Talk:Projection (linear algebra)

message: I have seen the word projection used for this, but never projector, until today.  Does someone use the former for linear transformations and the latter for matrices?  If so, it should say so.  Or do people in some scientific fields use one word while others use the other?  Michael Hardy 20:09, 17 Apr 2004 (UTC)
thread_title: Projection?

message: Should this article include affine-linear projections?  Or maybe they are covered somewhere else?  An example of what I mean in 2 dimensions is P((x,y)) = (2,y), which is a projection onto the vertical line through (2,0). McKay 16:07, 18 October 2006 (UTC)
replies: text: I don't think so, because this is an article about linear algrebra. Pdbailey 03:10, 24 October 2006 (UTC)
replies: text:  Affine transformations are a standard part of linear algebra.  McKay 01:31, 25 October 2006 (UTC)
thread_title:  Affine case 

message: I just tried to cleanup the abstract example, but it is so all over the place that I just can't. I wonder what the value of the section is at all.Pdbailey 01:30, 30 October 2006 (UTC)
replies: text: What does "k <= n column entries, each of length n" mean?  Just that there are k columns and n rows?  McKay 01:40, 30 October 2006 (UTC)
replies: text:  yes Pdbailey 05:47, 31 October 2006 (UTC)
replies: text:  So I fixed this point, and improved the typesetting, but I still don't understand it.  Both "span" and "kernel" are spaces, not matrices.  I vote to delete the section unless someone can clarify it. McKay 06:25, 31 October 2006 (UTC)
replies: text:  I think I understand this section. It's not really an example, but the general form for oblique projections, and it generalizes the formula mentioned at the end of "orthogonal projections". However, I agree that this section is not very clear. -- Jitse Niesen (talk) 08:20, 31 October 2006 (UTC)
thread_title:  abstract example makes no sense 

message: Anybody thinks it is worth merging this article with Projection operator (that is, redirecting that one to here)? Oleg Alexandrov (talk) 16:18, 31 October 2006 (UTC)
replies: text: I agree, so I did. -- Jitse Niesen (talk) 00:57, 1 November 2006 (UTC)
thread_title: Projection operator

message: There is an error in the content here.The image has the caption
"The transformation T is the projection along k onto m. The range of T is k and the null space is m".
It should be the other way around, null space k and range m. Similarly, the last sentence reads
"Only 0 and 1 can be an eigenvalue of a projection. The eigenspace corresponding to the eigenvalue 0 is the null space "U", and the eigenspace corresponding to 1 is the range "V"
Throughout the section he refers to the properties of the nullspace V, and range U. 
I have corrected these problems, they read correctly now.
:Thanks.  If someone gets the time, the notation in the image should be made consistent with the notation in the text (which could probably stand some reworking itself).  Lunch 21:57, 13 November 2006 (UTC)
thread_title:  Properties Section 

message: In my work, I use the terms "projection" and "projector" for these mathematical objects which project onto any set, and not only on linear or affine subspaces. I am sure there is some formal mathematical literature on that, although I only know references of this usage in more applied contexts. Shouldn't the definition be wider (that is, some non-linear transformations could also be called projections)? In metric spaces, one can further define a "distance-minimizing" projection, which is not only idempotent but also maps any point of a space to the point in a set that is closest to it (this can be seen as an extension of the "orthogonal" projection, I guess). Anyway, I just wanted to raise this point of view, I don't know to what extent this extended definition of projection is wide-spread enough to include it. Pierre T. 13:36, 19 March 2007 (UTC)
replies: text: seem like fair enough comments. if you, or some other knowledgable folks, got some references and are convinced that it is sufficiently common usage, i'd suggest add it to the article (or possible create a new article). Mct mht 13:59, 19 March 2007 (UTC)
replies: text: yup, there are other notions of projections, and if you add material about them, that'd be great.  this particular article is about the use of projections in linear algebra, though.  it'd probably be best to start another article; the addition would be very welcome.  cheers, Lunch 01:50, 20 March 2007 (UTC)
replies: text: Ok. I'll look into that. I now came to realize that there are no wiki entry on the convex feasibility problem... There is quite a large body of work on this, and people often use projections onto convex sets in the formulation of their algorithms. That might be the right place to start. Thanks for your encouragements.Pierre T. 13:35, 20 March 2007 (UTC)
thread_title:  Non-linear subspaces 

message: I think 'if x is a vector in the domain X of the projection', should say 'image/range of projection'; X is not the domain as described.  Also (P-P^2)=0 is not sufficient for orthogonal projection I don't think, just projection.  I think, transpose(P)*(1-P) should do.
:seems like it should just be domain, the "X" shouldn't be there. for your second point, article could be more clear there as well. the assumption throughout is, apparently, that P is self-adjoint, which is correct (notice self-adjointness is invoked in the first equation of the section). but perhaps that needs to be worded more explicitly. Mct mht 22:31, 25 March 2007 (UTC)
replies: text: what that paragraph is saying is that a Hermitian matrix P is a projection (P^2 = P) if and only if it is an orthogonal projection. that doesn't look very helpful. a more appropriate point to make would be "a projection is orthogonal if and only if it is Hermitian." Mct mht 22:45, 25 March 2007 (UTC)
replies: text: That is the point which I intended to make, though I wrote it in a terrible way. Thanks for fixing it. -- Jitse Niesen (talk) 12:07, 27 March 2007 (UTC)
thread_title: Orthogonal Projection Section

message: I didn't found the following pretty useful property of a projection. I'd like to know if people would like to add this property on the page and if there would be any needs for a proof.
Property: If P:V \longrightarrow V is a projection than tr(P) = dim(Im(P))
--Noud 18:46, 30 November 2007 (UTC)
thread_title:  Projection operator properties 

message: Start with vector-vector projections as people learn in physics and early on in linear algebra. This article is too technical for people looking for such basics, they serve to help build intuitions, and this would be better form for a general encyclopedia. The transformation view, while correct, is the wrong way to start the article.  —Preceding unsigned comment added by 71.111.195.53 (talk) 03:27, 30 April 2008 (UTC)
thread_title: Start with simpler examples

message: This article is nearly worthless.  It is a good example that stating a lot of facts is *not* the same as explaining something.
-- Bruce Jerrick  —Preceding unsigned comment added by 76.27.194.188 (talk) 00:22, 7 May 2008 (UTC)
thread_title:  Facts vs. explanations 

message: deleted
LearningQM (talk) 13:55, 5 September 2008 (UTC)
thread_title:  rewrite of Classification 

message: ignore - deleted
LearningQM (talk) 13:53, 5 September 2008 (UTC)
thread_title:  Classification 

message: I have removed the infobox added by User:SharkD (diff). In my opinion the infobox is only marginally related to this article as it discusses projections from a graphical or artist's point of view, whereas this article is theoretical. On the other hand the infobox detracts from the article by forcing the first figure to move down so that it is not visible while looking at the lead. The figure is a good intuitive explanation of the concept and ought to be the first thing seen when this page is opened. --Zvika (talk) 05:30, 15 January 2009 (UTC)
replies: text: I agree, I also don't like this box on top of all the other pages it is on - someone ought to convert it to a standard navigation box (at the bottom of the page) as is used on all other comparable pages instead. --Allefant (talk) 11:03, 15 January 2009 (UTC)
thread_title:  Infobox 

message: I think this article could be improved by adding a couple of examples in 2- and 3-d with pictures. The existing Simple Example section is a good start on this, but it lacks pictures, and could use some copy that takes more of a geometric or intuitive tone, than the math-term laden function which maps the point (x, y, z) in three-dimensional space R^3 to the point (x, y, 0) is a projection onto the x-y plane. Imagine being a newbie, and reading things like function which maps. Personally, as a college freshman (in engineering!), I would have been a little bit confused about that, let alone a layperson, who isn't familiar with the concept of a function as a map from space to another. I will make this change some day, but that day will likely never come, to be frank. User:!jimtalk  contribs 06:58, 20 October 2009 (UTC)
thread_title:  Could use better examples, with pictures 

message: There is a little error. The inverse (lambda I - P)^{-1} it is 1/lambda I + 1/lambda * 1/(lambda-1) P
Proof: If we call L=lambda:
(L I - P )^{-1} = 1/L (I-P/L)^{-1} = 1/L sum (P/L)^n = 1/L *I + 1/L *P/L + 1/L * P/L^2 + ... = 1/L I + 1/L (1/L + 1/L^2+ ... ) P = 1/L * I + 1/L(L-1) * P
Cheers.  — Preceding unsigned comment added by 95.120.230.142 (talk) 03:24, 8 January 2012 (UTC)
thread_title:  a little error 

message: I will delete the sentence "However, every continuous projection on a Banach space is an open mapping, by the open mapping theorem." This is simply wrong. Suppose that this is true. Then every continuous  projection P maps the whole space X onto an open set. But PX is a closed subspace and can only be open if it coincides with X. Hence, PX = X which implies P = I. In other words, the only continuous  projection is the identity (which is a contradiction whenever dim X is at least 2).  — Preceding unsigned comment added by 92.225.117.200 (talk) 14:39, 26 April 2014 (UTC)
thread_title:  Mistake in Banach Space Section 

message: The example shows an orthogonal projection onto a line and how the component orthogonal to u disappears. Shouldn't it be:
uu^\mathrm{T}x_\perp = u\cdot 0
not
\vec{0}
?
130.75.31.239 (talk) 19:40, 19 June 2015 (UTC)
replies: text:  <Px,(1-P)y>=<(1-P)Px,y> =0
LMSchmitt 22:59, 18 December 2020 (UTC)
thread_title:  Formulas subsection 

message: It is not clear what the "Proof of existence" is actually proving, because no statement is given. Is this a proof of the fact that an operator mapping a point in a Hilbert space to the nearest point within a closed linear subspace is an orthogonal projection as previously defined (e.g. linear, idempotent and self-adjoint)? If so, this statement must be formulated somewhere. Preferably, right before the proof.
AVM2019 (talk) 18:15, 17 December 2020 (UTC)
replies: text: shown: every closed subspace S in a Hilbert space H generates an orthogonal projection P onto S such that the orthogonal space S' in H corresponds to (1-P)=P'
LMSchmitt 18:42, 20 December 2020 (UTC)
thread_title:  Orthogonal projection. Proof of existence 

message: The current proof of existence of orthogonal projection claims
"For every \mathbf x the following set of non-negative norm-values \{\|\mathbf x - \mathbf u\| \mid \mathbf u \in U\} has an infimum, and due to the completeness of U it is a minimum."
Which confuses completeness and compactness; this is not sufficient to establish the existence of a minimum (or uniqueness). For an example of a Lipschitz-continuous function from a complete bounded metric space to \mathbb R which does not attain its infimum, consider d(-,A) : \bar B_{\ell^\infty}(0,1) \to \mathbb R with A = \{(1+2^{-i})e_i \mid i \in \mathbb N\} for e_i the standard Schauder basis.
(I do have a proof, but that uses approximations by projections to finite dimensional subspaces, and if there's a fix that doesn't require entirely changing the structure of the proof that would work better.)
Emlili (talk) 09:57, 27 April 2021 (UTC)
replies: text: WRONG. a) there is no use of compactness here. This works also in oo-dim HSs. b) The proof is as in Hilbert projection theorem as you found out. LMSchmitt 11:36, 28 April 2021 (UTC)
thread_title:  Error in orthogonal projection existence proof 

message: The article mentions that if we have a complete inner-product space V, then we can say a projection P\colon V\to V is orthogonal to mean \forall x,y\in V~~\langle Px,y\rangle=\langle x,Py\rangle. However, this definition does not require completeness. Does anything go wrong if we remove from the article the assumption of completeness? Thatsme314 (talk) 14:06, 17 October 2023 (UTC)
replies: text: I'd assume that typical sources only assume completeness to talk about adjoints, and they might define orthogonality as self-adjointness of P. Thatsme314 (talk) 14:09, 17 October 2023 (UTC)
thread_title:  Complete inner-product spaces