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There is a theorem that says: if $M=\{v_1,v_2,...,v_m\}$ are a set of vectors such that they are affine independent (i.e., $\{v_2-v_1,v_3-v_1,...,v_m-v_1\}$ are linearly independent), then for any vector $v$ that is in the convex hull of $M$, it can be written as a unique convex combination of the vectors in $M$.

My first question is: here the unique representation means $v=\sum_{j=1}^m K_jv_j$, where $\sum_{j=1}^mK_j=1$ and $K_j\geq 0,~\forall~1\leq j\leq m$. In particular, some of those $K_j$'s could be zero right?

My second question is: I couldn't feel the significance and importance of this theorem. Isn't that obvious? I mean, if we have a set of vectors, then we construct the convex hull of this set of vectors, which is taking all possible convex combinations of this set of vectors, then if you pick an arbitrary vector from this generated convex hull, it is for sure should be able to written as a unique representation of those vectors in the set that generates this convex hull. I guess my question is: what's the role played by this condition "$M$ is a set of affine independent vectors"? What if my set $M$ is not a set of affine independent vectors, then after I generate the convex hull (I can still do that by taking all possible convex combinations of the vectors in $M$), then if I pick an arbitrary vector from this convex hull, then why I cannot uniquely express this vector as a convex combination as those vectors in set $M$?

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Use the analogy to linear spaces:

If $b_1,...,b_n$ are linearly independent then $x \in \operatorname{sp} \{ b_k \}$ iff there are unique $\lambda_k$ such that $x =\sum_k \lambda_k b_k$.

Note that some of the $\lambda_k$ may be zero or negative.

Similarly, if $a_1,...,a_n$ are affinely independent then $x \in \operatorname{aff} \{ a_k \}$ iff there are unique $\mu_k$ such that $x =\sum_k \mu_k a_k$ and $\sum_k \mu_k = 1$.

Note that some (but not all) of the $\mu_k$ may be zero or negative.

In the latter case, if $x =\sum_k \mu_k a_k$, with $\sum_k \mu_k = 1$, then $x \in \operatorname{co} \{ a_k \}$ iff $\mu_k \ge 0$.

Notes:

A few related fact may help: (i) $\{ a_k \}_k$ are affinely independent iff $\{ \binom{1}{a_k} \}_k$ are linearly independent. (ii) $\{ a_k \}_k$ are affinely independent iff for any $k_0$ the collection $\{ a_k -a_{k_0} \}_{k \neq k_0} $are linearly independent

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    If I only have 2 different vectors in the set $M$, then by definition, they are affine independent right? Also, can you provide an example illustrating that if the set $M$ is not affine independent, then actually for a vector in $co\{M\}$, it can be written as two different convex combinations of the "pillar vectors" in $M$? – 2017-02-17
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    It is true, but not quite by definition, we have $\sum_{k=1}^2 \mu_k \binom{1}{ a_k} = 0$ **iff** $a_1 \neq a_2$. – 2017-02-17
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    Sure, take $M=\{1,2,3 \} \subset \mathbb{R}$, then $2 = {1 \over 2} 1 + {1 \over 2} 3 = 1 \cdot 2$. – 2017-02-17
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    Nice! I never thought points in R as vectors. But you are right. – 2017-02-17
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    Same applies in more complicated examples. Think of linear independence, you get some vector in the null space so you lose uniqueness of representation. The same sort of relationship holds with appropriate changes. – 2017-02-17
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    So if $M$ contains some "redudant" vectors in the sense of affine independence, then there is always a subset of vectors in $M$ denoted by $M'$ such that $M'$ contains only affine independent vectors and $co\{M'\}=co\{M\}$ and hence, we can uniquely express any vectors from $co\{M'\}$ (or equivalently $co\{M\}$) using convex combinations of the vectors in $M'$ right? – 2017-02-17
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    By analogy with the linear case, if you have $x = \sum_k \mu_k a_k \in \mathbb{R}^n$, with $\sum_k \mu_k = 1$, you can always write $x$ as an affine combination of at most $n+1$ of the $a_k$. This is given the fancy name of Carathéodory's theorem. – 2017-02-17
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    Hold on a second, you can always reduce $M$ so that the affine span is the same, but not necessarily the convex hull. Take the corners of a square in the plane. Any three are affinely independent, but you can't select three with the same convex hull. – 2017-02-17
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    I guess what I am trying to say is that the essential thing here is the affine nature rather than the convex nature. – 2017-02-17
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53786/discussion-between-kevinkim-and-copper-hat). – 2017-02-17
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    I checked the wiki for the Caratheodory Theorem and the picture on the wiki page. I have a question. So if $x\in R^d$ and $x\in co\{P\}$, then $x\in co\{P'\}$, where set $P'\subset P$ and only have $r\leq d+1$ elements. This $r$ is the Caratheodory number of $P$. Then my qeustion is, these $r$ vectors of $P'$ are not necessarily linear independent right? But are they always affine independent? The picture on Wiki looks confirm this claim. Also, if I choose a different $x'$, then it might corresponds to a different $P''$ which will likely generate a different $co\{P''\}$ that are not $co\{P'\}$ – 2017-02-17
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    You can always reduce the representation until the remaining vectors (that is, those with non zero multipliers) are affinely independent. Correct, the vectors are not necessarily linearly independent. Think of the corners of a square in the place, only if you are on the edges or diagonals will the representing points be linearly independent (and even there there are some caveats). – 2017-02-17
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    In the {(0,0),(1,0),(0,1),(1,1)} example, the convex hull is the unit square and these 4 vectors are not affine independent, but I feel that the point (0,0), which are in the convex hull (of course), but it does has unique convex representation of those 4 pillar vectors, right? which is putting all the weight to (0,0). Then does this contradict with the unique representation theorem? – 2017-02-17
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    No, the unique representation theorem that you are using just says that if $v_1,...,v_m$ are affinely independent, then any element of the convex hull (actually affine hull) has a unique representation in terms of the $v_k$. In the square example, you can have at most three affinely independent vectors (since you are in $\mathbb{R}^2$) and if the origin is in the convex hull of these vectors, it must, in this case, be one of the vectors, since the corners are extreme points. Hence the unique representation will be one times the origin plus zero times all the others. – 2017-02-17
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    So, the idea is, **first** fix the affinely independent set of vectors $v_1,...,v_m$ **then** compute the multipliers. These will be unique in terms of the $v_1,...,v_m$. – 2017-02-17
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    Think of affine as an 'origin free' version of linear. You can think of the multipliers are a probability measure on the supporting points. – 2017-02-17
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    I like your "probability measure" interpretation. It seems that "corner" point is special than "other" points (I am not sure what exactly I mean). I extend your previous example to M={0,1,2,3}, these 4 "vectors" are not affine independent, and $C=co\{M\}=[0,3]$, as in your example, $2\in C$ but 2=0.5*1+0.5*3=1*2. However, it seems that 0 only equal to 1*0. I am confused why 0 is so special, to me 0 is not very different from 2 in M. Why in a convex hull $C=co\{M\}$ generated by some set of nonaffine independent vectors M, then some points in C has unique convex representation of M, others not – 2017-02-17
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    In the $M$ example, the points $0,3$ are 'special' in that they are extreme. The corner points in the square example are also extreme. – 2017-02-17