Convex Geometry: Affine & Linear Dependency's Relationship

Question

The question is as follows:

Let $x_1, x_2, ..., x_n \in \mathbb{R}^d,$ and for every $i = 1, 2, . . . , n,$ let $y_i = (x_i , 1) \in \mathbb{R}^{d+1}$. Show that $x_1, x_2, ... , x_n$ are affinely dependent if and only if $y_1, y_2, ..., y_n$ are linearly dependent. What is the conclusion about the maximal number of affinely independent points in $\mathbb{R}^d$?

I think it is meant to be inherently simple, but I am having a hard time formulating a proof. We know that there exist scalars $\alpha_1, \alpha_2, ..., \alpha_n$ such that $$\alpha_1x_1 + \alpha_2x_2 + ... + \alpha_nx_n = 0$$ $$\sum^{n}_{i=1}\alpha_ix_i = 0, \sum^n_{i=1}\alpha_i = 0$$ Are these the same scalars we are using on each $y_i$ to eventually show linear dependency? This is only in reference to the first relation of the question.

Any help is appreciated! Much thanks.

Answer 1

Hint: If $\alpha_i\ne 0$, you can express $x_i$ as the affine combination of the other $x_j$'s. The converse will provide you nontrivial $\alpha_i$ satisfying your conditions.

Answer 2

What is the conclusion about the maximal number of affinely independent points in $\mathbb R^d$?

You can streamline this using matrices

$X := \bigg[\begin{array}{c|c|c|c|c} \mathbf x_1 & \mathbf x_2 &\cdots & \mathbf x_{n-1} & \mathbf x_{n} \end{array}\bigg] $

and
$Z := \begin{bmatrix} \mathbf 1_n^T \\ X\end{bmatrix}$
(you can cyclicaly permute the rows to put the ones on the bottom row if you want)

The problem is equivalent to asking about $\mathbf a \neq \mathbf 0$ such that

$Z\mathbf a = \mathbf 0$

since $\mathbf x_k \in \mathbb R^d$ this means $Z$ has $d+1$ rows. How many columns can it have before it must have linearly dependent columns / a nontrivial kernel? (note it could be useful to think of an example where $\mathbf x_k$ are distinct points on the moment curve and hence if $Z$ were a square matrix it would be Vandermonde)