3
votes
2answers
179 views

A version of the spectral theorem for group actions

Suppose $G$ is a sufficiently nice (maybe locally compact and abelian) group which acts on the separable Hilbert space $\mathcal{H}$ by unitary transformations. Is there a generalization of the ...
2
votes
0answers
177 views

Versions of the spectral theorem

Since any $C^*$-algebra can be represented as an algebra of bounded operators $\mathcal{B(H)}$ on a Hilbert space $\mathcal{H}$, the spectral theorem applies to all $C^*$-algebras: ($*$) ...
0
votes
1answer
158 views

Spectral decomposition function [closed]

Once I met a notation of "spectral decomposition function" (for a self-adjoint operator). No definition was given. Could someone give me a clue what can that be, cause I can't find this exact phrase ...
6
votes
0answers
219 views

Paving conjecture for Toeplitz matrices

Let me first recall what is the so-called paving conjecture: for any $\epsilon >0$, there exists $r\in \mathbb N$ such that for any bounded operator $A$ on $\ell^2(\mathbb Z)$, there exists a ...
4
votes
1answer
186 views

Well defined Tensoring of spectral triples

Hi, I have a misunderstanding that I am hoping is really quite trivial. I will give my question directly and context below for those that need/want it. Question: In connes standard model he takes ...
1
vote
0answers
211 views

Matrix conditions under which spectral radius is smaller than 1?

Hello everyone, I would like to find out which conditions are necessary so that the spectral radius $\rho(M)<1$ where $M$ represents the following matrix: $M = \left( \begin{array}{ccc} W & 0 ...
4
votes
2answers
288 views

Is independence meaningful for commutative $C^*$-algebras?

I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate. Let's say I have two self-adjoint operators on a Hilbert space and ...
4
votes
1answer
711 views

How “generalized eigenvalues” combine into producing the spectral measure?

Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of ...
9
votes
8answers
4k views

Can a self-adjoint operator have a continuous set of eigenvalues?

This should be a trivial question for mathematicians but not for typical physicists. I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" ...