本文为《Linear algebra and its applications》的读书笔记
Convex combinations
convex hull: 凸包
The convex hull of a single point v 1 \boldsymbol v_1 v1 is just the set { v 1 } \{\boldsymbol v_1\} { v1}, the same as the affine hull. In other cases, the convex hull is properly contained in the affine hull. Recall that the affine hull of distinct points v 1 \boldsymbol v_1 v1 and v 2 \boldsymbol v_2 v2 is the line
Because the weights in a convex combination are nonnegative, the points in c o n v { v 1 , v 2 } conv\{\boldsymbol v_1,\boldsymbol v_2\} conv{
v1,v2} is the line segment between v 1 \boldsymbol v_1 v1 and v 2 \boldsymbol v_2 v2, hereafter denoted by v 1 v 2 ‾ \overline{\boldsymbol v_1\boldsymbol v_2} v1v2.
If a set S S S is affinely independent and if p ∈ a f f S \boldsymbol p\in aff S p∈affS, then p ∈ c o n v S \boldsymbol p\in conv S p∈convS if and only if the barycentric coordinates of p \boldsymbol p p are nonnegative.
Example 1 shows a special situation in which S S S is much more than just affinely independent.
EXAMPLE 1
Let
and S = { v 1 , v 2 , v 3 } S=\{\boldsymbol v_1,\boldsymbol v_2,\boldsymbol v_3\} S={
v1,v2,v3}. Note that S S S is an orthogonal set. Determine whether p 1 \boldsymbol p_1 p1 is in S p a n S Span S SpanS, a f f S aff S affS, and c o n v S conv S convS.
SOLUTION
Let W W W be the subspace spanned by S S S.
The orthogonal projection of p 1 \boldsymbol p_1 p1 onto W W W is p 1 \boldsymbol p_1 p1 itself: This shows that p 1 \boldsymbol p_1 p1 is in S p a n S Span S SpanS. Also, since the coefficients sum to 1, p 1 \boldsymbol p_1 p1 is in a f f S aff S affS. In fact, p 1 \boldsymbol p_1 p1 is in c o n v S conv S convS, because the coefficients are also nonnegative.
Recall that a set S S S is affine if it contains all lines determined by pairs of points in S S S. When attention is restricted to convex combinations, the appropriate condition involves line segments rather than lines.
Intuitively, a set S S S is convex if every two points in the set can “see” each other without the line of sight leaving the set. Figure 1 illustrates this idea.
PROOF
The argument is similar to the proof of Theorem 2.
The intersection of any collection of subspaces is itself a subspace. A similar result holds for affine sets and convex sets.
PROOF
Let T T T denote the intersection of all the convex sets containing S S S. Since c o n v S conv S convS is a convex set containing S S S, it follows that T ⊂ c o n v S T\subset conv S T⊂convS. On the other hand, let C C C be any convex set containing S S S. Then C C C contains every convex combination of points of C C C, and hence also contains every convex combination of points of the subset S S S. That is, c o n v S ⊂ C conv S \subset C convS⊂C. Since this is true for every convex set C C C containing S S S, it is also true for the intersection of them all. That is, c o n v S ⊂ T conv S\subset T convS⊂T .
Theorem 9 shows that c o n v S conv S convS is in a natural sense the “smallest” convex set containing S S S. For example, consider a set S S S that lies inside some large rectangle in R 2 \R^2 R2, and imagine stretching a rubber band around the outside of S S S. As the rubber band contracts around S S S, it outlines the boundary of the convex hull of S S S. Or to use another analogy, the convex hull of S S S fills in all the holes in the inside of S S S and fills out all the dents in the boundary of S S S.
EXAMPLE 2
a. The convex hulls of sets S S S and T T T in R 2 \R^2 R2 are shown below.
b. Let S S S be the set consisting of the standard basis for R 3 \R^3 R3, S = { e 1 , e 2 , e 3 } S=\{\boldsymbol e_1,\boldsymbol e_2,\boldsymbol e_3\} S={
e1,e2,e3}. Then c o n v S conv S convS is a triangular surface in R 3 \R^3 R3, with vertices e 1 \boldsymbol e_1 e1, e 2 \boldsymbol e_2 e2, and e 3 \boldsymbol e_3 e3. See Figure 2.
EXAMPLE 3
Let
Show that the convex hull of S S S is
See Figure 3.
SOLUTION
Consider any point p \boldsymbol p p in the shaded region of Figure 3, say
The line through 0 \boldsymbol 0 0 and p \boldsymbol p p has the equation y = ( b / a ) t y =(b/a)t y=(b/a)t for t t t real. That line intersects S S S where t t t satisfies ( b / a ) t = t 2 (b/a)t=t^2 (b/a)t=t2, that is, when t = b / a t= b/a t=b/a. Thus, p \boldsymbol p p is on the line segment from 0 \boldsymbol 0 0 to [ b / a b 2 / a 2 ] \begin{bmatrix}b/a\\ b^2/a^2\end{bmatrix} [b/ab2/a2], which shows that Figure 3 is correct.
The following theorem is basic in the study of convex sets. It was first proved by Constantin Caratheodory in 1907. If p \boldsymbol p p is in the convex hull of S S S, then, by definition, p \boldsymbol p p must be a convex combination of points of S S S. But the definition makes no stipulation(规定) as to how many points of S S S are required to make the combination. Caratheodory’s remarkable theorem says that in an n n n-dimensional space, the number of points of S S S in the convex combination never has to be more than n + 1 n+ 1 n+1.
PROOF
Given p \boldsymbol p p in c o n v S conv S convS, one may write p = c 1 v 1 + . . . + c k v k \boldsymbol p = c_1\boldsymbol v_1+...+c_k\boldsymbol v_k p=c1v1+...+ckvk, where v i ∈ S \boldsymbol v_i\in S vi∈S, c 1 + . . . + c k = 1 c_1+...+ c_k= 1 c1+...+ck=1, and c i ≥ 0 c_i\geq 0 ci≥0, for some k k k and i = 1 , . . . , k i=1,..., k i=1,...,k. The goal is to show that such an expression exists for p \boldsymbol p p with k ≤ n + 1 k\leq n + 1 k≤n+1.
If k > n + 1 k > n + 1 k>n+1, then { v 1 , . . . , v k } \{\boldsymbol v_1,...,\boldsymbol v_k\} { v1,...,vk} is affinely dependent, by Exercise 12 in Section 8.2. Thus there exist scalars d 1 , . . . , d k d_1,..., d_k d1,...,dk, not all zero, such that
Consider the two equations
and
By subtracting an appropriate multiple of the second equation from the first, we now eliminate one of the v i \boldsymbol v_i vi terms and obtain a convex combination of fewer than k k k elements of S S S that is equal to p \boldsymbol p p.
Since not all of the d i d_i di coefficients are zero, we may assume (by reordering subscripts if if necessary) that d k > 0 d_k > 0 dk>0 and that c k / d k ≤ c i / d i c_k/d_k\leq c_i/d_i ck/dk≤ci/di for all those i i i for which d i > 0 d_i > 0 di>0. For i = 1 , . . . , k i= 1,..., k i=1,...,k, let b i = c i − ( c k / d k ) d i b_i=c_i-(c_k/d_k)d_i bi=ci−(ck/dk)di . Then b k = 0 b_k= 0 bk=0 and
Furthermore, each b i ≥ 0 b_i\geq 0 bi≥0. Indeed, if d i ≤ 0 d_i\leq 0 di≤0, then b i ≥ c i ≥ 0 b_i\geq c_i\geq 0 bi≥ci≥0. If d i > 0 d_i > 0 di>0, then b i = d i ( c i / d i − c k / d k ) ≥ 0 b_i=d_i(c_i/d_i-c_k/d_k)\geq 0 bi=di(ci/di−ck/dk)≥0. By construction,
Thus p \boldsymbol p p is now a convex combination of k − 1 k- 1 k−1 of the points v 1 , . . . , v k \boldsymbol v_1,...,\boldsymbol v_k v1,...,vk. This process may be repeated until p \boldsymbol p p is expressed as a convex combination of at most n + 1 n+1 n+1 of the points of S S S.
The following example illustrates the calculations in the proof above.
EXAMPLE 4
Let
and S = { v 1 , v 2 , v 3 , v 4 } S=\{\boldsymbol v_1,\boldsymbol v_2,\boldsymbol v_3,\boldsymbol v_4\} S={
v1,v2,v3,v4}. Then
Use the procedure in the proof of Caratheodory’s Theorem to express p \boldsymbol p p as a convex combination of three points of S S S.
SOLUTION
The set S S S is affinely dependent. Use the technique of Section 8.2 to obtain an affine dependence relation
Next, choose the points v 2 \boldsymbol v_2 v2 and v 4 \boldsymbol v_4 v4 in (3), whose coefficients are positive. For each point, compute the ratio of the coefficients in equations (2) and (3). The ratio for v 2 \boldsymbol v_2 v2 is 1 / 24 1/24 1/24, and that for v 4 \boldsymbol v_4 v4 is 1 / 48 1/48 1/48. The ratio for v 4 \boldsymbol v_4 v4 is smaller, so subtract 1 / 48 1/48 1/48 times equation (3) from equation (2) to eliminate v 4 \boldsymbol v_4 v4:
This result cannot, in general, be improved by decreasing the required number of points. Indeed, given any three non-collinear points in R 2 \R^2 R2, the centroid(质心) of the triangle formed by them is in the convex hull of all three, but is not in the convex hull of any two.