Talk:Spectrum (functional analysis)

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This page needs to be rewritten as it only discusses the spectrum of bdded operators. OoberMick 15:14, 18 Nov 2004 (UTC)

Huh? linas 6 July 2005 01:01 (UTC)

When I wrote that comment the remark about unbounded operators wasn't there, but still the article has a bias towards bounded operators. In fact the definition of the spectrum states "Let B be a complex Banach algebra containing a unit e". The definition of the spectrum does not require that B is a Banach algebra only a Banach space and such a definition, where B was a Banach Algebra would be equivelent to a definition on the Banach space B(X,X) of all bounded linear operators on a banach space X. OoberMick 11:09, 29 July 2005 (UTC)

Interesting. What is the definition of the spectrum of an element of a Banach space? Lupin 13:03, 29 July 2005 (UTC)

Appologies, what I've written isn't quite right. I should have wrote ...does not require that B is a Banach algebra only a space of operators on a Banach space and such a definition... But I wasn't completely wrong the dual space (space of linear operators) of a Banach space is indeed a Banach space :). With this is mind we can define the spectrum of a closed operator $T$ in our Banach space $X$. (If $T$ isn't closed, take it's closure... if $T$ isn't closable the whole complex plane is the spectrum). We define the spectrum as the complementary set of the resolvent set, where the resolvent set is the set of complex numbers $z$ such that $(T-z)^{-1}$ exists and lives in $B(X,X)$. So a complex number is an element of the spectrum if $T-z$ is not invertable or $(T-z)^{-1}$ is not densely defined or $(T-z)^{-1}$ is not bounded.

So are you advocating a definition for any complex unital algebra, not necessarily normed or complete? That sounds feasible, but I don't know what you can say in that generality. Maybe the algebraists already have it in an article somewhere. Lupin 16:15, 1 August 2005 (UTC)

What I'm looking for is a defintion of the spectrum which is general enough to allow discussion of the spectrum of differential operators such as the Laplacian and the Schrödinger equation. My personal preference would be to define the sepctrum in terms of operators (as opposed to an element of an algebra) since this would be more consistent with the definition in finite dim where we discuss matrices on a vector space. It would then be possible to show that if we consider only bounded operators then this is a Banach algebra. OoberMick 08:59, 2 August 2005 (UTC)

I agree with OoberMick that the definition is lacking a bit and can be generalized. Below is the definition I use, from notes taken from Kreysig.

Let $X$ be a non-trivial complex, normed linear space (not necessarily complete), and $T$ a linear operator (not necessarily bounded) that maps from a domain in $X$ into $X$. For a complex number $\lambda$, define $T_{\lambda }=T-\lambda I$ and the resolvent $R_{\lambda }=(T_{\lambda })^{-1}$. From now on, I'll write R to mean $R_{\lambda }$.

Then the regular valies of the operator are complex numbers such that (1) R exists, (2) R is bounded, and (3) R is defined on a dense set in $X$. Non-regular values are the spectrum, which are divided into the point spectrum [where (1) doesn't hold], the continuous spectrum [where (1) and (3) hold but (2) doesn't], and the residual spectrum [where (1) holds but (3) doesn't hold]. Lavaka 18:16, 7 November 2006 (UTC)

TeX vs. Non-Tex markup

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)

Spectrum categories are inconsistent

The names of the three categories of the spectrum are listed here as point spectrum, approximate point spectrum, and compression spectrum. But these categories have other names. I seem to recall that the point spectrum is sometimes called the eigenvalue spectrum (I am unsure, but I seem to recall this is the case). The approximate point spectrum is also called the continuous spectrum, and the compression spectrum is also called the essential spectrum or the residual spectrum. Given that the names used in the article aren't consistently followed. Further on, the wikipedia links to the spectrum categories are named differently. My recommendation is that we use the names "point spectrum", "continuous spectrum" (though this seems defined somewhat differently than the article specific to continuous spectrum), and "essential spectrum" to match the links to the relevant wikipedia articles, and mention in the brief description under each section in this article, the alternate names for these sets. -- KarlHallowell 04:54, 1 November 2005 (UTC)

Ok, I'm somewhat incorrect above. The essential spectrum is actually the combined approximate point spectrum and the compression spectrum. There may be other errors as well. I'll look this material up and see if I can at least find all the commonly used terms. -- KarlHallowell 17:23, 3 November 2005 (UTC)

all those types of spectra names above, and their definitions, can be encountered in the literature. they are not necessarily the same, if one accepts the standard definitions. for example, let's say a is in the approx. pt. spectrum if T - a is not bounded below. On the other hand, a is in the continuous spectrum if T - a is injective and have dense range. So the continuous spectrum would be contained in the approx. pt. spectrum but the converse need not be true in general. the continuous spectrum is by definition disjoint from the residual spectrum while parts of the approx. pt. spectrum may lie in the residual spectrum.

for shifts on l ^p, the approx. pt. and continuous spectra coincide, it's the unit circle. Mct mht 09:17, 15 August 2006 (UTC)

essential spectrum and residual spectrum do not agree in general. 10:17, 24 June 2010 (CET) —Preceding unsigned comment added by 134.60.67.134 (talk)

Operators on more general topological vector spaces

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)

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)

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)

Spectrum of an operator

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)

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)

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)

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)

Explanation

I have the right background to understand this topic, but this article isn't making things clear to me. I understand up to this point:

If X is a complex Banach space, then the set of all bounded linear operators on X forms a Banach algebra, denoted by B(X). The spectrum of a bounded linear operator is its spectrum when viewed as an element in this Banach algebra.

The first sentence seems understandable if I were fully familiar with Banach algebra. The second sentence seems almost like a recursive definition. I don't know what the spectrum of a bounded linear operator is, so I'm lost. I have more questions, but I'll stop there for now. —Ben FrantzDale 03:15, 1 May 2007 (UTC)

well, maybe the language can be made more explicit. the definition is simple: λ is in the spectrum of T if T - λ does not have a bounded inverse. the boundedness requirement is relevant in the infinite dimensional case. a special case is when T - λ has an algebraic inverse (i.e. is bijective). the open mapping theorem says that in that case, the algebraic inverse map is a bounded operator and so λ is not in the spectrum of T. Mct mht 04:45, 1 May 2007 (UTC)

I understand what you are saying, and that's about what my functional analysis text says, but it doesn't mean much to me in an intuitive sense. I think of eigenvectors of A as the directions that A leaves unchanged and the eigenvalues of A as the amounts that A can scale things by. This, of course, means that the eigenvalues with minimum and maximum magnitude tell you how much A can stretch the things it operates on.

Algebraically, it is easy to see that $Ax=\lambda x$ implies $(A-\lambda I)x=0$ and so $A-\lambda I$ has no inverse, but I never had an intuitive sense of how "$A-\lambda I$ has no inverse" is meaningful in its own right. Yet "spectrum" seems to be an extension of this definition of an eigenvalue. Perhaps someone could shed some light on the intuitive meaning of "$A-\lambda I$ has no inverse" for just the finite-dimensional case?

The best interpretation I have of that is "eigenvectors are the directions for which, if A didn't scale x by λ, A would be pretty much useless". I'm not sure if that helps much, and it depends on the idea of an eigenvector which clearly breaks down in infinite-dimensional cases such as the unilateral shift operator. —Ben FrantzDale 12:34, 1 May 2007 (UTC)

consider some special cases of the spectrum. in the finite-dimensional case, as you say it becomes the eigenvalues. in the commutative case, the spectrum of a continuous function f is its range. the resolvent function λ -> (λ - T)^-1 is analytic and gives a natural holomorphic functional calculus. it plays a somewhat similar role as the function (λ - z)^-1 in function theory (for example, in Cauchy's integral formula). so the spectrum, being set of "singularities" of the resolvent function, is important in determining the structure of an operator. even in the finite dimensional case, we see this in the Jordan form of a matrix. if T is a compact operator and σ ≠ 0 is an eigenvalue, then a contour integral around σ, ∫(λ - z)^-1 dλ, is the projection onto the generalized eigenspace corresponding to λ. Mct mht 14:03, 1 May 2007 (UTC)

I'm sorry, I don't know what you mean by some of this.

• "commutative case"
• holomorphic functional calculus
• resolvent function (resolvent formalism?)
• function theory, Cauchy's integral formula

I'll look up the ones with links and re-read your explanation later.

As for "the spectrum of a continuous function, f, is its range", that almost makes sense. That is, a matrix, A, can scale x by any amount in the range [λ_min,λ_max], depending on x, but the range of an $m\times n$ full-rank matrix A is the entire vector space R^m, so clearly its range isn't its set of eigenvalues, or even the range encompassed by its set of eigenvalues. What's up with that?

Again, I'll read over the links later and have another go at your explanation. Thanks. —Ben FrantzDale 19:25, 1 May 2007 (UTC)

re "commutative case": the family of continuous functions on, say, [0,1] is a commutative (Banach) algebra. in this algebra, f - λ is invertible iff λ does not lie in the range of f. Mct mht 03:21, 2 May 2007 (UTC)

Definition for Banach algebras matters! Perhaps split?

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)

Why Riesz' lemma to show that eigenvalues are in the approximate point spectrum?

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)

you're right, not quite sure what happened there. Mct mht 23:36, 30 September 2007 (UTC)

Definition for bounded operators

The article seems to say a couple of strange things about the spectrum for bounded linear operators. Firstly

• ... by the bounded inverse theorem, if λ I − T is bijective then λ does not lie in the spectrum of T

This is true of course, but the converse is notable by its absence. Then we see in the unbounded section:

• (Recall that bijectivity of T - λ is not implied by invertibility if T is bounded.)

So this is saying that there is an operator T - λ which is bounded, has inverse S in the Banach algebra B(X), but T - λ is not bijective as a function. This seems wrong to me. To cover the two possibilities separately:

We must have T - λ injective, since if not then the difference of any two inputs that are mapped to the same output is an eigenvector i.e. λ is an eigenvalue. I'm sure everyone agrees that eigenvalues are in the spectrum.

We must also have T - λ surjective, since we demand that S ∈ B(X), in particular that its domain is X so the range of T - λ is X. This is because S is in particular a right inverse (and a left one), so (T - λ)S=I where I is the identity of B(X); obviously the domain and range of I is X, so the range of (T - λ)S is X so the range of T - λ is X.

Even if we allowed the range of T - λ to only be a dense linear subspace of X, then we could extend S (which we at least ask be bounded, even in the case where T isn't) from having dense domain to the whole space by continuity.

Can anyone explain this? Am I confused about something, or is the article wrong? Quietbritishjim (talk) 12:11, 18 June 2009 (UTC)

Ok, after looking through the edit history, it seems the final point I mentioned (an operator with dense range) is the source of confusion. As I mentioned above it is not regular by definition, since we ask for the inverse to be a bounded operator on all of X. However, even if we relax the definition to include densely defined bounded operators, this case still doesn't occur because of the following:

Proposition If T:X→D is a bounded linear operator on a Banach space X, with range D which is dense in X, and S:D→X is a bounded linear two-side inverse for T, then D in fact equals X.

I could prove this directly with a limiting argument, but the easiest proof that I can think of relies on the following fact (really it just contains the same limiting argument in its own proof, of course). Hopefully it is well known, it appears in many operator theory textbooks, often as an exercise (we've had it in exams in previous years).

Lemma If T:X→X is a bounded linear operator, then

there exists a constant c such that for all x ∈ X: $\Vert Tx\Vert \geq c\Vert x\Vert$

if and only if

Ker(T)={} and Ran(T) is closed.

Proof of proposition First note that for any x ∈ X we have STx=x (since ST is the identity on X), and also

$\Vert STx\Vert \leq \Vert S\Vert \Vert Tx\Vert$

(since S is bounded), so the lemma holds with c=||S||^-1. Therefore D=Ran(S) is closed, and since it is also dense in X it equals X. Quietbritishjim (talk) 15:49, 20 June 2009 (UTC)

Lotsa stuff

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)

Right shift example

Doesn't the invertibility of this operator depend on how we define the codomain(it is clearly injective)? If we specify the codomain as the range itself, how does this effect the spectrum, for example. According to the given text, it would seem that the spectrum is still not empty or that the operator is then not bounded (continuous). Or are there other values of lambda that don't have easily fixed invertiblility? Forgive me if the questions sound ill-formed; I am just an undergrad trying to understand operator theory a little better, and the given text left some things open in my mind. —Preceding unsigned comment added by 79.235.179.181 (talk) 19:30, 3 July 2010 (UTC)

Very late reply. Not at all timely but maybe others would have the same question. If we do as you say, restrict the codomain to the range, then the resulting operator is now a unitary, which would have spectrum on the unit circle. So the spectrum shrinks from the disk to part of the circle

In fact a related construction happens in operator theory. The set of compact operators form an ideal in the ring of bounded operators. So we can take the quotient with respect to the compacts; this quotient is called the Calkin algebra. Given an operator T, the spectrum of its equivalence [T] in the Calkin algebra is called the essential spectrum of T. The essential spectrum are precisely those points in the spectrum of T that cannot be removed by adding a compact operator to T. Going back to your point, the essential spectrum is the unilateral shift is in fact the circle. Mct mht (talk) 00:58, 5 September 2012 (UTC)

Equivalently, the essential spectrum consists of all points of the spectrum that are not isolated eigenvalues of finite multiplicity. This is consistent with the analogy between the concept of spectrum for operators and the concept of poles in Complex Analysis, in which there are simple poles, poles of finite order and poles of infinite order (essential singularities) Tweet7 (talk) 04:27, 14 November 2012 (UTC)

Merged a section from Eigenvalues and eigenvectors article

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)

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)

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)

Fact tag

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)

Since nobody replied, I removed the tag. -- Jitse Niesen (talk) 12:23, 2 September 2013 (UTC)

Inconsistency in paragraph "Spectrum of the adjoint operator"

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)

"Bounded below" needs explanation or link

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)

Good catch. I added the definition. StrokeOfMidnight (talk) 23:54, 16 September 2022 (UTC)

Possible typo?

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)

Great catch! You're absolutely right. I was thinking about resolvent set when I was typing this. StrokeOfMidnight (talk) 23:47, 25 January 2023 (UTC)