Talk:Spectrum (functional analysis)

message: Re recent revert by User:Jitse Niesen of edits by User:MathKnight, titled :please see discussion on Wikipedia_talk:How to write a Wikipedia article on Mathematics." I actually liked MathKnight's tex-ification. -- linas 6 July 2005 01:01 (UTC)
: The God given inviolable rule is that there shall no be any worship to PNG idols unless on a separate line. The scripture at Wikipedia:How to write a Wikipedia article on Mathematics confirms that. Ask  God for MathML to be implemented in all browsers soon, that will solve the issue. Oleg Alexandrov 6 July 2005 03:24 (UTC)
:: Now that MathJAX/SVG seems to be working on all browsers, perhaps we can revise this position and use <math> more liberally.  (If only Wikipedia provided an easier markup for it, say [$...$] ) --Jorge Stolfi (talk) 02:34, 5 February 2013 (UTC)
thread_title: TeX vs. Non-Tex markup

message: What about operators on more general topological vector spaces (that are not Banach spaces)?  Certainly the requirement that T - lambda is not invertible makes sense in that context.
Crust 20:23, 15 November 2005 (UTC)
replies: text: Seems like you get a big mess from the topology. For example, an operator might be invertible in the TVS (topological vector space), but not in the completion of the space. Maybe the space of all differentiable functions would make a good example? -- KarlHallowell 01:16, 16 November 2005 (UTC)
replies: text: KarlHallowell, sounds like a good point.  (Just to make sure I understand you correctly since it's been some time since I did functional analysis: I think what you are considering a situtation with U,V TVS's with U \subset V, U has the induced topology and U dense in V.  Then your concern is that a continuous linear operator T: V->V might have a different spectrum than its restriction T:U->U.)  Yes, I was thinking primarily of locally convex TVS's given by a family seminorms, such as various spaces of differentiable functions.  I think my concern is similar to OoberMick's above.  Crust 20:25, 16 November 2005 (UTC)
thread_title:  Operators on more general topological vector spaces 

message: If we have a hermitian operator H that can be decomposed as 
:H = H_1 + H_2 = U_1^\dagger D_1 U_1  + U_2^\dagger D_2 U_2
with H1, H2 hermitian, U1, U2 unitary, is there a way of combining the Ds together to give the original spectrum of H? --HappyCamper 03:49, 18 September 2006 (UTC)
replies: text: in general, it sounds too much to hope for to me. if H_1 and H_2 commute, then the diagonalization of H is just D_1 + D_2. also, if we're talking about matrices, then there are perturbation-theoretic results from matrix analysis that give you some idea where the eigenvalues of H are located, in terms of eigenvalues of H_1 and H_2. Mct mht 05:40, 18 September 2006 (UTC)
replies: text: It does sound like wishful thinking. But going along the lines of your latter point, do you have good resources for that? Ideally, something with error bounds would be nice. I'd like to check up on that. --HappyCamper 15:21, 18 September 2006 (UTC)
replies: text: surely for matrices some estimates are possible. you might wanna look into books on matrix analysis. Horn and Johnson, both volumes, would certainly contain that kinda stuff. also, first chapters(chap. 1-2?) from Matrix Analysis by Bhatia (it's one of the GTM, i believe). Mct mht 18:36, 18 September 2006 (UTC)
thread_title:  Spectrum of an operator 

message: Looking at the history of this article I see that it previously defined the spectrum of an element in a Banach algebra and then gave the spectrum of a bounded operator as a special case. Someone then added the definition of the spectrum of an unbounded operator and the definition for a Banach algebra was taken out. However the Banach algebras definition is important in it's own right (not just for defining the operator case). It is one of the most important concepts in Banach algebra theory. 
Also there doesn't seem to be a purely algebraic article on the spectrum of an element in a unital algebra. Hence, I think the following split may be in order:
* One article called something like Spectrum of an element (or something less sucky if you can think of anything) which gives the most general (purely algebraic) definition and then has a section focussing on Banach algebras. There we have the information from here is basically Banach algebraic (such as the spectrum being compact and non-empty in that setting). It has some examples: C(X), B(E)... and the B(E) example points to the second article.
* An article at Spectrum of an operator which gives the definition of the spectrum of a (not necessarily bounded )linear operator, and then mentions that the spectrum of a bounded operator is the spectrum (in the previous sense) of that operator in B(E) (or in L(E) for that matter). All of the specifically operator theoretic stuff about decomposition goes here.
Lastly I think Spectrum (functional analysis) is a bad title for either  article because (for example) Spectrum of a C*-algebra is also a use of "spectrum" in functional analysis.
A Geek Tragedy 12:36, 28 May 2007 (UTC)
thread_title: Definition for [[Banach algebra]]s matters! Perhaps split?

message: From the article:
::The set of approximate eigenvalues is known as the approximate point spectrum. [...]
::"When T is bounded, then, by Riesz's lemma, the eigenvalues lie in the approximate point spectrum. [...]
::When T is unbounded, the definition of approximate point spectrum is slightly different. Continuity can no longer be used to show that that every eigenvalue is an approximate eigenvalue. So the approximate point spectrum of T is defined to be the union of eigenvalues and approximate eigenvalues."
But surely, if x is an eigenvector and (T-\lambda)x=0, then the constantly x sequence is an approximate eigenvector, whether T is bounded or not? Functor salad 18:11, 30 September 2007 (UTC)
replies: text: you're right, not quite sure what happened there. Mct mht 23:36, 30 September 2007 (UTC)
thread_title:  Why Riesz' lemma to show that eigenvalues are in the approximate point spectrum? 

message: I have made quite a few changes to this article; some are based on my observation in the previous section that this article had a major factual inaccuracy in it, and some are simply cleanup.
* I rearranged the leading section to be a bit clearer, in that each paragraph now has a clear separate purpose, whereas they ran on a bit before.  The first paragraph is purely introductory (and only covers bounded case), the 2nd contrasts the bounded inf. dim case with the finite dim. case with the aid of an example, the third discusses the unbounded case and the final the unital Banach algebra case.
* In the course of this I removed the factual inaccuracy in this the leading section: it didn't mention that we require the inverse to be bounded.  I left it out for the bounded case since its redundant, but I made it clear for the unbounded case.
* I removed all unital Banach algebra stuff from the bounded linear operator section.  Someone who is new to this idea shouldn't have to fight through the generalisation (especially since it is a choice of generalisation, the other being to the unbounded case).  It wasn't clear whether the "basic properties" section was about bounded op.s only or the general algebra case, so I changed it to be voiced in terms of just the bounded op case (hinting that something might be true is worse that not mentioning it at all).
* I gave the unital Banach algebra case its own section since it does deserve a mention. Since I haven't studied this case, the section is basically empty.  If some of the "basic properties" carry to this case then a simple reference to the fact that they do would suffice.  Another idea would be to move this section in between the current sections 1 and 1.1, and make basic properties explicit that it applies to the general Banach alegbra case.
* I removed all mention of the above fallacy that an invertible operator can exist without being bijective.  The exception is in the section on clasification of points in the spectrum.  I don't know enough to make this change, so I just tagged it.  Presumably the main article on the subject is full of problems too. In case it's not clear, the sort of problem I'm talking about is saying that an operator is in the spectrum because "it does not have dense range", when in fact dense range is not sufficient for invertibility.
* I removed the reasoning from the unbounded case that shows that bijectivity is required for invertibility, since this is true for any operator (indeed any function).  I think this was just a consequence of the confusion I've now removed from the first section.
An unrelated point: I was thinking of merging resolvent set into this article (when (if!) I get time), and moving resolvent formalism to resolvent function.  Any opinions? Quietbritishjim (talk) 14:12, 22 June 2009 (UTC)
thread_title:  Lotsa stuff 

message: There was a section in eigenvalues and eigenvectors about extensions to infinite-dimensional spaces.  Since it got too long and advanced for that article, I merged it into this article (in the "Definition" and "Spectrum and eigenvalues" subsections of "Spectrum of a bounded operator".  Also the "Example" subsection of "Classification of points in the spectrum of an operator". Since I am not familiar with the topic, could someone please check those sections for accuracy? Thanks... --Jorge Stolfi (talk) 02:30, 5 February 2013 (UTC)
replies: text: Decomposition of spectrum (functional analysis) is probably a better place for that example, where the self adjoint operators on Hilbert space case is discussed. Mct mht (talk) 10:21, 5 February 2013 (UTC)
replies: text:  OK. But conversely, perhaps some parts of that article (especially the definition of spectrum) should be here and not there? --Jorge Stolfi (talk) 17:58, 5 February 2013 (UTC)
thread_title:  Merged a section from Eigenvalues and eigenvectors article 

message: Could someone point out what the problem is with that section? Are we not happy with the definition? Mct mht (talk) 10:18, 5 February 2013 (UTC)
replies: text: Since nobody replied, I removed the tag. -- Jitse Niesen (talk) 12:23, 2 September 2013 (UTC)
thread_title:  Fact tag 

message: In paragraph "Spectrum of the adjoint operator" there is mismatch of terms - AFAIK there is firstly the concept of dual operator of an operator between normed spaces and then a standard theorem says that they have the same spectrum. Secondly, there is the notion of adjoint operator of an operator on a Hilbert space and there which is a similar notion, but the difference is that the identification of the Hilbert space and its dual comes into play and consequently, in this context, spectrum of the adjoint operator is the "conjugated" spectrum of the original operator. I think this should be clearly explained on that page.  A.j.rimmer.bdzp (talk) 15:14, 25 March 2021 (UTC)
thread_title:  Inconsistency in paragraph "Spectrum of the adjoint operator" 

message: The section Classification of points in the spectrum begins as follows:
"A bounded operator T on a Banach space is invertible, i.e. has a bounded inverse, if and only if T is bounded below and has dense range."
But there is neither a link nor an explanation for the term "bounded below".
Does this mean that the image of the unit sphere in Banach space does not contain the origin?  Something similar?
A brief explanation would be useful here. 2601:200:C000:1A0:243B:747:527E:21A1 (talk) 21:27, 11 September 2022 (UTC)
replies: text: Good catch. I added the definition. StrokeOfMidnight (talk) 23:54, 16 September 2022 (UTC)
thread_title:  "Bounded below" needs explanation or link 

message: Current wording is:
"By the closed graph theorem, \lambda is in the spectrum if and only if the bounded operator T - \lambda I: V\to V is bijective on V."
shouldn't it be the other way round? If T = I then for \lambda = 1 the operator T - \lambda I = 0 is not bijective on V (it is the zero operator). I really wouldn't want to introduce errors changing something. RadostW (talk) 23:14, 25 January 2023 (UTC)
thread_title: Possible typo?