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I'm really confused by the definition of general position at wikipedia.

I understand that the set of points/vectors in $\mathbb R^d$ is in general position iff every $(d+1)$ points are not in any possible hyperplane of dimension $d$.

However I found that this definition is equivalent to affine independence (according to wiki). Does general linear position mean something else?

Could you please explain that? It is extremely confusing since a lot of people omit "linear" and so on.

Anyway could you also please give some hints on the way of proving general position? The hyperplane definition is hard to use.

gebruiker
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petrbel
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2 Answers2

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$\def\R{\mathbb{R}}$General position is not equivalent to affine independence. For instance, in $\R^2$, you can have arbitrarily many points in general position, so long as no three of them are collinear. But you can have at most 3 affinely independent points.

Any random set of points in $\R^d$ is almost certain (with probability 1) to be in general position. The points have to be carefully arranged in order not to be in general position—for instance, by having three on one line, or four in one plane in $\mathbb{R}^3$ and above.

Now, if you have $d+1$ points in general position in $\R^d$, then they affinely span $\R^d$, so they are affinely independent; and conversely, any set of affinely independent points must be in general position. But you can have sets in general position with far more than $d+1$ points, and they won't be affinely independent.

Nick Matteo
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  • Or in other words, a set of $\ge n+1$ points in $\mathbb{R}^n$ are in general position if every $n+1$ subset of them are affinely independent. But for sets of size $ \le n$, you want the original definition i.e. forall $k \in [2, n+1]$, no k subset of them lie in a $(k-2)$ dimensional flat. – Anon Apr 25 '23 at 14:22
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Kundor's answer is correct, but I think this should be added for completeness.

Take a set of points $P \subset \mathbb R^d$. Then $P$ is in general linear position if and only if each k-tuple of points from $P$, $k=2,\dots,d+1$, is affinely independent.

But, as the OP notes, we only really need to check whether no $d+1$ points lie on a hyperplane of dimension $d$. This is equivalent with checking whether no $d+1$ points are affinely dependent.

Dahn
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  • I'm sorry for commenting after so many years, I don't even know if you'll know the answer to this, but where can I find a proof of this statement? I've been trying to find it with no success. – user8785084 Apr 03 '23 at 01:06
  • @user8785084 which of the statements in the answer? – Dahn Apr 03 '23 at 14:29
  • The one highlighted. "Take a ser of points $P\subset\mathbb{R}^d$. Then $P$ is in general position...$ – user8785084 Apr 03 '23 at 18:57