advanced27 min studyLesson 271 in Mathematics

Convex hulls and Carathéodory’s theorem

Compress a mixture using affine dependence while keeping its location and nonnegative weights unchanged.

On this page

What you’ll understand

  • Distinguish a convex hull from its closure and affine hull.
  • Prove the finite-dimensional support bound by eliminating a weight.
  • Construct and verify a shorter representation of a concrete mixture.

A mixture can be described by far more ingredients than its location actually requires. Imagine combining operating points whose coordinates record resource use and delivered output. Different nonnegative weight lists can produce the same average point. Convex geometry asks which averages are possible, then asks whether their descriptions can be shortened. Carathéodory’s theorem gives an exact answer depending on affine dimension rather than the number of available ingredients. The theorem is about representing a point, not about discovering an optimal point or preserving every property of its original mixture.

Three hulls answer different questions

Work in Euclidean space Rd\mathbb R^d. The convex hull of a nonempty set S consists of its finite convex combinations: coefficients are nonnegative and sum to one. Its affine hull permits negative coefficients but retains the sum condition. The closed convex hull also includes limits of convex combinations. These distinctions matter even in one dimension. For S=(0,1), its convex hull remains (0,1), while its closed convex hull contains both endpoints. No finite average of strictly positive numbers can equal zero.

The convex hull is the smallest convex set containing S. To verify this, combine two finite mixtures into one longer list to show convexity. Conversely, any convex set containing S contains every finite mixture by repeatedly applying the two-point definition. This argument does not require compactness or a finite set of generators. In finite dimensions, however, dimension imposes a powerful bound on the number of nonzero weights. A generator set lying in an affine plane uses that plane’s dimension, even if the surrounding ambient space is much larger.

conv⁡S={∑i=1mλisi:m≥1, si∈S, λi≥0, ∑i=1mλi=1}\operatorname{conv}S=\left\{\sum_{i=1}^{m}\lambda_i s_i:m\ge1,\ s_i\in S,\ \lambda_i\ge0,\ \sum_{i=1}^{m}\lambda_i=1\right\}
The combinations are finite. Taking a closure is a separate operation and can add boundary points.↔ Scroll wide equations horizontally.

Eliminate redundant weights

Suppose a representation uses more than d+1 points with strictly positive weights. The augmented vectors (si,1)(s_i,1) are linearly dependent in Rd+1\mathbb R^{d+1}. Consequently there is a nonzero coefficient list a with both weighted location sum zero and coefficient sum zero. The list a has a positive entry and a negative entry; otherwise its zero sum would force every entry to vanish. Subtract a small multiple of this dependence from the original weights. Both the represented point and the total weight remain unchanged.

Increase that multiple until the first positive weight reaches zero. More precisely, choose the minimum of λi/ai\lambda_i/a_i over positive a entries. Every new weight stays nonnegative, and at least one disappears. Repeating the procedure leaves at most d+1 points. Replacing d with the affine dimension gives the sharper bound. This is a complete existence proof and also an elimination algorithm when a dependence can be computed. It guarantees neither uniqueness of the final support nor a preference for particular generators.

∑iaisi=0,∑iai=0,t=min⁡ai>0λiai,λi′=λi−tai\sum_i a_i s_i=0,\quad\sum_i a_i=0,\qquad t=\min_{a_i>0}\frac{\lambda_i}{a_i},\quad\lambda_i'=\lambda_i-ta_i
The update preserves the average and weight sum, keeps all weights nonnegative, and removes at least one supported point.↔ Scroll wide equations horizontally.

What dimension does and does not buy

If S is compact and nonempty in finite dimensions, its convex hull is compact. Carathéodory reduces every point to d+1 generators, allowing repeated generators and zero weights. The hull is therefore the continuous image of the compact product of d+1 copies of S and the compact weight simplex. This compactness argument would fail without control over the number of generators. It gives existence results for maximizing a continuous objective over such a hull, but a nonlinear objective need not achieve its optimum at a generator.

The finite-dimensional hypothesis is substantive. In an infinite-dimensional Hilbert space, an infinite weighted limit can require arbitrarily large finite supports for accurate approximation, and need not belong to the finite convex hull at all. Even in finite dimensions the theorem preserves only the coordinates used to define the point. If a mixture must also preserve cost, variance, or another moment, include those quantities as extra coordinates before applying the bound. Adding constraints after compressing the original coordinates can destroy the feature one intended to preserve.

Remove one corner without moving the average

Use rectangle vertices v1=(0,0),v2=(3,0),v3=(3,2),v4=(0,2)v_1=(0,0),v_2=(3,0),v_3=(3,2),v_4=(0,2) with weights (0.3,0.3,0.1,0.3). Compress the representation.

  1. The represented point is (3(0.3+0.1),2(0.1+0.3))=(1.2,0.8)(3(0.3+0.1),2(0.1+0.3))=(1.2,0.8). All four weights are positive and sum to one.

  2. The affine dependence v1−v2+v3−v4=0v_1-v_2+v_3-v_4=0 has coefficient sum zero. For positive entries of a=(1,−1,1,−1), the ratios are 0.3 and 0.1, so choose t=0.1.

  3. Subtract ta to obtain (0.2,0.4,0,0.4). Its average is 0.4(3,0)+0.4(0,2)=(1.2,0.8)0.4(3,0)+0.4(0,2)=(1.2,0.8) and its weights still sum to one.

The result

Three vertices now suffice, exactly the general two-dimensional upper bound. This particular point may admit other representations; the procedure did not establish uniqueness.

Explore the model

Interactive lab

Interactive lab · matrix

Affine changes preserve convex mixtures

Pause and Reset, then set diagonal scale a=1 and shear b=1. Predict the four transformed unit-square corners and the image of its center: (1,0.5). Play to see the geometry interpolate, and use the final-matrix readouts to verify the selected transformation. For a fresh comparison Reset and set a=2,b=0; the center becomes (1,0.5) by a different map. Readouts describe the final matrix throughout playback. This instrument transforms a square; it does not perform weight elimination.

Affine changes preserve convex mixturesThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA linear image of a convex mixture is the same mixture of the transformed generators.

Check your understanding

A set in R¹⁰ lies in an affine plane of dimension two. How many generators always suffice for a point in its convex hull?

Try it yourself

Let S={(0,0),(2,0),(0,2),(2,2)}. Express (0.5,0.5) using two generators, and explain why the bound of three in the plane is still sharp for other sets.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Try the diagonal through the origin. For sharpness, examine the interior of a triangle using only its three vertices as generators.

Compare with the solution

The point equals 34(0,0)+14(2,2)\tfrac34(0,0)+\tfrac14(2,2), so two suffice here. For the three vertices (0,0),(1,0),(0,1), the point (1/3,1/3) has all three barycentric weights 1/3. Any two generators produce a triangle edge, whereas this point lies strictly inside. Thus three are necessary for that generator set, even though particular sets and points permit shorter descriptions.

Take it with you

  • Convex hulls use finite nonnegative mixtures; closure is additional.
  • Affine dependence removes weights without moving the represented point.
  • The exact support bound is affine dimension plus one, not a uniqueness statement.

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