This article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Mid‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles Mid This article has been rated as Mid-priority on the project's priority scale.

Text and/or other creative content from this version of Euclidean subspace was copied or moved into Linear subspace with this edit on April 2013. The former page's history now serves to provide attribution for that content in the latter page, and it must not be deleted as long as the latter page exists.

Archives

no archives yet ( create)

This page has archives. Sections older than 365 days may be automatically archived by Lowercase sigmabot III when more than 10 sections are present.

"To use this definition, we don't have to prove that all the properties of a vector space hold for W. Instead, we can prove a theorem that gives us an easier way to show that a subset of a vector space is a subspace."

This is confusing. The fact is that many of the axioms will necessarily hold for any subset of a vector space, so they don't have to be explicitly shown to still hold. But that is not explained here.

"Proof: Firstly, property 1 ensures W is nonempty. Looking at the definition of a vector space, we see that properties 2 and 3 above assure closure of W under addition and scalar multiplication, so the vector space operations are well defined. Since elements of W are necessarily elements of V, axioms 1, 2 and 5-8 of a vector space are satisfied. By the closure of W under scalar multiplication (specifically by 0 and -1), axioms 3 and 4 of a vector space are satisfied."

The axioms of a vector space have no inherent order, and should not be treated as if they do. If an order is to be imposed, it should be stated and not confusingly assumed by the writer. Not even on the Wikipedia article for vector space are the axioms numbered. The axioms should either be numbered by this article to facilitate a concise proof, or they should be described in the proof.

"Examples related to calculus"

This is a totally unnecessary subheading. Chesemonkyloma ( talk) 21:04, 25 July 2012 (UTC) reply

The article now contains too many stuff about linear equations (one piece which lied here initially, one written by me today, and one from user:Jim.belk’s fork). Now, someone may reduce it a bit. Also, Jim.belk’s portion may still contain some implicit dependencies on finite dimensions, as well as references to real numbers and “Euclideanness”, which I overlooked. Incnis Mrsi ( talk) 10:16, 30 April 2013 (UTC) reply

The article claimed "If V is an inner product space, then the orthogonal complement ⊥ of any subspace of V is again a subspace. This operation, understood as negation (¬), makes the lattice of subspaces a (possibly infinite) Boolean algebra."

However, it's not: distributivity fails. That is, the lattices of linear subspaces (with operations ${\displaystyle +}$ and ${\displaystyle \cap }$) is not distributive, and hence it cannot become a Boolean algebra. For instance, take ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$ 3 distinct lines in ${\displaystyle R^{2}}$. Then, ${\displaystyle a\cap (b+c)=a}$, while ${\displaystyle (a\cap b)+(a\cap c)=(0)}$.

I will delete the reference to Boolean algebra in the article, but if someone finds a mistake in what I said, let me know. 160.78.72.118 ( talk) 16:49, 10 February 2016 (UTC) reply

Where's the 'green' circle?I can't see it.Thank you. Warren Leywon ( talk) 23:49, 5 March 2017 (UTC) reply

The equivalent definition ("Equivalently, W is a subspace of V whenever...") allows W to be empty, so it is not an equivalent definition. Schubeda ( talk) 15:48, 16 April 2019 (UTC) reply

@ Schubeda: Indeed! Well-noticed. I have added the nonempty condition, which is included in the sources and should have been included here. -- JBL ( talk) 16:24, 16 April 2019 (UTC) reply

I understand that math encoding is a difficult subject, but I propose to change the Orthogonal Complements section at least for reasons of within-sentence consistency. The direction of consistency can go in more than one direction — it could also go towards more diverse templates or unicode. I propose that it go towards <math> because:

1. MathML has utmost support in the ecosystem of competing solutions for math rendering.
2. Latex is the utmost popular math markup language. Using Latex lowers the threshold to contribution.
3. Screen readers optimize for Latex but not for the myriad of Wikipedia markup.
4. Wikipedia offers the most technological fallback solutions for <math>
5. The paragraph is largely in <math> already.

If there are objections then I won't proceed.

PS. I have read the pages on Wikipedia consensus for math encoding and formula rendering. It is my reading that developing ad hoc consensus at the level of desired change is accordant with guidelines.

SirMeowMeow ( talk) 22:52, 17 February 2021 (UTC) reply

This section must be edited, as it contains three different encodings for the same symbol: ⊥, \perp, \bot. My opinion is that <math> and \perp must be used for formulas that contain this symbol, and {{ mvar}} for single letter formulas (generally, I recommend {{ math}} for very simple formulas without special symbols, but I have not seen any in this section). D.Lazard ( talk) 09:08, 18 February 2021 (UTC) reply

I'm reading this article alongside my linear algebra textbook and lectures from a college course, so I'm not going to edit this because I'm still learning about it. However...

In the four examples, the variables representing field "K" and subspace "V" are never brought up again in favor of just calling the fields and vector spaces what they are (saying R instead of referring it to it as K, for example). This makes it a bit confusing to read.

I like that it ties into the Definition section, but you could completely replace all lines such as "let K be the set of all real numbers R" with "Consider the set of all real numbers R", and it would increase readability while requiring zero changes to the examples themselves. Chersith ( talk) 16:00, 16 March 2023 (UTC) reply

@ Chersith: Well V is used again, but I agree with you that (since all examples are over the field R) it isn't necessary or obviously helpful to duplicate the notation with K in the first example. I've revised along those lines, let me know what you think. -- JBL ( talk) 00:19, 29 March 2023 (UTC) reply