drowsy bane
Let S = {P0,...,Pn} be a subet of A, an affine space (A,v, phi). Then show that if the subset L = { P0P1, P0P2, ... P0Pn} of the vector space V associated with the affine space A is LINEAR independent then S is affine independent (i.e. For any point P in S we have that P is not in the affine span of (S \ {P}) )
Here is my proof, but I am not sure if it's good. If anyone could correct me or give me a correct proof I would greatly appreacite it
Proof by contradiction:
Assume for the sake of contradiction that S is NOT affinr independent <=> There exists a point Pk in S such that Pk IS in the affine span of (S \ {Pk}) <=>
There exists a point Pk in S such that there exists a0, ... ,a(k-1), a(k+1), ... , an in K ( K is the field that the vector space V is defined on) such that their sum is 1 and Pk = a0*P0 + ... + a(k-1)*P(k-1) + a(k+1)P(k+1) + ... + anPn
Since Pk is an affine combination of the points P0,...,P(k-1),P(k+1),...Pn then if we take P0 as origin and move the Pk on the right of the equation we get :
0= a0* P0P0 + a1*P0P1 + ... + a(k-1)*P0P(k-1) - Pk + a(k+1)P0P(k+1) + ... + anP0Pn
We can ignore the a0*P0P0 since the vector from a point and itself is the null vector so that thing it's just 0
Since we know that L is linesr independent then that whole thing of sum of vectors of L = 0 must imply that a0 =...= ak =...=an = 0 but we can see that ak = -1 and this is a contradiction. Then S MUST be affine independent!
placid mountain