I cannot resist the temptation to explain another reason (maybe "the" reason) for this condition that, apparently, seemed so mysterious: namely, $\lambda_1 + \dots + \lambda_n = 0$.
In my previous answer, we have already shown two things:
- This condition is the one that makes the points $x_1, \dots , x_n$ to be affinely independent if and only if the vectors $\overrightarrow{x_1x_2 }, \dots ,\overrightarrow{x_1x_n} $ are linearly independent.
- Also, this condition makes the set $V = \left\{x = \lambda_1 x_1 + \dots + \lambda_n x_n \in \mathbb{R}^d \ \vert \ \lambda_1 + \dots + \lambda_n = 1 \right\}$ to be the smallest affine subspace containing all the points $x_i$.
But with just these two ideas in mind, it would be a legitimate doubt to ask ourselves: why do we need to define "affine independence" with that weird condition $\lambda_1 + \dots +\lambda_n = 0$ inside? Wouldn't be enough the common notion of "linear independence"? -Since, properly written, it's equivalent.
Indeed, one of the main reasons in mathematics to make a new definition is that it should save you time, not to lose it. So, let's see one natural next step in affine geometry where the notion of affine independence saves you time: it's the notion of barycentric coordinates.
So, assume your points $x_1, \dots , x_n$ are affinely independent. By definition, each point $x \in V$ can be written as
$ x = \lambda_1 x_1 + \dots + \lambda_n x_n \ \qquad \text{with} \qquad \lambda_1 + \dots + \lambda_n = 1 \ . $
But there is more. Because we have the following
Lemma. For each $x \in V$, these $(\lambda_1 , \dots , \lambda_n) \in \mathbb{R}^n$ are unique.
Proof. Assume we had
$ x = \lambda_1 x_1 + \dots + \lambda_n x_n \ \qquad \text{with} \qquad \lambda_1 + \dots + \lambda_n = 1 $
and also
x = \lambda'_1 x_1 + \dots + \lambda'_n x_n \ \qquad \text{with} \qquad \lambda'_1 + \dots + \lambda'_n = 1 \ .
Then, substracting both pairs of equations, we would get
0 = (\lambda_1- \lambda'_1) x_1 + \dots + (\lambda_n -\lambda'_n) x_n \ \qquad \text{with} \qquad (\lambda_1-\lambda'_1) + \dots + (\lambda_n-\lambda'_n) = 0 \ .
But our points $x_1, \dots , x_n$ are affinely independent. Hence, \lambda_1 = \lambda'_1, \dots , \lambda_n = \lambda'_n.
So, these $(\lambda_1 , \dots , \lambda_n)$ determine, and are determinated by, the point $x$ much in the same way as the coordinates of a vector with respect to a basis determine, and are determinated by, the vector. Indeed, these $(\lambda_1 , \dots , \lambda_n)$ are called the barycentric coordinates of the point $x$ with respect to the affine frame $x_1, \dots , x_n$ of $V$.
Of course, you have as a particular case $V = \mathbb{R}^d$. Then you would need $d+1$ affinely independent points to get an affine frame there.
It's good to have some intuition of how these barycentric coordinates work. For this, I suggest you the following
Exercises.
- Take $V = \mathbb{R}^1$. Then an affine frame is made out two different points, $p, q \in \mathbb{R}$ and every point $x \in \mathbb{R}$ can be written uniquely as
$ x = \lambda_1 p + \lambda_2 q \qquad \text{with} \qquad \lambda_1 + \lambda_2 = 1 \ . $
In this particular situation it's customary to write this simply as
$ x = t p + (1-t) q, \qquad t \in \mathbb{R} \ . $
So, the exercise is: assume for instance $p< q$ and place the points $x \in \mathbb{R}$ (on the left of $p$, between $p$ and $q$, on the right of $q$) according to its barycentric coordinates $(t, 1-t)$; i.e., the values of $t$.
- Take $V = \mathbb{R}^2$. Then an affine frame is made out of three points $p, q, r \in \mathbb{R}^2$ not lying in the same straight line and every point $x \in \mathbb{R}^2$ can be written uniquely as
$ x = \lambda_1 p + \lambda_2 q + \lambda_3 r \qquad \text{with} \qquad \lambda_1 + \lambda_2 +\lambda_3 = 1 \ . $
And the exercise is: place the points $x \in \mathbb{R}^2$ (inside the triangle determined by $p,q, r$, outside, lying in one of its edges) according to the values of its barycentric coordinates $(\lambda_1, \lambda_2, \lambda_3)$.