Affine hull and affine combinations equivalence

[Mr Davis 97]

Let $X = \{x_1 , \dots , x_n\}$. Then $\text{aff}(X) = \text{intersection of all affine spaces containing X}$. Let $C(X)$ be the set of all affine combinations of elements of $X$. We want to show that these two sets are equal. First we focus on the $\text{aff}(X) \subseteq C(X)$ inclusion. If we can show that $C(X)$ is affine and that it contains $X$ then this inclusion will hold. Let's just say that it's obvious that it contains $X$. So I want to prove that $C(X)$ is affine. Here is where I hit a roadblock. In my reference it only states that an affine space is a translate of a linear subspace. This is really the only definition I am given. Does this mean that I have to show that $C(X)$ is the translate of some linear subspace?

 

[fresh_42]

Let $X = \{x_1 , \dots , x_n\}$. Then $\text{aff}(X) = \text{intersection of all affine spaces containing X}$. Let $C(X)$ be the set of all affine combinations of elements of $X$. We want to show that these two sets are equal. First we focus on the $\text{aff}(X) \subseteq C(X)$ inclusion. If we can show that $C(X)$ is affine and that it contains $X$ then this inclusion will hold. Let's just say that it's obvious ...

or just note that $x_i=0+x_i$ is an affine combination

... that it contains $X$. So I want to prove that $C(X)$ is affine. Here is where I hit a roadblock. In my reference it only states that an affine space is a translate of a linear subspace. This is really the only definition I am given. Does this mean that I have to show that $C(X)$ is the translate of some linear subspace?

Yes, and again $0 + \operatorname{span}(X)$ is a linear and as such an affine space, too.

 

[Mr Davis 97]

or just note that $x_i=0+x_i$ is an affine combination
Yes, and again $0 + \operatorname{span}(X)$ is a linear and as such an affine space, too.

Are you saying that $C(X) = 0 + \text{span}(X)$?

 

[fresh_42]

Are you saying that $C(X) = 0 + \text{span}(X)$?

You're right, that wasn't a valid argument. But $0+\operatorname{span}(X) \subseteq C(X)$ whereas $0+\operatorname{span}(X) \nsubseteq \operatorname{aff}(X)$. What do I miss here? Are you sure the equation holds? I guess I'm currently a bit confused.

Edit: Assume we have two points in the plane. Then $\operatorname{aff}(\{x,y\})$ is the one straight line through $x$ and $y$. But any affine combination $\alpha x + \beta y = c$ yields the entire plane.

 

[member 587159]

Are you saying that $C(X) = 0 + \text{span}(X)$?

What is span(X)? X are elements in an affine space: writing span around them makes no sense.

Can you give your definition of affine space? I saw it as a set on which a vector space acts satisfying some axioms.

 

[fresh_42]

What is span(X)? X are elements in an affine space: writing span around them makes no sense.

It does. The linear span of all points in $X$ is still an affine space.

 

[member 587159]

It does. The linear span of all points in $X$ is still an affine space.

Span is something we write for vectors. The elements of X are not necessarily vectors if one treats the theory of affine geometry generally. It's why I asked for the OP's definition.

 

[fresh_42]

Span is something we write for vectors. The elements of X are not necessarily vectors if one treats the theory of affine geometry generally. It's why I asked for the OP's definition.

$\vec{0x_i}$ are the corresponding vectors. A space is affine if all $\vec{0x_i} - \vec{0x_j}$ are within. This is the case for $(0+\vec{0x_i}) - (0+\vec{0x_j})$. Nevertheless, I'm not sure what is meant by affine combination, other than all possible $\vec{c} + \sum_i \alpha_i \,\vec{0x_i}$ which is possibly wrong here. It should presumably be $\sum_{i,j} \alpha_{ij} \,(\vec{0x_i}-\vec{0x_j})\,.$