Why affinely independent points in $\mathbb{R^d}$ is $d+1$?

Question

Below text quoted from Jiff Matousek book:

Affine dependence of $a_1 ,\dots, a_n$ is equivalent to linear dependence of the $n-1$ vectors $a_1 - a_n, a_2 - a_n, \dots, a_{n-1}-a_n$ . Therefore, the maximum possible number of affinely independent points in $\mathbb{R^d}$ is $d+1$.

My question is why affinely independent points in $\mathbb{R^d}$ is $d+1$?

( We know that for linear dependence there are at least $d+1$ such points. The dimension of the space is $d$, so the maximum linearly independent set can have only $d$ elements. Hence, one of them is linearly dependent on the others.)

If use the logic of linear dependence, there should be $d-1$ points are independent in affine subspace in $\mathbb{R^d}$ because $d$(which $n$ points) points are dependent.

Answer 1

The $n$ vectors $a_1, \dots, a_n$ are affine dependent if the $n-1$ vectors $a_1 - a_n, \dots, a_{n-1} - a_n$ are linearly dependent.

In $\mathbb{R}^d$ we can have at most $d$ linearly independent vectors. Hence, we can have at most $d+1$ affinely independent vectors.

Answer 2

As you point out, there can be at most $d$ linearly independent elements in a subset of a $d$-dimensional vector space. If $\{a_0, \ldots, a_{n-1}\}$ are linearly independent, then the set $\{a_0, \ldots, a_{n-1}, a_n = 0\}$ is affinely indepnedent by the definition quoted in the question. Conversely, if $\{ a_0, \ldots, a_k \}$ is affinely independent, then $\{ a_0 - a_k, \ldots, a_{k-1} - a_k \}$ is linearly independent; there are $k-1$ such vectors, so $k-1 \le d$ and hence $k \le d+1$. Thus $d+1$ is the most possible such vectors in an affinely independent set.

Answer 3

Think of $a_1$ as a choice of origin. So it's $d$ linearly independent vectors and one additional vector to define the origin for a total of $d+1$.