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This is fine math, but no one has even defined a basis for a Vector Space, given the standard geometric interpretation of R3 as a Vector Space (a good concrete example.), etc. Lets try to fill in the more elementary material, before we soar to these heights...:-). I guess that includes me, too. RoseParks
I moved Rose's comment here, partly because there is at least entry for basis of a vector space now. DMD
I've seen what you call the "continuous dual" denoted by X* as well. Often the distinction is entirely based on context, if you're just working with vector spaces, X* is the algebraic dual, if you're workign with normed spaces, it's the continuous one. It might be worth noting this, because I for one have been confused by texts working with what they call X* when they mean what you call X'. cfp 22:37, 4 Apr 2004 (UTC)
I read elsewhere that a square matrix can be thought of as a tensor with one covariant index and one contravariant index. So it seems to me that row and column vectors correspond to tensors like . If dual spaces have to do with the relation between row and column vectors, does it have something to do with tensors?
The article claims that the continuous dual of a normed vector space is something stronger: a Banach space. It seems to me this cannot be so. For example, what if my normed vector space had the rationals as its underlying field? The dual will also have Q as its field, and so cannot be complete. - Lethe | Talk 17:40, 3 October 2005 (UTC)
Oleg, I think this should be an example of a linear map on a topological vector space which is not continuous. Let X be a space which is not locally compact and consider the space of real functions on it with the compact open topology. The Dirac delta functional send each function to its value at 0. Since X is not locally compact, this is not continuous. Since X is not compact, this is not a metric space. - lethe talk 15:12, 27 December 2005 (UTC)
I took the liberty of adding that transposing gives 'more or less' the same map, when you associate vectors in V with vectors in V** by the connection explained in the bidual part. However, in order to avoid confusion I had to place the transpose section after the bidual section.
A few questions : is what I said correct or is finiteness of dimension needed somewhere? Why does my expression at the end of the transpose part look 'not nice' Evilbu 22:38, 14 February 2006 (UTC)
I'm pulling this out since it may turn into a longer conversation.
What you said and what I said are the same thing. The dual of a coproduct is a product. The * indicates dual space. This article is about dual spaces, not about coproducts, which has its own article. - lethe talk + 19:47, 22 May 2006 (UTC)
I have added the assertion that dual of the coproduct is the product to the article coproduct and direct sum. Those are the right articles for this stuff. - lethe talk + 01:47, 23 May 2006 (UTC)
I've used my 3 reverts for the day. It's out of my hands. - lethe talk + 03:13, 23 May 2006 (UTC)
I have looked at the most recent versions. Although the difference is small, I prefer Lethe's version. I do not like category theory. Referring to it in this article is disturbing and makes me not want to read the article. JRSpriggs 08:02, 23 May 2006 (UTC)
OK, I have elaborated on the proof at coproduct in painful detail. I have added a brief indication of the idea in direct sum, and I have added a description to this article, of the fact that the dual space is the product, all infinite tuples. I have the feeling that there is too much discussion of this fact. I wonder what others think. - lethe talk + 05:31, 24 May 2006 (UTC)
I do not believe that a reader who is unfamiliar with the concept of dual space will find this article very helpful, even though its content as far as I can see is correct.
-- KYN 20:46, 15 August 2006 (UTC)
One of the authors seemed to believe that for dim(V) infinite dim(V*) > dim(V). The easiest example is in another part of the article where for any L^2 space V = V* and thus dim(V) = dim(V*). jbolden1517 Talk 19:35, 17 May 2006 (UTC)
I have rewritten your section. Comments welcome. I mostly added stuff, didn't remove anything you wrote, just shuffled and rephrased. The one thing I did remove was the examples of tuples from the product and coproduct. I'm not sure they're needed. - lethe talk + 03:54, 22 May 2006 (UTC)
I don't think this article is an appropriate place to provide a proof that the dual of the coproduct is the product. That should go in one of those articles. - lethe talk + 19:30, 22 May 2006 (UTC)
Similarly, we have the section:
Coming from a rather naive background, I find the above paragraphs utterly confusing. For example, the statement: the span of {σα} consists of all finite linear combinations of the dual vectors... ... Huh? Where is it required that the combination be finite? From the articles on Hamel basis and Schauder basis, I am typically used to using a Schauder basis for countably-dimensional vector spaces, so this appearance of a Hamel basis out of thin air is confusing. I suspect that this has something to do with the categorical notions of product and coproduct, based on the discussions above from May 2006, but if this is the case, it is quite opaque.
I am also unable to parse the sentence ... an infinite tuple of dual vectors evaluates to nonzero scalars only finitely many times. because I can't figure out what "evaluates" means in this context. Please fix. linas 01:08, 15 October 2006 (UTC)
The "examples" section currently states the following:
The statements about uncountablity is certainly wrong, or at least confusing. To make this more concrete, and easier to talk about, consider the space where the set with two elements. Then has a countable basis, and thus its dimension is countable. As a set, it is uncountable: taken as the set of all possible strings in two digits, it is homomorphic to the set of all reals on the unit interval, and thus clearly uncountable. The fact that there are an uncountable number of elements in the set should not be confused with the fact that, as a vector space, it has a countable basis, and thus a countable dimension. (Here, by "countable" I mean countably infinite; I do not mean "finite"). Yet the above example seems to be making exactly this confusion, with R standing in the place of . Can someone please fix this to say what should have been said? linas 00:09, 15 October 2006 (UTC)
Similarly, the set of strings with only a finite number of 1's in them is a countable set (its homomorphic to the rationals), and it has a countable dimension. Thus, I'd conclude that both W and both have countable dimension, but W has a countable number of elements, while has an uncountable number of elements. I think this is what the example is trying to say, right? linas 00:28, 15 October 2006 (UTC)
I suggested above that the article be split into continuous dual and algebraic dual, with dual space being used for an overview article. CSTAR agreed, but there has been no further discussion of this. I am prompted to make the suggestion again after finding that someone inserted the claim that Hilbert spaces are isomorphic to their duals in the section on algebraic duals. I think less confusion would arise if both concepts had their own entries. -- Zundark 09:51, 19 November 2006 (UTC)
I'm still not satisfied with the current introduction. I tried to rewrite it some time ago, but it was reverted on unclear grounds. Here are my concerns.
Instead I propose the following introduction:
-- KYN 16:27, 5 January 2007 (UTC)
My problem with the current introduction's presentation of the continuous dual space as a subspace of the algebraic dual space is that the context in which this is done suggests that "algebraic dual space" refers to a concept rather than (I believe?) to a specific instance of a dual space related to some specific topological space. Maybe it can be written something like:
-- KYN 20:12, 5 January 2007 (UTC)
Splitting the article into one part devoted to the algebraic dual space and one to the continuous dual space is either way to me. The two concepts appear to be reasonably well related, one is a specialization of the other, and to have them in the same article is not a problem. But since the stuff in the continuous dual space section is rather focused on the various properties of such a space it can also be in a separate article.
About the row and column idea, I agree that this can be used to illustrate how a practical implementation of a dual space may be set up for a particular choice of vector space. But I do not see the row-column idea as a vital thing in order to understand what a dual space is, in particular for the general case. I would like to remove the current wording in the intro since I believe that it does convey something that the "unfamiliar" reader may be able to grasp at that point. I'd rather see that this idea is further developed in the example section.
OK, let's remove tensor algebras from the intro but keep the tensors. Note that the current and proposed text does not exclude the possibility that tensor also may be defined on infinite dimensional spaces.
-- KYN 22:57, 8 January 2007 (UTC)
I was trying to prove that the canonical injection from a vector space to its double dual is indeed an injection. Seems to me that I couldn't do it without choosing a basis (and therefore invoking the axiom of choice). I don't like that. Is there a way to check the injectivity without a basis? - lethe talk + 03:54, 22 May 2006 (UTC)
But then is in every hyperplane of V, that is jbolden1517 Talk 04:49, 22 May 2006 (UTC)
I cannot understand this... The space which the gradient belongs to is the dual space relative to the space which or belong to.
The gradient of a function is a map from , not . So the gradient is not a linear functional and can not be in the dual space. Although you could map the gradient to the dual space by .
20:38, 21 September 2006 (UTC)
I was somewhat hesitant to edit this section, as I can see from the discussion and history that it has been the subject of lively debate in the past. However, I think I've been able to both clarify and shorten the treatment. There is still one thing missing here, however: although the dual space is seen to be "bigger", that does not prove it does not have a basis of the same cardinality (although the axiom of choice is needed to see that it has any basis at all!). It would be enough to show that the dual space itself has strictly larger cardinality than the original vector space, but I'm not able to see why that is. Geometry guy 17:39, 15 February 2007 (UTC)
I have reviewed the article having regard to the guidelines set out in Wikipedia:What is a good article. I have reached the following conclusions:
In summary, I have found the article to be good in many respects. However, the issues above are such that I am led to fail the GA nomination for now. I considered "On hold" status instead, but I find it hard to assume that the issues relating to the broadness of coverage on topological duals could be resolved in a week. One option would be to split the article into algebraic and topological duals, in which case the half on algebraic duals could be developed to the Good article level fairly quickly. However, the topic appears to have been discussed above, and reaching consensus on such a decision could take time and likely not be completed in a stable manner in one week either.
As a final note, I think this article is clearly capable of growing into a Good article status and beyond with some more editing.
Stca74 20:48, 11 May 2007 (UTC)
As far as I can see, we agree that the article needs a simple example which can introduce the reader to the concept of a dual space. However, this does not mean that reflects in an abstract way the relationship between row vectors and column vectors will make sense to the reader who is not already familiar with the concept. I'd rather see that this ambiguous and inprecise statement is replaced by an eample (early in the article) of a space and its dual space, possibly in the form of row and column vectors. Such an example can then provide the "key" which you want. -- KYN 23:00, 16 January 2007 (UTC)
I think the example near the beginning with the arrows and parallel lines is wrong: the result of counting lines crossed by the arrow is a natural number, and the natural numbers do not form a field. -- Vapniks ( talk) 17:11, 9 July 2008 (UTC)
I notice that the physicists' bracket notation is included, with the claim that it is a bilinear form. That's not quite true, as it is sesquilinear. More generally, I see no mention in the article of the picky issues that occur when working over C, e.g. the fact that a nondegenerate sesquilinear form yields an isomorphism of the dual with the conjugate space, not the space itself. Does anyone think that should be included, or at least mentioned? Or is that more suitable elsewhere? -- Spireguy ( talk) 03:27, 25 July 2008 (UTC)
The section called "Bilinear products and dual spaces" starts with the sentence. "If V is finite-dimensional, then V is isomorphic to V*." Further in the paragraph is reads: "If the bilinear form is assumed to be nondegenerate, then this is an isomorphism." not referencing the finite dimensionality anymore. I guess it's need. Should we change the first sentence to something like: "This paragraph concerns itself with finite-dimensional vectorspaces V. In this case V is isomorphic to V*. ...."? -- JanCK ( talk) 15:47, 7 August 2008 (UTC)
Somewhere above people have asked for more examples and I recall one person asking for details on calculating a dual basis. Assuming the following is mathematically correct, do people think it would be useful to add as a further example in the finite-dimensional case?
"Another basis of R2 is . From the biorthogonality conditions, it is clear that the dual basis of B is given by the rows of ."
digfarenough ( talk) 20:36, 22 December 2009 (UTC)
Why does a footnote say that the axiom of choice is needed to show that $\mathbb{R}^n$ has a basis? Infinite dimensional vector space, sure, but finite dimensional? That's wrong, isn't it? —Preceding unsigned comment added by 119.12.47.108 ( talk) 04:57, 20 December 2008 (UTC)
(Infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals.)
There should be more emphasis made that the double dual is only "interesting" in infinite dimensional spaces.... — Preceding unsigned comment added by 94.72.239.126 ( talk) 01:15, 25 December 2011 (UTC)
Presently, the article states that:
Moreover, if A and B are two subsets of V, then
and equality holds provided V is finite-dimensional.
It's tacitly implied that equality generally does not hold. However, it seems to be an easy exercise to prove equality always holds, by just using the fact that short exact sequences of vector spaces split, i.e., complements to subspaces always exist. In particular, one can decompose V by
where
Then given any linear function we can define functionals g and h by
and
Then it's clear that f=g + h and that 169.233.250.92 ( talk) 20:31, 14 May 2012 (UTC)
Excuse me, the term "continuous dual space", where is it from? In the theory of topological vector spaces people usually say just "dual space".
I would also like to insert here some facts about this notion. Eozhik ( talk) 13:43, 10 November 2012 (UTC)
I am sorry, I forgot about my question of November 2012 (and I did not see your answer of April 2013). I've just changed the title of this section "Continuous dual space" to "Dual space of a topological vector space", because this term -- continuous dual space -- is indeed very rarely used in this theory. I gave references (and I am a specialist in this field). Eozhik ( talk) 16:01, 24 July 2013 (UTC)
Really? I get no hits on Google books. — Quondum 15:25, 2 December 2013 (UTC)
In the context of an encyclopaedia, surely the full name "Dual vector space" should be used as the name of the article, and not the abbreviated form "Dual space"? In the context of a discussion of vector spaces, "dual space" is natural and a redirect with the shortened form should be given, but an encyclopaedia is not only about vector spaces. This is especially significant for this article, since " dual" and " space" individually have multiple meanings even when restricted to mathematics, so to combine them could produce multiple sensible meanings, and this article is about only one of these meanings. — Quondum 19:33, 14 January 2014 (UTC)
Subsection 2.5 "Topologies on the dual" of section 2 "Continuous dual space" has only one sentence, referring to another article, "dual space topology". Moreover its reference to using the topologies on the real or complex numbers seems in conflict with the relatively general setting of this article, which presumably concerns vector spaces in general and sometimes refers to properties of the scalar field.
Topologies on the dual space are discussed at more length above this subsection, in the main Section 2. This seems an appropriate place for the discussion because one might think that the dual of a topological vector space (TVS) should be another TVS.
I propose removing Subsection 2.5 and incorporating the link to the article "dual space topology" into the discussion of topologies on the dual in the main part of Section 2.
If someone believes that additional material on dual topologies beyond that already in the main Section 2 discussion, should be in this article under this subsection, that might be reasonable. I am not sufficiently expert to judge, but suspect that a more extensive discussion would most reasonably be left to the article "dual space topology".
MorphismOfDoom ( talk) 16:59, 19 March 2014 (UTC)
I disagree. If a man wants to know what a dual basis is, it will be not logical to send him to dual space, as dual basis is simpler to understand; only a link to dual space should be given at the beginning. I think I will write some more information about dual basis in the next month, before 2009-06-25, as I am a student myself. Q0k ( talk) 00:56, 29 May 2009 (UTC)
I agree with Stca74 here, though I guess it doesn't matter too much if there is duplication of subsections of articles as stand-alone articles. But if you want to know what a dual basis is, and get directed to a subsection of an article which does a perfectly fine job, what's the problem? It seems like you may see how your specific topic of interest fits into the broader picture.
MorphismOfDoom ( talk) 17:06, 19 March 2014 (UTC)
The lede states:"The dual space V* itself becomes a vector space over F when equipped with the following ..." In the article Linear_form {which is referenced in this lede as "linear functional"}, it is stated:"The set of all linear functionals from V to k, Homk(V,k), is itself a vector space over k. This space is called the dual space of V..." These are contradictory, as far as I can see. If additional structure is required in order for a dual space to be a vector space, then the Linear Form article is wrong, if the requirement that each function Φ:x→F be linear (ie Φ(x+y)=Φ(x)+Φ(y) and Φ(ax)=aΦ(x) somehow implies that for Φ,Θ:x→F, (Φ+Θ)(x) = Φ(x)+Θ(x) and (aΦ)(x) = a(Φ(x)) then no additional structure is needed and a dual space is a vector space.(x,y ∈ V; a ∈ F ) Could someone correct which ever article needs it? Abitslow ( talk) 21:10, 24 May 2014 (UTC)
The Kronecker delta symbol in the Finite-dimensional case subsection of the Algebraic dual space section, looks as if it should be instead of , in order to be consistent with the other indices in the equation, as in the dual basis article. — Anita5192 ( talk) 06:57, 28 July 2014 (UTC)
The article references Halmos as using the bracket notation phi(x) = [phi, x]. However Halmos actually uses the opposite notation phi(x) = [x, phi]. He gives an analogy that y(x)=x^2 is written as [x, x^2] (see section 14 "Brackets"). So I think some other reference is needed. — Preceding unsigned comment added by 188.120.148.184 ( talk) 09:59, 13 January 2015 (UTC)
I already have posted elsewhere, but I should do it mention it too here.
Annihilators on spaces, http://planetmath.org/encyclopedia/Annihilator4.html are very important when studying projective spaces, as they help you classify all correlations and thus also the polarities
Now I almost made an annihilator article (note there is already a disambiguation page linking to annihilator in ring theory etc) but then I wondered if it shouldn't be defined here?
Another thing that bothers me is my lack of inexperience with infinite dimensional vector spaces.
Let V be the vector space of all sequences of real numbers in which only a finite number of elements are non-zero. This vector space is not finite dimensional, and any injection into its dual space is a strict subspace (as we'd expect). That is, the dual space of V is isomorphic to the set of all sequences of reals, without the condition that almost all entries be zero. So what is its dual space, V^**? Why is this set strictly "larger" than V^*? The argument in the article relies on there being a basis of V. Can "V^* > V <=> V is not finite dimensional" be proved without using bases? Or does this boil down to some Zorn's lemma argument?
The first example given here is pretty much useless. It requires no thought and does not give a general method for finding the dual basis of a finite vector space. For anyone learning about Dual Spaces, the example could confuse them even more.
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 |
This is fine math, but no one has even defined a basis for a Vector Space, given the standard geometric interpretation of R3 as a Vector Space (a good concrete example.), etc. Lets try to fill in the more elementary material, before we soar to these heights...:-). I guess that includes me, too. RoseParks
I moved Rose's comment here, partly because there is at least entry for basis of a vector space now. DMD
I've seen what you call the "continuous dual" denoted by X* as well. Often the distinction is entirely based on context, if you're just working with vector spaces, X* is the algebraic dual, if you're workign with normed spaces, it's the continuous one. It might be worth noting this, because I for one have been confused by texts working with what they call X* when they mean what you call X'. cfp 22:37, 4 Apr 2004 (UTC)
I read elsewhere that a square matrix can be thought of as a tensor with one covariant index and one contravariant index. So it seems to me that row and column vectors correspond to tensors like . If dual spaces have to do with the relation between row and column vectors, does it have something to do with tensors?
The article claims that the continuous dual of a normed vector space is something stronger: a Banach space. It seems to me this cannot be so. For example, what if my normed vector space had the rationals as its underlying field? The dual will also have Q as its field, and so cannot be complete. - Lethe | Talk 17:40, 3 October 2005 (UTC)
Oleg, I think this should be an example of a linear map on a topological vector space which is not continuous. Let X be a space which is not locally compact and consider the space of real functions on it with the compact open topology. The Dirac delta functional send each function to its value at 0. Since X is not locally compact, this is not continuous. Since X is not compact, this is not a metric space. - lethe talk 15:12, 27 December 2005 (UTC)
I took the liberty of adding that transposing gives 'more or less' the same map, when you associate vectors in V with vectors in V** by the connection explained in the bidual part. However, in order to avoid confusion I had to place the transpose section after the bidual section.
A few questions : is what I said correct or is finiteness of dimension needed somewhere? Why does my expression at the end of the transpose part look 'not nice' Evilbu 22:38, 14 February 2006 (UTC)
I'm pulling this out since it may turn into a longer conversation.
What you said and what I said are the same thing. The dual of a coproduct is a product. The * indicates dual space. This article is about dual spaces, not about coproducts, which has its own article. - lethe talk + 19:47, 22 May 2006 (UTC)
I have added the assertion that dual of the coproduct is the product to the article coproduct and direct sum. Those are the right articles for this stuff. - lethe talk + 01:47, 23 May 2006 (UTC)
I've used my 3 reverts for the day. It's out of my hands. - lethe talk + 03:13, 23 May 2006 (UTC)
I have looked at the most recent versions. Although the difference is small, I prefer Lethe's version. I do not like category theory. Referring to it in this article is disturbing and makes me not want to read the article. JRSpriggs 08:02, 23 May 2006 (UTC)
OK, I have elaborated on the proof at coproduct in painful detail. I have added a brief indication of the idea in direct sum, and I have added a description to this article, of the fact that the dual space is the product, all infinite tuples. I have the feeling that there is too much discussion of this fact. I wonder what others think. - lethe talk + 05:31, 24 May 2006 (UTC)
I do not believe that a reader who is unfamiliar with the concept of dual space will find this article very helpful, even though its content as far as I can see is correct.
-- KYN 20:46, 15 August 2006 (UTC)
One of the authors seemed to believe that for dim(V) infinite dim(V*) > dim(V). The easiest example is in another part of the article where for any L^2 space V = V* and thus dim(V) = dim(V*). jbolden1517 Talk 19:35, 17 May 2006 (UTC)
I have rewritten your section. Comments welcome. I mostly added stuff, didn't remove anything you wrote, just shuffled and rephrased. The one thing I did remove was the examples of tuples from the product and coproduct. I'm not sure they're needed. - lethe talk + 03:54, 22 May 2006 (UTC)
I don't think this article is an appropriate place to provide a proof that the dual of the coproduct is the product. That should go in one of those articles. - lethe talk + 19:30, 22 May 2006 (UTC)
Similarly, we have the section:
Coming from a rather naive background, I find the above paragraphs utterly confusing. For example, the statement: the span of {σα} consists of all finite linear combinations of the dual vectors... ... Huh? Where is it required that the combination be finite? From the articles on Hamel basis and Schauder basis, I am typically used to using a Schauder basis for countably-dimensional vector spaces, so this appearance of a Hamel basis out of thin air is confusing. I suspect that this has something to do with the categorical notions of product and coproduct, based on the discussions above from May 2006, but if this is the case, it is quite opaque.
I am also unable to parse the sentence ... an infinite tuple of dual vectors evaluates to nonzero scalars only finitely many times. because I can't figure out what "evaluates" means in this context. Please fix. linas 01:08, 15 October 2006 (UTC)
The "examples" section currently states the following:
The statements about uncountablity is certainly wrong, or at least confusing. To make this more concrete, and easier to talk about, consider the space where the set with two elements. Then has a countable basis, and thus its dimension is countable. As a set, it is uncountable: taken as the set of all possible strings in two digits, it is homomorphic to the set of all reals on the unit interval, and thus clearly uncountable. The fact that there are an uncountable number of elements in the set should not be confused with the fact that, as a vector space, it has a countable basis, and thus a countable dimension. (Here, by "countable" I mean countably infinite; I do not mean "finite"). Yet the above example seems to be making exactly this confusion, with R standing in the place of . Can someone please fix this to say what should have been said? linas 00:09, 15 October 2006 (UTC)
Similarly, the set of strings with only a finite number of 1's in them is a countable set (its homomorphic to the rationals), and it has a countable dimension. Thus, I'd conclude that both W and both have countable dimension, but W has a countable number of elements, while has an uncountable number of elements. I think this is what the example is trying to say, right? linas 00:28, 15 October 2006 (UTC)
I suggested above that the article be split into continuous dual and algebraic dual, with dual space being used for an overview article. CSTAR agreed, but there has been no further discussion of this. I am prompted to make the suggestion again after finding that someone inserted the claim that Hilbert spaces are isomorphic to their duals in the section on algebraic duals. I think less confusion would arise if both concepts had their own entries. -- Zundark 09:51, 19 November 2006 (UTC)
I'm still not satisfied with the current introduction. I tried to rewrite it some time ago, but it was reverted on unclear grounds. Here are my concerns.
Instead I propose the following introduction:
-- KYN 16:27, 5 January 2007 (UTC)
My problem with the current introduction's presentation of the continuous dual space as a subspace of the algebraic dual space is that the context in which this is done suggests that "algebraic dual space" refers to a concept rather than (I believe?) to a specific instance of a dual space related to some specific topological space. Maybe it can be written something like:
-- KYN 20:12, 5 January 2007 (UTC)
Splitting the article into one part devoted to the algebraic dual space and one to the continuous dual space is either way to me. The two concepts appear to be reasonably well related, one is a specialization of the other, and to have them in the same article is not a problem. But since the stuff in the continuous dual space section is rather focused on the various properties of such a space it can also be in a separate article.
About the row and column idea, I agree that this can be used to illustrate how a practical implementation of a dual space may be set up for a particular choice of vector space. But I do not see the row-column idea as a vital thing in order to understand what a dual space is, in particular for the general case. I would like to remove the current wording in the intro since I believe that it does convey something that the "unfamiliar" reader may be able to grasp at that point. I'd rather see that this idea is further developed in the example section.
OK, let's remove tensor algebras from the intro but keep the tensors. Note that the current and proposed text does not exclude the possibility that tensor also may be defined on infinite dimensional spaces.
-- KYN 22:57, 8 January 2007 (UTC)
I was trying to prove that the canonical injection from a vector space to its double dual is indeed an injection. Seems to me that I couldn't do it without choosing a basis (and therefore invoking the axiom of choice). I don't like that. Is there a way to check the injectivity without a basis? - lethe talk + 03:54, 22 May 2006 (UTC)
But then is in every hyperplane of V, that is jbolden1517 Talk 04:49, 22 May 2006 (UTC)
I cannot understand this... The space which the gradient belongs to is the dual space relative to the space which or belong to.
The gradient of a function is a map from , not . So the gradient is not a linear functional and can not be in the dual space. Although you could map the gradient to the dual space by .
20:38, 21 September 2006 (UTC)
I was somewhat hesitant to edit this section, as I can see from the discussion and history that it has been the subject of lively debate in the past. However, I think I've been able to both clarify and shorten the treatment. There is still one thing missing here, however: although the dual space is seen to be "bigger", that does not prove it does not have a basis of the same cardinality (although the axiom of choice is needed to see that it has any basis at all!). It would be enough to show that the dual space itself has strictly larger cardinality than the original vector space, but I'm not able to see why that is. Geometry guy 17:39, 15 February 2007 (UTC)
I have reviewed the article having regard to the guidelines set out in Wikipedia:What is a good article. I have reached the following conclusions:
In summary, I have found the article to be good in many respects. However, the issues above are such that I am led to fail the GA nomination for now. I considered "On hold" status instead, but I find it hard to assume that the issues relating to the broadness of coverage on topological duals could be resolved in a week. One option would be to split the article into algebraic and topological duals, in which case the half on algebraic duals could be developed to the Good article level fairly quickly. However, the topic appears to have been discussed above, and reaching consensus on such a decision could take time and likely not be completed in a stable manner in one week either.
As a final note, I think this article is clearly capable of growing into a Good article status and beyond with some more editing.
Stca74 20:48, 11 May 2007 (UTC)
As far as I can see, we agree that the article needs a simple example which can introduce the reader to the concept of a dual space. However, this does not mean that reflects in an abstract way the relationship between row vectors and column vectors will make sense to the reader who is not already familiar with the concept. I'd rather see that this ambiguous and inprecise statement is replaced by an eample (early in the article) of a space and its dual space, possibly in the form of row and column vectors. Such an example can then provide the "key" which you want. -- KYN 23:00, 16 January 2007 (UTC)
I think the example near the beginning with the arrows and parallel lines is wrong: the result of counting lines crossed by the arrow is a natural number, and the natural numbers do not form a field. -- Vapniks ( talk) 17:11, 9 July 2008 (UTC)
I notice that the physicists' bracket notation is included, with the claim that it is a bilinear form. That's not quite true, as it is sesquilinear. More generally, I see no mention in the article of the picky issues that occur when working over C, e.g. the fact that a nondegenerate sesquilinear form yields an isomorphism of the dual with the conjugate space, not the space itself. Does anyone think that should be included, or at least mentioned? Or is that more suitable elsewhere? -- Spireguy ( talk) 03:27, 25 July 2008 (UTC)
The section called "Bilinear products and dual spaces" starts with the sentence. "If V is finite-dimensional, then V is isomorphic to V*." Further in the paragraph is reads: "If the bilinear form is assumed to be nondegenerate, then this is an isomorphism." not referencing the finite dimensionality anymore. I guess it's need. Should we change the first sentence to something like: "This paragraph concerns itself with finite-dimensional vectorspaces V. In this case V is isomorphic to V*. ...."? -- JanCK ( talk) 15:47, 7 August 2008 (UTC)
Somewhere above people have asked for more examples and I recall one person asking for details on calculating a dual basis. Assuming the following is mathematically correct, do people think it would be useful to add as a further example in the finite-dimensional case?
"Another basis of R2 is . From the biorthogonality conditions, it is clear that the dual basis of B is given by the rows of ."
digfarenough ( talk) 20:36, 22 December 2009 (UTC)
Why does a footnote say that the axiom of choice is needed to show that $\mathbb{R}^n$ has a basis? Infinite dimensional vector space, sure, but finite dimensional? That's wrong, isn't it? —Preceding unsigned comment added by 119.12.47.108 ( talk) 04:57, 20 December 2008 (UTC)
(Infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals.)
There should be more emphasis made that the double dual is only "interesting" in infinite dimensional spaces.... — Preceding unsigned comment added by 94.72.239.126 ( talk) 01:15, 25 December 2011 (UTC)
Presently, the article states that:
Moreover, if A and B are two subsets of V, then
and equality holds provided V is finite-dimensional.
It's tacitly implied that equality generally does not hold. However, it seems to be an easy exercise to prove equality always holds, by just using the fact that short exact sequences of vector spaces split, i.e., complements to subspaces always exist. In particular, one can decompose V by
where
Then given any linear function we can define functionals g and h by
and
Then it's clear that f=g + h and that 169.233.250.92 ( talk) 20:31, 14 May 2012 (UTC)
Excuse me, the term "continuous dual space", where is it from? In the theory of topological vector spaces people usually say just "dual space".
I would also like to insert here some facts about this notion. Eozhik ( talk) 13:43, 10 November 2012 (UTC)
I am sorry, I forgot about my question of November 2012 (and I did not see your answer of April 2013). I've just changed the title of this section "Continuous dual space" to "Dual space of a topological vector space", because this term -- continuous dual space -- is indeed very rarely used in this theory. I gave references (and I am a specialist in this field). Eozhik ( talk) 16:01, 24 July 2013 (UTC)
Really? I get no hits on Google books. — Quondum 15:25, 2 December 2013 (UTC)
In the context of an encyclopaedia, surely the full name "Dual vector space" should be used as the name of the article, and not the abbreviated form "Dual space"? In the context of a discussion of vector spaces, "dual space" is natural and a redirect with the shortened form should be given, but an encyclopaedia is not only about vector spaces. This is especially significant for this article, since " dual" and " space" individually have multiple meanings even when restricted to mathematics, so to combine them could produce multiple sensible meanings, and this article is about only one of these meanings. — Quondum 19:33, 14 January 2014 (UTC)
Subsection 2.5 "Topologies on the dual" of section 2 "Continuous dual space" has only one sentence, referring to another article, "dual space topology". Moreover its reference to using the topologies on the real or complex numbers seems in conflict with the relatively general setting of this article, which presumably concerns vector spaces in general and sometimes refers to properties of the scalar field.
Topologies on the dual space are discussed at more length above this subsection, in the main Section 2. This seems an appropriate place for the discussion because one might think that the dual of a topological vector space (TVS) should be another TVS.
I propose removing Subsection 2.5 and incorporating the link to the article "dual space topology" into the discussion of topologies on the dual in the main part of Section 2.
If someone believes that additional material on dual topologies beyond that already in the main Section 2 discussion, should be in this article under this subsection, that might be reasonable. I am not sufficiently expert to judge, but suspect that a more extensive discussion would most reasonably be left to the article "dual space topology".
MorphismOfDoom ( talk) 16:59, 19 March 2014 (UTC)
I disagree. If a man wants to know what a dual basis is, it will be not logical to send him to dual space, as dual basis is simpler to understand; only a link to dual space should be given at the beginning. I think I will write some more information about dual basis in the next month, before 2009-06-25, as I am a student myself. Q0k ( talk) 00:56, 29 May 2009 (UTC)
I agree with Stca74 here, though I guess it doesn't matter too much if there is duplication of subsections of articles as stand-alone articles. But if you want to know what a dual basis is, and get directed to a subsection of an article which does a perfectly fine job, what's the problem? It seems like you may see how your specific topic of interest fits into the broader picture.
MorphismOfDoom ( talk) 17:06, 19 March 2014 (UTC)
The lede states:"The dual space V* itself becomes a vector space over F when equipped with the following ..." In the article Linear_form {which is referenced in this lede as "linear functional"}, it is stated:"The set of all linear functionals from V to k, Homk(V,k), is itself a vector space over k. This space is called the dual space of V..." These are contradictory, as far as I can see. If additional structure is required in order for a dual space to be a vector space, then the Linear Form article is wrong, if the requirement that each function Φ:x→F be linear (ie Φ(x+y)=Φ(x)+Φ(y) and Φ(ax)=aΦ(x) somehow implies that for Φ,Θ:x→F, (Φ+Θ)(x) = Φ(x)+Θ(x) and (aΦ)(x) = a(Φ(x)) then no additional structure is needed and a dual space is a vector space.(x,y ∈ V; a ∈ F ) Could someone correct which ever article needs it? Abitslow ( talk) 21:10, 24 May 2014 (UTC)
The Kronecker delta symbol in the Finite-dimensional case subsection of the Algebraic dual space section, looks as if it should be instead of , in order to be consistent with the other indices in the equation, as in the dual basis article. — Anita5192 ( talk) 06:57, 28 July 2014 (UTC)
The article references Halmos as using the bracket notation phi(x) = [phi, x]. However Halmos actually uses the opposite notation phi(x) = [x, phi]. He gives an analogy that y(x)=x^2 is written as [x, x^2] (see section 14 "Brackets"). So I think some other reference is needed. — Preceding unsigned comment added by 188.120.148.184 ( talk) 09:59, 13 January 2015 (UTC)
I already have posted elsewhere, but I should do it mention it too here.
Annihilators on spaces, http://planetmath.org/encyclopedia/Annihilator4.html are very important when studying projective spaces, as they help you classify all correlations and thus also the polarities
Now I almost made an annihilator article (note there is already a disambiguation page linking to annihilator in ring theory etc) but then I wondered if it shouldn't be defined here?
Another thing that bothers me is my lack of inexperience with infinite dimensional vector spaces.
Let V be the vector space of all sequences of real numbers in which only a finite number of elements are non-zero. This vector space is not finite dimensional, and any injection into its dual space is a strict subspace (as we'd expect). That is, the dual space of V is isomorphic to the set of all sequences of reals, without the condition that almost all entries be zero. So what is its dual space, V^**? Why is this set strictly "larger" than V^*? The argument in the article relies on there being a basis of V. Can "V^* > V <=> V is not finite dimensional" be proved without using bases? Or does this boil down to some Zorn's lemma argument?
The first example given here is pretty much useless. It requires no thought and does not give a general method for finding the dual basis of a finite vector space. For anyone learning about Dual Spaces, the example could confuse them even more.