# Affine Independence $\iff$ 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?

• Hint: The affine hull is the set of linear combinations $\sum t_i \vec{x_i}$ with $\sum t_i = 1$. Now consider taking all linear combinations of the $(1,\vec{x_i})$ and seeing which lie in the affine subspace of points of $\mathbb{R}^{d+1}$ whose first coordinate is $1$. This is the same set of linear combinations! – Omar Antolín-Camarena Feb 7 '13 at 19:15

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.
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.