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I'm learning affine geometry and came across the following statement while reading the introductory paragraph on affine independence :

$\mathbb{R}^n$ contains at most $n+1$ affinely independent points.

Coming for linear algebra, the above statement is a little counterintuitive to me since $n+1$ vectors in $\mathbb{R}^n$ cannot be linearly independent. A proof of the later proposition can be found here.

I was wondering how one might prove the former statement. The definition I'm familiar for affine independence is the following :

A finite collection of points $x_0, x_1, \ldots, x_k$ are affinely independent if the vectors $x_1 - x_0, \ldots, x_k - x_0 \in \mathbb{R}^n$ are linearly independent.

glS
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2 Answers2

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Yes, so $n+1$ points can be affinely independent, since $n$ vectors can be linearly independent. For example, if $\{e_1,..,e_n\}$ denotes the basis vectors of $\mathbb R^n$, then the vectors $\{e_1,2e_1,e_2+e_1,...,e_n+e_1\}$ form an affinely independent subset, since if we take $x_0 = e_1$, then the vectors $e_1,e_2,...,e_n$ are generated after subtracting the first vector from every other vector, which are linearly independent. Hence, the given vectors are affinely independent.

But if we take more than $n+1$ points, then the number of vectors generated by the differences will be more than $n$, so cannot be linearly independent. Hence, atmost $n+1$ vectors can constitute an affinely independent set.

Note : The choice of $x_0$ does not affect affine independence i.e. if I had chosen some other vector instead of $e_1$ as $x_0$ and taken differences, it would not have affected linear independence. I leave you to see why, it is not so difficult.

Sarvesh Ravichandran Iyer
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Technically speaking, the answer is right there in the definition: if the $k+1$ vectors $x_0, \ldots, x_k$ are affinely independent then the $k$ vectors $x_1 - x_0, \ldots, x_k - x_0$ are linearly independent, so since we're in $\mathbb{R}^n$ we must have $k \leq n$ and hence $k+1 \leq n+1$.

Geometrically, $k$ vectors are linearly independent if and only if there is no hyperplane of dimension $d < k$ through the origin in $\mathbb{R}^n$ which contains them. On the other hand $k$ vectors are affinely independent if and only if there is no hyperplane of dimension $d < k$ (not necessarily passing through the origin) in $\mathbb{R}^n$ which contains them. This suggests that you can pass from a maximal linearly independent set to a maximal affinely independent set simply by adding the origin to the set: since no hyperplane containing the origin contained all of the points before adding the origin, no hyperplane whatsoever contains all of the points after adding the origin. This is intuitively why you get to have an extra point.

Paul Siegel
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