Convex Hull

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August 30th 2010, 01:22 AM
qspeechc

Convex Hull

Hello Everyone.

While I was reading along in a book I came across this: It says the convex hull is
$co(\{ x_n : n\in \mathbb{N}\}) = \{ \sum _{i=1}^m \alpha _i x_i : \sum |\alpha _i|\leq 1\}$
The $x_n$ are vectors in a vector space and the $\alpha _n$ complex scalars. But from my understanding (and wikipedia verifies this) the definition of a convex hull requires $\sum |\alpha _i|=1$ . Normally I'd just think this was a typo, but the text goes on to use the fact that $\sum |\alpha _i|\leq 1$ . So are the two things equivalent or what? I can't see how they are. (Crying). Help please!
August 30th 2010, 05:19 AM
Iondor

The definitions are only equivalent, if 0 is one of the $x_{n}$ .
I guess one could prove the following:
$co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$
August 31st 2010, 08:23 AM
qspeechc

No 0 is not one of them. I do not see how $co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$ helps...?
August 31st 2010, 11:17 AM
Iondor

Quote:

Originally Posted by qspeechc

No 0 is not one of them. I do not see how $co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$ helps...?

What exactly is your problem?
The two definitions are not equivalent.
The exact relationship between the definitions is expressed by above equality.
What more do you want?