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Characterizing Affine Independence

  1. Nov 9, 2011 #1
    Hello, I'm currently self-studying "An Introduction to Convex Polytopes" and I'm having some trouble understanding the different characterizations of affine independence. I understand that for an n-family (x_{1},...,x_{n}) of points from R^{d}, it is affinely independent if a linear combination [tex]\lambda_{1}x_{1} +...+\lambda_{n}x_{n}[/tex] with [tex]\lambda_{1} + ... + \lambda_{n} = 0[/tex] can only have the value of the zero vector when [tex]\lambda_{1} = ... = \lambda_{n}=0[/tex]. This makes sense to me and I understand how it is an analogue to the definition of linear independence, which I would like to think I understand quite well.

    It then states that, "affine independence of an n-family (x_{1},...,x_{n}) is equivalent to linear independence of one/all of the (n-1)-families:

    I'm not immediately seeing this connection, if someone would be able to help explain it to me or send me a link so that I could understand it better, that would help. I have checked Wikipedia, MathWorld, etc. already.

    Thank you very much,
  2. jcsd
  3. Nov 10, 2011 #2
    That's not right. Your definition of affine independence is just linear independence. You want the sum of the coefficients to be 1, not 0.

    Basically, affine independence is a version of linear independence that doesn't depend on basepoints. You have n+1 points. Take one of them as a reference point. That gives you n vectors. You want affine independence to mean that those vectors should be linearly independent.

    It helps to think of things in terms of centers of mass. You have a n+1 points. You can give each of them a mass, and imagine that the total mass has to be 1. When you vary the masses at the various points, the set of possible centers of mass forms a simplex. This is what happens when you allow only non-negative coefficents. Allowing negative coefficients gives you the smallest plane containing the n+1 points.

    Notice that if you have 3 points lying in a line, there will be more than one way (actually, a 1-parameter family of ways) to make a points in the segment containing them into a center of mass. And so on.

    So, it should be plausible that affine independence of n+1 points is equivalent to linear independence of the associated n vectors, taking one of the points as a reference point.
  4. Nov 10, 2011 #3
    Thank you very much, that clarifies a lot.
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