Relating Linear Independence to Affine Independence Question: (From an Introduction to Convex Polytopes)
Let $(x_{1},...,x_{n})$ be an $n$-family of points from $\mathbb{R}^d$, where $x_{i} = (\alpha_{1i},...,\alpha_{di})$, and $\bar{x_{i}} =(1,\alpha_{1i},...,\alpha_{di})$, where $i=1,...,n$. Show that the $n$-family $(x_{1},...,x_{n})$ is affinely independent if and only if the $n$-family $(\bar{x_{1}},...,\bar{x_{n}})$ of vectors from $\mathbb{R}^{d+1}$ is linearly independent.
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Here is what I have so far, it is mostly just writing out definitions, if you can give me some hints towards how I can start the problem that would be great.
$(\Rightarrow)$ Assume that for $x_{i} = (\alpha_{1i},...,\alpha_{di})$, the $n$-family $(x_{1},...,x_{n})$ is affinely independent. Then, a linear combination $\lambda_{1}x_{1} + ... + \lambda_{n}x_{n} = 0$ can only equal the zero vector when $\lambda_{1} + ... + \lambda_{n} = 0$. An equivalent characterization of affine independence is that the $(n-1)$-families $(x_{1}-x_{i},...,x_{i-1}-x_{i},x_{i+1}-x_{i},...,x_{n}-x_{i})$ are linearly independent. We want to prove that for $\bar{x_{i}}=(1,\alpha_{1i},...,\alpha_{di})$, the $n$-family $(\bar{x}_{1},...,\bar{x}_{n})$ of vectors from $\mathbb{R}^{d+1}$ is linearly independent.
 A: ($\Rightarrow$): Suppose $(\bar{x_1},\ldots,\bar{x_n})$ is linearly dependent, so we have $$\left(\sum\limits_{i=1}^n c_i,\sum\limits_{i=1}^n c_i\alpha_{1i},\ldots,\sum\limits_{i=1}^n c_i\alpha_{ni}\right) = \sum\limits_{i=1}^nc_i\bar{x_i} = 0$$
for some set of coefficients $c_i\in\mathbb{R}$, thus
$$\sum\limits_{i=1}^nc_ix_i = \left(\sum\limits_{i=1}^n c_i\alpha_{1i},\ldots,\sum\limits_{i=1}^n c_i\alpha_{ni}\right) = 0\text{ and }\sum\limits_{i=1}^n c_i = 0$$
so $(x_1,\ldots,x_n)$ is affinely dependent. Hence if $(x_1,\ldots,x_n)$ is affinely independent, $(\bar{x_1},\ldots,\bar{x_n})$ must be linearly independent.
($\Leftarrow$): Suppose $(x_1,\ldots,x_n)$ is affinely dependent, so we have $$\sum\limits_{i=1}^nc_ix_i = 0\text{ and }\sum\limits_{i=1}^nc_i=0$$ for some set of coefficients $c_i\in\mathbb{R}$. Then $$\sum\limits_{i=1}^nc_i\bar{x_i} = \left(\sum\limits_{i=1}^n c_i,\sum\limits_{i=1}^n c_i\alpha_{1i},\ldots,\sum\limits_{i=1}^n c_i\alpha_{ni}\right) = (0,0,\ldots,0) = 0$$ so $(\bar{x_1},\ldots,\bar{x_n})$ is linearly dependent. Hence if $(\bar{x_1},\ldots,\bar{x_n})$ is linearly independent, $(x_1,\ldots,x_n)$ must be affinely independent.
A: So, we want to prove that these two statements are equivalent:


*

*(a) The points $x_1, \dots , x_n \in \mathbb{R}^d$ are affinely independent.

*(b) The vectors $\overline{x}_1, \dots , \overline{x}_n \in \mathbb{R}^{d+1}$ are linearly independent.
Where $\overline{x}_i = (1, x_i),\ i = 1, \dots , n$.
Let's go.
$\mathbf{(a)\Longrightarrow (b)}$. Let $\lambda_1, \dots , \lambda_n \in \mathbb{R}$ be such that
$$
\lambda_1 \overline{x}_1 + \dots + \lambda_n \overline{x}_n = 0 \ .   \qquad \qquad \qquad [1]
$$
We have to show that $\lambda_1 = \dots = \lambda_n = 0$. But $[1]$ means
$$
\lambda_1 (1, x_1) + \dots + \lambda_n (1, x_n) = (0, 0) \ ,
$$
where $(0,0) \in \mathbb{R} \times \mathbb{R}^d$. And this is equivalent to
$$
\lambda_1 x_1 + \dots + \lambda_n x_n = 0  \qquad \text{and} \qquad \lambda_1 + \dots + \lambda_n = 0 \ .
$$
Now, $x_i = x_i - 0 = \overrightarrow{0x_i} , \ i = 1, \dots , n$. (Here, $0 \in \mathbb{R}^d$.) So, since we are assuming $(a)$, it follows that 
$$
\lambda_1 = \dots = \lambda_n = 0 \ .
$$
$\mathbf{(b)\Longrightarrow (a)}$. Let $p \in \mathbb{R}^d$ be any point. We have to show that
$$
\lambda_1 \overrightarrow{ px}_1 + \dots + \lambda_n \overrightarrow{ px}_n = 0  \qquad \text{and} \qquad \lambda_1 + \dots + \lambda_n = 0  \qquad \qquad \qquad [2]
$$
implies $\lambda_1 = \dots = \lambda_n = 0$.
If the point $p$ was $0 \in \mathbb{R}^d$, the conclusion should be clear because, in this case, $\overrightarrow{px_i} = x_i, \ i = 1, \dots , n$, and $[2]$ reads as follows:
$$
\lambda_1 x_1 + \dots + \lambda_n x_n = 0  \qquad \text{and} \qquad \lambda_1 + \dots + \lambda_n = 0 \ .  \qquad \qquad \qquad [3]
$$
From here, we do the same reasoning as in the previous proof, but backwars: these two things entail
$$
\lambda_1 (1, x_1) + \dots + \lambda_n (1, x_n) = (0, 0) \ .
$$
Which is the same as
$$
\lambda_1 \overline{x}_1 + \dots + \lambda_n \overline{x}_n = 0 \ .
$$
And this implies 
$$
\lambda_1 = \dots = \lambda_n = 0\ ,
$$
since we are assuming $(b)$. 
Hence, we have to show that the particular case $[3]$ already implies the general one $[2]$, for every $p\in \mathbb{R}^d$. But this is obvious:
$$
\lambda_1 \overrightarrow{ px}_1 + \dots + \lambda_n \overrightarrow{ px}_n = \lambda_1 (x_1 -p ) + \dots + \lambda_n (x_n - p) 
$$
Which is
$$
\lambda_1 x_1 + \dots + \lambda_n x_n - (\lambda_1 + \dots + \lambda_n)p = \lambda_1 x_1 + \dots + \lambda_n x_n = 0 \ .
$$
