Affine hull and affine combinations equivalence

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
Mentor
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.

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
Mentor
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.

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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
Mentor
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
Mentor
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})\,.##