Talk:Convex cone

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See Talk:Dual cone and polar cone. Oleg Alexandrov (talk) 03:54, 19 June 2007 (UTC)[reply]

Section on Convex Cone is lacking very much material. There is enough material in Dual Cone to warrant a separate entry. Dual cone should not be merged simply to make up for missing material on Convex Cone. —The preceding unsigned comment was added by Dattorro (talkcontribs).

Section Convex cones are linear cones should read:[edit]

If C is a convex cone, then for any positive scalar α and any x in C the vector αx = (α/2)x + (α/2)x is in C. It follows that a linear cone is a special case of a convex cone C.


Explanation: Since the scalar λ for the linear cone is arbitrary and the scalars α = β = λ/2 for the convex cone are special:

αx + βx= (λ/2)x + (λ/2)x = λx

Check: A blunt convex cone is necessarily salient?[edit]

I am not sure about this statement "A blunt convex cone is necessarily salient". It seems to be incompatible with this other source [1]. I am not an expert, but I think that maybe the right sentence is the opposite: "a salient cone is necessarily blunt", could someone check that?

Response: dunno if this has already been addressed but "a salient cone is necessarily blunt" is definitely wrong. Consider the set {(0,y):y≥0} in E2. This is a salient convex cone that is also pointed. on the other hand, if C is a blunt convex cone then it must be salient since if x,-x belong to C then x+(-x)=0 must belong to C (by convexity). This clearly contradicts the fact that C is blunt and therefore xC implies -x isnt in C, and hence C is salient.

Without 0 a set may no longer be a convex cone?[edit]

According to the above definition, if C is a convex cone, then C{0} and C{0} are convex cones, too.

Can somebody confirm the second part -- C{0}. This seems not to be the case for any flat convex cone, because one can get zero as a combination of opposite vectors with both coefficients equal 1. Wrwrwr (talk) 14:48, 7 May 2010 (UTC)[reply]

Quite right. I don't know the precise characterization of when C{0} is also a convex cone. — Arthur Rubin (talk) 17:35, 7 May 2010 (UTC)[reply]

Properties that are stated without condition and which fail in infinite dimensions[edit]

1) "any convex cone C that is not the whole space V must be contained in some closed half-space H of V."

2) the perfect half-spaces are the maximal salient convex cones (under the containment order). In fact, it can be proved that every pointed salient convex cone (independently of whether it is topologically open, closed, or mixed) is the intersection of all the perfect half-spaces that contain it. — Preceding unsigned comment added by 134.157.88.130 (talk) 14:25, 29 October 2015 (UTC)[reply]

Big cleanup May 2016[edit]

I propose to do the following

  • merge the good parts of cone (linear algebra) here (most of it is already here)
  • add a section/special case about polyhedral cones as they are vital to Polyhedral representation and I cannot find their information on wikipedia (done)
  • explicitly state the definition of cone, convex cone, polyhedral cone....so clean up the definition section. (done)
  • get picture examples of cones that are not convex, polyhedral cones, cones with infinite number of generators, cones that don't look like cones (linear subspaces, half spaces, etc). Basically, more examples with graphics. My graphic making skills are not that great, but I'm going to try. (done)
  • make the redirect on polyhedral cones map here and not to the cones page.(done)
  • add a bunch of references. (done)

DrWikiWikiShuttle (talk) 00:04, 24 May 2016 (UTC)[reply]

Equivalence of finitely generated and polyhedral cones[edit]

It says "Every finitely generated cone is a polyhedral cone". It seems to me that every polyhedral cone has nonempty interior: by Gaussian elimination we may assume it is in row echelon form, and then the cone is a product of half-spaces. While a f.g. cone may have empty interior. --nBarto (talk) 18:32, 2 September 2019 (UTC)[reply]

generating cone[edit]

"A cone C is said to be generating if is equal to the whole vector space." - Although this is cited correctly from the literature, it does not make sense to me. After all, is the empty set. I think it should be . --Tillmo (talk) 14:23, 8 February 2023 (UTC)[reply]

I believe that and I have edited the article in this sense. Otherwise, the term "generating" would be nonsensical. However, this is only my interpretation, and the source must be checked. D.Lazard (talk) 18:52, 8 February 2023 (UTC)[reply]

Positive scalar definition should depend only on the scalars[edit]

The section 'Definition includes this sentence:

"This concept is meaningful for any vector space that allows the concept of 'positive' scalar, such as spaces over the rational, algebraic, or (more commonly) the real numbers."

This is not a question of the vector space, but rather of the field over which the vector space is defined.

Shouldn't the quoted sentence refer to the field and not the vector space?

I hope someone knowledgeable about this subject can fix this.