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In Stephen Boyd's book on Convex optimization he points out that k+1 affinely independent points form a simplex with affine dimension k.

My understanding of affinely independent points is that no 3 points are in a line. So if I take 4 points no 3 of which are in a line in $R^2$ than I get a simplex of affine dimension 3.

How is it possible for a set to have dimension more than 2 in $R^2$?

Please correct me if I am wrong.

On further inspection I realized that Boyd says "affine dimension of simplex". Now simplex is a convex set and affine dimension should be defined for an affine set. Isn't that correct?

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They're affinely independent if none of them is in the affine space spanned by the others, i.e. the smallest affine space containing the others. Three points are in a plane, so the fourth point must not be in the same plane as the first three.

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  • $\begingroup$ This makes sense......Thank you..... $\endgroup$ – Shirin Feb 14 '13 at 16:14
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$n$ points are affinely independent iff no $3$ of them lie on a line, no $4$ of them lie on a plane, no $5$ of them lie on a 3d subspace, and so on..

Exactly, as you considered, it is possible to have $4$ points in one plane such that no $3$ of them lie on a line (e.g. a square). But they are still affinely dependent.

A precise definition of $P_0,P_1,\ldots,P_n$ points being affinely independent is that the collection of vectors $\vec{P_0P_1},\ldots,\vec{P_0P_n}$ is linearly independent.

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  • $\begingroup$ I still don't get what you mean when you say that the vertices of a square are still affinely dependent. Lets say I have 2 independent vectors in $R^2$. Its possible for me to express any other vector in $R^2$ using these 2 vectors by $\alpha*v_1 + \beta*v_2$. However if I take 2 points (1,1) and (2,3), they are affinely independent according to me. Now if i try to express a third point, lets say (3,1) using these 2 and satisfy the condition that $\alpha + \beta$ = 1, I can't....please point out where I am wrong..... $\endgroup$ – Shirin Feb 14 '13 at 15:08
  • $\begingroup$ Yes. But a square has $4$ points, call them $A,B,C,D$, $A$ opposite to $C$, then $B+D-A=C$, affinely dependents. (Because $\vec{AB}+\vec{AD}=\vec{AC}$, that is, $B-A+D-A=C-A$.) $\endgroup$ – Berci Feb 14 '13 at 15:10
  • $\begingroup$ So you are saying if I have n points I should rather think how many independent vectors can I make out of them to work out the affine dimension rather then going straight ahead and trying to check if I can express some points in terms of other points. $\endgroup$ – Shirin Feb 14 '13 at 15:20

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