I am wondering about affinely independent and just linearly independent. On Wikipedia it is explained that $u_i$ are affinely independent if $u_1 - u_0, ...,u_k -u_0$ are linearly independent. It is clear that if $u_i$ are linearly independent then $u_1 - u_0, ...,u_k -u_0$ are linearly independent. Is the other implication not also true? Then what is the difference between the two definitions?

  • $\begingroup$ What if $u_0$ itself is a linear combination of $u_i$'s? Then certainly, $\{u_i-u_0\}$ are not linearly independent. $\endgroup$ – Samrat Mukhopadhyay Jul 27 '13 at 11:06
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    $\begingroup$ affinely is for points whereas linearly is for vectors. $\endgroup$ – xavierm02 Jul 27 '13 at 11:09

The reverse implication is not true, consider the following three vectors in $\mathbb{R}^3$: $u_0=e_1, u_1=2e_1$ and $u_2=e_1+e_2$. The three vectors are affinely independent but not linearly independent.

The best way to understand the difference is from the picture on the wikipedia page. Think of $u_0$ as the "base vertex"and all the other $u_i$ as the positions of the other vertices. $u_i-u_0$ is then the position vector of all the other vertices relative to the base vertex, and we need these to be linearly independent (so that $u_0,\dots, u_k$ are affinely independent) so that we don't end up with three collinear vertices which would mess up our idea of what a simplex should be.

It should also be noted that the point $u_0$ is not special in the definition of affine indepenence, but in fact if $u_0,\dots,u_k$ are affinely independent then for anyfor any $j$, $u_0-u_j,\dots,u_k-u_j$ are all linearly independent. Check out this wikipedia page: http://en.wikipedia.org/wiki/Affine_space#Affine_combinations_and_affine_dependence

  • $\begingroup$ The thing is, those two are equivalent, so long as $u_i$ are in a vector space. In an abstract affine space, $u_i-u_0$ doesn't even make sense. $\endgroup$ – tomasz Jul 27 '13 at 11:33
  • $\begingroup$ @tomasz I'm a little confused, which two are equivalent? $\endgroup$ – Tom Oldfield Jul 27 '13 at 11:45
  • $\begingroup$ Linear independence of $u_1-u_0,\ldots,u_n-u_0$ and affine independence of $u_0,u_1,\ldots,u_n$. $\endgroup$ – tomasz Jul 27 '13 at 11:47
  • $\begingroup$ @tomasz I see, good point, I'll clarify the answer, thanks! (In fact, I just deleted the edit, I misinterpreted the "which means" part of the wiki article and as such the edit was unnecessary.) $\endgroup$ – Tom Oldfield Jul 27 '13 at 11:54
  • $\begingroup$ @TomOldfield If we only have 2 vectors, then are these two vectors affine independent by definition? $\endgroup$ – KevinKim Feb 17 '17 at 3:06

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