# Affine dimension of a simplex

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?

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.

• This makes sense......Thank you..... – Shirin Feb 14 '13 at 16:14

$n$ 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.

• 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..... – Shirin Feb 14 '13 at 15:08
• 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, $B-A+D-A=C-A$.) – Berci Feb 14 '13 at 15:10
• 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. – Shirin Feb 14 '13 at 15:20