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The contents of the cone (linear algebra) page were merged into Convex cone on 25 June 2016. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see its talk page. |
See Talk:Dual cone and polar cone. Oleg Alexandrov ( talk) 03:54, 19 June 2007 (UTC)
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
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.
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)
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)
I propose to do the following
DrWikiWikiShuttle ( talk) 00:04, 24 May 2016 (UTC)
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)
"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)
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.
This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
The contents of the cone (linear algebra) page were merged into Convex cone on 25 June 2016. For the contribution history and old versions of the redirected page, please see its history; for the discussion at that location, see its talk page. |
See Talk:Dual cone and polar cone. Oleg Alexandrov ( talk) 03:54, 19 June 2007 (UTC)
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
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.
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)
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)
I propose to do the following
DrWikiWikiShuttle ( talk) 00:04, 24 May 2016 (UTC)
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)
"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)
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.