Show the set is affine subspace.

Question

About 2 days ago I was learning stuff about affine geometry and yesterday I got stuck with the following problem

Suppose that $S$ is a subset of affine space A. Show the set:
$$
\langle S \rangle \stackrel{\text{def}}{=} a + \text{span}\{\vec{ax}: x \in S\}, \text{ for some $a \in S$ }
$$

Does not depend on $a$ and also is the minimal affine subspace containing the set S. And also

$$
\langle S \rangle=\big\{ {\sum_i \lambda_i x_i}: \sum_i \lambda_i=1, \space x_i \in S \big\}
$$

My attempt

I know how to prove the second task:
$$
v \in \langle S \rangle \implies v=a+ \sum_i \lambda_i \vec{ax_i} \\
=a + \sum_i {\lambda_i (x_i-a)} \\
=(1-\sum_i \lambda_i)a+\sum \lambda_i x_i
$$

Assuming that we proved the set $\langle S \rangle$ does not depend on $a$ it is clear that $v$ does not depend on $a$. Hence $\sum_i \lambda_i=1$. Since $v$ is an arbitrary point from $S$ then it implies that the whole set $\langle S \rangle$ consists of linear combinations of points from $S$ with sum of coefficients equal to $1$.

But how to prove the first one? E.g. to show that the set

$$
\langle S \rangle = a + \text{span}\{\vec{ax}: x \in S\}, \text{ for some $a \in S$ }
$$

does not depend on $a$? It doesn't look straightforward to me.

Have you got any ideas?

Answer 1

$\langle S\rangle$ does not depend on $a$ because $a=b+\overrightarrow{ba}$ for all $b\in S$. But $\overrightarrow{ba}$ is in $\operatorname{span}\{\overrightarrow{ax}~\vert~x\in S\}$ (which I will call $V_a$ from now on). So $$a+V_a=b+\overrightarrow{ba}+V_a=b+V_a=b+V_b.$$ For the last equality you need to show that $V_a=V_b$ for all $a,b\in S$. Note that $V_a$ is spanned by vectors of the form $\overrightarrow{ax}$, while $V_b$ is spanned by vectors of the form $\overrightarrow{bx}=\overrightarrow{ax}-\overrightarrow{ab}$.

It is also the minimal affine subspace containing $S$ because:

1. It's obviously an affine subspace.
2. It clearly contains $S$.
3. If all points of $S$ are to be contained in this affine space, then its spanning vector space must contain vectors pointing between these points. So all the vectors which span $V_a$ must be contained in the vector space spanning any affine space containing $S$ anyway.

Also, disregard my comment that you can't multiply $a$ by a number since it's not a vector. You seem to be working with a definition of affine space where the underlying set is still the vector space, so it's fine.