Affinely independent Points

Question

In my lecture notes I got this definition. But I don't see how, this last sum can be found.

The points $x_1,\ldots,x_n\in\Bbb R^d$ are said to be affinely independent if the relations $$\sum_{i=1}^n\lambda_ix_i=0,\quad\sum_{i=1}^n\lambda_i=0$$ imply that $\lambda_1=\ldots=\lambda_n=0$. For any $x_r$, $r\in\{1,\ldots,n\}$ and $\sum_{i=1}^n\lambda_i=0$ it follows that $$0=\sum_{i=1}^n\lambda_ix_i=\sum_{i=1\\i\neq r}^n\lambda_i(x_i-x_r).$$

Answer 1

If $\sum\limits_{i=1}^{n} \lambda_i=0$ then $\sum\limits_{i=1}^{n} \lambda_i x_r=0$

If also $\sum\limits_{i=1}^{n} \lambda_i x_i=0$ then by subtraction $\sum\limits_{i=1}^{n} \lambda_i(x_i- x_r)=0$

Clearly $\lambda_r(x_r-x_r)=0$ too, so these together imply $\sum\limits_{\substack{i=1\\i \not = r}}^{n} \lambda_i(x_i- x_r)=0$

Answer 2

If $\sum_{i=1}^n\lambda_ix_i=0$ and $\sum_{i=1}^n\lambda_i=0$, then

$$\begin{align*} \sum_{\substack{i=1\\i\ne r}}^n\lambda_i(x_i-x_r)&=\sum_{\substack{i=1\\i\ne r}}^n\lambda_ix_i-x_r\sum_{\substack{i=1\\i\ne r}}^n\lambda_i\\ &=\left(\sum_{i=1}^n\lambda_ix_i-\lambda_rx_r\right)-x_r\left(\sum_{i=1}^n\lambda_i-\lambda_r\right)\\ &=(0-\lambda_rx_r)-x_r(0-\lambda_r)\\ &=0\,. \end{align*}$$