𝐯_{𝑘}) =

aff {𝐯_{0}, \dots, 𝐯_{𝑘}}

The empty set has no affine combinations, because an empty list of coefficients adds up to $0$ , so $aff \emptyset = \emptyset$ .

Proposition 4.7.17 (The affine hull). Let $𝑆 \subseteq 𝑉$ be nonempty, and let $𝐯_{0}, \dots, 𝐯_{𝑘} \in 𝑉$ .

(a) $aff 𝑆$ is an affine subspace containing $𝑆$ , and it is contained in every affine subspace that contains $𝑆$ .
(b) $aff (𝐯_{0}, \dots, 𝐯_{𝑘}) =$ $𝐯_{0} +$ $span (𝐯_{1} - 𝐯_{0}, \dots, 𝐯_{𝑘} - 𝐯_{0})$ . In particular, $dim aff (𝐯_{0}, \dots, 𝐯_{𝑘}) \leq$ $𝑘$ .

Proof.standard

Uses. Theorem 4.7.14 (Affine subspaces are the sets closed under affine combinations), Lemma 4.7.13 (Affine combinations do not depend on the origin), Theorem 1.6.10 (Every spanning list contains a basis)

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. For (a), show that an affine combination of affine combinations of elements of

𝑆

is again one. List all the elements of

𝑆

involved once, padding with zero coefficients.

Hint 2. For (b), use

𝐯_{0}

as base point: Lemma 4.7.13(a) with

𝐩 = 𝐯_{0}

rewrites

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}

Sketch. (a) If

𝐰_{𝑗} =

\sum_{𝑖} 𝑎_{𝑖 𝑗} 𝐬_{𝑖}

with

\sum_{𝑖} 𝑎_{𝑖 𝑗} = 1

and

\sum_{𝑗} 𝑑_{𝑗} = 1

, then

\sum_{𝑗} 𝑑_{𝑗} 𝐰_{𝑗} =

\sum_{𝑖} (\sum_{𝑗} 𝑑_{𝑗} 𝑎_{𝑖 𝑗}) 𝐬_{𝑖}

, whose coefficients add up to

\sum_{𝑗} 𝑑_{𝑗} = 1

. So

aff 𝑆

is closed under affine combinations, hence affine by Theorem 4.7.14, and every affine subspace containing

𝑆

is closed under them, hence contains

aff 𝑆

. (b) By Lemma 4.7.13(a),

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} =

𝐯_{0} +

\sum_{𝑖 \geq 1} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐯_{0})

, and conversely

𝐯_{0} +

\sum_{𝑖 \geq 1} 𝑎_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =

(1 - \sum_{𝑖} 𝑎_{𝑖}) 𝐯_{0} +

\sum_{𝑖 \geq 1} 𝑎_{𝑖} 𝐯_{𝑖}

(a) Each $𝐬 \in 𝑆$ is the affine combination $1 𝐬$ , so $𝑆 \subseteq aff 𝑆$ , and $aff 𝑆$ is nonempty. Let $𝐰_{1}, \dots, 𝐰_{𝑚} \in aff 𝑆$ and $𝑑_{1} + \dots + 𝑑_{𝑚} = 1$ . Each $𝐰_{𝑗}$ is an affine combination of finitely many elements of $𝑆$ . Let $𝐬_{1}, \dots, 𝐬_{𝑁}$ list all elements of $𝑆$ that occur in any of these combinations, and pad with zero coefficients, so that $𝐰_{𝑗} =$ $\sum_{𝑖 = 1}^{𝑁} 𝑎_{𝑖 𝑗} 𝐬_{𝑖}$ with $\sum_{𝑖} 𝑎_{𝑖 𝑗} = 1$ for every $𝑗$ . Then

\sum_{𝑗} 𝑑_{𝑗} 𝐰_{𝑗} = \sum_{𝑖 = 1}^{𝑁} (\sum_{𝑗} 𝑑_{𝑗} 𝑎_{𝑖 𝑗}) 𝐬_{𝑖}, and \sum_{𝑖} \sum_{𝑗} 𝑑_{𝑗} 𝑎_{𝑖 𝑗} = \sum_{𝑗} 𝑑_{𝑗} \sum_{𝑖} 𝑎_{𝑖 𝑗} = \sum_{𝑗} 𝑑_{𝑗} = 1 .

So $aff 𝑆$ is closed under affine combinations, and it is an affine subspace by Theorem 4.7.14. If $𝐸$ is an affine subspace with $𝑆 \subseteq 𝐸$ , then $𝐸$ is closed under affine combinations by the same theorem, so it contains every affine combination of elements of $𝑆$ , that is, $aff 𝑆 \subseteq 𝐸$ .

(b) If $𝑐_{0} + \dots + 𝑐_{𝑘} = 1$ , then Lemma 4.7.13(a) with $𝐩 = 𝐯_{0}$ gives

\sum_{𝑖 = 0}^{𝑘} 𝑐_{𝑖} 𝐯_{𝑖} = 𝐯_{0} + \sum_{𝑖 = 1}^{𝑘} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐯_{0}),

because the term with $𝑖 = 0$ is $𝑐_{0} (𝐯_{0} - 𝐯_{0}) = 𝟎$ . So $aff (𝐯_{0}, \dots, 𝐯_{𝑘}) \subseteq$ $𝐯_{0} +$ $span (𝐯_{1} - 𝐯_{0}, \dots, 𝐯_{𝑘} - 𝐯_{0})$ . Conversely, for $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ ,

𝐯_{0} + \sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) = (1 - 𝑎_{1} - \dots - 𝑎_{𝑘}) 𝐯_{0} + 𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘},

an affine combination of $𝐯_{0}, \dots, 𝐯_{𝑘}$ . Hence the two sets are equal. The direction $span (𝐯_{1} - 𝐯_{0}, \dots, 𝐯_{𝑘} - 𝐯_{0})$ is spanned by a list of length $𝑘$ , which contains a basis (Theorem 1.6.10), so its dimension is at most $𝑘$ . □

For two points $𝐮 \neq 𝐯$ , (b) gives $aff (𝐮, 𝐯) =$ $𝐮 +$ $span (𝐯 - 𝐮) =$ ${(1 - 𝑡) 𝐮 + 𝑡 𝐯 | 𝑡 \in 𝐹}$ , the line through them. Over $𝔽_{2}$ this line consists of $𝐮$ and $𝐯$ alone. Three points of $ℝ^{3}$ that do not lie on one line have a plane as their affine hull.

4.7.3Affine independence and barycentric coordinates

Intuition (From linear to affine independence). By Proposition 4.7.17(b),

𝑘 + 1

points have an affine hull of dimension at most

𝑘

: two points span at most a line, three at most a plane. The points are in general position when they reach this bound, which by Proposition 4.7.17(b) means that the differences

𝐯_{𝑖} - 𝐯_{0}

are linearly independent. That description singles out

𝐯_{0}

. For a definition that treats all points alike, imitate linear independence, which asks that the only linear relation be the trivial one (Theorem 1.5.5). For points, only relations whose coefficients add up to

0

are meaningful (the intuition at the start of Section 4.7.2), so we test only those.

Definition 4.7.18 (Affine independence). A list $𝐯_{0}, \dots, 𝐯_{𝑘}$ in $𝑉$ is affinely independent if the only scalars $𝑐_{0}, \dots, 𝑐_{𝑘} \in 𝐹$ with

𝑐_{0} 𝐯_{0} + \dots + 𝑐_{𝑘} 𝐯_{𝑘} = 𝟎 and 𝑐_{0} + \dots + 𝑐_{𝑘} = 0

are $𝑐_{0} = \dots = 𝑐_{𝑘} = 0$ . Otherwise the list is affinely dependent, and a relation of this form with some $𝑐_{𝑖} \neq 0$ is an affine dependence.

Example 4.7.19 (Affine independence).

(a)A single point is affinely independent, since $𝑐_{0} = 0$ is forced by the coefficient sum. Two points $𝐮, 𝐯$ are affinely independent if and only if $𝐮 \neq 𝐯$ : a relation with coefficients $𝑐, - 𝑐$ says $𝑐 (𝐮 - 𝐯) = 𝟎$ .
(b)The points $(0, 0), (1, 0), (0, 1)$ of $ℝ^{2}$ are affinely independent: $𝑐_{0} (0, 0) +$ $𝑐_{1} (1, 0) +$ $𝑐_{2} (0, 1) =$ $(𝑐_{1}, 𝑐_{2})$ , so a relation has $𝑐_{1} = 𝑐_{2} = 0$ , and then $𝑐_{0} = 0$ . The list is linearly dependent, since it contains $𝟎$ .
(c)The points $(0, 0), (1, 1), (3, 3)$ are affinely dependent: $- 2 (0, 0) +$ $3 (1, 1) -$ $(3, 3) =$ $𝟎$ , and $- 2 + 3 - 1 = 0$ . They lie on one line.

An affine dependence is in particular a linear relation, so a linearly independent list is affinely independent; (b) shows that the converse fails. Unlike linear independence, affine independence does not change when all points are moved by the same vector $𝐰$ : by Lemma 4.7.13(b) with $𝐩 = - 𝐰$ , $\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖} + 𝐰) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ whenever $\sum_{𝑖} 𝑐_{𝑖} = 0$ .

Proposition 4.7.20 (Affine independence through differences). A list

𝐯_{0}, \dots, 𝐯_{𝑘}

𝑉

is affinely independent if and only if

𝐯_{1} -

𝐯_{0},

\dots,

𝐯_{𝑘} -

𝐯_{0}

is linearly independent.

Proof (translate the relations).routine

Uses. Definition 4.7.18 (Affine independence), Theorem 1.5.5 (Independence is decided at zero), Lemma 4.7.13 (Affine combinations do not depend on the origin)

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. A linear relation among the differences,

\sum_{𝑖 \geq 1} 𝑎_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =

𝟎

, expands to a relation among

𝐯_{0}, \dots, 𝐯_{𝑘}

. What are its coefficients, and what is their sum?

Hint 2. Conversely, an affine dependence has

𝑐_{0} = - (𝑐_{1} + \dots + 𝑐_{𝑘})

. Use Lemma 4.7.13(b) with

𝐩 = 𝐯_{0}

Sketch.

\sum_{𝑖 \geq 1} 𝑎_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =

(- \sum_{𝑖} 𝑎_{𝑖}) 𝐯_{0} +

\sum_{𝑖 \geq 1} 𝑎_{𝑖} 𝐯_{𝑖}

has coefficient sum

0

. So the linear relations among the differences are the affine relations among the points, read with

𝑐_{0} = - \sum_{𝑖 \geq 1} 𝑐_{𝑖}

, and trivial ones correspond to trivial ones.

( $\Rightarrow$ ) Suppose $𝐯_{0}, \dots, 𝐯_{𝑘}$ is affinely independent, and let $\sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =$ $𝟎$ . Expanding,

(- 𝑎_{1} - \dots - 𝑎_{𝑘}) 𝐯_{0} + 𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘} = 𝟎,

and the coefficients add up to $0$ . By affine independence they are all $0$ ; in particular $𝑎_{1} = \dots = 𝑎_{𝑘} = 0$ . So the differences are linearly independent by Theorem 1.5.5.

( $\Leftarrow$ ) Suppose the differences are linearly independent, and let $\sum_{𝑖 = 0}^{𝑘} 𝑐_{𝑖} 𝐯_{𝑖} =$ $𝟎$ with $\sum_{𝑖 = 0}^{𝑘} 𝑐_{𝑖} = 0$ . By Lemma 4.7.13(b) with $𝐩 = 𝐯_{0}$ , $\sum_{𝑖 = 0}^{𝑘} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =$ $𝟎$ , and the term with $𝑖 = 0$ is $𝟎$ , so $\sum_{𝑖 = 1}^{𝑘} 𝑐_{𝑖} (𝐯_{𝑖} - 𝐯_{0}) =$ $𝟎$ . Hence $𝑐_{1} = \dots = 𝑐_{𝑘} = 0$ , and $𝑐_{0} =$ $- (𝑐_{1} + \dots + 𝑐_{𝑘}) =$ $0$ . □

The first approach singles out $𝐯_{0}$ . The second treats all points alike, at the price of a larger space: it turns each point into a vector one dimension up, so that affine relations become linear relations. For a vector space $𝑉$ , the set $𝑉 \times 𝐹$ of pairs $(𝐯, 𝑎)$ with the componentwise operations $(𝐯, 𝑎) +$ $(𝐰, 𝑏) =$ $(𝐯 + 𝐰, 𝑎 + 𝑏)$ and $𝑐 (𝐯, 𝑎) =$ $(𝑐 𝐯, 𝑐 𝑎)$ is a vector space, because each axiom holds in each component separately. (Product spaces are studied in Section 4.6.) For $𝑉 = 𝐹^{𝑛}$ it is $𝐹^{𝑛 + 1}$ , with $(𝐯, 𝑎)$ the vector whose last entry is $𝑎$ .

Technique 4.7.21 (Homogenize: add a coordinate $1$ ). To turn an affine problem into a linear one without choosing a base point, replace each point $𝐯$ by its lift $\hat{𝐯} = (𝐯, 1)$ in $𝑉 \times 𝐹$ , or in $𝐹^{𝑛 + 1}$ if $𝑉 = 𝐹^{𝑛}$ . Then

\sum_{𝑖} 𝑐_{𝑖} {\hat{𝐯}}_{𝑖} = (\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}, \sum_{𝑖} 𝑐_{𝑖}),

so the last coordinate records the coefficient sum. Affine combinations of points are the linear combinations of their lifts that land in the hyperplane “last coordinate $1$ ”, and affine dependences are the linear relations among the lifts. Row reduction, rank, and matrices then apply directly.

Reach for it when you want to treat all points symmetrically, or when you want affine maps to become matrix multiplication. The canonical example is the second approach to Proposition 4.7.20. Homogeneous coordinates (Section 4.7.6) apply the same idea to maps.

Used in 3 places and 1 practice exercise

Proposition 4.7.20, Theorem 4.7.42, Exercise 4.7.58; 1 practice exercise.

Proof (lift to one dimension higher).standard

Uses. Definition 4.7.18 (Affine independence), Proposition 1.5.18 (Operations that preserve independence), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Homogenize: add a coordinate $1$ (Technique 4.7.21)

Hint 1. Compute

\sum_{𝑖} 𝑐_{𝑖} (𝐯_{𝑖}, 1)

𝑉 \times 𝐹

. What does it say about affine dependences?

Hint 2. The lifts are linearly independent exactly when the points are affinely independent. Now subtract the first lift from the others; Proposition 1.5.18 says that independence does not change.

Sketch.

\sum_{𝑖} 𝑐_{𝑖} {\hat{𝐯}}_{𝑖} =

(\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}, \sum_{𝑖} 𝑐_{𝑖})

, so the points are affinely independent if and only if their lifts are linearly independent, if and only if (by Proposition 1.5.18(c))

(𝐯_{0}, 1),

(𝐯_{1} - 𝐯_{0}, 0),

\dots,

(𝐯_{𝑘} - 𝐯_{0}, 0)

is. In a relation among these, the last coordinate forces the first coefficient to be

0

, so this list is independent if and only if the differences are.

Let ${\hat{𝐯}}_{𝑖} =$ $(𝐯_{𝑖}, 1) \in$ $𝑉 \times$ $𝐹$ . For $𝑐_{0}, \dots, 𝑐_{𝑘} \in 𝐹$ ,

\sum_{𝑖 = 0}^{𝑘} 𝑐_{𝑖} {\hat{𝐯}}_{𝑖} = (\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}, \sum_{𝑖} 𝑐_{𝑖}),

so $\sum_{𝑖} 𝑐_{𝑖} {\hat{𝐯}}_{𝑖} = 𝟎$ if and only if $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎$ and $\sum_{𝑖} 𝑐_{𝑖} = 0$ . Thus the affine dependences of $𝐯_{0}, \dots, 𝐯_{𝑘}$ are exactly the nontrivial linear relations of ${\hat{𝐯}}_{0}, \dots, {\hat{𝐯}}_{𝑘}$ , and the points are affinely independent if and only if their lifts are linearly independent.

Replacing ${\hat{𝐯}}_{𝑖}$ by ${\hat{𝐯}}_{𝑖} + (- 1) {\hat{𝐯}}_{0}$ for $𝑖 = 1, \dots, 𝑘$ , one at a time, does not change linear independence (Proposition 1.5.18(c)). The new list is

(𝐯_{0}, 1), (𝐮_{1}, 0), \dots, (𝐮_{𝑘}, 0), where 𝐮_{𝑖} = 𝐯_{𝑖} - 𝐯_{0} .

If $𝑎_{0} (𝐯_{0}, 1) +$ $\sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} (𝐮_{𝑖}, 0) =$ $𝟎$ , the last coordinate gives $𝑎_{0} = 0$ , and then $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} = 𝟎$ . So if $𝐮_{1}, \dots, 𝐮_{𝑘}$ is independent, every such relation is trivial. Conversely, a relation $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} = 𝟎$ gives the relation $0 (𝐯_{0}, 1) +$ $\sum_{𝑖} 𝑎_{𝑖} (𝐮_{𝑖}, 0) =$ $𝟎$ , so if the new list is independent, all $𝑎_{𝑖} = 0$ . Hence the new list is independent if and only if the differences are, and chaining the equivalences proves the proposition. □

The second approach also gives a mechanical test. For points of $𝐹^{𝑛}$ , $𝐯_{0}, \dots, 𝐯_{𝑘}$ is affinely independent if and only if the $(𝑛 + 1) \times (𝑘 + 1)$ matrix with columns ${\hat{𝐯}}_{0}, \dots, {\hat{𝐯}}_{𝑘}$ has linearly independent columns, which row reduction decides (Theorem 3.5.12).

Theorem 4.7.22 (Characterizations of affine independence). Let $𝐯_{0}, \dots, 𝐯_{𝑘} \in 𝑉$ . The following are equivalent.

(I) $𝐯_{0}, \dots, 𝐯_{𝑘}$ is affinely independent.
(II) $𝐯_{1} -$ $𝐯_{0},$ $\dots,$ $𝐯_{𝑘} -$ $𝐯_{0}$ is linearly independent.
(III) $dim aff (𝐯_{0}, \dots, 𝐯_{𝑘}) =$ $𝑘$ .
(IV)Every point of $aff (𝐯_{0}, \dots, 𝐯_{𝑘})$ is an affine combination of $𝐯_{0}, \dots, 𝐯_{𝑘}$ in exactly one way.
(V)No $𝐯_{𝑗}$ lies in the affine hull of the other entries of the list.

Proof.standard

Uses. Proposition 4.7.20 (Affine independence through differences), Proposition 4.7.17 (The affine hull), Proposition 1.6.2 (Bases are independent spanning lists), Corollary 4.1.11 (Spanning lists are long)

Techniques. Proving TFAE (Technique 2.7.1), Uniqueness via two candidates (Technique 1.1.8), Contrapositive (Technique 1.5.4)

Hint 1. (I)

\Leftrightarrow

(II) is Proposition 4.7.20. For (II)

\Leftrightarrow

(III), compute the direction of the affine hull with Proposition 4.7.17(b).

Hint 2. For (I)

\Leftrightarrow

(IV), subtract two representations of the same point. What do the coefficients of the difference add up to?

Hint 3. For (I)

\Leftrightarrow

(V), solve an affine dependence for a

𝐯_{𝑗}

with

𝑐_{𝑗} \neq 0

, and check that the new coefficients add up to

1

Sketch. The direction of the hull is spanned by the

𝑘

differences, so it has dimension

𝑘

exactly when they are a basis of it, that is, independent. Two representations of one point differ by an affine dependence, and conversely an affine dependence added to

𝐯_{0} = 1 𝐯_{0}

gives a second representation of

𝐯_{0}

. If

𝐯_{𝑗} =

\sum_{𝑖 \neq 𝑗} 𝑐_{𝑖} 𝐯_{𝑖}

with

\sum 𝑐_{𝑖} = 1

, moving

𝐯_{𝑗}

across gives an affine dependence; if

\sum 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎

\sum 𝑐_{𝑖} = 0

, and

𝑐_{𝑗} \neq 0

, then

𝐯_{𝑗} =

\sum_{𝑖 \neq 𝑗} (- 𝑐_{𝑖} / 𝑐_{𝑗}) 𝐯_{𝑖}

, and these coefficients add up to

1

We prove (I) $\Leftrightarrow$ (II), (II) $\Leftrightarrow$ (III), (I) $\Leftrightarrow$ (IV), and (I) $\Leftrightarrow$ (V).

(I) $\Leftrightarrow$ (II) is Proposition 4.7.20.

(II) $\Leftrightarrow$ (III). By Proposition 4.7.17(b), the direction of $aff (𝐯_{0}, \dots, 𝐯_{𝑘})$ is $𝐷 =$ $span (𝐯_{1} - 𝐯_{0}, \dots, 𝐯_{𝑘} - 𝐯_{0})$ , which is spanned by a list of length $𝑘$ . If this list is independent, it is a basis of $𝐷$ (Proposition 1.6.2), so $dim 𝐷 = 𝑘$ . If $dim 𝐷 = 𝑘$ , the spanning list of length $𝑘$ is a basis of $𝐷$ by Corollary 4.1.11(b), hence independent.

(I) $\Rightarrow$ (IV). Suppose $𝐩 =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} =$ $\sum_{𝑖} 𝑐_{𝑖}^{'} 𝐯_{𝑖}$ with $\sum_{𝑖} 𝑐_{𝑖} =$ $\sum_{𝑖} 𝑐_{𝑖}^{'} =$ $1$ . Subtracting, $\sum_{𝑖} (𝑐_{𝑖} - 𝑐_{𝑖}^{'}) 𝐯_{𝑖} =$ $𝟎$ and $\sum_{𝑖} (𝑐_{𝑖} - 𝑐_{𝑖}^{'}) = 0$ . By (I), $𝑐_{𝑖} = 𝑐_{𝑖}^{'}$ for all $𝑖$ .

(IV) $\Rightarrow$ (I). Suppose $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎$ with $\sum_{𝑖} 𝑐_{𝑖} = 0$ . Then

𝐯_{0} = 1 𝐯_{0} + 0 𝐯_{1} + \dots + 0 𝐯_{𝑘} = (1 + 𝑐_{0}) 𝐯_{0} + 𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}

are two affine combinations (their coefficients add up to $1$ and to $1 + 0 = 1$ ) that represent the point $𝐯_{0}$ of the affine hull. By (IV) they have the same coefficients, so $𝑐_{0} = \dots = 𝑐_{𝑘} = 0$ .

(I) $\Leftrightarrow$ (V). We show that (I) fails if and only if (V) fails. If $𝐯_{𝑗} =$ $\sum_{𝑖 \neq 𝑗} 𝑐_{𝑖} 𝐯_{𝑖}$ with $\sum_{𝑖 \neq 𝑗} 𝑐_{𝑖} = 1$ , then $\sum_{𝑖 \neq 𝑗} 𝑐_{𝑖} 𝐯_{𝑖} +$ $(- 1) 𝐯_{𝑗} =$ $𝟎$ is an affine dependence: its coefficients add up to $1 - 1 = 0$ , and the coefficient of $𝐯_{𝑗}$ is $- 1 \neq 0$ . Conversely, let $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎$ with $\sum_{𝑖} 𝑐_{𝑖} = 0$ and $𝑐_{𝑗} \neq 0$ . Then

𝐯_{𝑗} = \sum_{𝑖 \neq 𝑗} (- \frac{𝑐_{𝑖}}{𝑐_{𝑗}}) 𝐯_{𝑖}, and \sum_{𝑖 \neq 𝑗} (- \frac{𝑐_{𝑖}}{𝑐_{𝑗}}) = - \frac{1}{𝑐_{𝑗}} \sum_{𝑖 \neq 𝑗} 𝑐_{𝑖} = - \frac{1}{𝑐_{𝑗}} (- 𝑐_{𝑗}) = 1,

so $𝐯_{𝑗}$ lies in the affine hull of the other entries. □

Condition (V) is the affine form of the linear dependence lemma (Lemma 1.5.10), and (IV) is the affine form of unique coordinates. For $𝑘 = 0$ , (V) holds because the other entries form an empty list, whose affine hull is empty.

Corollary 4.7.23 (Affinely independent lists are short). If

dim 𝑉 = 𝑛

, then every affinely independent list in

𝑉

has at most

𝑛 + 1

entries. In particular, any

𝑛 + 2

points of

𝑉

are affinely dependent.

Proof.routine

Uses. Proposition 4.7.20 (Affine independence through differences), Theorem 4.1.10 (Independent lists are short)

Hint. Pass to the differences.

Sketch. If

𝐯_{0}, \dots, 𝐯_{𝑘}

is affinely independent, the

𝑘

differences

𝐯_{𝑖} - 𝐯_{0}

are linearly independent in an

𝑛

-dimensional space, so

𝑘 \leq 𝑛

𝐯_{0}, \dots, 𝐯_{𝑘}

is affinely independent, the

𝑘

vectors

𝐯_{𝑖} - 𝐯_{0}

are linearly independent by Proposition 4.7.20, so

𝑘 \leq 𝑛

by Theorem 4.1.10(a), and the list has

𝑘 + 1 \leq 𝑛 + 1

entries. A list of

𝑛 + 2

points has

𝑘 = 𝑛 + 1 > 𝑛

, so it is affinely dependent. □

Exercise 4.7.24.routine Decide whether each list is affinely independent. If it is not, find an affine dependence.

(a) $(1, 2, 3),$ $(2, 3, 4),$ $(4, 5, 6)$ in $ℝ^{3}$ .
(b) $(1, 0, 0),$ $(0, 1, 0),$ $(0, 0, 1),$ $(1, 1, 1)$ in $ℝ^{3}$ .
(c)The same four points, read in $𝔽_{2}^{3}$ .

Solution.

Uses. Proposition 4.7.20 (Affine independence through differences)

Hint 1. Subtract the first point from the others and test the differences for linear independence.

Hint 2. In (b), the relation

𝑎 (- 1, 1, 0) +

𝑏 (- 1, 0, 1) +

𝑐 (0, 1, 1) =

𝟎

leads to

- 2 𝑎 = 0

. What happens to this step over

𝔽_{2}

Sketch. (a) The differences

(1, 1, 1)

and

(3, 3, 3)

are dependent;

2 𝐯_{0} - 3 𝐯_{1} + 𝐯_{2} = 𝟎

. (b) The differences are independent over

ℝ

, so the points are the vertices of a tetrahedron. (c) Over

𝔽_{2}

, the four points add up to

𝟎

, and

1 + 1 + 1 + 1 = 0

(a) The differences from $(1, 2, 3)$ are $(1, 1, 1)$ and $(3, 3, 3) = 3 (1, 1, 1)$ , which are linearly dependent, so the list is affinely dependent (Proposition 4.7.20). Translating $3 (1, 1, 1) -$ $(3, 3, 3) =$ $𝟎$ back as in the proof of the proposition gives the affine dependence

2 (1, 2, 3) - 3 (2, 3, 4) + (4, 5, 6) = (2 - 6 + 4, 4 - 9 + 5, 6 - 12 + 6) = 𝟎, 2 - 3 + 1 = 0 .

(b) The differences from $(1, 0, 0)$ are $(- 1, 1, 0)$ , $(- 1, 0, 1)$ , $(0, 1, 1)$ . If $𝑎 (- 1, 1, 0) +$ $𝑏 (- 1, 0, 1) +$ $𝑐 (0, 1, 1) =$ $𝟎$ , then $- 𝑎 - 𝑏 = 0$ , $𝑎 + 𝑐 = 0$ , and $𝑏 + 𝑐 = 0$ . So $𝑏 = - 𝑎$ and $𝑐 = - 𝑎$ , and the last equation gives $- 2 𝑎 = 0$ , so $𝑎 = 0$ over $ℝ$ , and then $𝑏 = 𝑐 = 0$ . The differences are independent, and the four points are affinely independent.

(c) Over $𝔽_{2}$ the step $- 2 𝑎 = 0 \Rightarrow 𝑎 = 0$ fails, and $𝑎 = 𝑏 = 𝑐 = 1$ is a relation: $(1, 1, 0) +$ $(1, 0, 1) +$ $(0, 1, 1) =$ $(2, 2, 2) =$ $𝟎$ . Correspondingly,

(1, 0, 0) + (0, 1, 0) + (0, 0, 1) + (1, 1, 1) = (2, 2, 2) = 𝟎 and 1 + 1 + 1 + 1 = 0

in $𝔽_{2}$ , an affine dependence. The four points lie in the plane $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 1$ of $𝔽_{2}^{3}$ .

Definition 4.7.25 (Affine frame, barycentric coordinates). Let $𝐸$ be an affine subspace of $𝑉$ . An affine frame of $𝐸$ is an affinely independent list $𝐯_{0}, \dots, 𝐯_{𝑘}$ with $aff (𝐯_{0}, \dots, 𝐯_{𝑘}) =$ $𝐸$ . Given an affine frame, each $𝐩 \in 𝐸$ can be written as

𝐩 = 𝑡_{0} 𝐯_{0} + \dots + 𝑡_{𝑘} 𝐯_{𝑘}, 𝑡_{0} + \dots + 𝑡_{𝑘} = 1,

in exactly one way (Theorem 4.7.22(IV)). The scalars $(𝑡_{0}, \dots, 𝑡_{𝑘})$ are the barycentric coordinates of $𝐩$ with respect to $𝐯_{0}, \dots, 𝐯_{𝑘}$ .

The name comes from physics: if masses $𝑡_{0}, \dots, 𝑡_{𝑘} \geq 0$ with total $1$ sit at $𝐯_{0}, \dots, 𝐯_{𝑘}$ , their center of mass, or barycenter, is $\sum_{𝑖} 𝑡_{𝑖} 𝐯_{𝑖}$ .

Remark 4.7.26 (Frames exist, and coordinates are coordinates). Let $𝐸 = 𝐩 + 𝑈$ with $dim 𝑈 = 𝑘$ , and let $𝐮_{1}, \dots, 𝐮_{𝑘}$ be a basis of $𝑈$ . Then $𝐩,$ $𝐩 +$ $𝐮_{1},$ $\dots,$ $𝐩 +$ $𝐮_{𝑘}$ is an affine frame of $𝐸$ : its differences are the $𝐮_{𝑖}$ , which are independent, and its affine hull is $𝐩 +$ $span (𝐮_{1}, \dots, 𝐮_{𝑘}) =$ $𝐸$ by Proposition 4.7.17(b). Conversely, every affine frame of $𝐸$ has $dim 𝐸 + 1$ entries, by Theorem 4.7.22(III).

Barycentric coordinates are ordinary coordinates in disguise. By Lemma 4.7.13(a), $𝐩 =$ $\sum_{𝑖 = 0}^{𝑘} 𝑡_{𝑖} 𝐯_{𝑖}$ with $\sum_{𝑖} 𝑡_{𝑖} = 1$ if and only if

𝐩 - 𝐯_{0} = 𝑡_{1} (𝐯_{1} - 𝐯_{0}) + \dots + 𝑡_{𝑘} (𝐯_{𝑘} - 𝐯_{0}) and 𝑡_{0} = 1 - 𝑡_{1} - \dots - 𝑡_{𝑘} .

So $(𝑡_{1}, \dots, 𝑡_{𝑘}) =$ ${[𝐩 - 𝐯_{0}]}_{ℬ︀}$ for the basis $ℬ︀ =$ $𝐯_{1} -$ $𝐯_{0},$ $\dots,$ $𝐯_{𝑘} -$ $𝐯_{0}$ of $dir 𝐸$ , and $𝑡_{0}$ is determined by the others.

Example 4.7.27 (Barycentric coordinates in a triangle). Let $𝐯_{0} = (1, 1)$ , $𝐯_{1} = (5, 2)$ , $𝐯_{2} = (2, 5)$ in $ℝ^{2}$ . The differences $(4, 1)$ and $(1, 4)$ are independent, so the three points form an affine frame of $ℝ^{2}$ , a triangle. For $𝐩 = (3, 3)$ , Remark 4.7.26 asks for $𝐩 -$ $𝐯_{0} =$ $(2, 2) =$ $𝑡_{1} (4, 1) +$ $𝑡_{2} (1, 4)$ , that is, $4 𝑡_{1} + 𝑡_{2} = 2$ and $𝑡_{1} + 4 𝑡_{2} = 2$ . So $𝑡_{1} = 𝑡_{2} = \frac{2}{5}$ and $𝑡_{0} = \frac{1}{5}$ :

(3, 3) = \frac{1}{5} (1, 1) + \frac{2}{5} (5, 2) + \frac{2}{5} (2, 5) .

For $𝐪 = (6, 0)$ , the equations $4 𝑡_{1} + 𝑡_{2} = 5$ and $𝑡_{1} + 4 𝑡_{2} = - 1$ give $𝑡_{1} = \frac{7}{5}$ , $𝑡_{2} = - \frac{3}{5}$ , and $𝑡_{0} = \frac{1}{5}$ . Check: $\frac{1}{5} (1, 1) +$ $\frac{7}{5} (5, 2) -$ $\frac{3}{5} (2, 5) =$ $\frac{1}{5} (1 + 35 - 6, 1 + 14 - 15) =$ $(6, 0)$ .

Homogenizing (Technique 4.7.21) gives the same numbers in one step: $(𝑡_{0}, 𝑡_{1}, 𝑡_{2})$ is the unique solution of

(\begin{matrix} 1 & 5 & 2 \\ 1 & 2 & 5 \\ 1 & 1 & 1 \end{matrix}) (\begin{matrix} 𝑡_{0} \\ 𝑡_{1} \\ 𝑡_{2} \end{matrix}) = (\begin{matrix} 3 \\ 3 \\ 1 \end{matrix}),

whose columns are the lifts ${\hat{𝐯}}_{𝑖}$ .

Intuition (Reading barycentric coordinates off a picture). In a triangle in

ℝ^{2}

, the coordinate

𝑡_{0}

vanishes exactly on the line through

𝐯_{1}

and

𝐯_{2}

(the points whose representation does not use

𝐯_{0}

), equals

1

𝐯_{0}

, and changes at a constant rate as

𝐩

moves, because by Remark 4.7.26 it is an affine function of

𝐩

. So

𝑡_{0}

measures how far

𝐩

has traveled from the line

𝐯_{1} 𝐯_{2}

toward

𝐯_{0}

, as a fraction of the height of

𝐯_{0}

above that line. It is positive on the side of

𝐯_{0}

and negative on the other side, and the three edge lines cut the plane into seven regions with seven sign patterns (Figure 4.7.1, right). For

𝐩

inside the triangle, the triangles

𝐩 𝐯_{1} 𝐯_{2}

and

𝐯_{0} 𝐯_{1} 𝐯_{2}

share the base

𝐯_{1} 𝐯_{2}

, so the height description says that

𝑡_{0}

is the ratio of their areas (left). In Example 4.7.27,

𝐪

has

𝑡_{2} < 0

, and indeed it lies on the far side of the line

𝐯_{0} 𝐯_{1}

from

𝐯_{2}

Figure 4.7.1. Barycentric coordinates in a triangle. Left: the point

𝐩

with barycentric coordinates

(0.3, 0.4, 0.3)

; for a point inside the triangle,

𝑡_{𝑖}

is the fraction of the area taken by the small triangle opposite

𝐯_{𝑖}

. Right: the signs of

(𝑡_{0}, 𝑡_{1}, 𝑡_{2})

in the seven regions cut out by the lines through the edges.

When $𝐯_{0} = 𝟎$ , Remark 4.7.26 says that $(𝑡_{1}, 𝑡_{2})$ are the coordinates of $𝐩$ in the basis $𝐯_{1}, 𝐯_{2}$ , so the change-of-basis component of Section 4.5 also shows barycentric coordinates. Read its two basis vectors as two vertices of a triangle whose third vertex is $𝟎$ .

The triangle with vertices

𝟎

𝐛_{1} = (3, 1)

𝐛_{2} = (1, 2)

is half of a cell of the

ℬ︀

-grid. The point

𝐯 =

(1.5, 1) =

0.4 𝐛_{1} +

0.3 𝐛_{2}

has

ℬ︀

-coordinates

(𝑡_{1}, 𝑡_{2}) =

(0.4, 0.3)

, so its barycentric coordinates with respect to

𝟎, 𝐛_{1}, 𝐛_{2}

are

(0.3, 0.4, 0.3)

, with

𝑡_{0} = 1 - 𝑡_{1} - 𝑡_{2}

. In the interactive version, drag

𝐯

: it lies in the triangle exactly when

𝑡_{1} \geq 0

𝑡_{2} \geq 0

, and

𝑡_{1} + 𝑡_{2} \leq 1

. When

𝐛_{1}

and

𝐛_{2}

become parallel, the three points are affinely dependent and the coordinates are undefined.

If $(𝑘 + 1) \cdot 1 \neq 0$ in $𝐹$ , the point whose barycentric coordinates are all $1 / (𝑘 + 1)$ is the centroid $\frac{1}{𝑘 + 1} (𝐯_{0} + \dots + 𝐯_{𝑘})$ . Over $𝔽_{3}$ a triangle has no centroid, and Exercise 4.7.57 shows what its medians do instead.

4.7.4Convexity and simplices

Over $ℝ$ we can ask for barycentric coordinates to be nonnegative, which describes the solid triangle rather than the plane it spans. Throughout this subsection $𝑉$ is a real vector space; everything works verbatim over $ℚ$ .

Definition 4.7.28 (Convex combination, convex set, convex hull). Let

𝑉

be a real vector space. A convex combination of

𝐯_{1}, \dots, 𝐯_{𝑘}

is an affine combination

𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}

with

𝑐_{𝑖} \geq 0

for all

𝑖

. A subset

𝐶 \subseteq 𝑉

is convex if

(1 - 𝑡) 𝐮 + 𝑡 𝐯 \in 𝐶

for all

𝐮, 𝐯 \in 𝐶

and

𝑡 \in [0, 1]

, that is, if

𝐶

contains the segment

[𝐮, 𝐯] =

{(1 - 𝑡) 𝐮 + 𝑡 𝐯 | 𝑡 \in [0, 1]}

between any two of its points. The convex hull

conv 𝑆

𝑆 \subseteq 𝑉

is the set of all convex combinations of finitely many elements of

𝑆

Example 4.7.29 (Convex sets). Affine subspaces are convex, by Theorem 4.7.14. So is every half-space

𝐻 =

{𝐱 \in ℝ^{𝑛} | 𝑎_{1} 𝑥_{1} + \dots + 𝑎_{𝑛} 𝑥_{𝑛} \geq 𝑐}

: writing

𝛼 (𝐱) =

𝑎_{1} 𝑥_{1} +

\dots +

𝑎_{𝑛} 𝑥_{𝑛}

, if

𝛼 (𝐮) \geq 𝑐

and

𝛼 (𝐯) \geq 𝑐

, then

𝛼 ((1 - 𝑡) 𝐮 + 𝑡 𝐯) =

(1 - 𝑡) 𝛼 (𝐮) +

𝑡 𝛼 (𝐯) \geq

(1 - 𝑡) 𝑐 +

𝑡 𝑐 =

𝑐

for

𝑡 \in [0, 1]

. An intersection of convex sets is convex, so the solution set of any system of linear inequalities is convex; these are the feasible regions of linear programming. The union of the two coordinate axes in

ℝ^{2}

is not convex, since the midpoint of

(1, 0)

and

(0, 1)

is missing.

Proposition 4.7.30 (Convex sets and convex hulls). Let $𝑉$ be a real vector space, and let $𝑆 \subseteq 𝑉$ .

(a)A subset $𝐶 \subseteq 𝑉$ is convex if and only if every convex combination of elements of $𝐶$ lies in $𝐶$ .
(b) $conv 𝑆$ is convex, contains $𝑆$ , and is contained in every convex subset of $𝑉$ that contains $𝑆$ .

Proof.standard

Uses. Definition 4.7.28 (Convex combination, convex set, convex hull), Proposition 4.7.17 (The affine hull)

Techniques. Induction on the length of a list (Technique 1.6.9)

Hint 1. (

\Leftarrow

) of (a) is the case of two points. For (

\Rightarrow

), induct on the number

𝑘

of points, splitting off the last one.

Hint 2. If

𝑐_{𝑘} < 1

, write

\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} =

(1 - 𝑐_{𝑘}) 𝐪 +

𝑐_{𝑘} 𝐯_{𝑘}

, where

𝐪

is a convex combination of

𝐯_{1}, \dots, 𝐯_{𝑘 - 1}

Hint 3. For (b), copy the proof of Proposition 4.7.17(a): over a common list, a convex combination of two convex combinations is a convex combination.

Sketch. Induct on

𝑘

. For

𝑘 = 1

the combination is

𝐯_{1}

. If

𝑐_{𝑘} = 1

, the other coefficients are

0

and the point is

𝐯_{𝑘}

. Otherwise

𝐪 =

\sum_{𝑖 < 𝑘} (𝑐_{𝑖} / (1 - 𝑐_{𝑘})) 𝐯_{𝑖}

is a convex combination of

𝑘 - 1

points, so

𝐪 \in 𝐶

, and the point is

(1 - 𝑐_{𝑘}) 𝐪 +

𝑐_{𝑘} 𝐯_{𝑘} \in

𝐶

. For (b),

(1 - 𝑡) \sum_{𝑖} 𝑎_{𝑖} 𝐬_{𝑖} +

𝑡 \sum_{𝑖} 𝑏_{𝑖} 𝐬_{𝑖} =

\sum_{𝑖} ((1 - 𝑡) 𝑎_{𝑖} + 𝑡 𝑏_{𝑖}) 𝐬_{𝑖}

is a convex combination.

(a) If $𝐶$ contains every convex combination of its elements, then in particular it contains $(1 - 𝑡) 𝐮 + 𝑡 𝐯$ for $𝐮, 𝐯 \in 𝐶$ and $𝑡 \in [0, 1]$ , so $𝐶$ is convex. Conversely, let $𝐶$ be convex. We show by induction on $𝑘 \geq 1$ that every convex combination $𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}$ of elements of $𝐶$ lies in $𝐶$ . For $𝑘 = 1$ , $𝑐_{1} = 1$ and the combination is $𝐯_{1} \in 𝐶$ . Let $𝑘 \geq 2$ , and assume the claim for $𝑘 - 1$ . If $𝑐_{𝑘} = 1$ , then $𝑐_{1} + \dots + 𝑐_{𝑘 - 1} = 0$ with all $𝑐_{𝑖} \geq 0$ , so $𝑐_{1} = \dots = 𝑐_{𝑘 - 1} = 0$ and the combination is $𝐯_{𝑘} \in 𝐶$ . Otherwise $𝑠 =$ $1 -$ $𝑐_{𝑘} =$ $𝑐_{1} +$ $\dots +$ $𝑐_{𝑘 - 1} >$ $0$ , and

𝐪 = \frac{𝑐_{1}}{𝑠} 𝐯_{1} + \dots + \frac{𝑐_{𝑘 - 1}}{𝑠} 𝐯_{𝑘 - 1}

is a convex combination of $𝑘 - 1$ elements of $𝐶$ (its coefficients are nonnegative and add up to $𝑠 / 𝑠 = 1$ ), so $𝐪 \in 𝐶$ by the induction hypothesis. Then

𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘} = 𝑠 𝐪 + 𝑐_{𝑘} 𝐯_{𝑘} = (1 - 𝑐_{𝑘}) 𝐪 + 𝑐_{𝑘} 𝐯_{𝑘},

which lies in $𝐶$ because $𝐶$ is convex and $𝑐_{𝑘} \in [0, 1]$ .

(b) Each $𝐬 \in 𝑆$ is the convex combination $1 𝐬$ , so $𝑆 \subseteq conv 𝑆$ . Let $𝐮, 𝐯 \in conv 𝑆$ and $𝑡 \in [0, 1]$ . As in the proof of Proposition 4.7.17(a), write $𝐮 = \sum_{𝑖} 𝑎_{𝑖} 𝐬_{𝑖}$ and $𝐯 = \sum_{𝑖} 𝑏_{𝑖} 𝐬_{𝑖}$ over one list $𝐬_{1}, \dots, 𝐬_{𝑁}$ of elements of $𝑆$ , padding with zero coefficients. Then

(1 - 𝑡) 𝐮 + 𝑡 𝐯 = \sum_{𝑖} ((1 - 𝑡) 𝑎_{𝑖} + 𝑡 𝑏_{𝑖}) 𝐬_{𝑖},

and the coefficients are nonnegative and add up to $(1 - 𝑡) + 𝑡 = 1$ . So $conv 𝑆$ is convex. If $𝐶$ is convex and $𝑆 \subseteq 𝐶$ , then $𝐶$ contains every convex combination of elements of $𝑆$ by (a), that is, $conv 𝑆 \subseteq 𝐶$ . □

Definition 4.7.31 (Simplex). Let $𝐯_{0}, \dots, 𝐯_{𝑘}$ be affinely independent in a real vector space $𝑉$ . The $𝑘$ -simplex with vertices $𝐯_{0}, \dots, 𝐯_{𝑘}$ is

conv (𝐯_{0}, \dots, 𝐯_{𝑘}) = {𝑡_{0} 𝐯_{0} + \dots + 𝑡_{𝑘} 𝐯_{𝑘} | 𝑡_{𝑖} \geq 0 for all 𝑖, 𝑡_{0} + \dots + 𝑡_{𝑘} = 1} .

Indexing the coordinates of $ℝ^{𝑘 + 1}$ from $0$ , the standard $𝑘$ -simplex is

Δ^{𝑘} = {(𝑡_{0}, \dots, 𝑡_{𝑘}) \in ℝ^{𝑘 + 1} | 𝑡_{𝑖} \geq 0 for all 𝑖, 𝑡_{0} + \dots + 𝑡_{𝑘} = 1},

the simplex whose vertices are the standard basis vectors of $ℝ^{𝑘 + 1}$ .

A $0$ -simplex is a point, a $1$ -simplex is a segment, a $2$ -simplex is a solid triangle, and a $3$ -simplex is a solid tetrahedron. The points of $Δ^{𝑘}$ are the probability vectors with $𝑘 + 1$ entries. For instance, a classifier that assigns probabilities to $𝑘 + 1$ classes outputs a point of $Δ^{𝑘}$ , and stochastic matrices (Section 11.1) have columns in $Δ^{𝑛 - 1}$ .

Proposition 4.7.32 (Barycentric coordinates on a simplex). Let

𝜎

be the

𝑘

-simplex with vertices

𝐯_{0}, \dots, 𝐯_{𝑘}

. The map

𝛽 : Δ^{𝑘} \to 𝜎

𝛽 (𝑡_{0}, \dots, 𝑡_{𝑘}) =

𝑡_{0} 𝐯_{0} +

\dots +

𝑡_{𝑘} 𝐯_{𝑘}

, is a bijection. Equivalently, a point of

aff (𝐯_{0}, \dots, 𝐯_{𝑘})

lies in

𝜎

if and only if all of its barycentric coordinates are nonnegative, and

𝛽^{- 1}

sends a point of

𝜎

to its barycentric coordinates.

Proof.routine

Uses. Definition 4.7.31 (Simplex), Theorem 4.7.22 (Characterizations of affine independence), Definition 4.7.25 (Affine frame, barycentric coordinates)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Surjectivity is the definition of

𝜎

Hint 2. Injectivity is the uniqueness of barycentric coordinates, Theorem 4.7.22(IV).

Sketch.

𝛽

is onto

𝜎 =

conv (𝐯_{0}, \dots, 𝐯_{𝑘})

by definition, and two tuples with the same image are two affine combinations representing one point, so they are equal by Theorem 4.7.22(IV). A point lies in

𝜎

exactly when its unique coordinate tuple lies in

Δ^{𝑘}

By the definition of

𝜎

𝛽

is surjective. If

𝛽 (𝐭) = 𝛽 (𝐭^{'})

for

𝐭, 𝐭^{'} \in Δ^{𝑘}

, then

\sum_{𝑖} 𝑡_{𝑖} 𝐯_{𝑖}

and

\sum_{𝑖} 𝑡_{𝑖}^{'} 𝐯_{𝑖}

are affine combinations representing the same point, so

𝐭 = 𝐭^{'}

by Theorem 4.7.22(IV). Hence

𝛽

is a bijection. A point

𝐩 \in

aff (𝐯_{0}, \dots, 𝐯_{𝑘})

has unique barycentric coordinates

𝐭

, and

𝐩 \in 𝜎

if and only if

𝐩 = 𝛽 (𝐭^{'})

for some

𝐭^{'} \in Δ^{𝑘}

. By uniqueness this happens if and only if

𝐭 \in Δ^{𝑘}

, that is, if and only if all

𝑡_{𝑖} \geq 0

. □

In Example 4.7.27, $𝐩 = (3, 3)$ lies in the triangle, since its coordinates $(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})$ are positive, and $𝐪 = (6, 0)$ does not. Renderers test whether a pixel lies in a triangle exactly this way. They also use the coordinates to shade it: if the vertices carry colors $𝐜_{0}, 𝐜_{1}, 𝐜_{2} \in ℝ^{3}$ (red, green, and blue intensities), the pixel at $𝐩$ gets $𝑡_{0} 𝐜_{0} +$ $𝑡_{1} 𝐜_{1} +$ $𝑡_{2} 𝐜_{2}$ . This is the unique affine map $ℝ^{2} \to ℝ^{3}$ with the prescribed values at the vertices (Theorem 4.7.39 below), and the same formula interpolates depths and texture coordinates.

Carathéodory’s theorem

A point of $conv 𝑆$ may be given as a convex combination of many points of $𝑆$ . In $ℝ^{2}$ , though, every point of a convex polygon lies in a triangle formed by three of its vertices. In general a long convex combination can be shortened: when the points it uses are affinely dependent, the dependence gives a way to change the coefficients without changing the point or the coefficient sum, and we change them until one coefficient reaches $0$ .

Technique 4.7.33 (Slide along a dependence). Suppose $𝐩 = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ , where the coefficients are subject to constraints, such as $𝑐_{𝑖} \geq 0$ and $\sum_{𝑖} 𝑐_{𝑖} = 1$ , and the $𝐯_{𝑖}$ satisfy a relation $\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎$ that respects the equality constraints (here $\sum_{𝑖} 𝑑_{𝑖} = 0$ ). Then $𝐩 =$ $\sum_{𝑖} (𝑐_{𝑖} - 𝜆 𝑑_{𝑖}) 𝐯_{𝑖}$ for every scalar $𝜆$ , and the equality constraints still hold. Increase $𝜆$ from $0$ until the first inequality constraint becomes tight. At that extremal $𝜆$ some coefficient is $0$ , and its vector can be dropped. Repeat as long as a relation exists.

Reach for it to show that a representation can be chosen short, for instance using independent vectors only. It refines the argument of Theorem 1.6.10, where a redundant vector is simply discarded, to situations where the coefficients must keep their signs. The canonical example is Carathéodory’s theorem (Theorem 4.7.34).

Used in 2 places

Theorem 4.7.34, Exercise 4.7.59.

Theorem 4.7.34 (Carathéodory). Let

𝑉

be a real vector space with

dim 𝑉 = 𝑛

, and let

𝑆 \subseteq 𝑉

. Every point of

conv 𝑆

is a convex combination of at most

𝑛 + 1

elements of

𝑆

Proof.hard

Uses. Definition 4.7.28 (Convex combination, convex set, convex hull), Corollary 4.7.23 (Affinely independent lists are short), Definition 4.7.18 (Affine independence)

Techniques. Slide along a dependence (Technique 4.7.33), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. Take a representation

𝐩 =

\sum_{𝑖 = 1}^{𝑚} 𝑐_{𝑖} 𝐯_{𝑖}

with

𝑚

as small as possible. Then all

𝑐_{𝑖} > 0

. If

𝑚 \geq 𝑛 + 2

, what does Corollary 4.7.23 give?

Hint 2. An affine dependence

\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎

\sum_{𝑖} 𝑑_{𝑖} = 0

, has some

𝑑_{𝑖} > 0

. Subtract

𝜆

times it from the representation.

Hint 3. Take

𝜆 =

min {𝑐_{𝑖} / 𝑑_{𝑖} | 𝑑_{𝑖} > 0}

Sketch. Choose a representation

𝐩 =

\sum_{𝑖 = 1}^{𝑚} 𝑐_{𝑖} 𝐯_{𝑖}

with

𝑚

minimal, so every

𝑐_{𝑖} > 0

. If

𝑚 \geq 𝑛 + 2

, the

𝐯_{𝑖}

are affinely dependent:

\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎

\sum_{𝑖} 𝑑_{𝑖} = 0

, and some

𝑑_{𝑖} > 0

. With

𝜆 =

{min}_{𝑑_{𝑖} > 0} 𝑐_{𝑖} / 𝑑_{𝑖}

, the coefficients

𝑐_{𝑖} - 𝜆 𝑑_{𝑖}

are nonnegative, add up to

1

, represent

𝐩

, and one of them is

0

, contradicting minimality.

Let $𝐩 \in conv 𝑆$ . Among all ways of writing $𝐩$ as a convex combination of finitely many elements of $𝑆$ , choose one, $𝐩 =$ $𝑐_{1} 𝐯_{1} +$ $\dots +$ $𝑐_{𝑚} 𝐯_{𝑚}$ , with $𝑚$ as small as possible. Every $𝑐_{𝑖}$ is positive, since a term with $𝑐_{𝑗} = 0$ could be deleted. Suppose, for contradiction, that $𝑚 \geq 𝑛 + 2$ . By Corollary 4.7.23, the list $𝐯_{1}, \dots, 𝐯_{𝑚}$ is affinely dependent: there are $𝑑_{1}, \dots, 𝑑_{𝑚}$ , not all $0$ , with

𝑑_{1} 𝐯_{1} + \dots + 𝑑_{𝑚} 𝐯_{𝑚} = 𝟎, 𝑑_{1} + \dots + 𝑑_{𝑚} = 0 .

Since the $𝑑_{𝑖}$ add up to $0$ and are not all $0$ , at least one of them is positive. Let

𝜆 = min {\frac{𝑐_{𝑖}}{𝑑_{𝑖}} | 𝑑_{𝑖} > 0},

attained at $𝑖 = 𝑗$ . Then $𝜆 > 0$ . Put $𝑐_{𝑖}^{'} = 𝑐_{𝑖} - 𝜆 𝑑_{𝑖}$ . Then

\sum_{𝑖} 𝑐_{𝑖}^{'} 𝐯_{𝑖} = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} - 𝜆 \sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝐩, \sum_{𝑖} 𝑐_{𝑖}^{'} = \sum_{𝑖} 𝑐_{𝑖} - 𝜆 \sum_{𝑖} 𝑑_{𝑖} = 1 .

If $𝑑_{𝑖} \leq 0$ , then $𝑐_{𝑖}^{'} \geq 𝑐_{𝑖} > 0$ ; if $𝑑_{𝑖} > 0$ , then $𝑐_{𝑖} / 𝑑_{𝑖} \geq 𝜆$ , so $𝑐_{𝑖}^{'} =$ $𝑑_{𝑖} (𝑐_{𝑖} / 𝑑_{𝑖} - 𝜆) \geq$ $0$ . So $𝐩 = \sum_{𝑖} 𝑐_{𝑖}^{'} 𝐯_{𝑖}$ is a convex combination, and $𝑐_{𝑗}^{'} =$ $𝑐_{𝑗} -$ $(𝑐_{𝑗} / 𝑑_{𝑗}) 𝑑_{𝑗} =$ $0$ . Deleting the $𝑗$ th term represents $𝐩$ as a convex combination of $𝑚 - 1$ elements of $𝑆$ , contradicting the minimality of $𝑚$ . Hence $𝑚 \leq 𝑛 + 1$ . □

The bound $𝑛 + 1$ cannot be lowered. Let $𝑆$ be the vertex set of an $𝑛$ -simplex. Its centroid has all $𝑛 + 1$ barycentric coordinates positive, while a convex combination of fewer vertices has a zero coordinate, so by the uniqueness in Proposition 4.7.32 the centroid is not a convex combination of fewer than $𝑛 + 1$ points of $𝑆$ .

4.7.5Affine maps

In Section 2.1, the maps $𝐯 \mapsto 𝐿 (𝐯) + 𝐛$ were the basic non-examples of linear maps (Non-example 2.1.12), and Proposition 2.1.13 showed that, when $𝐛 \neq 𝟎$ , they respect exactly the affine combinations. They are the maps of affine geometry.

Definition 4.7.35 (Affine map). Let

𝑉

and

𝑊

be vector spaces over

𝐹

. A map

𝑓 : 𝑉 \to 𝑊

is affine if there are

𝐿 \in L (𝑉, 𝑊)

and

𝐛 \in 𝑊

with

𝑓 (𝐯) = 𝐿 (𝐯) + 𝐛

for all

𝐯 \in 𝑉

. We then write

𝑓 = 𝐿 + 𝐛

The pair is determined by $𝑓$ : $𝑓 (𝟎) =$ $𝐿 (𝟎) +$ $𝐛 =$ $𝐛$ , and then $𝐿 (𝐯) =$ $𝑓 (𝐯) -$ $𝑓 (𝟎)$ . We call $𝐿$ the linear part of $𝑓$ .

Example 4.7.36 (Affine maps).

(a)Linear maps ( $𝐛 = 𝟎$ ), constant maps ( $𝐿 = 0$ ), and translations $𝜏_{𝐛} (𝐯) = 𝐯 + 𝐛$ (with $𝐿 = {id}_{𝑉}$ ) are affine.
(b)The rotation of $ℝ^{2}$ by the angle $𝜃$ about a point $𝐚$ moves $𝐚$ to the origin, rotates, and moves back:
$𝑓 (𝐯) = 𝐚 + 𝑅_{𝜃} (𝐯 - 𝐚) = 𝑅_{𝜃} 𝐯 + (𝐚 - 𝑅_{𝜃} 𝐚),$
where $𝑅_{𝜃}$ is the rotation about $𝟎$ (Section 2.1). Its linear part is $𝑅_{𝜃}$ wherever the center is. Likewise the scaling by $𝑐$ about $𝐚$ is $𝐯 \mapsto$ $𝐚 +$ $𝑐 (𝐯 - 𝐚) =$ $𝑐 𝐯 +$ $(1 - 𝑐) 𝐚$ .
(c)The map $𝑥 \mapsto 𝑥^{2}$ on $ℝ$ is not affine, since it sends $0, 1, 2$ to $0, 1, 4$ , and $1$ is the midpoint of $0$ and $2$ while $1$ is not the midpoint of $0$ and $4$ . (Affine maps preserve midpoints, by the next theorem.)

Theorem 4.7.37 (Affine maps preserve affine combinations). Let $𝑓 : 𝑉 \to 𝑊$ be a map. The following are equivalent.

(I) $𝑓$ is affine.
(II) $𝑓 (𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}) =$ $𝑐_{1} 𝑓 (𝐯_{1}) +$ $\dots +$ $𝑐_{𝑘} 𝑓 (𝐯_{𝑘})$ for all $𝑘 \geq 1$ , all $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ , and all $𝑐_{1}, \dots, 𝑐_{𝑘} \in 𝐹$ with $𝑐_{1} + \dots + 𝑐_{𝑘} = 1$ .
(III)For every $𝐩 \in 𝑉$ , the map $𝑔_{𝐩} : 𝑉 \to 𝑊$ , $𝑔_{𝐩} (𝐯) =$ $𝑓 (𝐩 + 𝐯) -$ $𝑓 (𝐩)$ , is linear.
(IV)For some $𝐩 \in 𝑉$ , the map $𝑔_{𝐩}$ is linear.

In that case $𝑔_{𝐩}$ is the linear part of $𝑓$ for every $𝐩$ .

Intuition. Condition (II) says that

𝑓

commutes with weighted averages. Condition (III) says that

𝑓

becomes linear once positions are measured from

𝐩

𝑉

and from

𝑓 (𝐩)

𝑊

. The definition measures from

𝟎

; the theorem says that

𝟎

is not special. To get from (II) to (III), express

𝐩 + 𝐮 + 𝐯

and

𝐩 + 𝑐 𝐯

as affine combinations of

𝐩

𝐩 + 𝐮

, and

𝐩 + 𝐯

Proof.standard

Uses. Proposition 2.1.13 (Which combinations an affine map respects), Theorem 2.1.3 (Checking linearity)

Techniques. Proving TFAE (Technique 2.7.1), Choose a base point (Technique 4.7.4)

Hint 1. (I)

\Rightarrow

(II) is Proposition 2.1.13.

Hint 2. For (II)

\Rightarrow

(III):

𝐩 +

𝐮 +

𝐯 =

(𝐩 + 𝐮) +

(𝐩 + 𝐯) -

𝐩

and

𝐩 +

𝑐 𝐯 =

𝑐 (𝐩 + 𝐯) +

(1 - 𝑐) 𝐩

Hint 3. For (IV)

\Rightarrow

(I), write

𝑓 (𝐯) =

𝑓 (𝐩 + (𝐯 - 𝐩))

Sketch. Applying

𝑓

𝐩 +

𝐮 +

𝐯 =

(𝐩 + 𝐮) +

(𝐩 + 𝐯) -

𝐩

and to

𝐩 +

𝑐 𝐯 =

𝑐 (𝐩 + 𝐯) +

(1 - 𝑐) 𝐩

and subtracting

𝑓 (𝐩)

gives additivity and homogeneity of

𝑔_{𝐩}

. Conversely,

𝑓 (𝐯) =

𝑔_{𝐩} (𝐯 - 𝐩) +

𝑓 (𝐩) =

𝑔_{𝐩} (𝐯) +

(𝑓 (𝐩) - 𝑔_{𝐩} (𝐩))

. If

𝑓 = 𝐿 + 𝐛

, then

𝑔_{𝐩} (𝐯) =

𝐿 (𝐩 + 𝐯) -

𝐿 (𝐩) =

𝐿 (𝐯)

We prove (I) $\Rightarrow$ (II) $\Rightarrow$ (III) $\Rightarrow$ (IV) $\Rightarrow$ (I).

(I) $\Rightarrow$ (II). If $𝑓 = 𝐿 + 𝐛$ , then by Proposition 2.1.13, $𝑓 (\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}) -$ $\sum_{𝑖} 𝑐_{𝑖} 𝑓 (𝐯_{𝑖}) =$ $(1 - \sum_{𝑖} 𝑐_{𝑖}) 𝐛 =$ $𝟎$ when $\sum_{𝑖} 𝑐_{𝑖} = 1$ .

(II) $\Rightarrow$ (III). Fix $𝐩$ and write $𝑔 = 𝑔_{𝐩}$ . For $𝐮, 𝐯 \in 𝑉$ , the vector $𝐩 +$ $𝐮 +$ $𝐯 =$ $1 (𝐩 + 𝐮) +$ $1 (𝐩 + 𝐯) +$ $(- 1) 𝐩$ is an affine combination, as $1 + 1 - 1 = 1$ . By (II), $𝑓 (𝐩 + 𝐮 + 𝐯) =$ $𝑓 (𝐩 + 𝐮) +$ $𝑓 (𝐩 + 𝐯) -$ $𝑓 (𝐩)$ , and subtracting $𝑓 (𝐩)$ from both sides gives

𝑔 (𝐮 + 𝐯) = (𝑓 (𝐩 + 𝐮) - 𝑓 (𝐩)) + (𝑓 (𝐩 + 𝐯) - 𝑓 (𝐩)) = 𝑔 (𝐮) + 𝑔 (𝐯) .

For $𝑐 \in 𝐹$ , $𝐩 +$ $𝑐 𝐯 =$ $𝑐 (𝐩 + 𝐯) +$ $(1 - 𝑐) 𝐩$ , so by (II), $𝑓 (𝐩 + 𝑐 𝐯) =$ $𝑐 𝑓 (𝐩 + 𝐯) +$ $(1 - 𝑐) 𝑓 (𝐩)$ , and

𝑔 (𝑐 𝐯) = 𝑐 𝑓 (𝐩 + 𝐯) - 𝑐 𝑓 (𝐩) = 𝑐 𝑔 (𝐯) .

So $𝑔$ is linear by Theorem 2.1.3.

(III) $\Rightarrow$ (IV) holds because $𝑉$ is nonempty.

(IV) $\Rightarrow$ (I). Let $𝑔 = 𝑔_{𝐩}$ be linear. For $𝐯 \in 𝑉$ ,

𝑓 (𝐯) = 𝑓 (𝐩 + (𝐯 - 𝐩)) = 𝑔 (𝐯 - 𝐩) + 𝑓 (𝐩) = 𝑔 (𝐯) + (𝑓 (𝐩) - 𝑔 (𝐩)) .

So $𝑓 = 𝑔 + 𝐛$ with $𝐛 = 𝑓 (𝐩) - 𝑔 (𝐩)$ , and $𝑓$ is affine.

Finally, if $𝑓 = 𝐿 + 𝐛$ , then for every $𝐩$ and $𝐯$ , $𝑔_{𝐩} (𝐯) =$ $(𝐿 (𝐩 + 𝐯) + 𝐛) -$ $(𝐿 (𝐩) + 𝐛) =$ $𝐿 (𝐯)$ , so $𝑔_{𝐩} = 𝐿$ . □

Compare Exercise 2.1.39, which assumed only that $𝑓$ respects affine combinations of two points and needed $1 + 1 \neq 0$ to recover additivity. The three-point combination with coefficients $1, 1, - 1$ makes the division by $2$ unnecessary, exactly as in the proof of Theorem 4.7.14.

Proposition 4.7.38 (Affine maps and affine subspaces). Let $𝑓 = 𝐿 + 𝐛 : 𝑉 \to 𝑊$ be affine.

(a)If $𝑔 = 𝑀 + 𝐜 : 𝑊 \to 𝑋$ is affine, then $𝑔 \circ 𝑓$ is affine, and $(𝑔 \circ 𝑓) (𝐯) =$ $𝑀 𝐿 (𝐯) +$ $(𝑀 (𝐛) + 𝐜)$ .
(b) $𝑓$ is bijective if and only if $𝐿$ is. In that case $𝑓^{- 1}$ is affine, and $𝑓^{- 1} (𝐰) =$ $𝐿^{- 1} (𝐰) -$ $𝐿^{- 1} (𝐛)$ .
(c)If $𝐸$ is an affine subspace of $𝑉$ , then $𝑓 (𝐸)$ is an affine subspace of $𝑊$ with direction $im (𝐿 ↾_{dir 𝐸})$ . If $𝐸$ is finite-dimensional, then $dim 𝑓 (𝐸) \leq dim 𝐸$ .
(d)If $𝐸^{'}$ is an affine subspace of $𝑊$ , then $𝑓^{- 1} (𝐸^{'})$ is empty or an affine subspace of $𝑉$ with direction ${𝐮 \in 𝑉 | 𝐿 (𝐮) \in dir 𝐸^{'}}$ .

Proof.standard

Uses. Definition 4.7.35 (Affine map), Proposition 2.1.17 (Compositions of linear maps are linear), Proposition 2.7.7 (Bijective linear maps are invertible), Proposition 4.7.3 (Direction and base points), Proposition 2.6.8 (Kernels and images are subspaces), Theorem 4.2.11 (Rank–nullity), Theorem 1.4.3 (Subspace criterion)

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. (a) is a computation. For (b), write

𝑓 = 𝜏_{𝐛} \circ 𝐿

, where

𝜏_{𝐛}

is the translation by

𝐛

, a bijection.

Hint 2. For (c) and (d), choose base points:

𝐩 \in 𝐸

for (c), and a point

𝐪 \in 𝑓^{- 1} (𝐸^{'})

, if there is one, for (d). The key identity is

𝑓 (𝐩 + 𝐮) =

𝑓 (𝐩) +

𝐿 (𝐮)

Sketch. (a)

𝑀 (𝐿 (𝐯) + 𝐛) + 𝐜

. (b)

𝜏_{𝐛}

is a bijection with inverse

𝜏_{- 𝐛}

, so

𝑓 = 𝜏_{𝐛} \circ 𝐿

is bijective if and only if

𝐿

is, and then

𝑓^{- 1} = 𝐿^{- 1} \circ 𝜏_{- 𝐛}

. (c)

𝑓 (𝐩 + 𝑈) =

𝑓 (𝐩) +

𝐿 (𝑈)

, and

dim 𝐿 (𝑈) \leq dim 𝑈

by rank–nullity. (d) With

𝐪 \in 𝑓^{- 1} (𝐸^{'})

𝑓 (𝐯) \in 𝐸^{'}

if and only if

𝐿 (𝐯 - 𝐪) =

𝑓 (𝐯) -

𝑓 (𝐪) \in

dir 𝐸^{'}

(a) $𝑔 (𝑓 (𝐯)) =$ $𝑀 (𝐿 (𝐯) + 𝐛) +$ $𝐜 =$ $𝑀 𝐿 (𝐯) +$ $(𝑀 (𝐛) + 𝐜)$ , and $𝑀 𝐿$ is linear by Proposition 2.1.17.

(b) The translation $𝜏_{𝐛} : 𝑊 \to 𝑊$ , $𝐰 \mapsto 𝐰 + 𝐛$ , is a bijection with inverse $𝜏_{- 𝐛}$ , and $𝑓 = 𝜏_{𝐛} \circ 𝐿$ . If $𝐿$ is bijective, so is $𝑓$ ; if $𝑓$ is bijective, so is $𝐿 = 𝜏_{- 𝐛} \circ 𝑓$ . In that case $𝑓^{- 1} = 𝐿^{- 1} \circ 𝜏_{- 𝐛}$ , that is, $𝑓^{- 1} (𝐰) =$ $𝐿^{- 1} (𝐰 - 𝐛) =$ $𝐿^{- 1} (𝐰) -$ $𝐿^{- 1} (𝐛)$ , and $𝐿^{- 1}$ is linear by Proposition 2.7.7. So $𝑓^{- 1}$ is affine.

(c) Let $𝐸 = 𝐩 + 𝑈$ . For $𝐮 \in 𝑈$ , $𝑓 (𝐩 + 𝐮) =$ $𝐿 (𝐩) +$ $𝐿 (𝐮) +$ $𝐛 =$ $𝑓 (𝐩) +$ $𝐿 (𝐮)$ . Hence $𝑓 (𝐸) =$ $𝑓 (𝐩) +$ $𝐿 (𝑈)$ , where $𝐿 (𝑈) = im (𝐿 ↾_{𝑈})$ is a subspace of $𝑊$ (Proposition 2.6.8, applied to $𝐿 ↾_{𝑈}$ ). So $𝑓 (𝐸)$ is an affine subspace with direction $im (𝐿 ↾_{𝑈})$ . If $𝑈$ is finite-dimensional, rank–nullity for $𝐿 ↾_{𝑈}$ (Theorem 4.2.11) gives $dim im (𝐿 ↾_{𝑈}) \leq$ $dim 𝑈$ .

(d) Let $𝑈^{'} = dir 𝐸^{'}$ and $𝑈^{″} =$ ${𝐮 \in 𝑉 | 𝐿 (𝐮) \in 𝑈^{'}}$ . Then $𝑈^{″}$ is a subspace: $𝐿 (𝟎) = 𝟎 \in 𝑈^{'}$ , and if $𝐿 (𝐮),$ $𝐿 (𝐮^{'}) \in$ $𝑈^{'}$ and $𝑐 \in 𝐹$ , then $𝐿 (𝐮 + 𝐮^{'}) =$ $𝐿 (𝐮) +$ $𝐿 (𝐮^{'}) \in$ $𝑈^{'}$ and $𝐿 (𝑐 𝐮) =$ $𝑐 𝐿 (𝐮) \in$ $𝑈^{'}$ . Suppose $𝑓^{- 1} (𝐸^{'})$ is nonempty, and let $𝐪 \in 𝑓^{- 1} (𝐸^{'})$ . By Proposition 4.7.3(b), $𝐸^{'} = 𝑓 (𝐪) + 𝑈^{'}$ , so for $𝐯 \in 𝑉$ we have $𝑓 (𝐯) \in 𝐸^{'}$ if and only if $𝑓 (𝐯) - 𝑓 (𝐪) \in 𝑈^{'}$ . Since $𝑓 (𝐯) -$ $𝑓 (𝐪) =$ $𝐿 (𝐯 - 𝐪)$ , this holds if and only if $𝐯 - 𝐪 \in 𝑈^{″}$ . Hence $𝑓^{- 1} (𝐸^{'}) = 𝐪 + 𝑈^{″}$ . □

Over $ℝ$ , condition (II) of Theorem 4.7.37 applied to convex combinations shows that an affine map sends segments onto segments, convex sets to convex sets, and $conv 𝑆$ onto $conv 𝑓 (𝑆)$ . It also preserves ratios along a line: if $𝐩 = (1 - 𝑡) 𝐚 + 𝑡 𝐛$ , then $𝑓 (𝐩) =$ $(1 - 𝑡) 𝑓 (𝐚) +$ $𝑡 𝑓 (𝐛)$ , so midpoints go to midpoints. Lengths and angles are not preserved (think of a shear), which is why affine geometry has parallel lines and ratios along a line but no distances.

Theorem 4.7.39 (Affine maps are determined by a frame). Let $𝐯_{0}, \dots, 𝐯_{𝑛}$ be an affine frame of $𝑉$ , and let $𝐰_{0}, \dots, 𝐰_{𝑛} \in 𝑊$ be arbitrary. There is exactly one affine map $𝑓 : 𝑉 \to 𝑊$ with $𝑓 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for $𝑖 = 0, \dots, 𝑛$ , namely

𝑓 (𝑡_{0} 𝐯_{0} + \dots + 𝑡_{𝑛} 𝐯_{𝑛}) = 𝑡_{0} 𝐰_{0} + \dots + 𝑡_{𝑛} 𝐰_{𝑛} whenever 𝑡_{0} + \dots + 𝑡_{𝑛} = 1 .

Proof.standard

Uses. Theorem 4.7.37 (Affine maps preserve affine combinations), Definition 4.7.25 (Affine frame, barycentric coordinates), Proposition 4.7.17 (The affine hull), Theorem 4.7.22 (Characterizations of affine independence), Theorem 2.2.1 (Linear maps are determined by a basis)

Techniques. Choose a base point (Technique 4.7.4), Construct a map by choosing it on a basis (Technique 2.2.4)

Hint 1. Uniqueness: every point is an affine combination of the

𝐯_{𝑖}

, and affine maps preserve affine combinations.

Hint 2. Existence: use

𝐯_{0}

as base point. The differences

𝐯_{𝑖} - 𝐯_{0}

form a basis of

𝑉

. Which linear map should be the linear part?

Sketch. Every point has barycentric coordinates, and Theorem 4.7.37(II) forces the formula, which gives uniqueness. For existence, the differences

𝐯_{𝑖} - 𝐯_{0}

are independent and span

𝑉

, so there is a linear

𝐿

with

𝐿 (𝐯_{𝑖} - 𝐯_{0}) =

𝐰_{𝑖} -

𝐰_{0}

; then

𝑓 (𝐯) =

𝐰_{0} +

𝐿 (𝐯 - 𝐯_{0})

is affine and sends

𝐯_{𝑖}

𝐰_{𝑖}

Uniqueness and the formula. Let $𝑓$ be affine with $𝑓 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for all $𝑖$ . Every $𝐯 \in$ $𝑉 =$ $aff (𝐯_{0}, \dots, 𝐯_{𝑛})$ has barycentric coordinates, $𝐯 = \sum_{𝑖} 𝑡_{𝑖} 𝐯_{𝑖}$ with $\sum_{𝑖} 𝑡_{𝑖} = 1$ , and by Theorem 4.7.37(II), $𝑓 (𝐯) =$ $\sum_{𝑖} 𝑡_{𝑖} 𝑓 (𝐯_{𝑖}) =$ $\sum_{𝑖} 𝑡_{𝑖} 𝐰_{𝑖}$ . So $𝑓$ is determined by the $𝐰_{𝑖}$ and is given by the formula.

Existence. By Proposition 4.7.17(b), $𝑉 = 𝐯_{0} + 𝐷$ with $𝐷 =$ $span (𝐯_{1} - 𝐯_{0}, \dots, 𝐯_{𝑛} - 𝐯_{0})$ . For every $𝐱 \in 𝑉$ , $𝐯_{0} + 𝐱 \in 𝑉$ , so $𝐱 \in 𝐷$ ; hence $𝐷 = 𝑉$ . By Theorem 4.7.22(II), the differences are linearly independent, so they form a basis of $𝑉$ . By Theorem 2.2.1 there is a linear map $𝐿 : 𝑉 \to 𝑊$ with $𝐿 (𝐯_{𝑖} - 𝐯_{0}) =$ $𝐰_{𝑖} -$ $𝐰_{0}$ for $𝑖 = 1, \dots, 𝑛$ . Define

𝑓 (𝐯) = 𝐰_{0} + 𝐿 (𝐯 - 𝐯_{0}) = 𝐿 (𝐯) + (𝐰_{0} - 𝐿 (𝐯_{0})),

an affine map. Then $𝑓 (𝐯_{0}) = 𝐰_{0}$ , and $𝑓 (𝐯_{𝑖}) =$ $𝐰_{0} +$ $(𝐰_{𝑖} - 𝐰_{0}) =$ $𝐰_{𝑖}$ for $𝑖 \geq 1$ . □

Example 4.7.40 (Mapping one triangle onto another). The affine map $𝑓 : ℝ^{2} \to ℝ^{2}$ with $𝑓 (0, 0) = (1, 1)$ , $𝑓 (1, 0) = (5, 2)$ , $𝑓 (0, 1) = (2, 5)$ has translation part $𝑓 (𝟎) = (1, 1)$ and linear part $𝐿$ with $𝐿 (𝐞_{1}) = (4, 1)$ and $𝐿 (𝐞_{2}) = (1, 4)$ :

𝑓 (𝐱) = (\begin{matrix} 4 & 1 \\ 1 & 4 \end{matrix}) 𝐱 + (\begin{matrix} 1 \\ 1 \end{matrix}) .

By Theorem 4.7.39, $𝑓$ sends the point with barycentric coordinates $(𝑡_{0}, 𝑡_{1}, 𝑡_{2})$ with respect to $𝟎, 𝐞_{1}, 𝐞_{2}$ , which is $(𝑡_{1}, 𝑡_{2})$ , to the point with the same barycentric coordinates with respect to the triangle of Example 4.7.27. For instance $𝑓 (\frac{2}{5}, \frac{2}{5}) =$ $(\frac{8}{5} + \frac{2}{5} + 1, \frac{2}{5} + \frac{8}{5} + 1) =$ $(3, 3)$ , matching the coordinates $(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})$ of $(3, 3)$ found there. Texture mapping works this way: the renderer finds the affine map from a triangle in the texture image to a triangle on the screen.

4.7.6Homogeneous coordinates

A graphics program moves objects by rotations, scalings, and translations, composed in long chains, and applies the result to every vertex of the scene. Composites of linear maps are products of matrices, which can be computed once and reused. A translation is not linear, so a chain such as “rotate about a point, then translate” is not a single matrix in $𝐹^{𝑛 \times 𝑛}$ . Homogenizing (Technique 4.7.21) repairs this.

Definition 4.7.41 (Homogeneous coordinates). For

𝐱 \in 𝐹^{𝑛}

, let

\hat{𝐱} = (\begin{matrix} 𝐱 \\ 1 \end{matrix}) \in 𝐹^{𝑛 + 1}

. A vector of homogeneous coordinates of

𝐱

is any vector

𝑐 \hat{𝐱} = (𝑐 𝐱, 𝑐)

with

𝑐 \in 𝐹

𝑐 \neq 0

. Conversely, a vector

(𝐳, 𝑐) \in 𝐹^{𝑛 + 1}

with

𝐳 \in 𝐹^{𝑛}

and

𝑐 \neq 0

is a vector of homogeneous coordinates of exactly one point of

𝐹^{𝑛}

, namely

𝑐^{- 1} 𝐳

The last claim holds because $(𝐳, 𝑐) = 𝑐^{'} (𝐱, 1)$ forces $𝑐^{'} = 𝑐$ and $𝐱 = 𝑐^{- 1} 𝐳$ . For example, $(6, - 3, 9, 3)$ and $(2, - 1, 3, 1)$ are both vectors of homogeneous coordinates of $(2, - 1, 3) \in ℝ^{3}$ . The vectors of homogeneous coordinates of $𝐱$ are the nonzero vectors on the line $span (\hat{𝐱})$ through $𝟎$ in $𝐹^{𝑛 + 1}$ , so $𝐱 \mapsto span (\hat{𝐱})$ matches the points of $𝐹^{𝑛}$ with the lines through $𝟎$ in $𝐹^{𝑛 + 1}$ that are not contained in the hyperplane $𝑥_{𝑛 + 1} = 0$ . Vectors with last coordinate $0$ represent no point of $𝐹^{𝑛}$ ; the aside at the end of the section says what they represent instead.

Theorem 4.7.42 (Affine maps as matrices). Let $𝑓 : 𝐹^{𝑛} \to 𝐹^{𝑚}$ be affine, $𝑓 (𝐱) = 𝐴 𝐱 + 𝐛$ with $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐛 \in 𝐹^{𝑚}$ , and let

𝑀_{𝑓} = (\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix}) \in 𝐹^{(𝑚 + 1) \times (𝑛 + 1)},

where $0$ is the $1 \times 𝑛$ zero row.

(a) $𝑀_{𝑓} \hat{𝐱} = \hat{𝑓 (𝐱)}$ for every $𝐱 \in 𝐹^{𝑛}$ , and $𝑀_{𝑓}$ is the only matrix in $𝐹^{(𝑚 + 1) \times (𝑛 + 1)}$ with this property.
(b) $𝑓 \mapsto 𝑀_{𝑓}$ is a bijection from the set of affine maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ onto the set of matrices in $𝐹^{(𝑚 + 1) \times (𝑛 + 1)}$ whose last row is $(0, \dots, 0, 1)$ .
(c)If $𝑔 : 𝐹^{𝑚} \to 𝐹^{𝑙}$ is affine, then $𝑀_{𝑔 \circ 𝑓} = 𝑀_{𝑔} 𝑀_{𝑓}$ .
(d)If $𝑚 = 𝑛$ , then $𝑓$ is bijective if and only if $𝑀_{𝑓}$ is invertible, and in that case $𝑀_{𝑓^{- 1}} = 𝑀_{𝑓}^{- 1}$ .

Proof.standard

Uses. Theorem 2.4.34 (Block multiplication), Proposition 4.7.38 (Affine maps and affine subspaces), Proposition 2.3.17 (Every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is some $𝑇_{𝐴}$ ), Corollary 2.2.6 (Agreeing on a spanning list), Theorem 2.3.20 (Linear maps correspond to matrices), Corollary 3.5.25 (The invertible matrix theorem)

Techniques. Homogenize: add a coordinate $1$ (Technique 4.7.21), Block decomposition (Technique 2.4.35)

Hint 1. Multiply in blocks (Theorem 2.4.34), splitting

𝑛 + 1

𝑛

and

1

, and

𝑚 + 1

𝑚

and

1

Hint 2. For uniqueness in (a), the vectors

\hat{𝐱}

span

𝐹^{𝑛 + 1}

: look at

\hat{𝟎}

and the

{\hat{𝐞}}_{𝑗}

Hint 3. (c) is block multiplication together with Proposition 4.7.38(a). For (d), apply (c) to

𝑓

and

𝑓^{- 1}

; for the converse, apply

𝑀_{𝑓}

(𝐚, 0)

with

𝐴 𝐚 = 𝟎

Sketch. Block multiplication gives

𝑀_{𝑓} \hat{𝐱} =

(𝐴 𝐱 + 𝐛, 1)

. Since

\hat{𝟎} = 𝐞_{𝑛 + 1}

and

{\hat{𝐞}}_{𝑗} - \hat{𝟎} = 𝐞_{𝑗}

, the

\hat{𝐱}

span

𝐹^{𝑛 + 1}

, so a matrix is determined by its products with them.

𝑀_{𝑔} 𝑀_{𝑓} =

(\begin{matrix} 𝐵 𝐴 & 𝐵 𝐛 + 𝐜 \\ 0 & 1 \end{matrix}) =

𝑀_{𝑔 \circ 𝑓}

. If

𝑓

is bijective,

𝑀_{𝑓^{- 1}} 𝑀_{𝑓} =

𝑀_{id} =

𝐼_{𝑛 + 1}

and likewise in the other order. If

𝑀_{𝑓}

is invertible and

𝐴 𝐚 = 𝟎

, then

𝑀_{𝑓} (𝐚, 0) = 𝟎

, so

𝐚 = 𝟎

and

𝐴

is invertible.

(a) By Theorem 2.4.34, with blocks of sizes $𝑚, 1$ and $𝑛, 1$ ,

𝑀_{𝑓} \hat{𝐱} = (\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix}) (\begin{matrix} 𝐱 \\ 1 \end{matrix}) = (\begin{matrix} 𝐴 𝐱 + 𝐛 \\ 0 𝐱 + 1 \end{matrix}) = (\begin{matrix} 𝑓 (𝐱) \\ 1 \end{matrix}) = \hat{𝑓 (𝐱)} .

For uniqueness, note that $\hat{𝟎} = 𝐞_{𝑛 + 1}$ and ${\hat{𝐞}}_{𝑗} - \hat{𝟎} = 𝐞_{𝑗}$ for $𝑗 \leq 𝑛$ , so the vectors $\hat{𝐱}$ span $𝐹^{𝑛 + 1}$ . If $𝑁 \hat{𝐱} = 𝑀_{𝑓} \hat{𝐱}$ for all $𝐱$ , then the linear maps $𝑇_{𝑁}$ and $𝑇_{𝑀_{𝑓}}$ agree on a spanning list, so they are equal (Corollary 2.2.6), and $𝑁 = 𝑀_{𝑓}$ (Theorem 2.3.20).

(b) By Proposition 2.3.17, the linear part of an affine map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is $𝑇_{𝐴}$ for exactly one $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ , so each affine map is $𝐱 \mapsto 𝐴 𝐱 + 𝐛$ for exactly one pair $(𝐴, 𝐛)$ , since the linear part and $𝐛 = 𝑓 (𝟎)$ are determined by $𝑓$ . The matrices with last row $(0, \dots, 0, 1)$ are exactly the $(\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix})$ with $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐛 \in 𝐹^{𝑚}$ , and each of them determines $(𝐴, 𝐛)$ . So $𝑓 \mapsto 𝑀_{𝑓}$ is a bijection.

(c) Let $𝑔 (𝐲) = 𝐵 𝐲 + 𝐜$ . By Theorem 2.4.34,

𝑀_{𝑔} 𝑀_{𝑓} = (\begin{matrix} 𝐵 & 𝐜 \\ 0 & 1 \end{matrix}) (\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 𝐵 𝐴 & 𝐵 𝐛 + 𝐜 \\ 0 & 1 \end{matrix}),

and by Proposition 4.7.38(a), $𝑔 \circ 𝑓$ is $𝐱 \mapsto$ $𝐵 𝐴 𝐱 +$ $(𝐵 𝐛 + 𝐜)$ . So $𝑀_{𝑔} 𝑀_{𝑓} = 𝑀_{𝑔 \circ 𝑓}$ .

(d) Note that $𝑀_{id} =$ $(\begin{matrix} 𝐼_{𝑛} & 𝟎 \\ 0 & 1 \end{matrix}) =$ $𝐼_{𝑛 + 1}$ . If $𝑓$ is bijective, then $𝑓^{- 1}$ is affine (Proposition 4.7.38(b)), and by (c), $𝑀_{𝑓^{- 1}} 𝑀_{𝑓} =$ $𝑀_{𝑓^{- 1} \circ 𝑓} =$ $𝐼_{𝑛 + 1}$ and $𝑀_{𝑓} 𝑀_{𝑓^{- 1}} = 𝐼_{𝑛 + 1}$ . So $𝑀_{𝑓}$ is invertible with inverse $𝑀_{𝑓^{- 1}}$ . Conversely, suppose $𝑀_{𝑓}$ is invertible, and let $𝐴 𝐚 = 𝟎$ with $𝐚 \in 𝐹^{𝑛}$ . Then $𝑀_{𝑓} (\begin{matrix} 𝐚 \\ 0 \end{matrix}) =$ $(\begin{matrix} 𝐴 𝐚 \\ 0 \end{matrix}) =$ $𝟎$ , so $(𝐚, 0) = 𝟎$ and $𝐚 = 𝟎$ . Hence $ker 𝑇_{𝐴} = {𝟎}$ , $𝐴$ is invertible by Corollary 3.5.25, and $𝑓$ is bijective by Proposition 4.7.38(b). □

The translation by $𝐚$ has matrix $(\begin{matrix} 𝐼_{𝑛} & 𝐚 \\ 0 & 1 \end{matrix})$ . On $𝐹^{𝑛 + 1}$ it is a linear map, a shear that slides the hyperplane $𝑥_{𝑛 + 1} = 1$ along itself by $𝐚$ and fixes the hyperplane $𝑥_{𝑛 + 1} = 0$ . For $𝑛 = 1$ : the map $(𝑥, 𝑦) \mapsto$ $(𝑥 + 𝑎 𝑦, 𝑦)$ of $𝐹^{2}$ is linear and moves the line $𝑦 = 1$ , a copy of $𝐹$ , by $𝑎$ . So translations, which are not linear on $𝐹^{𝑛}$ , become linear one dimension up.

Exercise 4.7.43.routine Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ be invertible and $𝐛 \in 𝐹^{𝑛}$ . Verify by block multiplication that

{(\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} 𝐴^{- 1} & - 𝐴^{- 1} 𝐛 \\ 0 & 1 \end{matrix}) .

Use it to invert the matrix of the translation by $𝐚$ , and the matrix $(\begin{matrix} 0 & - 1 & 3 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix})$ .

Solution.

Uses. Theorem 2.4.34 (Block multiplication), Theorem 2.5.10 (Left and right inverses agree)

Hint 1. Multiply the two block matrices in both orders.

Hint 2. For the

3 \times 3

matrix,

𝐴

is the rotation

(\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})

, whose inverse is

(\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})

Sketch. Both products are

(\begin{matrix} 𝐼_{𝑛} & 𝟎 \\ 0 & 1 \end{matrix}) = 𝐼_{𝑛 + 1}

. The translation by

𝐚

inverts to the translation by

- 𝐚

. For the

3 \times 3

matrix,

- 𝐴^{- 1} (3, 1) =

- (1, - 3) =

(- 1, 3)

, giving

(\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 3 \\ 0 & 0 & 1 \end{matrix})

By Theorem 2.4.34,

(\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix}) (\begin{matrix} 𝐴^{- 1} & - 𝐴^{- 1} 𝐛 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 𝐴 𝐴^{- 1} & - 𝐴 𝐴^{- 1} 𝐛 + 𝐛 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 𝐼_{𝑛} & 𝟎 \\ 0 & 1 \end{matrix}) = 𝐼_{𝑛 + 1},

and

(\begin{matrix} 𝐴^{- 1} & - 𝐴^{- 1} 𝐛 \\ 0 & 1 \end{matrix}) (\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 𝐴^{- 1} 𝐴 & 𝐴^{- 1} 𝐛 - 𝐴^{- 1} 𝐛 \\ 0 & 1 \end{matrix}) = 𝐼_{𝑛 + 1} .

For the translation, $𝐴 = 𝐼_{𝑛}$ and $𝐛 = 𝐚$ , so the inverse is $(\begin{matrix} 𝐼_{𝑛} & - 𝐚 \\ 0 & 1 \end{matrix})$ , the matrix of the translation by $- 𝐚$ .

For the $3 \times 3$ matrix, $𝐴 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ and $𝐛 = (3, 1)$ . Since $𝐴 (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) = 𝐼_{2}$ , $𝐴^{- 1} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$ , and $- 𝐴^{- 1} 𝐛 =$ $- (1, - 3) =$ $(- 1, 3)$ . So the inverse is $(\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 3 \\ 0 & 0 & 1 \end{matrix})$ . Check: the product of the first row of the inverse with the third column of the original is $0 \cdot 3 + 1 \cdot 1 + (- 1) \cdot 1 = 0$ , as it should be.

Example 4.7.44 (Rotation about a point). Let $𝑓$ be the rotation of $ℝ^{2}$ by $90 °$ counterclockwise about $𝐚 = (1, 2)$ , and $𝑅 = 𝑅_{𝜋 / 2} = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ . By Example 4.7.36, $𝑓 = 𝜏_{𝐚} \circ 𝑅 \circ 𝜏_{- 𝐚}$ , so by Theorem 4.7.42(c),

𝑀_{𝑓} = (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 0 & - 1 & 3 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}) .

The last column is $𝐚 -$ $𝑅 𝐚 =$ $(1, 2) -$ $(- 2, 1) =$ $(3, 1)$ . As a check, $𝑀_{𝑓} \hat{(1, 2)} =$ $(0 - 2 + 3, 1 + 0 + 1, 1)$ , so the center is fixed, and $𝑀_{𝑓} \hat{(2, 2)} =$ $(1, 3, 1)$ : the point one unit to the right of $𝐚$ goes to the point one unit above it.

The order of the factors matters. The product

(\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 0 & - 1 & - 2 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix})

is the matrix of “translate by $𝐚$ , then rotate about $𝟎$ ”, which is $𝐱 \mapsto 𝑅 𝐱 + 𝑅 𝐚$ , a rotation about a different center.

In three dimensions the same matrices are $4 \times 4$ . A renderer stores each object once, in coordinates convenient for that object, and places it in the scene by a product $𝑀 = 𝑀_{𝑘} \dots 𝑀_{1}$ of rotations, scalings, and translations, computed once per object. Each vertex then costs one $4 \times 4$ matrix–vector product, which is why graphics processors are built around $4 \times 4$ matrix arithmetic. In the product, $𝑀_{1}$ acts first.

Perspective projection

The projection $(𝑥, 𝑦, 𝑧) \mapsto$ $(𝑥, 𝑦, 0)$ onto the $𝑥 𝑦$ -plane is linear, but it draws distant objects as large as near ones. A camera, or an eye, projects through a point instead. Fix $𝑑 \neq 0$ , the center $𝐜 = (0, 0, 𝑑)$ , and the image plane $𝑧 = 0$ . The perspective projection $𝜋$ sends $𝐯 = (𝑥, 𝑦, 𝑧)$ to the point where the line through $𝐜$ and $𝐯$ meets the image plane (Figure 4.7.2).

Figure 4.7.2. The perspective projection seen from the side, with the image plane

𝑧 = 0

drawn as the horizontal line. The point

𝐯

between the plane and the center is projected outward; the point

𝐰

behind the plane is projected inward, so distant points come closer together.

The points of the line through $𝐜$ and $𝐯$ are $𝐜 + 𝑠 (𝐯 - 𝐜)$ for $𝑠 \in ℝ$ . The third coordinate of such a point is $𝑑 + 𝑠 (𝑧 - 𝑑)$ , which is $0$ if and only if $𝑠 (𝑑 - 𝑧) = 𝑑$ . If $𝑧 = 𝑑$ , there is no such $𝑠$ : the line is parallel to the image plane, or $𝐯 = 𝐜$ , and $𝐯$ has no image. Otherwise $𝑠 = 𝑑 / (𝑑 - 𝑧)$ , and the first two coordinates are $𝑠 𝑥$ and $𝑠 𝑦$ , so

𝜋 (𝑥, 𝑦, 𝑧) = (\frac{𝑥}{1 - 𝑧 / 𝑑}, \frac{𝑦}{1 - 𝑧 / 𝑑}, 0) (𝑧 \neq 𝑑) .

The same formula covers points on the far side of the image plane from the center, and points on the far side of the center from the image plane, where $𝑠 < 0$ .

The map $𝜋$ is not affine. With $𝑑 = 1$ , $𝜋 (0, 0, 0) = 𝟎$ and $𝜋 (1, 0, \frac{1}{2}) =$ $(2, 0, 0)$ , but the midpoint $(\frac{1}{2}, 0, \frac{1}{4})$ goes to $(\frac{2}{3}, 0, 0)$ , not to the midpoint $(1, 0, 0)$ of the images. So by Theorem 4.7.37 no matrix $𝑀_{𝑓}$ represents $𝜋$ . Even so:

Proposition 4.7.45 (Perspective projection in homogeneous coordinates). Let $𝑑 \neq 0$ and

𝑃 = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 / 𝑑 & 1 \end{matrix}) .

For every $𝐯 = (𝑥, 𝑦, 𝑧) \in ℝ^{3}$ with $𝑧 \neq 𝑑$ , $𝑃 \hat{𝐯} =$ $(𝑥, 𝑦, 0, 1 - 𝑧 / 𝑑)$ , and this is a vector of homogeneous coordinates of $𝜋 (𝐯)$ .

Proof.routine

Uses. Definition 2.3.8 (Matrix–vector product), Definition 4.7.41 (Homogeneous coordinates)

Hint 1. Compute

𝑃 \hat{𝐯}

row by row.

Hint 2. Which point does a vector

(𝐳, 𝑐)

with

𝑐 \neq 0

represent?

Sketch. The rows of

𝑃

give

𝑥

𝑦

0

, and

- 𝑧 / 𝑑 + 1

. The last entry is nonzero because

𝑧 \neq 𝑑

, and dividing by it gives

𝜋 (𝐯)

Multiplying the rows of

𝑃

\hat{𝐯} = (𝑥, 𝑦, 𝑧, 1)

gives

𝑥

𝑦

0

, and

- 𝑧 / 𝑑 + 1

. Since

𝑧 \neq 𝑑

, the last entry

𝑐 = 1 - 𝑧 / 𝑑

is nonzero, and by Definition 4.7.41 the vector represents the point

𝑐^{- 1} (𝑥, 𝑦, 0) =

𝜋 (𝐯)

. □

The nonlinearity has moved into the freedom of Definition 4.7.41: $𝑃 \hat{𝐯}$ does not have last coordinate $1$ , and the division by the last coordinate, performed once at the very end, supplies it. The matrix $𝑃$ has rank $3$ , so the depth $𝑧$ is lost. Renderers therefore use the third row $(0, 0, 1, 0)$ instead of $(0, 0, 0, 0)$ , which keeps the number $𝑧 / (1 - 𝑧 / 𝑑) =$ $𝑑 𝑧 / (𝑑 - 𝑧)$ as a third coordinate. Its derivative in $𝑧$ is $𝑑^{2} / {(𝑑 - 𝑧)}^{2} > 0$ , so on each side of the plane $𝑧 = 𝑑$ it increases with $𝑧$ , and it still tells which of two points that project to the same pixel is nearer to the camera.

For a center off the $𝑧$ -axis, or another image plane, conjugate by translations and rotations as in Example 4.7.44; Exercise 4.7.49 carries this out. Under $𝜋$ , parallel lines in space usually stop being parallel, as the rails of a track do in a photograph, and Exercise 4.7.50 finds where their images meet.

Aside (Projective space). By Definition 4.7.41, a point $𝐱$ of $𝐹^{𝑛}$ is the same thing as a line through $𝟎$ in $𝐹^{𝑛 + 1}$ , namely $span (\hat{𝐱})$ , and every line through $𝟎$ that is not contained in the hyperplane $𝐻 = {𝑥_{𝑛 + 1} = 0}$ arises this way exactly once. The projective space $ℙ^{𝑛} (𝐹)$ is the set of all lines through $𝟎$ in $𝐹^{𝑛 + 1}$ , that is, of all $1$ -dimensional subspaces. Its elements are the points of $𝐹^{𝑛}$ together with the lines inside $𝐻$ , called points at infinity. A point at infinity $span ((𝐰, 0))$ , with $𝐰 \neq 𝟎$ , corresponds to a direction in $𝐹^{𝑛}$ . Over $ℝ$ , the points $𝐱 + 𝑠 𝐰$ of a line have the vectors of homogeneous coordinates $(𝐱 + 𝑠 𝐰, 1)$ and $(𝑠^{- 1} 𝐱 + 𝐰, 𝑠^{- 1})$ for $𝑠 \neq 0$ , and the latter tend to $(𝐰, 0)$ as $𝑠 \to \infty$ . So all lines with direction $span (𝐰)$ share one point at infinity, and in Exercise 4.7.50 the vanishing point of a family of parallel lines is the image of their common point at infinity under $𝑃$ .

A projective line is a $2$ -dimensional subspace of $𝐹^{𝑛 + 1}$ , viewed as the set of lines through $𝟎$ that it contains. In the projective plane $ℙ^{2} (𝐹)$ , two distinct projective lines $𝑈 \neq 𝑊$ meet in exactly one point: $𝑊 ⊈ 𝑈$ by Proposition 4.1.21(b), so $𝑈 + 𝑊$ is larger than $𝑈$ and equals $𝐹^{3}$ , and Theorem 4.4.19 gives $dim (𝑈 \cap 𝑊) =$ $2 +$ $2 -$ $3 =$ $1$ . In particular, two parallel lines of $𝐹^{2}$ , completed by their common point at infinity, meet there. An invertible matrix in $𝐹^{(𝑛 + 1) \times (𝑛 + 1)}$ permutes the $1$ -dimensional subspaces of $𝐹^{𝑛 + 1}$ , so it acts on $ℙ^{𝑛} (𝐹)$ , and nonzero multiples of a matrix act in the same way. These are the projective transformations. They send projective lines to projective lines, but unlike affine maps they can move points at infinity into $𝐹^{𝑛}$ , which is how a photograph can show a horizon. Over a field with $𝑞$ elements, $ℙ^{2}$ has $(𝑞^{3} - 1) / (𝑞 - 1) =$ $𝑞^{2} +$ $𝑞 +$ $1$ points; for $𝑞 = 2$ it is the seven-point Fano plane.

Exercises

Exercise 4.7.46.routine Decide which of the following sets are affine subspaces. For each one that is, find its direction and, if it is finite, its dimension.

(a)The solution set in $ℝ^{3}$ of $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 1$ and $𝑥_{1} - 𝑥_{3} = 2$ .
(b)The set of $𝑝 \in ℝ [𝑥]$ with $𝑝 (0) = 1$ and $𝑝 (1) = 3$ .
(c)The set of $𝐱 \in ℝ^{2}$ with $𝑥_{1} 𝑥_{2} = 0$ .
(d)The set of $(𝑥_{1}, 𝑥_{2}) \in 𝔽_{3}^{2}$ with $𝑥_{2} = 𝑥_{1}^{2}$ .

Solution.

Uses. Theorem 4.7.8 (Affine subspaces of $𝐹^{𝑛}$ are solution sets), Theorem 4.7.14 (Affine subspaces are the sets closed under affine combinations), Proposition 4.7.3 (Direction and base points), Corollary 2.6.31 (Fibers are translates of the kernel)

Hint 1. (a) and (b) are solution sets of linear equations, in

ℝ^{3}

and in

ℝ [𝑥]

. Find one solution and the homogeneous solutions.

Hint 2. For (c) and (d), look for an affine combination of three points of the set, such as

𝐮 + 𝐯 - 𝐰

, that leaves the set.

Sketch. (a)

(2, - 1, 0) +

span ((1, - 2, 1))

, a line. (b) The fiber of

𝑝 \mapsto (𝑝 (0), 𝑝 (1))

over

(1, 3)

1 +

2 𝑥 +

{𝑝 | 𝑝 (0) = 𝑝 (1) = 0}

, an affine subspace with infinite-dimensional direction. (c) No:

(1, 0) +

(0, 1) -

(0, 0) =

(1, 1)

. (d) No: the set is

{(0, 0), (1, 1), (2, 1)}

, and

(1, 1) +

(2, 1) -

(0, 0) =

(0, 2)

is missing.

(a) Yes. The homogeneous equations $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ , $𝑥_{1} = 𝑥_{3}$ give $𝑥_{2} = - 2 𝑥_{3}$ , so their solutions are $span ((1, - 2, 1))$ . One solution of the system is $(2, - 1, 0)$ : $2 - 1 + 0 = 1$ and $2 - 0 = 2$ . By Theorem 4.7.8(a), the set is $(2, - 1, 0) +$ $span ((1, - 2, 1))$ , a line, with direction $span ((1, - 2, 1))$ and dimension $1$ .

(b) Yes. The map $𝑇 : ℝ [𝑥] \to ℝ^{2}$ , $𝑇 (𝑝) =$ $(𝑝 (0), 𝑝 (1))$ , is linear, and the set is the fiber $𝑇^{- 1} ((1, 3))$ , which contains $1 + 2 𝑥$ . By Corollary 2.6.31 it is $1 + 2 𝑥 + ker 𝑇$ , an affine subspace with direction $ker 𝑇 =$ ${𝑝 | 𝑝 (0) = 𝑝 (1) = 0}$ . This direction contains $𝑥^{𝑗} (𝑥 - 1)$ for every $𝑗 \geq 1$ , and these polynomials have distinct degrees, so $ker 𝑇$ is infinite-dimensional and the dimension is not defined.

(c) No. The points $(1, 0)$ , $(0, 1)$ , $(0, 0)$ lie in the set, and the affine combination $(1, 0) +$ $(0, 1) -$ $(0, 0) =$ $(1, 1)$ does not, since $1 \cdot 1 \neq 0$ . By Theorem 4.7.14 the set is not affine.

(d) No. In $𝔽_{3}$ , $0^{2} = 0$ , $1^{2} = 1$ , and $2^{2} = 4 = 1$ , so the set is ${(0, 0), (1, 1), (2, 1)}$ . The affine combination $(1, 1) +$ $(2, 1) -$ $(0, 0) =$ $(3, 2) =$ $(0, 2)$ is not in it, so it is not affine by Theorem 4.7.14.

Exercise 4.7.47.routine Let

𝐯_{0} = (0, 0)

𝐯_{1} = (4, 0)

𝐯_{2} = (1, 3)

ℝ^{2}

. Find the barycentric coordinates of

𝐩 = (2, 1)

𝐪 = (3, 2)

, and

𝐫 = (- 1, 1)

with respect to

𝐯_{0}, 𝐯_{1}, 𝐯_{2}

, and decide which of the three points lie in the triangle

conv (𝐯_{0}, 𝐯_{1}, 𝐯_{2})

. For each point outside, name the edge line it lies beyond.

Solution.

Uses. Remark 4.7.26 (Frames exist, and coordinates are coordinates), Proposition 4.7.32 (Barycentric coordinates on a simplex)

Hint 1. Since

𝐯_{0} = 𝟎

(𝑡_{1}, 𝑡_{2})

are the coordinates of the point in the basis

(4, 0), (1, 3)

, and

𝑡_{0} = 1 - 𝑡_{1} - 𝑡_{2}

Hint 2. Solve

4 𝑡_{1} + 𝑡_{2} = 𝑥

and

3 𝑡_{2} = 𝑦

. A negative

𝑡_{𝑖}

puts the point beyond the edge opposite

𝐯_{𝑖}

Sketch.

𝑡_{2} = 𝑦 / 3

and

𝑡_{1} = (𝑥 - 𝑦 / 3) / 4

. This gives

𝐩 : (\frac{1}{4}, \frac{5}{12}, \frac{1}{3})

, inside;

𝐪 : (- \frac{1}{4}, \frac{7}{12}, \frac{2}{3})

, beyond the edge

𝐯_{1} 𝐯_{2}

;

𝐫 : (1, - \frac{1}{3}, \frac{1}{3})

, beyond the edge

𝐯_{0} 𝐯_{2}

The differences $(4, 0)$ and $(1, 3)$ are independent, so the points form an affine frame of $ℝ^{2}$ . By Remark 4.7.26, since $𝐯_{0} = 𝟎$ , a point $(𝑥, 𝑦)$ has $(𝑥, 𝑦) =$ $𝑡_{1} (4, 0) +$ $𝑡_{2} (1, 3)$ , that is, $4 𝑡_{1} + 𝑡_{2} = 𝑥$ and $3 𝑡_{2} = 𝑦$ , and $𝑡_{0} = 1 - 𝑡_{1} - 𝑡_{2}$ . So $𝑡_{2} = 𝑦 / 3$ and $𝑡_{1} = (𝑥 - 𝑦 / 3) / 4$ .

For $𝐩 = (2, 1)$ : $𝑡_{2} = \frac{1}{3}$ , $𝑡_{1} = (2 - \frac{1}{3}) / 4 = \frac{5}{12}$ , $𝑡_{0} = 1 - \frac{5}{12} - \frac{4}{12} = \frac{1}{4}$ . All are positive, so $𝐩$ lies in the triangle (Proposition 4.7.32). Check: $\frac{5}{12} (4, 0) +$ $\frac{1}{3} (1, 3) =$ $(\frac{5}{3} + \frac{1}{3}, 1) =$ $(2, 1)$ .

For $𝐪 = (3, 2)$ : $𝑡_{2} = \frac{2}{3}$ , $𝑡_{1} = (3 - \frac{2}{3}) / 4 = \frac{7}{12}$ , $𝑡_{0} = 1 - \frac{7}{12} - \frac{8}{12} = - \frac{1}{4}$ . So $𝐪$ is outside, beyond the edge $𝐯_{1} 𝐯_{2}$ opposite $𝐯_{0}$ . Check: $\frac{7}{12} (4, 0) +$ $\frac{2}{3} (1, 3) =$ $(\frac{7}{3} + \frac{2}{3}, 2) =$ $(3, 2)$ .

For $𝐫 = (- 1, 1)$ : $𝑡_{2} = \frac{1}{3}$ , $𝑡_{1} = (- 1 - \frac{1}{3}) / 4 = - \frac{1}{3}$ , $𝑡_{0} = 1 + \frac{1}{3} - \frac{1}{3} = 1$ . So $𝐫$ is outside, beyond the edge $𝐯_{0} 𝐯_{2}$ opposite $𝐯_{1}$ . Check: $- \frac{1}{3} (4, 0) +$ $\frac{1}{3} (1, 3) =$ $(- 1, 1)$ .

Exercise 4.7.48.routine Work in $ℝ^{2}$ with homogeneous coordinates.

(a)Write the $3 \times 3$ matrices of the translation $𝑇$ by $(2, - 1)$ , the rotation $𝑅$ by $90 °$ counterclockwise about $𝟎$ , and the scaling $𝑆$ by the factor $3$ about $(1, 1)$ .
(b)Compute $𝑀_{𝑇} 𝑀_{𝑅}$ and $𝑀_{𝑅} 𝑀_{𝑇}$ , and describe each product as a map of $ℝ^{2}$ .
(c)Which point of $ℝ^{2}$ has $(6, - 3, 3)$ as a vector of homogeneous coordinates? Give a vector of homogeneous coordinates of the same point with last entry $2$ .

Solution.

Uses. Theorem 4.7.42 (Affine maps as matrices), Example 4.7.36 (Affine maps), Definition 4.7.41 (Homogeneous coordinates)

Hint 1. Each matrix is

(\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix})

with

𝐴

the linear part and

𝐛

the image of

𝟎

. The scaling about

(1, 1)

𝐯 \mapsto 3 𝐯 - 2 (1, 1)

Hint 2. By Theorem 4.7.42(c),

𝑀_{𝑇} 𝑀_{𝑅}

is the matrix of

𝑇 \circ 𝑅

, which rotates first.

Sketch.

𝑀_{𝑇} = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix})

𝑀_{𝑅} = (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix})

𝑀_{𝑆} =

(\begin{matrix} 3 & 0 & - 2 \\ 0 & 3 & - 2 \\ 0 & 0 & 1 \end{matrix})

𝑀_{𝑇} 𝑀_{𝑅} =

(\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & - 1 \\ 0 & 0 & 1 \end{matrix})

𝐯 \mapsto 𝑅 𝐯 + (2, - 1)

, and

𝑀_{𝑅} 𝑀_{𝑇} =

(\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & 2 \\ 0 & 0 & 1 \end{matrix})

𝐯 \mapsto 𝑅 𝐯 + (1, 2)

. (c)

(2, - 1)

, and

(4, - 2, 2)

(a) By Theorem 4.7.42, the matrix of $𝐯 \mapsto 𝐴 𝐯 + 𝐛$ is $(\begin{matrix} 𝐴 & 𝐛 \\ 0 & 1 \end{matrix})$ . The scaling about $(1, 1)$ is $𝐯 \mapsto$ $(1, 1) +$ $3 (𝐯 - (1, 1)) =$ $3 𝐯 -$ $(2, 2)$ (Example 4.7.36). So

𝑀_{𝑇} = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix}), 𝑀_{𝑅} = (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}), 𝑀_{𝑆} = (\begin{matrix} 3 & 0 & - 2 \\ 0 & 3 & - 2 \\ 0 & 0 & 1 \end{matrix}) .

(b) Multiplying,

𝑀_{𝑇} 𝑀_{𝑅} = (\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & - 1 \\ 0 & 0 & 1 \end{matrix}), 𝑀_{𝑅} 𝑀_{𝑇} = (\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & 2 \\ 0 & 0 & 1 \end{matrix}) .

By Theorem 4.7.42(c), the first is the matrix of $𝑇 \circ 𝑅$ , which rotates and then translates: $𝐯 \mapsto 𝑅 𝐯 + (2, - 1)$ . The second is $𝑅 \circ 𝑇$ , which translates and then rotates: $𝐯 \mapsto$ $𝑅 (𝐯 + (2, - 1)) =$ $𝑅 𝐯 +$ $(1, 2)$ , since $𝑅 (2, - 1) = (1, 2)$ . The two maps have the same linear part but differ by a translation, so the order matters.

(c) By Definition 4.7.41, $(6, - 3, 3)$ represents $3^{- 1} (6, - 3) = (2, - 1)$ . Multiplying $\hat{(2, - 1)} = (2, - 1, 1)$ by $2$ gives $(4, - 2, 2)$ .

Exercise 4.7.49.standard Let $𝑑 \neq 0$ and $𝑐_{1}, 𝑐_{2} \in ℝ$ , and let $𝜋_{𝐜}$ be the perspective projection onto the plane $𝑧 = 0$ with center $𝐜 = (𝑐_{1}, 𝑐_{2}, 𝑑)$ .

(a)Show that for $𝐯 = (𝑥, 𝑦, 𝑧)$ with $𝑧 \neq 𝑑$ ,
$𝜋_{𝐜} (𝐯) = (\frac{𝑑 𝑥 - 𝑐_{1} 𝑧}{𝑑 - 𝑧}, \frac{𝑑 𝑦 - 𝑐_{2} 𝑧}{𝑑 - 𝑧}, 0) .$
(b)Let $𝐚 = (𝑐_{1}, 𝑐_{2}, 0)$ and let $𝜋$ be the projection with center $(0, 0, 𝑑)$ . Show that $𝜋_{𝐜} =$ $𝜏_{𝐚} \circ$ $𝜋 \circ$ $𝜏_{- 𝐚}$ .
(c)Compute the $4 \times 4$ matrix $𝑀_{𝜏_{𝐚}} 𝑃 𝑀_{𝜏_{- 𝐚}}$ , with $𝑃$ as in Proposition 4.7.45, and check that it sends $\hat{𝐯}$ to a vector of homogeneous coordinates of $𝜋_{𝐜} (𝐯)$ .

Solution.

Uses. Proposition 4.7.45 (Perspective projection in homogeneous coordinates), Theorem 4.7.42 (Affine maps as matrices), Definition 4.7.41 (Homogeneous coordinates)

Hint 1. Parametrize the line as

𝐜 + 𝑠 (𝐯 - 𝐜)

and solve for the

𝑠

that makes the third coordinate

0

Hint 2. For (b), compute

𝜏_{𝐚} (𝜋 (𝜏_{- 𝐚} (𝐯)))

with the formula for

𝜋

; note that

𝜏_{- 𝐚}

does not change

𝑧

Hint 3. In (c), the translation matrices are

(\begin{matrix} 𝐼_{3} & \pm 𝐚 \\ 0 & 1 \end{matrix})

. Multiply

𝑃 𝑀_{𝜏_{- 𝐚}}

first.

Sketch. (a)

𝑠 = 𝑑 / (𝑑 - 𝑧)

, and the first coordinate is

𝑐_{1} +

𝑠 (𝑥 - 𝑐_{1}) =

(𝑑 𝑥 - 𝑐_{1} 𝑧) / (𝑑 - 𝑧)

. (b)

𝜋 (𝑥 - 𝑐_{1}, 𝑦 - 𝑐_{2}, 𝑧) +

𝐚

has first coordinate

𝑑 (𝑥 - 𝑐_{1}) / (𝑑 - 𝑧) +

𝑐_{1}

, the same. (c) The product is

(\begin{matrix} 1 & 0 & - 𝑐_{1} / 𝑑 & 0 \\ 0 & 1 & - 𝑐_{2} / 𝑑 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 / 𝑑 & 1 \end{matrix})

, which sends

\hat{𝐯}

(𝑥 - 𝑐_{1} 𝑧 / 𝑑, 𝑦 - 𝑐_{2} 𝑧 / 𝑑, 0, 1 - 𝑧 / 𝑑)

(a) The line through $𝐜$ and $𝐯$ consists of the points $𝐜 + 𝑠 (𝐯 - 𝐜)$ , whose third coordinate is $𝑑 + 𝑠 (𝑧 - 𝑑)$ . It vanishes exactly for $𝑠 = 𝑑 / (𝑑 - 𝑧)$ , which exists since $𝑧 \neq 𝑑$ . The first coordinate is then

𝑐_{1} + \frac{𝑑}{𝑑 - 𝑧} (𝑥 - 𝑐_{1}) = \frac{𝑐_{1} (𝑑 - 𝑧) + 𝑑 𝑥 - 𝑑 𝑐_{1}}{𝑑 - 𝑧} = \frac{𝑑 𝑥 - 𝑐_{1} 𝑧}{𝑑 - 𝑧},

and the second is $(𝑑 𝑦 - 𝑐_{2} 𝑧) / (𝑑 - 𝑧)$ in the same way.

(b) $𝜏_{- 𝐚} (𝐯) =$ $(𝑥 - 𝑐_{1}, 𝑦 - 𝑐_{2}, 𝑧)$ has third coordinate $𝑧 \neq 𝑑$ , so by the formula for $𝜋$ (the case $𝑐_{1} = 𝑐_{2} = 0$ of (a)),

𝜋 (𝜏_{- 𝐚} (𝐯)) = (\frac{𝑑 (𝑥 - 𝑐_{1})}{𝑑 - 𝑧}, \frac{𝑑 (𝑦 - 𝑐_{2})}{𝑑 - 𝑧}, 0) .

Adding $𝐚$ , the first coordinate becomes $(𝑑 𝑥 - 𝑑 𝑐_{1} + 𝑐_{1} (𝑑 - 𝑧)) / (𝑑 - 𝑧) =$ $(𝑑 𝑥 - 𝑐_{1} 𝑧) / (𝑑 - 𝑧)$ , and likewise the second. By (a), this is $𝜋_{𝐜} (𝐯)$ .

(c) With $𝑀_{𝜏_{\pm 𝐚}} =$ $(\begin{matrix} 𝐼_{3} & \pm 𝐚 \\ 0 & 1 \end{matrix})$ ,

𝑃 𝑀_{𝜏_{- 𝐚}} = (\begin{matrix} 1 & 0 & 0 & - 𝑐_{1} \\ 0 & 1 & 0 & - 𝑐_{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 / 𝑑 & 1 \end{matrix}), 𝑀_{𝜏_{𝐚}} 𝑃 𝑀_{𝜏_{- 𝐚}} = (\begin{matrix} 1 & 0 & - 𝑐_{1} / 𝑑 & 0 \\ 0 & 1 & - 𝑐_{2} / 𝑑 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 / 𝑑 & 1 \end{matrix}) .

(In the second product, row $1$ is row $1$ plus $𝑐_{1}$ times row $4$ of $𝑃 𝑀_{𝜏_{- 𝐚}}$ , and similarly for row $2$ .) This matrix sends $\hat{𝐯} = (𝑥, 𝑦, 𝑧, 1)$ to $(𝑥 - 𝑐_{1} 𝑧 / 𝑑, 𝑦 - 𝑐_{2} 𝑧 / 𝑑, 0, 1 - 𝑧 / 𝑑)$ . The last entry is nonzero, and dividing by it gives first coordinate $(𝑥 - 𝑐_{1} 𝑧 / 𝑑) / (1 - 𝑧 / 𝑑) =$ $(𝑑 𝑥 - 𝑐_{1} 𝑧) / (𝑑 - 𝑧)$ , and similarly for the second, which is $𝜋_{𝐜} (𝐯)$ by (a).

Exercise 4.7.50.standard Let $𝑑 \neq 0$ , and let $𝜋$ and $𝑃$ be as in Proposition 4.7.45. Let $𝐯 = (𝑣_{1}, 𝑣_{2}, 𝑣_{3})$ with $𝑣_{3} \neq 𝑑$ , and let $𝐰 =$ $(𝑤_{1}, 𝑤_{2}, 𝑤_{3}) \neq$ $𝟎$ .

(a)Suppose $𝑤_{3} \neq 0$ , and let $𝐪_{𝐰} =$ $(- 𝑑 𝑤_{1} / 𝑤_{3}, - 𝑑 𝑤_{2} / 𝑤_{3}, 0)$ . Show that $𝑃 (\begin{matrix} 𝐰 \\ 0 \end{matrix})$ is a vector of homogeneous coordinates of $𝐪_{𝐰}$ , and that for every $𝑠 \in ℝ$ with $𝑣_{3} + 𝑠 𝑤_{3} \neq 𝑑$ , the point $𝜋 (𝐯 + 𝑠 𝐰)$ lies on $aff (𝜋 (𝐯), 𝐪_{𝐰})$ . Conclude that the images of all lines with direction $span (𝐰)$ lie on lines (or points) through the single point $𝐪_{𝐰}$ , the vanishing point of the direction $𝐰$ .
(b)Suppose $𝑤_{3} = 0$ . Show that $𝜋$ maps the line $𝐯 + span (𝐰)$ onto the line $𝜋 (𝐯) +$ $span ((𝑤_{1}, 𝑤_{2}, 0))$ . So lines parallel to the image plane have parallel images.

Solution.

Uses. Proposition 4.7.45 (Perspective projection in homogeneous coordinates), Definition 4.7.41 (Homogeneous coordinates), Proposition 4.7.17 (The affine hull)

Hint 1.

𝑃

is linear and

\hat{𝐯 + 𝑠 𝐰} =

\hat{𝐯} +

𝑠 (\begin{matrix} 𝐰 \\ 0 \end{matrix})

, so

𝑃 \hat{𝐯 + 𝑠 𝐰} =

𝐡_{1} +

𝑠 𝐡_{2}

with

𝐡_{1} = 𝑃 \hat{𝐯}

and

𝐡_{2} = 𝑃 (\begin{matrix} 𝐰 \\ 0 \end{matrix})

Hint 2. Let

𝛼

and

𝛽

be the last entries of

𝐡_{1}

and

𝐡_{2}

. Dividing

𝐡_{1} + 𝑠 𝐡_{2}

𝛼 + 𝑠 𝛽

writes the image point as an affine combination of

𝜋 (𝐯)

and

𝐪_{𝐰}

with coefficients

𝛼 / (𝛼 + 𝑠 𝛽)

and

𝑠 𝛽 / (𝛼 + 𝑠 𝛽)

Sketch.

𝐡_{2} =

(𝑤_{1}, 𝑤_{2}, 0, - 𝑤_{3} / 𝑑)

represents

𝐪_{𝐰}

. With

𝛼 = 1 - 𝑣_{3} / 𝑑

and

𝛽 = - 𝑤_{3} / 𝑑

, the point

𝜋 (𝐯 + 𝑠 𝐰)

is the first three entries of

𝐡_{1} + 𝑠 𝐡_{2}

divided by

𝛼 + 𝑠 𝛽

, which equals

𝛼 / (𝛼 + 𝑠 𝛽) \cdot

𝜋 (𝐯) +

𝑠 𝛽 / (𝛼 + 𝑠 𝛽) \cdot

𝐪_{𝐰}

, an affine combination. In (b),

𝑧

is constant along the line, so

𝜋 (𝐯 + 𝑠 𝐰) =

𝜋 (𝐯) +

(𝑠 / 𝛼) (𝑤_{1}, 𝑤_{2}, 0)

(a) $𝑃 (\begin{matrix} 𝐰 \\ 0 \end{matrix}) =$ $(𝑤_{1}, 𝑤_{2}, 0, - 𝑤_{3} / 𝑑) =$ $𝐡_{2}$ , whose last entry $𝛽 = - 𝑤_{3} / 𝑑$ is nonzero, so it represents $𝛽^{- 1} (𝑤_{1}, 𝑤_{2}, 0) =$ $𝐪_{𝐰}$ . Let $𝐡_{1} =$ $𝑃 \hat{𝐯} =$ $(𝑣_{1}, 𝑣_{2}, 0, 𝛼)$ with $𝛼 = 1 - 𝑣_{3} / 𝑑 \neq 0$ ; it represents $𝜋 (𝐯)$ by Proposition 4.7.45. Since $\hat{𝐯 + 𝑠 𝐰} =$ $\hat{𝐯} +$ $𝑠 (\begin{matrix} 𝐰 \\ 0 \end{matrix})$ and $𝑃$ is linear, $𝑃 \hat{𝐯 + 𝑠 𝐰} =$ $𝐡_{1} +$ $𝑠 𝐡_{2}$ . If $𝑣_{3} + 𝑠 𝑤_{3} \neq 𝑑$ , its last entry $𝛼 +$ $𝑠 𝛽 =$ $1 -$ $(𝑣_{3} + 𝑠 𝑤_{3}) / 𝑑$ is nonzero, and by Proposition 4.7.45,

𝜋 (𝐯 + 𝑠 𝐰) = \frac{1}{𝛼 + 𝑠 𝛽} ((𝑣_{1}, 𝑣_{2}, 0) + 𝑠 (𝑤_{1}, 𝑤_{2}, 0)) = \frac{𝛼}{𝛼 + 𝑠 𝛽} 𝜋 (𝐯) + \frac{𝑠 𝛽}{𝛼 + 𝑠 𝛽} 𝐪_{𝐰},

using $(𝑣_{1}, 𝑣_{2}, 0) =$ $𝛼 𝜋 (𝐯)$ and $(𝑤_{1}, 𝑤_{2}, 0) =$ $𝛽 𝐪_{𝐰}$ . The coefficients add up to $1$ , so the point lies in $aff (𝜋 (𝐯), 𝐪_{𝐰})$ (Proposition 4.7.17), which is the line through $𝜋 (𝐯)$ and $𝐪_{𝐰}$ if these differ, and the single point $𝐪_{𝐰}$ otherwise. The point $𝐪_{𝐰}$ depends only on $𝐰$ , so this holds for every line $𝐯 + span (𝐰)$ with $𝑣_{3} \neq 𝑑$ . (A line in this direction with $𝑣_{3} = 𝑑$ contains a point $𝐯 + 𝑠 𝐰$ with $𝑣_{3} + 𝑠 𝑤_{3} \neq 𝑑$ , and we may start from that point instead.)

(b) If $𝑤_{3} = 0$ , then $(𝑤_{1}, 𝑤_{2}) \neq (0, 0)$ , and every point of the line has third coordinate $𝑣_{3} \neq 𝑑$ . So

𝜋 (𝐯 + 𝑠 𝐰) = \frac{1}{𝛼} (𝑣_{1} + 𝑠 𝑤_{1}, 𝑣_{2} + 𝑠 𝑤_{2}, 0) = 𝜋 (𝐯) + \frac{𝑠}{𝛼} (𝑤_{1}, 𝑤_{2}, 0) .

As $𝑠$ runs over $ℝ$ , so does $𝑠 / 𝛼$ , and the image is the line $𝜋 (𝐯) +$ $span ((𝑤_{1}, 𝑤_{2}, 0))$ .

Exercise 4.7.51.standard Prove or disprove: if

𝐸_{1}

and

𝐸_{2}

are affine subspaces of

𝑉

and

𝐸_{1} \cup 𝐸_{2}

is an affine subspace, then

𝐸_{1} \subseteq 𝐸_{2}

𝐸_{2} \subseteq 𝐸_{1}

. Does the answer depend on

𝐹

Solution.

Uses. Theorem 4.7.14 (Affine subspaces are the sets closed under affine combinations), Proposition 4.7.17 (The affine hull)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Try the smallest field first.

Hint 2. If

𝐹

has an element

𝑡 \notin {0, 1}

, take

𝐚 \in 𝐸_{1} ∖ 𝐸_{2}

and

𝐛 \in 𝐸_{2} ∖ 𝐸_{1}

, and ask where

(1 - 𝑡) 𝐚 + 𝑡 𝐛

lies.

Hint 3. If

𝐜 =

(1 - 𝑡) 𝐚 +

𝑡 𝐛 \in

𝐸_{1}

, then

𝐛

is an affine combination of

𝐚

and

𝐜

Sketch. Over

𝔽_{2}

it is false: in

𝑉 = 𝔽_{2}

{0} \cup {1} = 𝔽_{2}

. If

| 𝐹 | \geq 3

it is true: with

𝐚, 𝐛

as in the hint and

𝑡 \notin {0, 1}

𝐜 =

(1 - 𝑡) 𝐚 +

𝑡 𝐛 \in

𝐸_{1} \cup

𝐸_{2}

. If

𝐜 \in 𝐸_{1}

, then

𝐛 =

(1 - 1 / 𝑡) 𝐚 +

(1 / 𝑡) 𝐜 \in

𝐸_{1}

; if

𝐜 \in 𝐸_{2}

, then

𝐚 =

(1 / (1 - 𝑡)) 𝐜 -

(𝑡 / (1 - 𝑡)) 𝐛 \in

𝐸_{2}

. Both are contradictions.

The answer depends on $𝐹$ . Over $𝔽_{2}$ the statement is false: in $𝑉 = 𝔽_{2}$ , the points ${0}$ and ${1}$ are affine subspaces, neither contains the other, and their union $𝔽_{2}$ is a subspace.

Now suppose $𝐹$ has more than two elements, and choose $𝑡 \in 𝐹$ with $𝑡 \neq 0$ and $𝑡 \neq 1$ . Suppose $𝐸_{1} \cup 𝐸_{2}$ is affine but neither of $𝐸_{1}, 𝐸_{2}$ contains the other. Pick $𝐚 \in 𝐸_{1}$ with $𝐚 \notin 𝐸_{2}$ , and $𝐛 \in 𝐸_{2}$ with $𝐛 \notin 𝐸_{1}$ . The point $𝐜 = (1 - 𝑡) 𝐚 + 𝑡 𝐛$ is an affine combination of elements of $𝐸_{1} \cup 𝐸_{2}$ , so it lies in $𝐸_{1} \cup 𝐸_{2}$ by Theorem 4.7.14.

If $𝐜 \in 𝐸_{1}$ , then

(1 - \frac{1}{𝑡}) 𝐚 + \frac{1}{𝑡} 𝐜 = (1 - \frac{1}{𝑡} + \frac{1 - 𝑡}{𝑡}) 𝐚 + 𝐛 = 𝐛,

an affine combination of $𝐚, 𝐜 \in 𝐸_{1}$ , so $𝐛 \in 𝐸_{1}$ , a contradiction. If $𝐜 \in 𝐸_{2}$ , then

\frac{1}{1 - 𝑡} 𝐜 - \frac{𝑡}{1 - 𝑡} 𝐛 = 𝐚 + \frac{𝑡}{1 - 𝑡} 𝐛 - \frac{𝑡}{1 - 𝑡} 𝐛 = 𝐚,

an affine combination of $𝐜, 𝐛 \in 𝐸_{2}$ (the coefficients add up to $(1 - 𝑡) / (1 - 𝑡) = 1$ ), so $𝐚 \in 𝐸_{2}$ , a contradiction. Hence, when $| 𝐹 | \geq 3$ , the statement is true. The argument breaks down over $𝔽_{2}$ exactly because the line through $𝐚$ and $𝐛$ has no third point.

Exercise 4.7.52.standard Let

dim 𝑉 = 𝑛 \geq 2

, and let

𝐻

and

𝐻^{'}

be hyperplanes in

𝑉

. Show that either

𝐻

and

𝐻^{'}

are parallel, or

𝐻 \cap 𝐻^{'}

is an affine subspace of dimension

𝑛 - 2

. Deduce that two lines in a plane are parallel or meet in exactly one point.

Solution.

Uses. Proposition 4.7.10 (Intersections of affine subspaces), Theorem 4.4.19 (Dimension of a sum), Proposition 4.1.21 (Dimensions of subspaces), Corollary 4.7.6 (Parallel affine subspaces are equal or disjoint)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Let

𝑈 = dir 𝐻

and

𝑈^{'} = dir 𝐻^{'}

. If

𝑈 \neq 𝑈^{'}

, what is

𝑈 + 𝑈^{'}

Hint 2. To find a common point, write

𝐡^{'} - 𝐡 = 𝐮 + 𝐮^{'}

with

𝐡 \in 𝐻

𝐡^{'} \in 𝐻^{'}

𝐮 \in 𝑈

𝐮^{'} \in 𝑈^{'}

Sketch. If

𝑈 \neq 𝑈^{'}

, then

𝑈^{'} ⊈ 𝑈

(equal dimensions), so

𝑈 + 𝑈^{'}

strictly contains

𝑈

and equals

𝑉

. Writing

𝐡^{'} - 𝐡 = 𝐮 + 𝐮^{'}

gives the common point

𝐡 + 𝐮 = 𝐡^{'} - 𝐮^{'}

. By Proposition 4.7.10 the intersection has direction

𝑈 \cap 𝑈^{'}

, of dimension

2 (𝑛 - 1) - 𝑛 = 𝑛 - 2

Let $𝑈 = dir 𝐻$ and $𝑈^{'} = dir 𝐻^{'}$ , both of dimension $𝑛 - 1$ , and suppose $𝐻$ and $𝐻^{'}$ are not parallel, so $𝑈 \neq 𝑈^{'}$ . If $𝑈^{'} \subseteq 𝑈$ , then $𝑈^{'} = 𝑈$ by Proposition 4.1.21(b), which is not the case. So some vector of $𝑈^{'}$ lies outside $𝑈$ , and $𝑈 ⊊ 𝑈 + 𝑈^{'}$ . Then $𝑛 -$ $1 <$ $dim (𝑈 + 𝑈^{'}) \leq$ $𝑛$ , so $𝑈 + 𝑈^{'} = 𝑉$ by Proposition 4.1.21(b).

Pick $𝐡 \in 𝐻$ and $𝐡^{'} \in 𝐻^{'}$ , and write $𝐡^{'} - 𝐡 = 𝐮 + 𝐮^{'}$ with $𝐮 \in 𝑈$ and $𝐮^{'} \in 𝑈^{'}$ . Then $𝐡 + 𝐮 = 𝐡^{'} - 𝐮^{'}$ lies in $𝐻$ and in $𝐻^{'}$ . By Proposition 4.7.10, $𝐻 \cap 𝐻^{'}$ is an affine subspace with direction $𝑈 \cap 𝑈^{'}$ , and by Theorem 4.4.19,

dim (𝑈 \cap 𝑈^{'}) = dim 𝑈 + dim 𝑈^{'} - dim (𝑈 + 𝑈^{'}) = 2 (𝑛 - 1) - 𝑛 = 𝑛 - 2 .

For $𝑛 = 2$ , the hyperplanes are lines. Two lines in a plane are either parallel, or they meet in an affine subspace of dimension $0$ , a single point.

Exercise 4.7.53.standard Let

𝐮, 𝐰 \in 𝑉

be nonzero, let

𝐩, 𝐪 \in 𝑉

, and let

ℓ_{1} = 𝐩 + span (𝐮)

and

ℓ_{2} = 𝐪 + span (𝐰)

. Show that

ℓ_{1}

and

ℓ_{2}

are disjoint and not parallel if and only if

𝐮, 𝐰, 𝐪 - 𝐩

is linearly independent. Deduce that if

dim 𝑉 = 2

, there are no skew lines in

𝑉

Solution.

Uses. Proposition 1.5.14 (Appending a vector), Theorem 4.1.10 (Independent lists are short), Definition 4.7.5 (Direction, dimension, parallel)

Hint 1. The lines meet if and only if

𝐩 + 𝑎 𝐮 = 𝐪 + 𝑏 𝐰

for some scalars. Rewrite this as a statement about

𝐪 - 𝐩

Hint 2. The lines are parallel if and only if

span (𝐮) = span (𝐰)

, that is, if and only if

𝐮, 𝐰

is dependent. Then use Proposition 1.5.14.

Sketch.

ℓ_{1} \cap ℓ_{2} \neq \emptyset

if and only if

𝐪 - 𝐩 \in span (𝐮, 𝐰)

, and the lines are parallel if and only if

𝐮, 𝐰

is dependent. So “disjoint and not parallel” means “

𝐮, 𝐰

independent and

𝐪 - 𝐩 \notin span (𝐮, 𝐰)

”, which is the independence of

𝐮, 𝐰, 𝐪 - 𝐩

by Proposition 1.5.14. In dimension

2

, three vectors are dependent.

The lines meet if and only if $𝐩 + 𝑎 𝐮 = 𝐪 + 𝑏 𝐰$ for some $𝑎, 𝑏 \in 𝐹$ , that is, if and only if $𝐪 - 𝐩 = 𝑎 𝐮 - 𝑏 𝐰$ for some $𝑎, 𝑏$ , that is, if and only if $𝐪 - 𝐩 \in span (𝐮, 𝐰)$ .

The lines are parallel if and only if $span (𝐮) = span (𝐰)$ . Since $𝐮, 𝐰 \neq 𝟎$ , this holds if and only if $𝐰 = 𝑐 𝐮$ for some $𝑐$ (necessarily nonzero), that is, if and only if $𝐮, 𝐰$ is linearly dependent: a nontrivial relation $𝑎 𝐮 + 𝑏 𝐰 = 𝟎$ must have $𝑎, 𝑏 \neq 0$ , as both vectors are nonzero.

So $ℓ_{1}, ℓ_{2}$ are disjoint and not parallel if and only if $𝐮, 𝐰$ is independent and $𝐪 - 𝐩 \notin span (𝐮, 𝐰)$ . By Proposition 1.5.14, this is equivalent to the independence of $𝐮, 𝐰, 𝐪 - 𝐩$ . (If $𝐮, 𝐰$ is dependent, so is the longer list, and both sides fail.)

If $dim 𝑉 = 2$ , no list of three vectors is independent (Theorem 4.1.10), so two lines are never disjoint and non-parallel.

Exercise 4.7.54.routine Find a system of linear equations whose solution set is

aff ((1, 0, 2, 1), (2, 1, 1, 1), (0, 1, 3, 2))

ℝ^{4}

. How many equations are needed?

Solution.

Uses. Proposition 4.7.17 (The affine hull), Theorem 4.7.8 (Affine subspaces of $𝐹^{𝑛}$ are solution sets), Proposition 4.1.21 (Dimensions of subspaces)

Hint 1. The direction is spanned by

(1, 1, - 1, 0)

and

(- 1, 1, 1, 1)

. Find the dimension, then the number of equations from Theorem 4.7.8(b).

Hint 2. An equation

𝑎_{1} 𝑥_{1} +

\dots +

𝑎_{4} 𝑥_{4} =

𝑐

works for the direction if

(𝑎_{1}, \dots, 𝑎_{4})

kills both direction vectors. Find two independent such rows, then compute the right-hand sides at

(1, 0, 2, 1)

Sketch. The two differences are independent, so the hull is a plane and needs

4 - 2 = 2

equations. The rows

(1, 0, 1, 0)

and

(1, - 1, 0, 2)

kill both differences, and the system

𝑥_{1} + 𝑥_{3} = 3

𝑥_{1} - 𝑥_{2} + 2 𝑥_{4} = 3

works.

By Proposition 4.7.17(b), the hull is $(1, 0, 2, 1) +$ $span (𝐝_{1}, 𝐝_{2})$ with $𝐝_{1} = (1, 1, - 1, 0)$ and $𝐝_{2} = (- 1, 1, 1, 1)$ . These are independent (the last entries force the coefficient of $𝐝_{2}$ to be $0$ , and then that of $𝐝_{1}$ ), so the hull is a plane, and by Theorem 4.7.8(b) it takes $4 - 2 = 2$ equations, and no fewer.

A row $𝐚 =$ $(𝑎_{1}, 𝑎_{2}, 𝑎_{3}, 𝑎_{4})$ with $𝐴 𝐝_{1} = 𝐴 𝐝_{2} = 0$ satisfies $𝑎_{1} + 𝑎_{2} - 𝑎_{3} = 0$ and $- 𝑎_{1} +$ $𝑎_{2} +$ $𝑎_{3} +$ $𝑎_{4} =$ $0$ . Taking $(𝑎_{3}, 𝑎_{4}) = (1, 0)$ gives $(1, 0, 1, 0)$ , and $(𝑎_{3}, 𝑎_{4}) = (0, 2)$ gives $(1, - 1, 0, 2)$ . With $𝐴 = (\begin{matrix} 1 & 0 & 1 & 0 \\ 1 & - 1 & 0 & 2 \end{matrix})$ , of rank $2$ , and $𝐴 (1, 0, 2, 1) = (3, 3)$ , the system is

𝑥_{1} + 𝑥_{3} = 3, 𝑥_{1} - 𝑥_{2} + 2 𝑥_{4} = 3 .

Its solution set is an affine subspace containing $(1, 0, 2, 1)$ with direction $ker 𝑇_{𝐴}$ of dimension $4 - 2 = 2$ , which contains $𝐝_{1}, 𝐝_{2}$ ; so the direction is $span (𝐝_{1}, 𝐝_{2})$ by Proposition 4.1.21(b), and the solution set is the hull. Check the other points: $2 + 1 = 3$ , $2 - 1 + 2 = 3$ , and $0 + 3 = 3$ , $0 - 1 + 4 = 3$ .

Exercise 4.7.55.hard Let $𝑓 : ℚ^{𝑛} \to ℚ^{𝑚}$ satisfy

𝑓 (\frac{1}{2} 𝐮 + \frac{1}{2} 𝐯) = \frac{1}{2} 𝑓 (𝐮) + \frac{1}{2} 𝑓 (𝐯) for all 𝐮, 𝐯 \in ℚ^{𝑛} .

Show that $𝑓$ is affine. (Over $ℝ$ the same hypothesis does not imply that $𝑓$ is affine; the counterexamples need a basis of $ℝ$ over $ℚ$ , see Section 15.1.)

Solution.

Uses. Theorem 2.1.3 (Checking linearity), Definition 4.7.35 (Affine map)

Hint 1. Replace

𝑓

𝑔 = 𝑓 - 𝑓 (𝟎)

. It satisfies the same identity and

𝑔 (𝟎) = 𝟎

Hint 2. Take

𝐯 = 𝟎

to get

𝑔 (\frac{1}{2} 𝐮) = \frac{1}{2} 𝑔 (𝐮)

. Then

𝑔 (𝐮 + 𝐯) =

2 𝑔 (\frac{1}{2} (𝐮 + 𝐯))

Hint 3. Additivity gives

𝑔 (𝑘 𝐮) = 𝑘 𝑔 (𝐮)

for integers

𝑘

. For

𝑟 = 𝑘 / 𝑚

, compute

𝑚 𝑔 (𝑟 𝐮)

Sketch.

𝑔 = 𝑓 - 𝑓 (𝟎)

satisfies the midpoint identity and

𝑔 (𝟎) = 𝟎

, so

𝑔 (𝐮 / 2) = 𝑔 (𝐮) / 2

and

𝑔 (𝐮 + 𝐯) =

2 𝑔 ((𝐮 + 𝐯) / 2) =

𝑔 (𝐮) +

𝑔 (𝐯)

. By induction

𝑔 (𝑘 𝐮) = 𝑘 𝑔 (𝐮)

for

𝑘 \geq 0

, also for

𝑘 < 0

since

𝑔 (- 𝐮) = - 𝑔 (𝐮)

, and

𝑚 𝑔 ((𝑘 / 𝑚) 𝐮) =

𝑔 (𝑘 𝐮) =

𝑘 𝑔 (𝐮)

. So

𝑔

is linear over

ℚ

and

𝑓 = 𝑔 + 𝑓 (𝟎)

Let $𝑔 (𝐱) =$ $𝑓 (𝐱) -$ $𝑓 (𝟎)$ . Then $𝑔 (𝟎) = 𝟎$ , and subtracting $𝑓 (𝟎) =$ $\frac{1}{2} 𝑓 (𝟎) +$ $\frac{1}{2} 𝑓 (𝟎)$ from the hypothesis gives

𝑔 (\frac{1}{2} 𝐮 + \frac{1}{2} 𝐯) = \frac{1}{2} 𝑔 (𝐮) + \frac{1}{2} 𝑔 (𝐯) for all 𝐮, 𝐯 .

With $𝐯 = 𝟎$ this says $𝑔 (\frac{1}{2} 𝐮) = \frac{1}{2} 𝑔 (𝐮)$ . Applying that to $𝐮 + 𝐯$ ,

𝑔 (𝐮 + 𝐯) = 2 𝑔 (\frac{1}{2} (𝐮 + 𝐯)) = 2 (\frac{1}{2} 𝑔 (𝐮) + \frac{1}{2} 𝑔 (𝐯)) = 𝑔 (𝐮) + 𝑔 (𝐯) .

So $𝑔$ is additive. By induction on $𝑘$ , $𝑔 (𝑘 𝐮) = 𝑘 𝑔 (𝐮)$ for integers $𝑘 \geq 0$ (the case $𝑘 = 0$ is $𝑔 (𝟎) = 𝟎$ , and $𝑔 ((𝑘 + 1) 𝐮) =$ $𝑔 (𝑘 𝐮) +$ $𝑔 (𝐮)$ ). Since $𝑔 (𝐮) +$ $𝑔 (- 𝐮) =$ $𝑔 (𝟎) =$ $𝟎$ , the same holds for negative $𝑘$ . For $𝑟 = 𝑘 / 𝑚$ with $𝑘, 𝑚 \in ℤ$ and $𝑚 > 0$ ,

𝑚 𝑔 (𝑟 𝐮) = 𝑔 (𝑚 𝑟 𝐮) = 𝑔 (𝑘 𝐮) = 𝑘 𝑔 (𝐮),

so $𝑔 (𝑟 𝐮) = 𝑟 𝑔 (𝐮)$ . Hence $𝑔$ is additive and homogeneous over $ℚ$ , so it is linear by Theorem 2.1.3, and $𝑓 = 𝑔 + 𝑓 (𝟎)$ is affine.

Exercise 4.7.56.routine Find the error in the following argument, and decide which half of the claim is true.

Claim. A list $𝐯_{0}, \dots, 𝐯_{𝑘}$ in $𝑉$ is affinely independent if and only if it is linearly independent.

Proof. An affine dependence is a linear relation, so if the list is linearly independent, it is affinely independent. Conversely, let $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎$ be a linear relation, and put $𝑠 = \sum_{𝑖} 𝑐_{𝑖}$ . Replacing $𝑐_{0}$ by $𝑐_{0} - 𝑠$ gives coefficients that add up to $0$ , so an affinely independent list has $𝑐_{0} -$ $𝑠 =$ $𝑐_{1} =$ $\dots =$ $𝑐_{𝑘} =$ $0$ . Then $𝑠 = 𝑐_{0}$ , so $𝑐_{0} =$ $𝑐_{0} +$ $𝑐_{1} +$ $\dots +$ $𝑐_{𝑘}$ is consistent only if all $𝑐_{𝑖} = 0$ .

Solution.

Uses. Definition 4.7.18 (Affine independence), Example 4.7.19 (Affine independence)

Hint 1. Changing a coefficient changes the vector that the combination represents. Is the new combination still

𝟎

Hint 2. Test the claim on a list containing

𝟎

Sketch. Replacing

𝑐_{0}

𝑐_{0} - 𝑠

changes the combination to

- 𝑠 𝐯_{0}

, which need not be

𝟎

, so the new coefficients need not form an affine dependence. The forward half is true. The converse is false:

𝟎, 𝐞_{1}

is affinely independent but linearly dependent.

The first half is correct: an affine dependence is a nontrivial linear relation, so a linearly independent list has none.

The error is in the converse. After $𝑐_{0}$ is replaced by $𝑐_{0} - 𝑠$ , the combination represents

(𝑐_{0} - 𝑠) 𝐯_{0} + 𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘} = 𝟎 - 𝑠 𝐯_{0} = - 𝑠 𝐯_{0},

which is not $𝟎$ unless $𝑠 = 0$ or $𝐯_{0} = 𝟎$ . So the new coefficients are not an affine dependence, and affine independence says nothing about them. (The last sentence of the argument is also confused, but the claim fails before it.)

The converse is false. The list $𝟎, 𝐞_{1}$ in $ℝ^{2}$ consists of two distinct points, so it is affinely independent (Example 4.7.19(a)), but it contains $𝟎$ , so it is linearly dependent. Likewise $(0, 0), (1, 0), (0, 1)$ is affinely independent and linearly dependent (Example 4.7.19(b)).

Exercise 4.7.57.hard Let $𝐯_{0}, 𝐯_{1}, 𝐯_{2}$ be affinely independent in $𝑉$ . When $1 + 1 \neq 0$ , the median from $𝐯_{𝑖}$ is the line through $𝐯_{𝑖}$ and the midpoint $𝐦_{𝑖}$ of the other two vertices.

(a)Suppose $2 \neq 0$ and $3 \neq 0$ in $𝐹$ . Show that the three medians are distinct lines that all pass through the centroid $𝐠 =$ $\frac{1}{3} (𝐯_{0} + 𝐯_{1} + 𝐯_{2})$ , and that $𝐠$ is their only common point.
(b)Now let $𝐹 = 𝔽_{3}$ and $𝑉 = 𝔽_{3}^{2}$ . Show that the three medians are parallel and pairwise disjoint, and that together they contain every point of $𝔽_{3}^{2}$ .

Solution.

Uses. Theorem 4.7.22 (Characterizations of affine independence), Proposition 4.7.17 (The affine hull), Corollary 4.7.6 (Parallel affine subspaces are equal or disjoint), Proposition 4.7.3 (Direction and base points), Proposition 4.1.34 (Size of a vector space over a finite field)

Hint 1. For (a), write

𝐠

as an affine combination of

𝐯_{𝑖}

and

𝐦_{𝑖}

. To see that two medians differ, show that if the median from

𝐯_{0}

contained

𝐯_{1}

, then

𝐯_{2}

would lie in

aff (𝐯_{0}, 𝐯_{1})

Hint 2. For (b), in

𝔽_{3}

we have

1 / 2 = 2 = - 1

, so

𝐦_{𝑖} = - (𝐯_{𝑗} + 𝐯_{𝑘})

. Compute

𝐦_{𝑖} - 𝐯_{𝑖}

Hint 3. Let

𝐬 = 𝐯_{0} + 𝐯_{1} + 𝐯_{2}

. Show

𝐬 \neq 𝟎

, and that

𝐯_{1} - 𝐯_{0} \in span (𝐬)

would give an affine dependence, using

3 = 0

Sketch. (a)

𝐠 = \frac{1}{3} 𝐯_{𝑖} + \frac{2}{3} 𝐦_{𝑖}

lies on every median. If two medians were equal, the median from

𝐯_{0}

would contain

𝐯_{1}

, so

𝐦_{0} \in aff (𝐯_{0}, 𝐯_{1})

and

𝐯_{2} = 2 𝐦_{0} - 𝐯_{1}

would be in it too. Distinct lines share at most one point. (b)

𝐦_{𝑖} - 𝐯_{𝑖} = - 𝐬

for every

𝑖

, and

𝐬 \neq 𝟎

, so each median is

𝐯_{𝑖} + span (𝐬)

. If

𝐯_{𝑗} - 𝐯_{𝑖} = 𝑐 𝐬

, the coefficients of the resulting relation add up to

- 3 𝑐 = 0

, giving an affine dependence. So the medians are three distinct parallel lines, disjoint by Corollary 4.7.6, with

3

points each, covering all

9

points.

(a) Write $𝐦_{0} = \frac{1}{2} (𝐯_{1} + 𝐯_{2})$ , and similarly $𝐦_{1}, 𝐦_{2}$ . Then

\frac{1}{3} 𝐯_{0} + \frac{2}{3} 𝐦_{0} = \frac{1}{3} (𝐯_{0} + 𝐯_{1} + 𝐯_{2}) = 𝐠,

an affine combination of $𝐯_{0}$ and $𝐦_{0}$ , so $𝐠$ lies on the median from $𝐯_{0}$ (Proposition 4.7.17), and likewise on the other two. The median from $𝐯_{0}$ is a line, since $𝐦_{0} \neq 𝐯_{0}$ : otherwise $𝐯_{0}$ would be an affine combination of $𝐯_{1}, 𝐯_{2}$ , contradicting Theorem 4.7.22(V).

Suppose two medians coincide, say those from $𝐯_{0}$ and $𝐯_{1}$ (the other cases are the same up to renaming). Then $𝐯_{1}$ lies on the median from $𝐯_{0}$ , so this median is $aff (𝐯_{0}, 𝐯_{1})$ , and it contains $𝐦_{0}$ . Then $𝐯_{2} = 2 𝐦_{0} - 𝐯_{1}$ is an affine combination of points of $aff (𝐯_{0}, 𝐯_{1})$ , so it lies there, contradicting Theorem 4.7.22(V). So the medians are three distinct lines. Two distinct lines meet in at most one point: their intersection is empty or an affine subspace of dimension at most $1$ (Proposition 4.7.10), and if it had dimension $1$ it would equal both lines (Proposition 4.1.21(b) applied to the directions, and Proposition 4.7.3). So $𝐠$ is the only common point.

(b) In $𝔽_{3}$ , $2 \cdot 2 = 4 = 1$ , so $1 / 2 = 2 = - 1$ , and $𝐦_{𝑖} = - (𝐯_{𝑗} + 𝐯_{𝑘})$ , where ${𝑖, 𝑗, 𝑘} =$ ${0, 1, 2}$ . Let $𝐬 = 𝐯_{0} + 𝐯_{1} + 𝐯_{2}$ . Then $𝐦_{𝑖} - 𝐯_{𝑖} = - 𝐬$ for every $𝑖$ . If $𝐬 = 𝟎$ , then $𝐯_{2} =$ $- 𝐯_{0} -$ $𝐯_{1} =$ $2 𝐯_{0} +$ $2 𝐯_{1}$ with $2 + 2 = 4 = 1$ , an affine combination of $𝐯_{0}, 𝐯_{1}$ , contradicting affine independence. So $𝐬 \neq 𝟎$ , and the median from $𝐯_{𝑖}$ is $𝐯_{𝑖} + span (𝐬)$ . All three medians have direction $span (𝐬)$ , so they are parallel.

Suppose two of them coincide, say $𝐯_{1} - 𝐯_{0} = 𝑐 𝐬$ for some $𝑐 \in 𝔽_{3}$ (by Proposition 4.7.3(c)). Then

(- 1 - 𝑐) 𝐯_{0} + (1 - 𝑐) 𝐯_{1} - 𝑐 𝐯_{2} = 𝟎,

with coefficient sum $- 3 𝑐 = 0$ , and the coefficients are not all $0$ (that would need $𝑐 = - 1$ , $𝑐 = 1$ , and $𝑐 = 0$ ). This is an affine dependence, a contradiction. So the three medians are distinct parallel lines, hence pairwise disjoint (Corollary 4.7.6). Each has $| span (𝐬) | = 3$ points, so together they have $9 = | 𝔽_{3}^{2} |$ points (Proposition 4.1.34), and they cover $𝔽_{3}^{2}$ .

Exercise 4.7.58.standard

(a)Find the affine map $𝑓 : ℝ^{2} \to ℝ^{2}$ with $𝑓 (0, 0) = (2, 1)$ , $𝑓 (1, 0) = (4, 2)$ , $𝑓 (0, 1) = (1, 3)$ , and its matrix $𝑀_{𝑓}$ .
(b)Find the affine map $𝑔 : ℝ^{2} \to ℝ^{2}$ with $𝑔 (1, 1) = (2, 1)$ , $𝑔 (3, 1) = (4, 2)$ , $𝑔 (1, 2) = (1, 3)$ .
(c)Let $𝐚_{0}, 𝐚_{1}, 𝐚_{2}$ be affinely independent in $ℝ^{2}$ , let $𝐛_{0}, 𝐛_{1}, 𝐛_{2} \in ℝ^{2}$ , and let $\hat{𝐴}, \hat{𝐵} \in ℝ^{3 \times 3}$ have columns ${\hat{𝐚}}_{𝑖}$ and ${\hat{𝐛}}_{𝑖}$ . Show that $\hat{𝐴}$ is invertible and that the affine map with $𝐚_{𝑖} \mapsto 𝐛_{𝑖}$ has matrix $\hat{𝐵} {\hat{𝐴}}^{- 1}$ . Check (b) this way.

Solution.

Uses. Theorem 4.7.39 (Affine maps are determined by a frame), Theorem 4.7.42 (Affine maps as matrices), Proposition 4.7.20 (Affine independence through differences), Corollary 3.5.25 (The invertible matrix theorem), Proposition 2.4.8 (Four views of the product)

Techniques. Homogenize: add a coordinate $1$ (Technique 4.7.21)

Hint 1. In (a),

𝑓 (𝟎)

is the translation part, and

𝑓 (𝐞_{𝑗}) - 𝑓 (𝟎)

is column

𝑗

of the linear part.

Hint 2. In (b), compose: if

ℎ

sends

𝟎, 𝐞_{1}, 𝐞_{2}

(1, 1), (3, 1), (1, 2)

, then

𝑔 = 𝑓 \circ ℎ^{- 1}

Hint 3. In (c),

𝑀 {\hat{𝐚}}_{𝑖} = {\hat{𝐛}}_{𝑖}

for all

𝑖

says

𝑀 \hat{𝐴} = \hat{𝐵}

. The columns of

\hat{𝐴}

are independent by the second approach to Proposition 4.7.20.

Sketch. (a)

𝑓 (𝐱) =

(\begin{matrix} 2 & - 1 \\ 1 & 2 \end{matrix}) 𝐱 +

(2, 1)

. (b)

ℎ (𝐱) =

(\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}) 𝐱 +

(1, 1)

, and

𝑔 = 𝑓 \circ ℎ^{- 1}

𝑔 (𝐲) =

(\begin{matrix} 1 & - 1 \\ \frac{1}{2} & 2 \end{matrix}) 𝐲 +

(2, - \frac{3}{2})

. (c)

𝑀_{𝑔} \hat{𝐴} = \hat{𝐵}

column by column, and

\hat{𝐴}

has independent columns, so it is invertible.

(a) By Theorem 4.7.39 there is exactly one such map. Its translation part is $𝑓 (𝟎) = (2, 1)$ , and its linear part $𝐿$ has $𝐿 (𝐞_{1}) =$ $(4, 2) -$ $(2, 1) =$ $(2, 1)$ and $𝐿 (𝐞_{2}) =$ $(1, 3) -$ $(2, 1) =$ $(- 1, 2)$ . So

𝑓 (𝐱) = (\begin{matrix} 2 & - 1 \\ 1 & 2 \end{matrix}) 𝐱 + (\begin{matrix} 2 \\ 1 \end{matrix}), 𝑀_{𝑓} = (\begin{matrix} 2 & - 1 & 2 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}) .

(b) The affine map $ℎ (𝐱) =$ $(\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}) 𝐱 +$ $(1, 1)$ sends $𝟎, 𝐞_{1}, 𝐞_{2}$ to $(1, 1), (3, 1), (1, 2)$ , and it is bijective with $ℎ^{- 1} (𝐲) =$ $(\begin{matrix} \frac{1}{2} & 0 \\ 0 & 1 \end{matrix}) (𝐲 - (1, 1))$ . The map $𝑔 = 𝑓 \circ ℎ^{- 1}$ is affine and sends $(1, 1), (3, 1), (1, 2)$ to $(2, 1), (4, 2), (1, 3)$ , so it is the unique such map (Theorem 4.7.39). Explicitly,

𝑔 (𝐲) = (\begin{matrix} 2 & - 1 \\ 1 & 2 \end{matrix}) (\begin{matrix} \frac{1}{2} & 0 \\ 0 & 1 \end{matrix}) (𝐲 - (\begin{matrix} 1 \\ 1 \end{matrix})) + (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 1 & - 1 \\ \frac{1}{2} & 2 \end{matrix}) 𝐲 + (\begin{matrix} 2 \\ - \frac{3}{2} \end{matrix}) .

Check: $𝑔 (1, 1) =$ $(0, \frac{5}{2}) +$ $(2, - \frac{3}{2}) =$ $(2, 1)$ , $𝑔 (3, 1) =$ $(2, \frac{7}{2}) +$ $(2, - \frac{3}{2}) =$ $(4, 2)$ , $𝑔 (1, 2) =$ $(- 1, \frac{9}{2}) +$ $(2, - \frac{3}{2}) =$ $(1, 3)$ .

(c) The columns of $\hat{𝐴}$ are the lifts ${\hat{𝐚}}_{𝑖}$ , which are linearly independent by the second approach to Proposition 4.7.20. So $\hat{𝐴}$ is invertible (Corollary 3.5.25). Let $𝑔$ be the affine map with $𝑔 (𝐚_{𝑖}) = 𝐛_{𝑖}$ (Theorem 4.7.39). By Theorem 4.7.42(a), $𝑀_{𝑔} {\hat{𝐚}}_{𝑖} = {\hat{𝐛}}_{𝑖}$ for each $𝑖$ , which says that $𝑀_{𝑔} \hat{𝐴} = \hat{𝐵}$ column by column (Proposition 2.4.8). Hence $𝑀_{𝑔} = \hat{𝐵} {\hat{𝐴}}^{- 1}$ . For (b),

\hat{𝐴} = (\begin{matrix} 1 & 3 & 1 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{matrix}), {\hat{𝐴}}^{- 1} = (\begin{matrix} - \frac{1}{2} & - 1 & \frac{5}{2} \\ \frac{1}{2} & 0 & - \frac{1}{2} \\ 0 & 1 & - 1 \end{matrix}), \hat{𝐵} {\hat{𝐴}}^{- 1} = (\begin{matrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{matrix}) {\hat{𝐴}}^{- 1} = (\begin{matrix} 1 & - 1 & 2 \\ \frac{1}{2} & 2 & - \frac{3}{2} \\ 0 & 0 & 1 \end{matrix}),

in agreement with (b).

Exercise 4.7.59.hard (Radon’s theorem.) Let $𝐯_{1}, \dots, 𝐯_{𝑛 + 2} \in ℝ^{𝑛}$ . Show that the index set ${1, \dots, 𝑛 + 2}$ can be split into two disjoint nonempty parts $𝐼$ and $𝐽$ with

conv {𝐯_{𝑖} | 𝑖 \in 𝐼} \cap conv {𝐯_{𝑗} | 𝑗 \in 𝐽} \neq \emptyset .

Find such a split for the four points $(0, 0),$ $(2, 0),$ $(0, 2),$ $(1, 1)$ of $ℝ^{2}$ , and for $(0, 0),$ $(4, 0),$ $(0, 4),$ $(1, 1)$ .

Solution.

Uses. Corollary 4.7.23 (Affinely independent lists are short), Definition 4.7.28 (Convex combination, convex set, convex hull)

Techniques. Slide along a dependence (Technique 4.7.33)

Hint 1. The

𝑛 + 2

points are affinely dependent. Take an affine dependence

\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎

\sum_{𝑖} 𝑑_{𝑖} = 0

Hint 2. Split the indices by the sign of

𝑑_{𝑖}

, and move the negative terms to the other side.

Sketch. Let

\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎

\sum_{𝑖} 𝑑_{𝑖} = 0

, not all

𝑑_{𝑖} = 0

. Put

𝐼 = {𝑖 | 𝑑_{𝑖} > 0}

and

𝐽

the rest; both are nonempty, and

𝑠 =

\sum_{𝑖 \in 𝐼} 𝑑_{𝑖} =

- \sum_{𝑗 \in 𝐽} 𝑑_{𝑗} >

0

. Then

\sum_{𝑖 \in 𝐼} (𝑑_{𝑖} / 𝑠) 𝐯_{𝑖} =

\sum_{𝑗 \in 𝐽} (- 𝑑_{𝑗} / 𝑠) 𝐯_{𝑗}

is a common point. In the examples,

(1, 1)

is the midpoint of

(2, 0)

and

(0, 2)

, and lies inside the triangle

(0, 0), (4, 0), (0, 4)

By Corollary 4.7.23, the $𝑛 + 2$ points are affinely dependent: there are $𝑑_{1}, \dots, 𝑑_{𝑛 + 2}$ , not all $0$ , with $\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖} = 𝟎$ and $\sum_{𝑖} 𝑑_{𝑖} = 0$ . Let $𝐼 = {𝑖 | 𝑑_{𝑖} > 0}$ and $𝐽 = {𝑗 | 𝑑_{𝑗} \leq 0}$ . Since the $𝑑_{𝑖}$ add up to $0$ and are not all $0$ , some $𝑑_{𝑖}$ is positive and some is negative, so $𝐼$ and $𝐽$ are nonempty. Let $𝑠 = \sum_{𝑖 \in 𝐼} 𝑑_{𝑖} > 0$ ; then $\sum_{𝑗 \in 𝐽} 𝑑_{𝑗} = - 𝑠$ . Moving the terms with $𝑗 \in 𝐽$ to the other side of the dependence and dividing by $𝑠$ ,

𝐩 = \sum_{𝑖 \in 𝐼} \frac{𝑑_{𝑖}}{𝑠} 𝐯_{𝑖} = \sum_{𝑗 \in 𝐽} \frac{- 𝑑_{𝑗}}{𝑠} 𝐯_{𝑗} .

On the left, the coefficients are positive and add up to $1$ ; on the right, they are nonnegative and add up to $(- (- 𝑠)) / 𝑠 = 1$ . So $𝐩$ lies in both convex hulls.

For $(0, 0),$ $(2, 0),$ $(0, 2),$ $(1, 1)$ : the dependence $(2, 0) +$ $(0, 2) -$ $2 (1, 1) =$ $𝟎$ (coefficients $0, 1, 1, - 2$ , sum $0$ ) gives $𝐼 = {2, 3}$ and $𝐽 = {1, 4}$ , with common point $(1, 1) =$ $\frac{1}{2} (2, 0) +$ $\frac{1}{2} (0, 2)$ , which is also the endpoint $(1, 1)$ of the segment $conv {(0, 0), (1, 1)}$ . For $(0, 0),$ $(4, 0),$ $(0, 4),$ $(1, 1)$ : $2 (0, 0) +$ $(4, 0) +$ $(0, 4) -$ $4 (1, 1) =$ $𝟎$ , with coefficients $2, 1, 1, - 4$ adding up to $0$ , gives $𝐼 = {1, 2, 3}$ and $𝐽 = {4}$ : the point $(1, 1) =$ $\frac{1}{2} (0, 0) +$ $\frac{1}{4} (4, 0) +$ $\frac{1}{4} (0, 4)$ lies in the triangle.

Exercise 4.7.60.standard A function $ℎ : ℝ^{𝑛} \to ℝ$ of the form $ℎ (𝐱) =$ $𝑤_{1} 𝑥_{1} +$ $\dots +$ $𝑤_{𝑛} 𝑥_{𝑛} +$ $𝑏$ is an affine map, and the rule “label $𝐱$ by the sign of $ℎ (𝐱)$ ” is a linear classifier, a single neuron of a neural network (Section 16.6).

(a)Show that no affine $ℎ : ℝ^{2} \to ℝ$ has $ℎ (0, 0) > 0$ , $ℎ (1, 1) > 0$ , $ℎ (1, 0) < 0$ , and $ℎ (0, 1) < 0$ . (The labels of these four points form the “exclusive or” pattern.)
(b)Show that after adding the feature $𝑥_{1} 𝑥_{2}$ , that is, after replacing each point $𝐱$ by $𝜑 (𝐱) =$ $(𝑥_{1}, 𝑥_{2}, 𝑥_{1} 𝑥_{2}) \in$ $ℝ^{3}$ , there is an affine $ℎ^{'} : ℝ^{3} \to ℝ$ with $ℎ^{'} \circ 𝜑$ positive at $(0, 0), (1, 1)$ and negative at $(1, 0), (0, 1)$ .

Solution.

Uses. Theorem 4.7.37 (Affine maps preserve affine combinations)

Hint 1. The point

(\frac{1}{2}, \frac{1}{2})

is the midpoint of

(0, 0)

and

(1, 1)

, and also of

(1, 0)

and

(0, 1)

Hint 2. For (b), try

ℎ^{'} (𝐲) =

𝑐 -

𝑦_{1} -

𝑦_{2} +

2 𝑦_{3}

and choose

𝑐

Sketch. (a) Affine maps preserve midpoints, so

ℎ (\frac{1}{2}, \frac{1}{2})

is the average of two positive numbers and also of two negative numbers. (b)

ℎ^{'} (𝐲) =

\frac{1}{2} -

𝑦_{1} -

𝑦_{2} +

2 𝑦_{3}

gives

\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

(a) Suppose such an affine $ℎ$ exists. The point $𝐦 = (\frac{1}{2}, \frac{1}{2})$ is the affine combination $\frac{1}{2} (0, 0) + \frac{1}{2} (1, 1)$ , and also $\frac{1}{2} (1, 0) + \frac{1}{2} (0, 1)$ . By Theorem 4.7.37(II),

ℎ (𝐦) = \frac{1}{2} ℎ (0, 0) + \frac{1}{2} ℎ (1, 1) > 0 and ℎ (𝐦) = \frac{1}{2} ℎ (1, 0) + \frac{1}{2} ℎ (0, 1) < 0,

a contradiction. In words, the segment between the two positive points crosses the segment between the two negative points, and at the crossing point an affine function would have to be both positive and negative.

(b) The four points become $𝜑 (0, 0) = (0, 0, 0)$ , $𝜑 (1, 1) = (1, 1, 1)$ , $𝜑 (1, 0) = (1, 0, 0)$ , and $𝜑 (0, 1) = (0, 1, 0)$ . Let $ℎ^{'} (𝐲) =$ $\frac{1}{2} -$ $𝑦_{1} -$ $𝑦_{2} +$ $2 𝑦_{3}$ . Then $ℎ^{'}$ takes the values $\frac{1}{2}$ , $\frac{1}{2} - 2 + 2 = \frac{1}{2}$ , $- \frac{1}{2}$ , and $- \frac{1}{2}$ at these points, as required. The midpoint argument of (a) no longer applies, because $𝜑$ is not affine: the midpoint of $𝜑 (0, 0)$ and $𝜑 (1, 1)$ is $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ , while that of $𝜑 (1, 0)$ and $𝜑 (0, 1)$ is $(\frac{1}{2}, \frac{1}{2}, 0)$ .

Exercise 4.7.61.standard Show that a nonempty subset

𝑆 \subseteq 𝔽_{2}^{𝑛}

is an affine subspace if and only if

𝐮 + 𝐯 + 𝐰 \in 𝑆

for all

𝐮, 𝐯, 𝐰 \in 𝑆

. Use this to decide whether

{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)}

and

{(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)}

are affine subspaces of

𝔽_{2}^{3}

Solution.

Uses. Theorem 4.7.14 (Affine subspaces are the sets closed under affine combinations)

Techniques. Induction on the length of a list (Technique 1.6.9)

Hint 1. Over

𝔽_{2}

, an affine combination of

𝐬_{1}, \dots, 𝐬_{𝑘}

is the sum of the

𝐬_{𝑖}

whose coefficient is

1

, and there is an odd number of them.

Hint 2. Show by induction that the sum of any odd number of elements of

𝑆

lies in

𝑆

: replace the first three summands by their sum.

Sketch. (

\Rightarrow

)

𝐮 + 𝐯 + 𝐰

is an affine combination, as

1 + 1 + 1 = 1

𝔽_{2}

. (

\Leftarrow

) An affine combination over

𝔽_{2}

is a sum of an odd number

2 𝑗 + 1

of elements of

𝑆

; replacing three of them by their sum, which is in

𝑆

, lowers

𝑗

, so by induction the sum lies in

𝑆

. Then Theorem 4.7.14 applies. The first set is the subspace

𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0

; the second is the plane

𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 1

, and both pass the test.

( $\Rightarrow$ ) If $𝑆$ is affine, then $𝐮 +$ $𝐯 +$ $𝐰 =$ $1 𝐮 +$ $1 𝐯 +$ $1 𝐰$ is an affine combination, since $1 + 1 + 1 = 1$ in $𝔽_{2}$ , so it lies in $𝑆$ by Theorem 4.7.14.

( $\Leftarrow$ ) Every coefficient in $𝔽_{2}$ is $0$ or $1$ , so an affine combination $\sum_{𝑖} 𝑐_{𝑖} 𝐬_{𝑖}$ of elements of $𝑆$ is the sum of those $𝐬_{𝑖}$ with $𝑐_{𝑖} = 1$ , and their number is odd, because the coefficients add up to $1$ in $𝔽_{2}$ . We show by induction on $𝑗 \geq 0$ that a sum of $2 𝑗 + 1$ elements of $𝑆$ (repetitions allowed) lies in $𝑆$ . For $𝑗 = 0$ it is a single element of $𝑆$ . For $𝑗 \geq 1$ , the sum $𝐬_{1} +$ $𝐬_{2} +$ $𝐬_{3} +$ $𝐬_{4} +$ $\dots +$ $𝐬_{2 𝑗 + 1}$ equals $𝐬^{'} + 𝐬_{4} + \dots + 𝐬_{2 𝑗 + 1}$ with $𝐬^{'} =$ $𝐬_{1} +$ $𝐬_{2} +$ $𝐬_{3} \in$ $𝑆$ by hypothesis, a sum of $2 (𝑗 - 1) + 1$ elements of $𝑆$ , which lies in $𝑆$ by induction. So $𝑆$ is closed under affine combinations, and it is affine by Theorem 4.7.14.

The first set consists of the vectors with $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ , and each of its triples sums to a vector with the same property, so it passes the test; it is in fact a subspace. The second consists of the vectors with $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 1$ , and the sum of three such vectors has coordinate sum $1 + 1 + 1 = 1$ , so it passes too: it is the affine plane $(1, 0, 0) +$ ${𝐱 | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0}$ . (It is the plane of Exercise 4.7.24(c).)

Exercise 4.7.62.standard

(a)Using $3 \times 3$ matrices, compute the composite of the rotation by $90 °$ counterclockwise about $(0, 0)$ followed by the rotation by $90 °$ counterclockwise about $(1, 0)$ . Show that it is a rotation by $180 °$ , and find its center.
(b)More generally, let $𝑓 (𝐯) = 𝑅_{𝜃} 𝐯 + 𝐛$ be an affine map of $ℝ^{2}$ whose linear part is the rotation by an angle $𝜃$ that is not a multiple of $2 𝜋$ . Show that $𝑓$ has exactly one fixed point $𝐜$ , and that $𝑓$ is the rotation by $𝜃$ about $𝐜$ .

Solution.

Uses. Theorem 4.7.42 (Affine maps as matrices), Example 4.7.36 (Affine maps), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix), Proposition 3.1.19 (Solution sets are translates of the kernel)

Hint 1. The rotation about

𝐚

has matrix

(\begin{matrix} 𝑅 & 𝐚 - 𝑅 𝐚 \\ 0 & 1 \end{matrix})

. Multiply in the right order.

Hint 2. A fixed point solves

(𝐼_{2} - 𝑅_{𝜃}) 𝐜 = 𝐛

. Compute

𝑎 𝑑 - 𝑏 𝑐

for

𝐼_{2} - 𝑅_{𝜃}

Hint 3. If

𝐜 = 𝑅_{𝜃} 𝐜 + 𝐛

, then

𝐛 = 𝐜 - 𝑅_{𝜃} 𝐜

; substitute into

𝑓

Sketch. (a)

(\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) =

(\begin{matrix} - 1 & 0 & 1 \\ 0 & - 1 & - 1 \\ 0 & 0 & 1 \end{matrix})

, the map

𝐯 \mapsto - 𝐯 + (1, - 1)

, with fixed point

(\frac{1}{2}, - \frac{1}{2})

; it is

𝐯 \mapsto 𝐜 - (𝐯 - 𝐜)

. (b)

𝐼_{2} - 𝑅_{𝜃}

has

𝑎 𝑑 -

𝑏 𝑐 =

{(1 - cos 𝜃)}^{2} +

{sin}^{2} 𝜃 =

2 -

2 cos 𝜃 \neq

0

, so the fixed point exists and is unique, and

𝑓 (𝐯) =

𝑅_{𝜃} 𝐯 +

𝐜 -

𝑅_{𝜃} 𝐜 =

𝐜 +

𝑅_{𝜃} (𝐯 - 𝐜)

(a) Let $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ . The rotation about $(0, 0)$ has matrix $𝑀_{1} = (\begin{matrix} 𝑅 & 𝟎 \\ 0 & 1 \end{matrix})$ . The rotation about $𝐚 = (1, 0)$ is $𝐯 \mapsto 𝑅 𝐯 + (𝐚 - 𝑅 𝐚)$ (Example 4.7.36), with $𝐚 -$ $𝑅 𝐚 =$ $(1, 0) -$ $(0, 1) =$ $(1, - 1)$ . The composite applies $𝑀_{1}$ first, so by Theorem 4.7.42(c) its matrix is

(\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} - 1 & 0 & 1 \\ 0 & - 1 & - 1 \\ 0 & 0 & 1 \end{matrix}),

the map $𝑔 (𝐯) = - 𝐯 + (1, - 1)$ . Its fixed points solve $2 𝐯 = (1, - 1)$ , so $𝐜 = (\frac{1}{2}, - \frac{1}{2})$ , and $𝑔 (𝐯) =$ $- 𝐯 +$ $2 𝐜 =$ $𝐜 -$ $(𝐯 - 𝐜)$ . This is the rotation by $180 °$ about $𝐜$ , since $𝑅_{𝜋} = - 𝐼_{2}$ .

(b) A fixed point is a solution of $𝑅_{𝜃} 𝐜 + 𝐛 = 𝐜$ , that is, of $(𝐼_{2} - 𝑅_{𝜃}) 𝐜 = 𝐛$ , where

𝐼_{2} - 𝑅_{𝜃} = (\begin{matrix} 1 - cos 𝜃 & sin 𝜃 \\ - sin 𝜃 & 1 - cos 𝜃 \end{matrix}) .

Here $𝑎 𝑑 -$ $𝑏 𝑐 =$ ${(1 - cos 𝜃)}^{2} +$ ${sin}^{2} 𝜃 =$ $2 -$ $2 cos 𝜃$ , which is nonzero because $cos 𝜃 \neq 1$ . By Proposition 2.5.28, $𝐼_{2} - 𝑅_{𝜃}$ is invertible, so the fixed point $𝐜 = {(𝐼_{2} - 𝑅_{𝜃})}^{- 1} 𝐛$ exists and is unique. Then $𝐛 = 𝐜 - 𝑅_{𝜃} 𝐜$ , and

𝑓 (𝐯) = 𝑅_{𝜃} 𝐯 + 𝐜 - 𝑅_{𝜃} 𝐜 = 𝐜 + 𝑅_{𝜃} (𝐯 - 𝐜),

the rotation by $𝜃$ about $𝐜$ . Part (a) is the case $𝜃 = 𝜋$ , $𝐛 = (1, - 1)$ .

Exercise 4.7.63.standard Let $𝐹$ be a field with $𝑞$ elements, and let $0 \leq 𝑘 \leq 𝑛$ .

(a)Show that the number of $𝑘$ -dimensional affine subspaces of $𝐹^{𝑛}$ is $𝑞^{𝑛 - 𝑘}$ times the number of $𝑘$ -dimensional subspaces of $𝐹^{𝑛}$ .
(b)Count the lines in $𝔽_{3}^{2}$ and the planes in $𝔽_{2}^{3}$ .

Solution.

Uses. Proposition 4.7.3 (Direction and base points), Corollary 4.7.6 (Parallel affine subspaces are equal or disjoint), Proposition 4.1.34 (Size of a vector space over a finite field), Proposition 4.1.37 (Counting subspaces)

Techniques. Double counting (Technique 4.1.36)

Hint 1. Group the affine subspaces by their direction. How many translates does a

𝑘

-dimensional subspace

𝑈

have?

Hint 2. The translates of

𝑈

are disjoint and cover

𝐹^{𝑛}

, and each has

| 𝑈 | = 𝑞^{𝑘}

elements.

Sketch. Each

𝑘

-dimensional affine subspace has exactly one direction, a

𝑘

-dimensional subspace

𝑈

. The distinct translates of

𝑈

partition

𝐹^{𝑛}

(Corollary 4.7.6), and each has

𝑞^{𝑘}

elements, so there are

𝑞^{𝑛} / 𝑞^{𝑘} = 𝑞^{𝑛 - 𝑘}

of them. (b)

𝔽_{3}^{2}

has

4

one-dimensional subspaces, so

4 \cdot 3 = 12

lines;

𝔽_{2}^{3}

has

7

two-dimensional subspaces, so

7 \cdot 2 = 14

planes.

(a) By Proposition 4.7.3(a), each affine subspace has exactly one direction, so the $𝑘$ -dimensional affine subspaces are grouped by their directions, which are the $𝑘$ -dimensional subspaces $𝑈$ of $𝐹^{𝑛}$ . Fix such a $𝑈$ . By Corollary 4.7.6, the distinct translates of $𝑈$ are pairwise disjoint and every vector lies in one of them. Each translate $𝐯 + 𝑈$ is in bijection with $𝑈$ via $𝐮 \mapsto 𝐯 + 𝐮$ , so it has $| 𝑈 | = 𝑞^{𝑘}$ elements (Proposition 4.1.34). Since $| 𝐹^{𝑛} | = 𝑞^{𝑛}$ , there are exactly $𝑞^{𝑛} / 𝑞^{𝑘} = 𝑞^{𝑛 - 𝑘}$ translates of $𝑈$ . Summing over $𝑈$ gives the claim.

(b) By Proposition 4.1.37, $𝔽_{3}^{2}$ has $(9 - 1) / (3 - 1) = 4$ one-dimensional subspaces, so it has $4 \cdot 3^{2 - 1} = 12$ lines. And $𝔽_{2}^{3}$ has $((8 - 1) (8 - 2)) / ((4 - 1) (4 - 2)) =$ $42 / 6 =$ $7$ two-dimensional subspaces, so it has $7 \cdot 2^{3 - 2} = 14$ planes.

Exercise 4.7.64.hard Let $𝑉$ and $𝑊$ be real vector spaces, and let $𝑓 : 𝑉 \to 𝑊$ satisfy

𝑓 ((1 - 𝑡) 𝐮 + 𝑡 𝐯) = (1 - 𝑡) 𝑓 (𝐮) + 𝑡 𝑓 (𝐯) for all 𝐮, 𝐯 \in 𝑉 and 𝑡 \in [0, 1] .

That is, $𝑓$ maps each segment onto a segment, at constant speed. Show that $𝑓$ is affine.

Solution.

Uses. Exercise 2.1.39, Definition 4.7.35 (Affine map)

Hint 1. First extend the identity to all

𝑡 \in ℝ

. For

𝑡 > 1

, the point

𝐯

lies on the segment from

𝐮

𝐰 = (1 - 𝑡) 𝐮 + 𝑡 𝐯

Hint 2. For

𝑡 < 0

, swap the roles of

𝐮

and

𝐯

Hint 3. Now

𝑔 = 𝑓 - 𝑓 (𝟎)

respects all affine combinations of two points and fixes

𝟎

. Use Exercise 2.1.39(a).

Sketch. For

𝑡 > 1

𝐯 =

(1 - 1 / 𝑡) 𝐮 +

(1 / 𝑡) 𝐰

with

1 / 𝑡 \in (0, 1)

, so

𝑓 (𝐯) =

(1 - 1 / 𝑡) 𝑓 (𝐮) +

(1 / 𝑡) 𝑓 (𝐰)

, which solves to

𝑓 (𝐰) =

(1 - 𝑡) 𝑓 (𝐮) +

𝑡 𝑓 (𝐯)

. For

𝑡 < 0

, write the point as

(1 - 𝑡^{'}) 𝐯 + 𝑡^{'} 𝐮

with

𝑡^{'} = 1 - 𝑡 > 1

. So

𝑔 = 𝑓 - 𝑓 (𝟎)

satisfies

𝑔 (𝑎 𝐮 + 𝑏 𝐯) =

𝑎 𝑔 (𝐮) +

𝑏 𝑔 (𝐯)

whenever

𝑎 + 𝑏 = 1

, and

𝑔 (𝟎) = 𝟎

; since

1 + 1 \neq 0

ℝ

𝑔

is linear.

We first show that the identity holds for all $𝑡 \in ℝ$ . Let $𝑡 > 1$ and $𝐰 = (1 - 𝑡) 𝐮 + 𝑡 𝐯$ . Then

(1 - \frac{1}{𝑡}) 𝐮 + \frac{1}{𝑡} 𝐰 = (1 - \frac{1}{𝑡} + \frac{1 - 𝑡}{𝑡}) 𝐮 + 𝐯 = 𝐯,

and $1 / 𝑡 \in (0, 1)$ , so the hypothesis gives $𝑓 (𝐯) =$ $(1 - \frac{1}{𝑡}) 𝑓 (𝐮) +$ $\frac{1}{𝑡} 𝑓 (𝐰)$ . Multiplying by $𝑡$ and solving,

𝑓 (𝐰) = 𝑡 𝑓 (𝐯) - (𝑡 - 1) 𝑓 (𝐮) = (1 - 𝑡) 𝑓 (𝐮) + 𝑡 𝑓 (𝐯) .

If $𝑡 < 0$ , then $(1 - 𝑡) 𝐮 +$ $𝑡 𝐯 =$ $(1 - 𝑡^{'}) 𝐯 +$ $𝑡^{'} 𝐮$ with $𝑡^{'} = 1 - 𝑡 > 1$ , and the case just proved, with $𝐮$ and $𝐯$ exchanged, gives $𝑓 ((1 - 𝑡^{'}) 𝐯 + 𝑡^{'} 𝐮) =$ $(1 - 𝑡^{'}) 𝑓 (𝐯) +$ $𝑡^{'} 𝑓 (𝐮) =$ $(1 - 𝑡) 𝑓 (𝐮) +$ $𝑡 𝑓 (𝐯)$ .

Now let $𝑔 (𝐱) =$ $𝑓 (𝐱) -$ $𝑓 (𝟎)$ . Then $𝑔 (𝟎) = 𝟎$ , and for $𝑎 + 𝑏 = 1$ ,

𝑔 (𝑎 𝐮 + 𝑏 𝐯) = 𝑎 𝑓 (𝐮) + 𝑏 𝑓 (𝐯) - (𝑎 + 𝑏) 𝑓 (𝟎) = 𝑎 𝑔 (𝐮) + 𝑏 𝑔 (𝐯) .

Since $1 + 1 \neq 0$ in $ℝ$ , Exercise 2.1.39(a) shows that $𝑔$ is linear. Hence $𝑓 = 𝑔 + 𝑓 (𝟎)$ is affine.

Exercise 4.7.65.standard A point

𝐩

of a convex set

𝐶

is an extreme point of

𝐶

𝐩 = (1 - 𝑡) 𝐚 + 𝑡 𝐛

with

𝐚, 𝐛 \in 𝐶

and

𝑡 \in (0, 1)

implies

𝐚 = 𝐛 = 𝐩

. Show that the extreme points of a simplex are exactly its vertices. Conclude that a simplex determines its set of vertices.

Solution.

Uses. Proposition 4.7.32 (Barycentric coordinates on a simplex), Theorem 4.7.22 (Characterizations of affine independence)

Hint 1. Work in barycentric coordinates (Proposition 4.7.32). A vertex

𝐯_{0}

has coordinates

(1, 0, \dots, 0)

Hint 2. If

𝐯_{0} = (1 - 𝑡) 𝐚 + 𝑡 𝐛

, compare coordinates

𝑖 \geq 1

: a sum of two nonnegative numbers is

0

Hint 3. If

𝐩

is not a vertex, two of its coordinates are positive, say

𝑡_{0}, 𝑡_{1} > 0

. Move

𝐩

a little in the direction

\pm (𝐯_{0} - 𝐯_{1})

Sketch. If

𝐯_{0} = (1 - 𝑡) 𝐚 + 𝑡 𝐛

, the barycentric coordinates satisfy

0 =

(1 - 𝑡) 𝛼_{𝑖} +

𝑡 𝛽_{𝑖}

for

𝑖 \geq 1

with both terms nonnegative, so

𝛼_{𝑖} = 𝛽_{𝑖} = 0

and

𝐚 = 𝐛 = 𝐯_{0}

. If

𝐩

has

𝑡_{0}, 𝑡_{1} > 0

, then with

𝛿 = min (𝑡_{0}, 𝑡_{1})

, the points

𝐩 \pm 𝛿 (𝐯_{0} - 𝐯_{1})

lie in the simplex, differ, and have midpoint

𝐩

Let $𝜎$ have vertices $𝐯_{0}, \dots, 𝐯_{𝑘}$ , and write points of $𝜎$ in barycentric coordinates (Proposition 4.7.32). The vertex $𝐯_{𝑗}$ has coordinates $𝐞_{𝑗}$ , with $1$ in place $𝑗$ and $0$ elsewhere.

Vertices are extreme. Suppose $𝐯_{𝑗} =$ $(1 - 𝑡) 𝐚 +$ $𝑡 𝐛$ with $𝐚, 𝐛 \in 𝜎$ and $𝑡 \in (0, 1)$ , and let $𝐚, 𝐛$ have barycentric coordinates $(𝛼_{𝑖})$ and $(𝛽_{𝑖})$ . Then $\sum_{𝑖} ((1 - 𝑡) 𝛼_{𝑖} + 𝑡 𝛽_{𝑖}) 𝐯_{𝑖}$ is an affine combination representing $𝐯_{𝑗}$ , so by uniqueness (Theorem 4.7.22(IV)), $(1 - 𝑡) 𝛼_{𝑖} +$ $𝑡 𝛽_{𝑖} =$ $0$ for $𝑖 \neq 𝑗$ . Both terms are nonnegative and $1 - 𝑡, 𝑡 > 0$ , so $𝛼_{𝑖} = 𝛽_{𝑖} = 0$ for $𝑖 \neq 𝑗$ , and then $𝛼_{𝑗} = 𝛽_{𝑗} = 1$ . Hence $𝐚 = 𝐛 = 𝐯_{𝑗}$ .

Other points are not extreme. Let $𝐩 \in 𝜎$ have coordinates $(𝑡_{𝑖})$ and not be a vertex. The $𝑡_{𝑖}$ are nonnegative and add up to $1$ , and they are not a standard basis vector, so at least two of them are positive, say $𝑡_{𝑟}, 𝑡_{𝑠} > 0$ with $𝑟 \neq 𝑠$ . Let $𝛿 =$ $min (𝑡_{𝑟}, 𝑡_{𝑠}) >$ $0$ , and $𝐚 =$ $𝐩 +$ $𝛿 (𝐯_{𝑟} - 𝐯_{𝑠})$ , $𝐛 =$ $𝐩 -$ $𝛿 (𝐯_{𝑟} - 𝐯_{𝑠})$ . Their coordinates are those of $𝐩$ with $𝑡_{𝑟}, 𝑡_{𝑠}$ replaced by $𝑡_{𝑟} + 𝛿, 𝑡_{𝑠} - 𝛿$ and by $𝑡_{𝑟} - 𝛿, 𝑡_{𝑠} + 𝛿$ ; these are nonnegative and add up to $1$ , so $𝐚, 𝐛 \in 𝜎$ . They differ, since $𝐯_{𝑟} \neq 𝐯_{𝑠}$ , and $𝐩 = \frac{1}{2} 𝐚 + \frac{1}{2} 𝐛$ . So $𝐩$ is not extreme.

The extreme points are defined in terms of the set $𝜎$ alone, so $𝜎$ determines its vertex set. In particular, a $𝑘$ -simplex is not a $𝑘^{'}$ -simplex for $𝑘^{'} \neq 𝑘$ .

More practice

The practice bank has 16 exercises on this section’s topics.