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affine combination

Defines: 
affine independence, affinely independent, affine hull
Synonym: 
affine independent
Type of Math Object: 
Definition
Major Section: 
Reference

Mathematics Subject Classification

51A15 no label found

Comments

If anyone can help me with this proof, that would be great.

I need to show that if v_0,...,v_n are n+1 affine-independent points in R^n, and if there exits x,y in R^n such that |x - v_i| = |y - v_i| for i = 0,...,n, then x = y.

Start with the identity x-v_i = (x-y)+(y-v_i). Square it (scalar product by itself), after simplification you get:
|x-y|^2 = -2(x-y).(y-v_i) for all i, so (x-y).(v_i-v_j) = 0 for all i and j.
Since the n+1 vectors v_i are affine independent, v_i-v_j generate the whole space R^n. That means that x-y is orhogonal to every vector in R^n.

Hi there, I follow the argument until you say

> so (x-y).(v_i-v_j) = 0 for all i and j

how does this follow from

> |x-y|^2 = -2(x-y).(y-v_i) for all i

? Thanks for all the help!

Since |x-y|^2 = -2(x-y).(y-v_i) for all i, 0 = |x-y|^2 - |x-y|^2 = -2(x-y).(y-v_i) - (-2(x-y).(y-v_j)) = 2(x-y)(v_i-v_j).

wow ok that's rather clever...thanks so much; this was a big help!!!

I need to prove two things

1) If x_0,...x_n are n+1 affine-independent points in R^n and if |x_i - x_j| = |y_i - y_j| for all i,j then y_0,...,y_n are also affine independent.

2) If x_0,...x_n are n+1 affine-independent points in R^n and A:R^n -->R^n is an affine tranformation such that for all i,j

|A(x_i) - A(x_j)| = |x_i - x_j|

then A is a Euclidean isometry.

Thanks so much for the help!

I figured out (2); since A is affine, there is a linear transformation L and a vector v such that A(x) = L(x) + v for all x. It suffices to show that L is orthogonal which is to say that
L(x).L(y) = x.y for all x,y. Now notice that by hypothesis we have

(x_i - x_j).(x_i - x_j) = |x_i - x_j|^2
= |A(x_i) - A(x_j)|^2
= |L(x_i) - L(x_j)|^2
= |L(x_i - x_j)|^2
= L(x_i - x_j).L(x_i - x_j)

since x_0,...,x_n are affine-ind., the vectors x_0-x_i form a basis for R^n so we can expand x = sum_i(a_i(x_0 - x_i)) and y = sum_j(b_j(x_0 - x_j)). Using the above computation we have

L(x).L(y) = L(sum_i(a_i(x_0 - x_i))).L(sum_j(b_j(x_0 - x_j)))
= sum_ij(a_ib_jL(x_0 - x_i).L(x_0 - x_j))
= sum_ij(a_ib_j(x_0 - x_i).(x_0 - x_j))
= (sum_i(a_i(x_0 - x_i))).(sum_j(b_j(x_0 - x_j)))
= x.y

I also figured out (1) with the help of a professor. It's not hard to show that the set v_i - v_0 has the same Gramian matrix as w_i - w_0 by using the constraints |v_i - v_j| = |w_i - w_j|. It follows that since the set v_i - v_0 is linearly independent, so is the set w_i - w_0 and thus w_0,...,w_n are affine independent.

If anyone finds any errors in the proofs of either of the two parts of the original post, then I'd be grateful to know about them.

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