# Talk:Vector space

As I understand the word "generalization" in math, one says that objects of type A generalize objects of type B if certain A-objects are B-objects. So:

it is perfectly correct to say that modules generalize vector spaces.
Affine spaces do not generalize vector spaces; any vector space defines an affine space, but it is not the case that certain affine spaces are vector spaces.
One could consider those certain vector bundles in which the base space is a point (I note this is not even mentioned in the article), and to then identify the total space with a vector space. It may be overly pedantic to say that the extra specification of the particular point matters, but I think it is simply true. It may be more correct to say that "vector space" is generalized by the concept of total space of a vector bundle, and not by vector bundle itself.
Sorry for the pedantry, but the article seems to suggest that all three concepts are equally well generalizations of vector space, and maybe this is not so good. Gumshoe2 (talk) 07:23, 15 February 2022 (UTC)

 Fixed D.Lazard (talk) 10:08, 15 February 2022 (UTC)

The redirect Abstract vector space has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2023 June 24 § Abstract vector space until a consensus is reached.  Hildeoc (talk) 00:59, 24 June 2023 (UTC)

They are supposed to be axioms. Given the amount of time vector spaces have been defined, I would assume there would be a counterexample showing some mathematical object satisfying all axioms except one. 
Counterexample help understanding just as much as examples. Sometimes more. It certainly would be helpful for me. Ndhananj (talk) 22:57, 20 August 2023 (UTC)

The page would gain comprehension and usefulness if a sub-section in the respective operations sections is also devoted to using the operation using component vectors. For example, in the addition of two vectors, using |R| = √((Ax + Bx)² + (Ay + By)²) and Angle between the resultant and base vector = tan^(-1) ((Ay + By)/(Ax + Bx)). R0ck$ (talk) 05:46, 30 October 2023 (UTC)

@R0ck$: I'm not sure what you mean. This is an article about general vector spaces that might have no inner product associated with them, and which might be multidimensional. There is an example (ordered pairs of numbers) that showcases components, but in no way are the definitions dependent on the existence of a basis.--Jasper Deng (talk) 06:11, 30 October 2023 (UTC)
@Jasper Deng apologies, I am not well-versed at all about "inner product" as a terminology. Will educate myself about the same and try to understand what you are trying to say.
All I was concerned about was that frequently it is useful to perform a vector sum in a manner that separates the magnitude and direction, and the formula for the same does not seem to be there in this particular article. R0ck$ (talk) 11:01, 30 October 2023 (UTC)
@R0ck$: I don't think you understand. "Magnitude and direction" are only meaningful in inner product spaces, like for Euclidean vectors. Adding and subtracting vectors is always possible using components in a basis, but it is nontrivial to show that a basis always exists and certainly not part of the definition. For many infinite-dimensional vector spaces, you are not going to be adding componentwise: the set of all real-valued functions on a given set forms a vector space, but any basis is going to be of uncountably large dimension. If anything, the magnitude and direction formulation is emphatically not what this article is getting at.--Jasper Deng (talk) 11:03, 30 October 2023 (UTC)

The definition erroneously said there are two binary operations, one of which is scalar multiplication. In fact, the group of units of the field produces a group action on the vectors. Inclusion of zero for scalar multiplication annihilates the vector space to the zero vector. Binary operations require one set, but scalar multiplication starts with two: F and V. Precision is mathematics is expected, and this hitch of imprecision might have led to confusion.  Reference has been made to binary function since introduction of group action at this level assumes much of the reader. — Rgdboer (talk) 02:41, 22 December 2023 (UTC)

Please, read the third paragraph of Binary operation. If you find it erroneous, you must discuss it on that talk page. In any case, most textbooks call scalar multiplication a binary operation. 09:49, 22 December 2023 (UTC) D.Lazard (talk) 09:49, 22 December 2023 (UTC)
That paragraph in Binary operation conflicts with the definition given in the article proper. In usage it seems Scalar multiplication is a legacy exception, or carveout, as the usage perpetuates a misnomer. The paragraph seems appropriate, given usage, particularly as Binary function is mentioned as a valid alternative. — Rgdboer (talk) 00:57, 27 December 2023 (UTC)

As far as I know, most textbooks on vector spaces use "binary operation" for the sclar multiplication, and not "binary function". So, Wikipedia must follow the common usage. D.Lazard (talk) 09:01, 27 December 2023 (UTC)
Composition is the word used by Emil Artin in his Geometric Algebra (book), available via page 4, Internet Archive. — Rgdboer (talk) 00:56, 4 January 2024 (UTC)


### Vector space
@Dedhert.Jr Thank you very much for your efforts on this article! What exactly though is the rationale behind the rearrangement of the second section? It seems to me to be a lot more difficult to understand in the new, compressed form, and for instance the definition of linear independence as "the linear combination that is equal to zero" has little resemblance to the usual definition, according to which a set of vectors is linearly independent if there is no non-trivial linear combination of those vectors that equal 0. Felix QW (talk) 19:03, 3 February 2024 (UTC)

@Felix QW My opinion about rearrangement is that there are some relation between basis and linear combination, and this could be explained in one single paragraph rather than described in list. Dedhert.Jr (talk) 06:16, 4 February 2024 (UTC)
I partially reverted the change for now, as I think it is important to have correct and clear definitions of the basic concepts in our vector space article. While I think the structured pairs of concept and definition work well, I would be fine with any other layout, as long as the definitions themselves are preserved. Felix QW (talk) 08:28, 4 February 2024 (UTC)
@Felix QW You partially reverted the edit, but it could also mean that you deleted more citations to be added. Can you please add them up? Dedhert.Jr (talk) 09:25, 4 February 2024 (UTC)
@Felix QW Nevermind. I will added it later. Dedhert.Jr (talk) 09:38, 4 February 2024 (UTC)

The presentation of the definition is not very good. A clear definition should IMHO mention all the components used. Furthermore, writing scalar multiplication not explicitly might be okay when working with vector spaces daily, but a definition should make this explicit.
Also in the table of "axiom" the header "meaning" is a bit misleading, like that there is some room for interpretation, but actually whats given in the column is the definition. 132.176.73.161 (talk) 06:56, 19 April 2024 (UTC)

I do not understand your second concern, since the word "meaning' does not appear in the article. I do not understand either your first concern. By "component", I suppose that you mean coordinates on a basis. This cannnot appear in the definition since not all bases are finite, and many vector spaces do not have a given basis. For example, the real valued functions with the reals as domain form a vector space for which no basis can be explicitly described. D.Lazard (talk) 13:20, 19 April 2024 (UTC)

I suggest removing 1*v = v as a defining axiom of a vector space and instead show that it is a property that follows from the other axioms. Note that once we have shown that 0*v = 0 (zero vector) and -v = (-1)*v, both of which follow from the other axioms, we have that:
0 = (1 + (-1))*v = 1*v + (-1)*v <=> 0 + v = 1*v + -v + v <=> v = 1*v. 68.193.34.74 (talk) 11:09, 2 October 2024 (UTC)

You wrote "-v = (-1)*v, which [...] follows from the other axioms". This is true, but this cannot be used to prove 1*v = v, since 1*v = v is used for proving -v = (-1)*v (circular reasoning). D.Lazard (talk) 14:11, 2 October 2024 (UTC)
Good point. However, after further consideration, I still conclude we do not need to assume 1•v=v as an axiom of a vector space but rather can prove it from the other axioms, alone.
By the axiom of the compatability of scalar and field multiplication in vector spaces, we have that a•(b•v) = (ab)•v, for all a and b in our base field and all v in V (a vector space). To be clear, I am using • to symbolize scalar multiplication (i.e., • : F x V --> V) in an F-vector space, and juxtaposition to indicate field multiplication (i.e., F x F --> F) in the base field F.
Since the above axiom holds for all values a in F, set a = 1, i.e., the number 1 (multiplicative identity) from our base field F. Then:
1•(c•v) = (1c)•v (axiom of compatability of scalar product and field multiplication)
= c•v (multiplicative identity axiom, i.e., 1c=c for all c in our base field F)
By the axiom of closure of scalar product in V, c•v is an element of V for all c in F and all v in V, so we may define x := c•v where x is an element in V. Then, substituting above and reading the extreme left and right hand sides, we have 1•x = x.
I checked carefully but did not find any circular reasoning problems with my argument. In particular, the axiom of the existence of a multiplicative identity 1 in F such that 1c=c for all c in F is part of our assumption that V is an F-vector space over F, a field.
What am I missing? 68.193.34.74 (talk) 13:45, 11 November 2024 (UTC)
You prove that 1•x = x for every x such that there exist c and v such that x = c•v, but you do not prove that every element of the vector space can be factored this way. D.Lazard (talk) 22:10, 11 November 2024 (UTC)
If you consider the scalar multiplication c•v = 0 for every c and every v, then every abelian group satisfies all axioms of a vector space except that 1•v = 0 ≠ v. This shows that the axiom 1•v = v is really needed. D.Lazard (talk) 22:48, 11 November 2024 (UTC)
Thanks for your response. I fully agree that 1•v = v is indeed necessary for the reasons you indicated and that it cannot be proven from the other axioms.
After spending (way too much) time thinking about this, I realized many other problems arise if we do not assume 1•v = v.
First, with your proposed definition of scalar multiplication sending everything to the zero vector, there is no field F that is a vector space over itself. In particular, if • : F x F --> F is defined as you indicated, then for all field elements a, b (both nonzero), we have a•b = 0 = 0, meaning that a and b are zero divisors in F, a contradiction because F is a field (hence an integral domain, hence has no zero divisors). I guess this argument assumes that scalar multiplication IS field multiplication, but I think that is the usual assumption when considering F as a vector space over itself.
Another thing I noticed is that the operation of • "looks like" a left group action, albeit of F* (nonzero field elements), viewed as a group under field multiplication with identity 1, on V. Moreover, the distributivity axioms of a vector space further imply that 0•v = 0 so that • is also a well-defined map of all of F (and not just F*) on V. However, we no longer have a group action if 1•s = s is not assumed. This seems to cause all sorts of problems I had not even considered. Among other things, a group action of G on another group H is supposed to permute the elements of H, which we do not get if everything is mapped to 0.
I am curious if it is indeed possible to reconstruct vector spaces by starting with V as an abelian group and then defining the scalar product as being a left action along the lines above. In any event thank you for disabusing me of my idea to discard 1•v = v as an axiom! 68.193.34.74 (talk) 14:31, 20 November 2024 (UTC)