Talk:Norm (mathematics)

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modulus of an element of an arbitrary field

Does anyone know what the modulus of an element of an arbitrary field is? Does the notion of modulus exist in every field?

You can't do it for all fields. The article as it stands right now is wrong. AxelBoldt 19:28, 22 Apr 2005 (UTC)

Elaborate $p \geq 0$, simple consequence?

It is not obvious for me why we have the implication that $p$ is positive, in none of my references (or the book cited) is the norm defined as anything but positive. Could you please elaborate, why does it follow that it is positive through this definition? See the article on normed vector space for example.

Link to an explanation of the || notation?

I don't know what the || notation means. From context, I think I get that it is analagous to absolute value or cardinality, but for vectors. Will someone please add something explictly introducing the || notation either here or on it's own page?

Kal Culus

I do not understand your question. || is a function on a vector space. We write |.| to indicate the function has to be feed a vector. We could also write f(x), with f the norm and x an element of the vector space V. The properties of the norm function are given in the article. I do not know how to make this more explicit in the article. MathMartin 13:33, 29 Oct 2004 (UTC)

Merging norm (mathematics) and normed vector space

Would it be agreeable to merge the material from here back into normed vector space? I can see a couple of reasons for this:

1. all examples for norms given here are also examples of normed vector spaces, so we could have all examples in one spot.
2. norms and normed vector spaces are essentially the same concept; in all likelihood, the reader who wants to learn about one wants to learn about the other as well.
3. the topology induced by a norm is currently explained in both articles
4. there are several other concepts called "norm" in mathematics, so norm (mathematics) should really be a disambiguation page. See norm.

AxelBoldt 19:59, 22 Apr 2005 (UTC)

I did not separate norm (mathematics) and normed vector space but I did separate metric (mathematics) and metric space recently and I think most of the arguments are valid for those pages as well. See Talk:Metric_space#Split_metric_space_into_metric_.28mathematics.29_and_metric_space

1. The examples should be separated better.
2. Although the concepts are closely related I think it is clearer to discuss them separately. The norm (mathematics) should focus in the norm as a special function on vector spaces whereas normed vector space should focus on the properties of a normed vector space.
3. The topology stuff should be discussed mainly on normed vector space
4. I have no problem with moving this article to norm (vector space) and putting a disambiguation page here instead.

MathMartin 20:53, 22 Apr 2005 (UTC)

If you are not convinced by my arguments feel free to merge the pages again. MathMartin 13:52, 23 Apr 2005 (UTC)

I'd support this merge. Neither definition can be made without making both and neither can be understood without reference to the other. If we want to separate out examples the articles should be something like normed vector space and examples of normed vector spaces —Preceding unsigned comment added by A Geek Tragedy (talkcontribs)
Ummm, this section is a discussion from two years ago. If you want to merge the articles, I think you need to make a fresh argument. I think you also need to make a sound refutation of MathMartin's arguments above. Lunch 16:44, 5 April 2007 (UTC)

Any field?

This article defines norms for any field F, but there is a problem with axiom 2 since it relies on a real-valued function |a| defined in the field. What does that mean for a finite field, for example? --Zero 06:28, 23 Apr 2005 (UTC)

I really don't know whether it's possible to make the definition of norm general enough for more fields. I guess for the norm to exist, we need some kind of map from our base field to the reals which will itself satisfy the axioms of a norm. a sort of subadditive field homomorphism to R. Do such things exist for finite fields? I have no idea. Certainly I've never seen anything like that. -Lethe | Talk 17:25, July 11, 2005 (UTC)

I agree something does not make sense. If F=C are the complex numbers, then axiom 2 has no meaning, as C is not an ordered field, so we cant say that p(u+v) <= p(u) + p(v) — Preceding unsigned comment added by 186.18.76.220 (talk) 01:53, 28 September 2011 (UTC)

Disambiguation???

On the Gauge disambiguation page, I tried to come up with a short one-line description, and wrote:

"*Gauge (mathematics), a semi norm, a concept related to convex sets.

Unfortunately, I don't understand this stuff well enough to know if I butchered the description or not. Could somebody please check it out and correct if needed? Thanks. RoySmith 8 July 2005 11:58 (UTC)

turning a seminorm into a norm

Somehow I seem to remember there being a person's name associated with the process of turning a seminorm into a norm by modding out by the vectors of norm zero, in just the same way that you turn the space of Lebesgue square integrable functions into a Banach space by identifying functions that agree almost everywhere. I came to wikipedia to look up this name, but couldn't find it. Am I imagining this? -Lethe | Talk 02:46, July 11, 2005 (UTC)

It works exactly like for Lebesgue case (which is in fact a special case of this): You just consider the quotient space modulo the subspace of vectors of zero seminorm, on which the (induced) seminorm (the seminorm is independent of the representative of a class) is a norm. (But I don't know whose name could be associated to this -- maybe you think of the associated Hausdorff space, where this is a special case of.) — MFH:Talk 21:59, 28 February 2006 (UTC)
For a finite field, I think there is only one. Define |x|=1 if x is not zero, and |0|=0. Not terribly useful! --Zero 17:02, 13 July 2005 (UTC)

L2 and energy norms?

In my Finite Element Analysis course, I keep running into the L2 norm,

$\|u-u_h\|_0 = \sqrt{\int_a^b |u-u_h|^2\,dx}$

and the "energy norm" (which apparently, for most solid mechanics applications, denotes energy):

$\|u-u_h\|_m = \sqrt{\int_a^b\sum_{i=0}^m \left| \frac{d^i u}{dx^i} - \frac{d^i u_h}{dx^i}\right|^2\,dx}$

Can someone who knows more than I on the subject add information on these? —BenFrantzDale 03:49, 27 September 2005 (UTC)

In the section p-norm in this article, there is a reference to Lp space where that kind of norms are discussed. Oleg Alexandrov 05:16, 27 September 2005 (UTC)
That article is not very well suited for explaining the L2 norm. IMO it's worth having a separate article, as it's used quite commonly. Info from Mathworld could be adapted. 142.103.235.1 19:30, 21 February 2006 (UTC)
You want L2 norm, then visit the article on L2 norm. :) Oleg Alexandrov (talk) 00:30, 22 February 2006 (UTC)

infinity norm

We need to generalize max to sup for infinity norm. MathStatWoman 09:26, 22 January 2006 (UTC)

I don't think that's right. If a sequence doesn't have a largest element, then its l-∞ norm is not finite. -lethe talk 10:24, 22 January 2006 (UTC)

This article has been very helpful, but I don't understand this about the infinity norm: "The set of vectors whose ∞-norm is a given constant forms the surface of a hypercube of dimension equivalent to that of the norm minus 1." I understand why it form a hypercube, but what is the part about the dimension? Oh never mind, I get it now. I just didn't know a norm had a dimension and as far as I can tell that is never mentioned in the article. I thought of the dimension of the hypercube as the size or edge length or something like that. Which I guess would be equal to two times the norm. 150.135.159.218 (talk) 18:47, 15 October 2008 (UTC)

Unless your "Oh never mind" is something I'm missing, the wording is ambiguous. A norm is always a scalar, so if you take "dimension" in the number-of-components sense, then the phrasing is wrong. Taking the other meaning -- edge length -- the article is just wrong. I just rephrased it. —Ben FrantzDale (talk) 10:52, 16 October 2008 (UTC)

I see this portion as inconsistent since $\infty \notin R$: Let p ≥ 1 be a real number. ...and for p = $\infty$ we get the infinity norm --Kmatzen (talk) 15:10, 21 October 2011 (UTC)

If we have an article entitled "Norm(mathematics)", should we not include the concepts of Banach spaces, norms based on the general Lebesgue integrals as well as countable sums, and sup norms? To quote from Banach spaces: "The Banach space l∞ consists of all bounded sequences of elements in K; the norm of such a sequence is defined to be the supremum of the absolute values of the sequence's members." MathStatWoman 13:04, 22 January 2006 (UTC)

If you want to add more stuff to this article, please go ahead. But don't do much integration work by removing stuff from other articles. That seldom works. Oleg Alexandrov (talk) 17:30, 22 January 2006 (UTC)
• JA: Yes, that's kinda what I meant by "co-ordination", interlinking and such. Jon Awbrey 17:33, 22 January 2006 (UTC)
We are on the same wavelenth. When I hear people taking big plans about reorganizing things I assume the worst. :) Happily not this time. :) Oleg Alexandrov (talk) 01:10, 23 January 2006 (UTC)

Someone should add some links (in this article) to other articles explaining norms, and the sup norm as an example, since it is so important in other areas of mathematics. Is that all right with everyone out there? MathStatWoman 18:33, 23 January 2006 (UTC)

Go ahead. :) Oleg Alexandrov (talk) 02:40, 24 January 2006 (UTC)

Weak norms

I found a weak norm being used at http://www.sm.luth.se/~johanb/applmath/chap3en/part3.htm. Should this be in the main article? --Sunnyside 10:40, 9 February 2006 (UTC)

I guess one needs to have a section in this article discussing norms on function spaces. That seems to be missing now. Then, that norm would be most welcome, but I never heard it being called weak norm though, some references for that (except the web page) would be needed. Oleg Alexandrov (talk) 23:02, 9 February 2006 (UTC)

Streamlined definition of a semi-norm

I prefer to define a semi-norm as in the latest version just posted today. Positive homogeneity and the triangle inequality are the only two properties you need, and it seems silly to keep carrying around positive definiteness when it follows. Perhaps I am missing something as I have not seen this anywhere. Jenny Harrison 12 July 2006

I reverted that change. It may be more streamlined but it is harder to read. I prefer the classical definition, and I added a brief sentence saying that first property is actually not necessary. Oleg Alexandrov (talk) 20:14, 12 July 2006 (UTC)
Let's think about this again. When we define a linear transformation T of a vector space, we do not usually list the fact that T(0) = 0 in the definition. We would appear ignorant if we did so. Similary, there is no need to include positive definitness. Since everyone has memorized it this way, most don't realize it is not needed. It becomes a mantra that may be comforting but is pointless to carry it around. The lean definition actually reads very well, for it looks a lot like the definition of a linear transformation, a kind of sublinear functional, as it were. So I am going to try again to make this read well for you. Jenny Harrison (talk) 16 July 2006 (UTC)
I see your point. However, T(0)=0 is trival to derive and not that essential for linear transformations. For seminorms, the positivity is crucial, and is important enough to actually state it as a definition. It is also rather clumsy to derive, and people better spend their attention on something else than that derivation when reading the article. So I disagree. Oleg Alexandrov (talk) 17:28, 17 July 2006 (UTC)
Leave the discussion of which definition to be followed officially. First of all, Did you people check completely the if and only if condition?? I think you are talking only one way i.e. "||v||=0 if v=0" , but not "||v||=0 only if v=0". which makes the condition of "positive definiteness" must to be mentioned, though "positivity" (as you people said) could be omitted. More over, I didnot understand at all why you people are thinking of "linear transformation" here!. my word is simply that we need all 3 properties.. positive definiteness, subadditivity and homogenity, compulsarily but not positivity to define NORM.

Hi, on a possibly related note, I noticed this under Notes:

A useful consequence of the norm axioms is the inequality ||u ± v|| ≥ | ||u|| − ||v|| | for all u and v ∈ K.

Would it be okay to move this along with positive definiteness as another property that follows from the definitions? The Absolute value article reads much better, witha section of fundamental properties and another for derived ones. MisterSheik 03:03, 17 July 2006 (UTC)

||u ± v|| ≥ | ||u|| − ||v|| | is indeed a derived property and followw very easily from the triangle inequality. Positivity is much more crucial and not as easy to derive and I think it should stay in the definition. Oleg Alexandrov (talk) 17:28, 17 July 2006 (UTC)
I still like the the way "absolute value" reads, and think that this article could benefit from immitating that model. My view is that all of the derived properties worth mentioning should be mentioned together. Also, I think that the positive definiteness is not that hard to derive :) (from 1, p(0)=p(0*u)=0*p(u)=0, and then let u=-v in 2, where p(-v)=p(v) from 1). MisterSheik 18:59, 17 July 2006 (UTC)
It is not worth the trouble spending time on the derivation, I think. Also, the absolute value article is different. There all the properties are consequences. Here are are talking about how to define an abstract concent like norm. Oleg Alexandrov (talk) 02:35, 18 July 2006 (UTC)
As MisterSheik points out, the derivation of positivity is rather trivial. The adjective "crucial" is subjective. All properties are crucial when you need them. The question comes down to this: what is fundamental and what is derived? If a derived property is included in a definition, then it must be verified each time one needs to check that a definition holds. It might be helpful to compare the definition of a linear transformation with the definition of a seminorm, with and without positivity. To show a linear transformation T is linear, we do not have to verify that T(0) = 0. Jenny Harrison (talk) 15:37, 18 July 2006 (UTC)
Leave the discussion of which definition to be followed officially. First of all, Did you people check completely the if and only if condition?? I think you are talking only one way i.e. "||v||=0 if v=0" , but not "||v||=0 only if v=0". so positive definiteness is compulsary!!
A similar discussion is going on at talk:metric space about the need for the positivity axiom. Oleg Alexandrov (talk) 00:50, 19 July 2006 (UTC)
I would have done it already if this discussion hadn't taken place, but since it has, I'll put it up for comment here first. In the definition, I propose to replace "3. p(v) = 0 if and only if v is the zero vector (positive definiteness)" with "3. if p(v) = 0 then v is the zero vector (separates points)", since the other direction follows from axiom 1. JumpDiscont (talk) 01:10, 1 October 2010 (UTC)
Since there were no comments, I made the change. JumpDiscont (talk) 05:39, 16 October 2010 (UTC)

The terminology "positive homogeneity" is misleading, especially since the article then links to a different and much more common definition of 'positive homogeneity'. This property should be called 'symmetry' or possibly 'symmetric homogeneity'. Cerberus (talk) 13:15, 26 July 2013 (UTC)

Apparently the preferred term is "absolute homogeneity" ([1]). I will make the edit. Cerberus (talk) 18:20, 28 August 2013 (UTC)

L_-inf

Perhaps there is a L_(-infinity), with a unit circle of a cross (0 width)? —The preceding unsigned comment was added by 67.183.154.231 (talk) 04:01, 5 December 2006 (UTC).

what quite sure what you mean by inf there. if you consider the Lp norm on say R2, with p < 1. this is not really a norm since the unit sphere is not convex. as p tends to 0, the unit circle probably looks more and more like a cross. Mct mht 09:03, 5 December 2006 (UTC)

Taxicab, 1-norm, and hypercubes.

Regarding edits. I didn't confuse hamming distance with the taxicab norm. My interpretation is not incorrect, hamming distance and 1-norm are equivalent reductions(HAMMING 1-norm). You could consider the hamming distance to be the 1-norm of the binary vertex representing the difference between two vertices on a Hypercube graph. (hamming(A, B) = "1-norm"(A-B) or "1-norm"(AB)). In addition Hamming_distance states "and the Hamming distance of the strings is equivalent to the Manhattan_distance between the vertices." and you can see that "Manhattan distance" is redirected to Taxicab_geometry. These two concepts are very similar and indeed they are reducibly equivalent. --ANONYMOUS COWARD0xC0DE 10:01, 3 January 2007 (UTC)

you're right, on the hypercube the hamming distance and the l1 distance are the same. however, the rest of the article is using vector spaces over the reals as examples. i think introducing the hamming distance at that point in the article confuses the reader. perhaps you could make a breakout section on norms for vector spaces over other fields ({0,1} in the case of the hypercube). Lunch 19:31, 3 January 2007 (UTC)
Meh I thought I would add it, but it is a little bit off topic I agree. --ANONYMOUS COWARD0xC0DE 00:53, 10 January 2007 (UTC)

Don't understand norm of W + v

In the part where it talks about how every seminormed vector space induces a normed vector space V/W, I don't understand the formula (||W+v|| = p(v)) that is given for the induced norm on V/W. For one thing, what does it means to add the vector space W to the vector v?

W + v denotes an element of the quotient space V/W; it's the equivalence class [v]. say we're in R^3 and W is a 2-d subspace, then W + v is the affine plane obtained by translating W by v. we should also check that the definition (||W+v|| = p(v)) is independent of the representative v. Mct mht 03:49, 6 July 2007 (UTC)

Minor cleanup needed

There are a lot of names for various norms. We need some cleanup to clarify this. There's "Euclidean norm", L2, taxicab, etc. It should be made clear in the introduction which of these is which and some of the redirects need to be looked into. For example L1 norm redirects neither here nor to Lp space. —Ben FrantzDale 00:53, 11 October 2007 (UTC)

Rhomboid → square?

The L1 circle is described as a rhomboid when it is a square. Any objections to fixing this? --Vaughan Pratt (talk) 19:22, 26 June 2008 (UTC)

Proposal: Merge Uniform norm to here

Just as Euclidean distance has an article and Euclidean norm redirects to a section in Norm (mathematics), so should Chebyshev distance continue to have an article and Uniform norm redirect to a section here that's prsently missing. Let's merge it in. This is an alternative to the proposal to merge Chebyshev distance into Uniform norm, which seems backwards to me. Dicklyon (talk) 20:33, 21 September 2008 (UTC)

Disagree. The uniform norm is a big concept of mathematical analysis. I would support merging Chebyshev distance here, since it is just a finitary special case, under the general principle of making abstract concepts central. Charles Matthews (talk) 20:39, 21 September 2008 (UTC)
You would merge Chebyshev distance into Norm (mathematics)??? And where do you get this general concept of making astract concepts central? Seems like a bad idea to me. Dicklyon (talk) 20:49, 21 September 2008 (UTC)
As Charles Matthews says, the uniform norm is a very important concept in analysis, and deserves its own article.
(Among concepts it refers to are uniform convergence and L space.)
It should certainly have a section here, but it should also have a longer main article.
I don’t think separating “norm” and “distance” helps – they’re very closely linked, and, unlike for Euclidean distance and norm (which are very common concepts), few people will care about the one and not the other.
As for whether the finite and infinite case should be separate articles or merged into one article, I have no strong feelings – I think they’re very close and should go together, but one can reasonably argue that the R2 case and the function space case are conceptually pretty far.
Nils von Barth (nbarth) (talk) 00:19, 22 September 2008 (UTC)
• Don't merge. I've noticed that making single, large articles that try to "say everything" are daunting for the reader (lots and lots to read, which can be discouraging), and are difficult for editors to maintain (since large articles get frequent revisions, and as a result, little but important parts of the article get chipped away, and are lost, because no one notices, because they're hidden by the large changes.). Smaller articles are more easily monitored for deleterious changes.
Also, norms are not 'almost the same as' distances. Metrics have to satisfy the triangle inequality, there are plenty of norms that fail do so, and thus cannot be used to define a metric. These are not interchangeable concepts. Silly me, what was I thinking of? Something else, clearly.linas (talk) 01:33, 9 November 2008 (UTC)

Semi-* vs. Pseudo-*

Note that in some contexts, the term "semi" is used when the triangle equality doesn't hold. In some cases, it is used synonym to "pseudo" (non-zero vectors of length 0 allowed). However these don't necessarily fall together (at least not for distance functions), do they? E.g. euclidean-like distance in a torus space (using the shortest path on the torus) doesn't fulfill the triangle equality, but the condition d(x,x)==0 <=> x=x holds. Unfortunately these terms are totally mixed up in all the related articles here on wikipedia. 138.246.7.155 (talk) 16:35, 24 November 2008 (UTC)

The p-norm formula for 0 < p < 1 is "also valid" in what sense???

The article defines the p-norm of a vector x in the usual way -- as the pth root of the sum of the pth powers of its components.

Two lines later it states, referring to this definition:

"This formula is also valid for 0 < p < 1, but the resulting function does not define a norm, because it violates the triangle inequality."

Agreed that it violates the triangle inequality. But I do not understand what "the formula is also valid" means here. Simply that the formula can be evaluated, i.e., that it makes mathematical sense? In that case, why exclude negative values of p ? (And why not also mention what happens when p -> 0 and when p -> -oo ?)Daqu (talk) 06:55, 26 December 2008 (UTC)

Weighted norm

I could not find an article or any mention of the following norm,

$\|x\|_A=(Ax,x),$

where A is positive definite matrix, and () is the inner product. Does this norm have a name in English? (Igny (talk) 15:38, 18 January 2009 (UTC))

That's not a norm. Do you mean $\sqrt{(Ax,x)}$? If so, I don't know a name for it. Algebraist 20:51, 18 January 2009 (UTC)
Err, yes that is what I meant. I just thought to improve the expression as well as provide a relevant link for the last formula in the GMM. (Igny (talk) 03:07, 19 January 2009 (UTC))
It is sometimes called the Energy norm. I think it is in my Springer Functional Analysis graduate text. —Ben FrantzDale (talk) 21:51, 19 January 2009 (UTC)

Examples

To add to the overall clarification in the examples section, it might be a good idea to add some numerical examples in with all of the symbolic examples. —Preceding unsigned comment added by 132.235.19.87 (talk) 13:03, 28 April 2009 (UTC)

Length and the norm

regarding this in the introduction: The Euclidean norm assigns to each vector the length of its arrow. Because of this, the Euclidean norm is often known as the magnitude. Isn't the "length" of the arrow determined by the norm? Is there anything really more fundamental about "length"? The statement seems circular. 76.175.72.51 (talk) 16:26, 20 September 2009 (UTC)

It is trying to be friendly to readers who already know what "length" means but came here to find out what "norm" means. So the point is "you already know one norm under the name 'length', it is called the Euclidean norm". Zerotalk 01:08, 21 September 2009 (UTC)

"Almost norm"

I find it instructive to link to related concepts that answer the question "What if you remove axioms?". In particular, I am curious about what happens when you drop axiom 1 from the definition of norm? Do you get a function that defines a metric? I think so, but am not sure...

Metric requires:

For all x, y, z in X, this function is required to satisfy the following axioms:
1. d(x, y) ≥ 0     (non-negativity)
2. d(x, y) = 0   if and only if   x = y     (identity of indiscernibles. Note that condition 1 and 2 together produce positive definiteness)
3. d(x, y) = d(y, x)     (symmetry)
4. d(x, z) ≤ d(x, y) + d(y, z)     (subadditivity / triangle inequality).
The first condition is implied by the others.

Norm requires:

For all a in F and all u and v in V,
1. p(a v) = |a| p(v), (positive homogeneity or positive scalability)
2. p(u + v) ≤ p(u) + p(v) (triangle inequality or subadditivity).
3. p(v) ≥ 0, with p(v) = 0 if and only if v is the zero vector (positive definiteness).
A simple consequence of the first two axioms, positive homogeneity and the triangle inequality, is p(0) = 0 and thus
p(v) ≥ 0 (positivity).

The thing that throws me is the note at the end of the "norm" axiom. It seems that criterion 3 is prescribing positive-definiteness. But the note says positivity and non-degeneracy (in the sense that $p(x)=0\,\Rightarrow\, x=0$) is implied by the first two "norm" axioms. If so, then all axiom 3 adds is definiteness. Is that right?

If all of axiom 3 is required, then dropping axiom 1 seems to leave you with a function, p, still defining a metric, $d(x,y) = p(x-y)$. However, if criterion 3 is really a note, implied by axioms 1 and 2, but not an axiom defining "norm", then criterion 3 would fail if we remove axiom 1 and so "a norm without criterion 1" wouldn't define a metric without criterion 3.

Why is axiom 3 so long if much of it is implied by axioms 1 and 2?

Which axioms can be removed from "norm" to still get something that defines a metric? —Ben FrantzDale (talk) 15:13, 28 January 2010 (UTC)

equivalence of norms missing from article

One of the key properties of norms is that they are all equivalent up to constant factors. In particular, if $\Vert x \Vert_a$ and $\Vert x \Vert_b$ are two norms, then there exist some positive constants C1 and C2 such that:

$C_1 \Vert x \Vert_b \leq \Vert x \Vert_a \leq C_2 \Vert x \Vert_b$

for all x in a vector space (for the same constants). As a consequence, it doesn't matter what norm you pick if you don't care about constant factors, which is very convenient for proving lots of asymptotic bounds, talking about stability of algorithms, etcetera. This is discussed e.g. in Trefethen and Bau, Numerical Linear Algebra.

(At least, this is true for finite-dimensional vector spaces; I'm not sure about the infinite-dimensional case.)

(Proof in brief: prove that any norm is continuous, reduce to considering the unit ball by dividing by $\Vert x \Vert_b$, then invoke fact that continuous functions achieve their maxima and minima on a compact set to obtain C1 and C2.)

Anyway, it seems like this should be mentioned somewhere.

— Steven G. Johnson (talk) 02:58, 15 March 2010 (UTC)

Look in the section "Properties". Zerotalk 03:59, 15 March 2010 (UTC)

Norm of a complex number

It's possibly a simple oversight, but complex numbers all have norms, or at least that was what Euler called $\sqrt{x^2 +y^2}$ for the complex number x + iy. And yet this article has no section on complex number norms. The complex number wiki has a section on how to compute this value for complex numbers, but doesn't CALL it a "norm," but rather modulus, absolute value, or magnitude. But if a complex number is considered a vector, then obviously its magnitude is the norm. Has the Euler usage entirely gone by the wayside?

It's not obvious how to compute a complex norm, as you multiply by the complex conjugate, which means you essentially disregard the quantity i. If you didn't, and included it, you'd get a different kind of norm (the Minkowski norm). Is there something in the terminology here that I'm missing? Obviously I'm no mathematician. SBHarris 01:55, 8 April 2010 (UTC)

Nevermind, I found my answer in the article on absolute value. The Euclidian norm of a complex number is indeed the absolute value of it, if the complex plane is identified with the Euclidean plane R2. I'll stick that in here, someplace.SBHarris 22:16, 20 April 2010 (UTC)

Imaginary numbers a vector space?

Regarding this edit [1] I don't believe 0 is an imaginary number, so the imaginary numbers are not an abelian group. much less a vector space. — Carl (CBM · talk) 10:44, 27 October 2010 (UTC)

Arguable, but I suppose the ambiguity doesn't help. I'll revert, although the real numbers really aren't a good example of a vector space over a field, if the field is also the real numbers. — Arthur Rubin (talk) 14:17, 27 October 2010 (UTC)

plus or minus

A moment ago I wrote in an edit summary

an n-sphere is the surface of an (n-1)-ball.

And after clicking, of course, I noticed that I meant (n+1)-ball. Oh well. —Tamfang (talk) 21:25, 11 July 2011 (UTC)

row sum? column sum?

The one norm is simply the maximum column sum of absolute values.
The infinity norm is simply the maximum row sum of absolute values.

Uh, what? I guess these definitions concern something fancier than the plain old vectors that appear to be the context of the rest of the article. —Tamfang (talk) 01:20, 12 December 2011 (UTC)

I haven't checked the article, but those are quick ways to calculate the operator norm of a matrix acting on vectors with the p-norm for p=1 and p=∞. |A| = max( |Av|/|v| : v ≠ 0 ). JackSchmidt (talk) 17:38, 12 December 2011 (UTC)

Pseudonorm

Why does Pseudonorm redirect to this page, when the page doesn't mention "Pseudonorm" anywhere? I have seen pseudonorm used in the sense of the Hamming-distance from 0, but there maybe other conventions. It would be nice to mention it, even if only to say that there is no consensus on the terminology. Lavaka (talk) 14:35, 20 April 2012 (UTC)

Notation

I improved the Notation section and put it after Definition section which is a more logical place for it.

I removed the claim that set cardinality is a norm and should be noted by double vertical bars because I cannot see how sets would make up a normed vector space (if there is a natural way to do it, it would be interesting to know) and single-bar notation is also much more common for cardinality.

I removed the claim that denoting norm by single bars is generally discouraged. Actually, the the only case where I have encountered the single-bar notation is the case of Euclidean spaces, and for them, such usage is widespread, so a statement that it has been discouraged should be backed up by an authoritative source. I also removed the suggestion of denoting matrix norm with double vertical line since matrix spaces are special cases of vector spaces and I haven't seen their norms being denoted otherwise. (I personally do not see a problem with norm being confused with absolute value in case of Euclidean spaces, since the Euclidean norm in 1-dimensional space is exactly the absolute value. For matrices, I would rather blame the notation of determinant with single bars as a source of confusion, since for the usual matrix norms, the norm of a 1-by-1 matrix is the absolute value of its only element but it may differ from the determinant of the matrix by sign.)

If someone knows some sources where norms on generic vector spaces or matrix spaces or some other specific spaces are denoted by single bars and can find a source that such notation is discouraged, then such suggestions can be added back to the Notation section. Jaan Vajakas (talk) 16:30, 14 September 2012 (UTC)

All algebraic vector spaces over subfields of $\mathbb C$ are normable

I removed the claim "Although every vector space is seminormed (e.g., with the trivial seminorm in the Examples section below), it may not be normed." since it is not true (for vector spaces over subfields of $\mathbb C$). Indeed, for a field F, algebraic vector spaces over F are characterized by the cardinalities of their Hamel bases. For a vector space $V$ Hamel basis $e_\alpha$, $\alpha \in A$, where A is an arbitrary (possibly infinite) set, we can consider any p-norm w. r. t. the basis, e. g. the 1-norm $\left\|\sum_{i=1}^k a_i e_{\alpha_i}\right\| = \sum_{i=1}^k |a_i|$. Jaan Vajakas (talk) 17:07, 14 September 2012 (UTC)

Derivative

It wasn't my idea to include it...

But the more formal equation is:

$\frac {\partial} {\partial x_k} \|\mathbf{x}\|_p = \frac{ x_k |x_k|^{p-2} }{ \| \mathbf{x} \|_{p}^{p-1} }.$Arthur Rubin (talk) 21:34, 7 January 2013 (UTC)

It was my idea. I agree on the correctness of your formal equation. Maybe it is better to use it to avoid confusion about the power that has to be applied component-wise. --User:Bugmenot10 —Preceding undated comment added 13:29, 8 January 2013 (UTC)
There is the presentation question of whether the expression I used (only clearly usable for a zero component if p ≥ 2), or:
$\frac {\partial} {\partial x_k} \|\mathbf{x}\|_p = \frac{ |x_k|^{p-1}\operatorname{sign}x_k }{ \| \mathbf{x} \|_{p}^{p-1} },$
usable for a zero component where p > 1. Also, in theory, this is a little complicated for WP:CALC, so it would need a source. — Arthur Rubin (talk) 15:10, 8 January 2013 (UTC)
• ^ Advanced Statistics: Volume 1: Description of Populations By Shelby J. Haberman, p.46