8.1 Affine combinations (仿射组合)

本文为《Linear algebra and its applications》的读书笔记

This chapter begins with geometric ideas that are already familiar to you. Affine combinations of vectors generalize the concepts of lines and planes to higher dimensions and describe the mathematical objects formed when subspaces are shifted away from the origin. Hyperplanes are a generalization of planes, and polytopes are a generalization of polygons.

The geometric objects studied in this chapter include line segments, solid triangles and polygons, specials types of curves and surfaces, and generalizations to sets in more than three dimensions. These objects form the building blocks for applications to computer graphics and to linear programming.

Affine combinations

In this section, vectors are viewed as points in ${\mathbb{R}}^n$.

Given vectors (or “points”) ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_p$ in ${\mathbb{R}}^n$ and scalars $c_1,...,c_p$, an affine combination of ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_p$ is a linear combination

such that the weights satisfy $c_1+...+c_p=1$.

仿射包 / 仿射生成集

The affine hull of a single point ${\mathbf{v}}_1$ is just the set { v 1 } \{\boldsymbol v_1\} { v1​}. The affine hull of two distinct points is often written in a special way. Suppose $\mathbf{y}=c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2$ with $c_1+c_2=1$. Write $t$ in place of $c_2$, so that $c_1=1-t$ . Then the affine hull of { v 1 , v 2 } \{\boldsymbol v_1, \boldsymbol v_2\} { v1​,v2​} is the set

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This set of points includes ${\mathbf{v}}_1$ (when $t=0$) and ${\mathbf{v}}_2$ (when $t=1$). If ${\mathbf{v}}_2={\mathbf{v}}_1$, then (1) again describes just one point. Otherwise, (1) describes the line through ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$. To see this, rewrite (1) in the form

where $\mathbf{p}$ is ${\mathbf{v}}_1$ and $\mathbf{u}$ is ${\mathbf{v}}_2-{\mathbf{v}}_1$.

Figure 2 uses the original points ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$, and displays a f f { v 1 , v 2 } aff\{\boldsymbol v_1,\boldsymbol v_2\} aff{ v1​,v2​} as the line through ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$.

PROOF
[Hint: $\mathbf{y}=(1-c_2-...-c_p){\mathbf{v}}_1+c_2{\mathbf{v}}_2+...+c_p{\mathbf{v}}_p$]

In the statement of Theorem 1, the point ${\mathbf{v}}_1$ could be replaced by any of the other points in the list ${\mathbf{v}}_1,...,{\mathbf{v}}_p$: Only the notation in the proof would change.

We can judge whether an arbitrary vector $\mathbf{y}$ is an affine combination of { v 1 , . . . , v n } \{\boldsymbol v_1,...,\boldsymbol v_n\} { v1​,...,vn​} by using Theorem 1. But the question can be answered more directly if the chosen points ${\mathbf{v}}_i$ are a basis for ${\mathbb{R}}^n$. For example, let B = { b 1 , . . . , b n } \mathcal B = \{\boldsymbol b_1,...,\boldsymbol b_n\} B={ b1​,...,bn​} be such a basis. Then any $\mathbf{y}$ in ${\mathbb{R}}^n$ is a unique linear combination of ${\mathbf{b}}_1,...,{\mathbf{b}}_n$. This combination is an affine combination of the $\mathbf{b}$’s if and only if the weights sum to 1.

EXAMPLE 2
Let

The set B = { b 1 , b 2 , b 3 } \mathcal B =\{\boldsymbol b_1, \boldsymbol b_2,\boldsymbol b_3\} B={ b1​,b2​,b3​} is a basis for ${\mathbb{R}}^3$. Determine whether the points ${\mathbf{p}}_1$ and ${\mathbf{p}}_2$ are affine combinations of the points in $\mathcal{B}$.
SOLUTION
Find the $\mathcal{B}$-coordinates of ${\mathbf{p}}_1$ and ${\mathbf{p}}_2$. These two calculations can be combined by row reducing the matrix $[{\mathbf{b}}_1{\mathbf{b}}_2{\mathbf{b}}_3{\mathbf{p}}_1{\mathbf{p}}_2]$:

Thus,

${\mathbf{p}}_1$ is not an affine combination of the $\mathbf{b}$’s and ${\mathbf{p}}_2$ is an affine combination of the $\mathbf{b}$’s.

集合 $S$ 是仿射的

Geometrically, a set is affine if whenever two points are in the set, the entire line through these points is in the set. (If $S$ contains only one point, $\mathbf{p}$, then the line through $\mathbf{p}$ and $\mathbf{p}$ is just a point, a “degenerate” line.)

Algebraically, for a set $S$ to be affine, the definition requires that every affine combination of two points of $S$ belong to $S$. Remarkably, this is equivalent to requiring that $S$ contain every affine combination of an arbitrary number of points of $S$.

PROOF
Suppose that $S$ is affine and use induction(数学归纳法) on the number $m$ of points of $S$ occurring in an affine combination.
(i). When $m$ is 1 or 2, an affine combination of $m$ points of $S$ lies in $S$, by the definition of an affine set.
(ii). Now, assume that every affine combination of $k$ or fewer points of $S$ yields a point in $S$, and consider a combination of $k+1$ points. Take ${\mathbf{v}}_i$ in $S$ for $i=1,...,k+1$, and let $\mathbf{y}=c_1{\mathbf{v}}_1+...+c_k{\mathbf{v}}_k+c_{k+1}{\mathbf{v}}_{k+1}$, where $c_1+...+c_{k+1}=1$. Since the $c_i$ ’s sum to 1, at least one of them must not be equal to 1. By reindexing the ${\mathbf{v}}_i$ and $c_i$ , if necessary, we may assume that $c_{k+1}\ne 1$. Let $t=c_1+...+c_k$. Then $t=1-c_{k+1}\ne 0$, and

By the induction hypothesis, the point $\mathbf{z}=(c_1/t){\mathbf{v}}_1+...+(c_k/t){\mathbf{v}}_k$ is in $S$. Thus (6) displays $\mathbf{y}$ as an affine combination of two points in $S$. By the principle of induction, every affine combination of such points lies in $S$. That is, $affS\subset S$. The reverse inclusion, $S\subset affS$, always applies.

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The next definition provides terminology for affine sets that emphasizes their close connection with subspaces of ${\mathbb{R}}^n$.

translate: 平移
flat: 平面

For example, in ${\mathbb{R}}^3$, the proper subspaces consist of the origin $0$, the set of all lines through $0$, and the set of all planes through $0$. Thus the proper flats in ${\mathbb{R}}^3$ are points (zero-dimensional), lines (one-dimensional), and planes (two-dimensional), which may or may not pass through the origin.

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The next theorem shows that these geometric descriptions of lines and planes in ${\mathbb{R}}^3$ (as translates of subspaces) actually coincide with their earlier algebraic descriptions as sets of all affine combinations of two or three points, respectively.

Remark: Notice the key role that definitions play in this proof. For example, the first part assumes that $S$ is affine and seeks to show that $S$ is a flat. By definition, a flat is a translate of a subspace. By choosing $\mathbf{p}$ in $S$ and defining $W=S+(-\mathbf{p})$, the set $S$ is translated to the origin and $S=W+\mathbf{p}$. It remains to show that $W$ is a subspace, for then $S$ will be a translate of a subspace and hence a flat.

PROOF
Suppose that $S$ is affine. Let $\mathbf{p}$ be any fixed point in $S$ and let $W=S+(-\mathbf{p})$, so that $S=W+\mathbf{p}$. To show that $S$ is a flat, it suffices to show that $W$ is a subspace of ${\mathbb{R}}^n$.
Since $\mathbf{p}$ is in $S$, the zero vector is in $W$. To show that $W$ is closed under sums and scalar multiples, it suffices to show that if ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ are elements of $W$, then ${\mathbf{u}}_1+t{\mathbf{u}}_2$ is in $W$ for every real $t$. Since ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ are in $W$ , there exist ${\mathbf{s}}_1$ and ${\mathbf{s}}_2$ in $S$ such that

Let $\mathbf{y}=s_1+s_2-\mathbf{p}$. Then $\mathbf{y}$ is an affine combination of points in $S$. Since $S$ is affine, $\mathbf{y}$ is in $S$. But then $(1-t){\mathbf{s}}_1+t\mathbf{y}$ is also in $S$. So ${\mathbf{u}}_1+t{\mathbf{u}}_2$ is in $-\mathbf{p}+S=W$. This shows that $W$ is a subspace of ${\mathbb{R}}^n$. Thus $S$ is a flat, because $S=W+\mathbf{p}$.

Conversely, suppose $S$ is a flat. That is, $S=W+\mathbf{p}$ for some $\mathbf{p}\in {\mathbb{R}}^n$ and some subspace $W$ . To show that $S$ is affine, it suffices to show that for any pair ${\mathbf{s}}_1$ and ${\mathbf{s}}_2$ of points in $S$, the line through ${\mathbf{s}}_1$ and ${\mathbf{s}}_2$ lies in $S$. By definition of $W$ , there exist ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ in $W$ such that ${\mathbf{s}}_1={\mathbf{u}}_1+\mathbf{p}$ and ${\mathbf{s}}_2={\mathbf{u}}_2+\mathbf{p}$. So, for each real $t$,

Since $W$ is a subspace, $(1-t){\mathbf{u}}_1+t{\mathbf{u}}_2\in W$ and so $(1-t)s_1+t{s}_2\in W+\mathbf{p}=S$. Thus $S$ is affine.

Theorem 3 provides a geometric way to view the affine hull of a set: it is the flat that consists of all the affine combinations of points in the set. For instance, Figure 3 shows the points studied in Example 2. Although the set of all linear combinations of ${\mathbf{b}}_1,{\mathbf{b}}_2$, and ${\mathbf{b}}_3$ is all of ${\mathbb{R}}^3$, the set of all affine combinations is only the plane through ${\mathbf{b}}_1,{\mathbf{b}}_2$, and ${\mathbf{b}}_3$. Note that ${\mathbf{p}}_2$ (from Example 2) is in the plane through ${\mathbf{b}}_1,{\mathbf{b}}_2$, and ${\mathbf{b}}_3$, while ${\mathbf{p}}_1$ is not in that plane.

The affine combination of two linearly independent vectors (points) is just the line through these two points; the affine combination of three linearly independent vectors (points) is the plane through these three points.

The next example takes a fresh look at a familiar set—the set of all solutions of a system $A\mathbf{x}=\mathbf{b}$.

EXAMPLE 3
Suppose that the solutions of an equation $A\mathbf{x}=\mathbf{b}$ are all of the form $\mathbf{x}=x_3\mathbf{u}+\mathbf{p}$, (通解 + 特解 的形式)

Find points ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ such that the solution set of $A\mathbf{x}=\mathbf{b}$ is a f f { v 1 , v 2 } aff\{\boldsymbol v_1, \boldsymbol v_2\} aff{ v1​,v2​}.
SOLUTION
The solution set is a line through $\mathbf{p}$ in the direction of $\mathbf{u}$. Since
a f f { v 1 , v 2 } aff\{\boldsymbol v_1,\boldsymbol v_2\} aff{ v1​,v2​} is a line through ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$, identify two points on the line $\mathbf{x}=x_3\mathbf{u}+\mathbf{p}$. Two simple choices appear when $x_3=0$ and $x_3=1$. That is, take ${\mathbf{v}}_1=\mathbf{p}$ and ${\mathbf{v}}_2=\mathbf{u}+\mathbf{p}$, so that

In this case, the solution set is described as the set of all affine combinations of the form

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The next theorem provides another view of affine combinations, which for ${\mathbb{R}}^2$ and ${\mathbb{R}}^3$ is closely connected to applications in computer graphics, discussed in the next section (and in Section 2.7).

EXERCISE 13
Suppose { v 1 , v 2 , v 3 } \{\boldsymbol v_1, \boldsymbol v_2,\boldsymbol v_3\} { v1​,v2​,v3​} is a basis for ${\mathbb{R}}^3$. Show that S p a n { v 2 − v 1 , v 3 − v 1 } Span\{\boldsymbol v_2-\boldsymbol v_1 ,\boldsymbol v_3-\boldsymbol v_1\} Span{ v2​−v1​,v3​−v1​} is a plane in ${\mathbb{R}}^3$.
SOLUTION
[Hint: S p a n { v 2 − v 1 , v 3 − v 1 } Span\{\boldsymbol v_2-\boldsymbol v_1 ,\boldsymbol v_3-\boldsymbol v_1\} Span{ v2​−v1​,v3​−v1​} is a plane if and only if { v 2 − v 1 , v 3 − v 1 } \{\boldsymbol v_2-\boldsymbol v_1 ,\boldsymbol v_3-\boldsymbol v_1\} { v2​−v1​,v3​−v1​} is linearly independent.]

EXERCISE 14
Show that if { v 1 , v 2 , v 3 } \{\boldsymbol v_1, \boldsymbol v_2,\boldsymbol v_3\} { v1​,v2​,v3​} is a basis for ${\mathbb{R}}^3$, then a f f { v 1 , v 2 , v 3 } aff\{\boldsymbol v_1, \boldsymbol v_2,\boldsymbol v_3\} aff{ v1​,v2​,v3​} is the plane through ${\mathbf{v}}_1,{\mathbf{v}}_2,$ and ${\mathbf{v}}_3$.
SOLUTION
It suffices to show that a f f { v 1 , v 2 , v 3 } aff\{\boldsymbol v_1, \boldsymbol v_2,\boldsymbol v_3\} aff{ v1​,v2​,v3​} is the translation of a subspace. if y ∈ a f f { v 1 , v 2 , v 3 } \boldsymbol y\in aff\{\boldsymbol v_1, \boldsymbol v_2,\boldsymbol v_3\} y∈aff{ v1​,v2​,v3​}, then y − v 1 ∈ S p a n { v 2 − v 1 , v 3 − v 1 } \boldsymbol y-\boldsymbol v_1\in Span\{\boldsymbol v_2-\boldsymbol v_1,\boldsymbol v_3-\boldsymbol v_1\} y−v1​∈Span{ v2​−v1​,v3​−v1​}. Thus y ∈ v 1 + S p a n { v 2 − v 1 , v 3 − v 1 } \boldsymbol y\in \boldsymbol v_1+Span\{\boldsymbol v_2-\boldsymbol v_1,\boldsymbol v_3-\boldsymbol v_1\} y∈v1​+Span{ v2​−v1​,v3​−v1​}.

EXERCISE 16
Let $\mathbf{v}\in {\mathbb{R}}^n$ and let $k\in \mathbb{R}$. Prove that S = { x ∈ R n : x ⋅ v = k } S =\{\boldsymbol x\in \R^n:\boldsymbol x\cdot\boldsymbol v=k\} S={ x∈Rn:x⋅v=k} is an affine subset of ${\mathbb{R}}^n$.
SOLUTION
W = { x ∈ R n : x ⋅ v = 0 } W =\{\boldsymbol x\in \R^n:\boldsymbol x\cdot\boldsymbol v=0\} W={ x∈Rn:x⋅v=0} is a subspace of ${\mathbb{R}}^n$. Suppose $\mathbf{p}\cdot \mathbf{v}=k$, then $S=W+\mathbf{p}$.

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EXAMPLE 4
Let

Write $\mathbf{p}$ as an affine combination of ${\mathbf{v}}_1,{\mathbf{v}}_2$, and ${\mathbf{v}}_3$, if possible.
SOLUTION
Row reduce the augmented matrix for the equation

To simplify the arithmetic, move the fourth row of 1’s to the top (equivalent to three row interchanges). After this, the number of arithmetic operations here is basically the same as the number needed for the method using Theorem 1.

Thus