The

proof of

Radon's theorem is little more than an imaginative

exercise in

linear algebra, but (like the

theorem) it always seems to me to reveal some of the

structure of

**R**^{d}. In any case, make sure you read

what we're proving before reading the proof...

So suppose we're given d+2 points x_{0},...,x_{d+1}∈**R**^{d} (the theorem is true for *any* n>=d+2 points, but obviously adding more points just makes it easier to prove). Then these points have some affine dependence, and WLOG (maybe after reordering the points) we may write

x_{d+1} = ∑_{i=0}^{d} c_{i}x_{i},

∑_{i=0}^{d} c_{i} = 1.

Since their sum is

positive,

*some* of the c

_{i}'s must be positive, and WLOG (again, after reordering the points) we may say c

_{0},...,c

_{k} > 0 >= c

_{k+1},...,c

_{d}.

Rewrite the affine dependence all coefficients separated by their signs:

x_{d+1} + ∑_{i=k+1}^{d} (-c_{i})x_{i} =
∑_{i=0}^{k} c_{i}x_{i}.

The point here is that

*all coefficients appearing are positive*!

Now let S = ∑_{i=0}^{k} c_{i} = 1 - ∑_{i=k+1}^{d} c_{i} __> 0__. Write b_{i}=c_{i}/S for 0<=i<=k, b_{i}=-c_{i}/S for k<i<=d, and b_{d+1}=1/S. Then all the b_{i}'s are positive, ∑_{i=0}^{k}b_{i} = ∑_{i=k+1}^{d+1}b_{i} = 1, and we have the equal convex combinations

∑_{i=0}^{k} b_{i}x_{i} = ∑_{i=k+1}^{d+1} b_{i}x_{i}

Q.E.D.!