Prove that $\{\vec x_i\}\subset\mathbb R^d$ is affinely independent iff $\{(1,\vec x_i)\}$ is linearly independent I guess I'm having some trouble getting my head around the notion of affine independence.  As I've been taught, a set of vectors $\{\vec{x_1},\ldots,\vec{x_n}\}\subset \mathbb{R}^d$ is affinely independent if its affine hull has dimension $n-1$. I want instead to think about linearity in $\mathbb{R}^{d+1}$, and I'm told I can.
The claim is that my set of vectors is affinely independent if and only if $\{\hat{x_i}\}$ is linearly independent, where $\hat{x_i}=(1,\vec{x_i})$.  Any tips for proving such a thing?
 A: The points $x_k$ are affine independent when 
$$
 \sum \lambda_k x_k = 0 \text{ with }\sum \lambda_k =0
$$
implies all $\lambda_k = 0$.
The vectors $\hat x_i = (1, x_i)$ are linearly independent if
$$
  \sum \lambda_k \hat x_k = 0
$$
implies $\lambda_k=0$ for all $k$. But since the first component of $\hat x_k$ is always $1$ the sum of the first components is $\sum \lambda_k$ which needs to be zero. So the same coefficients can be used to prove/disprove affine independency.
edit
To understand the definition of affine independency: suppose that there exist $\lambda_k$ such that $\sum_k \lambda_k = 0$ and 
$$
  \sum_k \lambda_k v_k = 0.
$$
If the coefficients $\lambda_k$ are not all equal to zero there exists one which is different from $0$. Suppose $\lambda_1 \neq 0$. By dividing all by $-\lambda_1$ you can also suppose that $\lambda_1 = -1$. This means that $\lambda_2+\dots+\lambda_n=1$ and that 
$$
  v_1 = \lambda_2 v_2 + \dots + \lambda_n v_n
$$
i.e. $v_1$ is in the affine hull of $\lambda_2,\dots, \lambda_n$. So the hull cannot be $n-1$ dimensional.
