The intuition behind this explicit form of an affine hull? I have come across an explicit formula of an affine hull from here. I'm trying to prove that formula in 2. below. However, my approach is quite muscular and not natural, i.e. I have to know the formula beforehand in order to prove it.
Could you elaborate on an intuition behind such formula and another approach to obtain that it?

Let $V$ be a vector space. A set $A \subset V$ is called an affine subspace if $A = S+a$ for some vector subspace $S\subset V$ and $a\in V$.


*

*Let $(A_i)_{i\in I}$ be a collection of affine subspaces of $V$. Then $\bigcap_{i\in I} A_i$ is also an affine subspace.


Assume $A_i = S_i + a_i$ for some vector subspace $S_i \subset V$ and $a_i\in V$. If $\bigcap_{i\in I} A_i = \emptyset$. Then we are done. If not, there is $x \in \bigcap_{i\in I} A_i$. It's clear that $A_i = x+S_i$. Then $\bigcap_{i\in I} A_i = \bigcap_{i\in I} (x+S_i) = x+\bigcap_{i\in I} S_i$. Because $\bigcap_{i\in I} S_i$ is a vector subspace, the claim then follows.



*Let $B \subset V$. The affine hull of $B$, denoted by $\operatorname{aff} (B)$ is defined as the smallest affine subspace containing $B$. Then
$$
\operatorname{aff}(B)= C:=\left\{\sum_{i=1}^{k} a_{i} x_{i} \,\middle\vert\, k \in \mathbb N^*, x_{i} \in B, a_{i} \in \mathbb{R}, \sum_{i=1}^{k} a_{i}=1\right\}.
$$


*

*First, we show that $C$ is indeed an affine subspace.

We fix some $b \in B$. We will show that $C-b$ is a vector subspace. Let $c_1, \ldots, c_n \in C$, i.e.,
$$
c_j = \sum_{i=1}^{k_j} a_{j,i} x_{j,i}
$$
with $k_j \in \mathbb N^*, x_{j,i} \in B, a_{j,i} \in \mathbb{R}, \sum_{i=1}^{k_j} a_{j,i}=1$.
It suffices to show that $\sum_{j=1}^n \lambda_j (c_j-b)  \in C-b$ for any $\lambda_1, \ldots, \lambda_n \in \mathbb R$. This is equivalent to showing that $\sum_{j=1}^n \lambda_j (c_j-b) +b \in C$. In fact, we have
$$
\begin{align}
\sum_{j=1}^n \lambda_j (c_j-b) +b &= \sum_{j=1}^n \lambda_j \left ( \sum_{i=1}^{k_j} a_{j,i} x_{j,i} - b \right) +b \\
&= \sum_{j=1}^n \lambda_j \sum_{i=1}^{k_j}  a_{j,i} x_{j,i} + \left (1- \sum_{j=1}^n \lambda_j\right)b.
\end{align}
$$
Notice that $\sum_{i=1}^{k_j} a_{j,i}=1$, so the sum of coefficients is
$$
\sum_{j=1}^n \lambda_j \sum_{i=1}^{k_j}  a_{j,i} + \left (1- \sum_{j=1}^n \lambda_j\right) = \sum_{j=1}^n \lambda_j + \left (1- \sum_{j=1}^n \lambda_j\right) = 1.
$$

*

*Second, we show that if $A$ is a affine subspace containing $B$, then $A$ contains $C$.

Assume $A = S+a$ for some vector subspace $S\subset V$ and $a\in V$. Let $c\in C$, i.e., $c=\sum_{i=1}^{k} a_{i} x_{i}$ with $k \in \mathbb N^*, x_{i} \in B, a_{i} \in \mathbb{R}, \sum_{i=1}^{k} a_{i}=1$. To prove that $c\in A$, we will show that $c-a \in S$. We have
$$
c-a = \sum_{i=1}^{k} a_{i} x_{i}-a = \sum_{i=1}^{k} a_{i} (x_{i}-a).
$$
We have $B \subset A = S+a$, so $B-a \subset S$. Hence $x_i-a \in S$. The claim then follows.
 A: Thanks to @dxiv's comment, the idea gets clearer to me. It lies in the characterization of an affine subspace by affine combination.

Let $V$ be a vector space. A set $A \subset V$ is called affine (subspace) if $A = S+a$ for some vector subspace $S\subset V$ and $a\in V$. The following statements are equivalent.

*

*$A$ is affine

*$\forall x,y\in A, \forall t\in \mathbb R: tx+(1-t)y \in A$.

*$\forall x_i \in A, \forall t_i \in \mathbb R \text{ s.t. } \sum_i t_i=1: \sum_i t_i x_i \in A$.

The proof $(2) \iff (3)$ is given here.

a. $(1) \implies (2)$.

Assume $A$ is affine, i.e., $A = S+a$ for some vector subspace $S\subset V$ and $a\in V$. Let $x,y\in A$. Then $x=s_1+a$ and $y=s_2+a$ for some $s_1, s_2\in S$. For all $t\in \mathbb R$, we get
$$
tx+(1-t)y = t(s_1+a)+(1-t)(s_2+a) = [ts_1+(1-t)s_2] +a \in S+a=A.
$$

b. $(2) \implies (1)$.

Assume $A$ such that
$$
\forall x,y\in A, \forall t\in \mathbb R: tx+(1-t)y \in A.
$$
We fix some $a\in A$. We will prove that $S:=A-a$ is a vector subspace. For all $t_1,t_2 \in \mathbb R$ and $a_1, a_2\in A$, we have
$$
t_1(a_1-a)+t_2(a_2-a) = t_1a_1+t_2a_2+[1-(t_1+t_2)]a-a \in A-a
$$
This is because $t_1+t_2 + [1-(t_1+t_2)]=1$ and thus $t_1a_1+t_2a_2+[1-(t_1+t_2)]a\in A$.
