Definition: Let $v_0, v_1.. v_k$ be points in $\mathbb{R}^d$. These points are called affinely independent if there do not exist real numbers $\alpha_0, \alpha_1…\alpha_k$ that are not all zero such that $\sum_{i=0}^k \alpha_i v_i = 0$ and $\sum_{i=0}^k \alpha_i = 0$.
We need to prove the following:
The points $v_0, v_1… v_k$ are affinely independent if and only if the vectors $v_1 – v_0, v_2 -v_0… v_k – v_0$ are linearly independent.
Thank you so much.
Best Answer
Statement A: $v_0,v_1,...,v_k$ are affinely independent.
Statement B: $v_1-v_0,v_2-v_0,...,v_k-v_0$ are linearly independent.
First let us prove that $A \implies B$.
Consider some $\lambda_1,...,\lambda_k$, such that:
$$\sum_{i=1}^{k} \lambda_i (v_i - v_0) = 0 \tag{1}$$ We have to show that, if affine independence holds, all the cofficients ($\lambda$'s) must be zero.
Now, consider some $\lambda_0 \in \mathbb{R}$, such that: $$\sum_{i=0}^{k} \lambda_i = 0\tag{2}$$ Also, we have: $$\sum_{i=0}^{k} \lambda_i v_i = \sum_{i=1}^{k} \lambda_i (v_i - v_0) + (\sum_{i=0}^{k} \lambda_i)v_0 \tag{3}$$ Using equations 1 and 2, we observe that both terms on the RHS of (3) are zero. This means that: $$\sum_{i=0}^{k} \lambda_i v_i = 0 \tag{4}$$
From equations 2 and 4, we can say that $\lambda_i = 0$, $\forall i$ due to affine independence. Therefore, since all coefficients must be zero, this implies that B is true.
Now, let us prove the converse, that $B \implies A$.
Consider some $\lambda_0,\lambda_1,...,\lambda_k$, such that: $\sum_{i=0}^{k} \lambda_i v_i = 0$ and $\sum_{i=0}^{k} \lambda_k = 0$.
We have to show that all these coefficients must be zero under the condition of linear independence.
Using equation 3, and the above two conditions, we can conclude that $\sum_{i=1}^{k} \lambda_i (v_i - v_0) = 0$.
Therefore, due to linear independence of $(v_i - v_0)$, we conclude that: $$\lambda_1 = \lambda_2 = .... = \lambda_k = 0$$ Also, $\sum_{i=0}^{k} \lambda_k = 0 \implies \lambda_0 = 0$.
This proves that they are affinely independent (B is true).
We have shown $A \implies B$ and $B \implies A$.
$\therefore A \iff B$.