Spectral theory is a study of generalized notions of eigenvalues and eigenvectors.

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Characteristic Function of a Spextral Density Function

I am struggling with understanding the link between the Spectral Density Function and the Characteristic Function. In particular, can you find the Characteristic function when only the SPX and ...
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49 views

Fredholm index for 1-d Schroedinger operator

if I look at a Schroedinger-operator $-\frac{d^2}{dx^2} +V$ on a compact intervall $[a,b] \subset \mathbb{R}$ and I take boundary conditions that this operator is self-adjoint (for example periodic ...
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Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom.

I have the following two questions: The Fourier transform defines a unitary (provided that it is normalized properly) map $\hat{\cdot}:L^2(\mathbf{R})\rightarrow L^2(\mathbf{R})$. I figured out its ...
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1answer
48 views

Exercise about spectrum of selfadjoint operator.

I'm stuck on an exercise about the spectrum of a selfadjoint operator on a Hilbert space. The problem is the following: Let $(X,\langle \cdot, \cdot\rangle)$ a Hilbert space and let $A \in B(H)$ a ...
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1answer
120 views

Spectral theory

I have absolutely no idea about Spectral theory and want to ask some fundamental questions. 1.) What does it mean that the resolvent of an operator is Hilbert-Schmidt? (Cause I saw a theorem that was ...
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28 views

logarithm of projection

I want to prove what's used in the fourth line below the "Proof" section here: http://en.wikipedia.org/wiki/Quantum_relative_entropy#The_result The statement is: Let $\rho$ be a density operator on a ...
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1answer
48 views

Is it possible to consider an approximation to a (non-self adjoint) operator with a self adjoint one?

In operator theory it's wonderful if we have a self-adjoint operator (non necessarily bounded) due to all the work that has been done using their symmetry,... etc. I.e there are many powerful tools. ...
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1answer
54 views

To What Extent is the Fourier Inversion Theorem Due to the Self-Adjointedness of the Laplacian

I've tried looking this up (I looked at various spectral theorems) but couldn't find anything that talks about the connection between Fourier transforms and the eigenfunctions of the Laplacian (we may ...
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93 views

Spectrum of normal elements in C*-algebras

Let $\mathcal{A}$ be a C*-algebra and $x \in \mathcal{A}$ a normal element. Can you show that $\left\{ \phi(x) : \phi \text{ is a state on } \mathcal{A} \right\}$ is the closed convex hull of the ...
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1answer
91 views

eigenvalues to Laplacian

Let $\Omega$ be a bounded subset of $\mathbb R^d$ and let $g\in L^2(\Omega )$. Let $(\lambda_n)_{n\in \mathbb{N}}$ be the eigenvalues of the Laplacian operator $\Delta$, and $(e_n)_n$ the ...
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Reference for simplicity of the principal eigenvalue of the Laplacian

i'm currently searching for a proper reference or proof to see that the first eigenvalue $\lambda \in \mathbb{R}$ of \begin{equation*} - \Delta u = \lambda u \text{ in } \Omega, \\ u \in ...
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What is the difference between Singular Value and Eigenvalue?

I am trying to prove some statements about singular value decomposition, but I am not sure what the difference between singular value and eigenvalue is. Is singular value just another name for ...
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1answer
37 views

Lower bound for the spectralradius of a matrix

Any submultiplicative norm (for example the row-sum-norm) is an upper bound for the spectralradius of a matrix A. But is there a way to get a suitable LOWER bound for the spectralradius ? ...
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1answer
59 views

Spectrum of the operator $A(f,g)=(g,\Delta f-f)$

Let $\Omega$ be an open set in $\mathbb{R^n}$. We consider the product Hilbert space $H=H^1_0(\Omega)\times L^2(\Omega)$ with the norm $$|(f,g)|^2=\int_\Omega (|\nabla f|^2+|f|^2+|g|^2 ) dx$$ We ...
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1answer
126 views

Proving the spectral theorem for unbounded self-adjoint operators

Let $A$ be (densely-defined) self-adjoint operator on a (complex) Hilbert space $H$. Then, the Cayley transform $U=(A-i)(A+i)^{-1}$ is a unitary operator, and $(A\pm i)^{-1} \in B(H)$. Using the ...
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2answers
56 views

Spectrum of a bounded operator $T$ satisfying $T^n=I$

Let $\mathcal{H}$ be an infinite dimensional Hilbert space, suppose $T\in \mathcal{B}(\mathcal{H})$ is a bounded operator and suppose that $n$ is the smallest natural number so that $T^n=I$. Let ...
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Pseudospectrum of an $n\times n$ matrix has at most n connected components

For $\epsilon>0$, the $\epsilon$ pseudospectrum of an $n\times n$ matrix $A$ is given by $\sigma_{\epsilon}(A)=\{z\in \mathbb{C}:\|(z-A)^{-1}\|>\epsilon^{-1}\}$, with the convention that ...
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94 views

Mathieu differential equation

Given the operator $T (\psi)(x):= \psi''(x)-2q \cos(2x)\psi(x)$ with $T : D(T) \subset L^2[0,2\pi] $ I was wondering: What is the right domain $D(T)$ for this operator if we want to solve the ...
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64 views

What is the right domain for this Hamiltonian

I want to define a proper domain $D(H) \subset L^2$ for this Hamiltonian ( $\theta$, $\phi$ are the standard angles in spherical coordinates). Furthermore, the wave function is supposed to satisfy ...
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1answer
81 views

Eigenvalues of polynomials of a matrix and its inverse up to summation by identity

There is a paper that I am reading and the following has been considered without proof: (Suppose $\lambda(.)$ defines the spectrum of a matrix and one can define a random variable on this spectrum say ...
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1answer
25 views

Proof that solution of $\lambda$-affine, linear ODE is entire in $\lambda$

Suppose $F(\lambda)~(\lambda\in\mathbb{C})$ is a linear ordinary differential operator (with, say, domain $D$ dense in some Hilbert space), and is also affine-linear in $\lambda$. Is there a proof ...
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107 views

Prove that if $\lambda$ is an isolated eigen-value of $T=T^*$, then $\ker(T-\lambda)=E_{\{\lambda\}}H$

Here we have a self-adjoint, densely-defined operator $T$ on a Hilbert space $H$, and $E_M$ is the usual spectral projector for any Borel set $M$, i.e., $E_M=\int_M\text{d}E_t$ (this means, by ...
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18 views

Question about Joint spectral radius.

Given a bounded set $\mathcal A\subset \Bbb R^{n x n}$. The joint spectral radius is given by: $\sigma(\mathcal A)$=$limsup_{m\to\infty}(sup_{A\in\mathcal A^m} \rho(A))$ where $\rho$ is the normal ...
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26 views

Diagonalisation spectral theorem

For the proof of the spectral theorem for complex numbers I know that the proof follows that, as T is normal then the algebraic and geometric multiplicities coincide. This means that there will be n ...
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1answer
26 views

A little question about using spectral mapping theorem for polynomials

A question from Kreyszig: Let $X$ be complex Banach space with $T\in B(X,X)$ and a $p$ a polynomial. Show that the equation $p(T)x=y$ has a unique solution $x$ for every $y \in X$ if and only if ...
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61 views

Spectrum of a bounded operator and Liouville's theorem

Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We know that the spectrum of $A$ is not empty, because otherwise we find a contradiction by using the holomorphy of the function ...
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1answer
54 views

Spectral radius of an operator equals its norm

Let $X$ be a Banach space and $A:X\to X$ a bounded operator. We know that the spectrum of $A$ is always included in the ball $B(0,|A|)$ and the spectral radious $r(A)$ is the smalest radius such that ...
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26 views

Spectrum of the sum on a tensor product?

I have the following problem. Consider the operator $R= H\otimes 1 + 1 \otimes K$ on the tensor product $\mathcal H \otimes \mathcal K$ where $H$ and $K$ are self-adjoint. I know that $R$ has a ...
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68 views

A decomposition of Hilbert space via self-adjoint operator

Let $H$ be a complex Hilbert space and $A:H\to H$ self-adjoint. Show that one can decompose $H$ into two $A$-invariant closed subspaces as $H=H_{p} \bigoplus H_{c}$ such that the spectrum of ...
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What is spectrum for Laplacian in $\mathbb{R}^n$?

I know very well that Laplacian in bounded domain has a discrete spectrum. How about Laplacian in $\mathbb{R}^n$?(not in some fancy-shaped unbounded domain, but the whole domain) Where can I find ...
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Skew adjoint operator with uncountable spectrum

Let $H$ be a Hilbert space. I just want an example of a skew adjoint operator $(A^*=-A)$ with uncountable spectrum. I also want an example for unbounded differential operators. The only example I ...
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502 views

Spectrum of an Orthogonal Projection Operator

I want to show that $ \sigma(p) = \{ 0,1 \} $ for any orthogonal projection operator $ p \notin \{ 0,I \} $ on a Hilbert space $ \mathcal{H} $. Recall that an orthogonal projection operator $ p $ on $ ...
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Integral representation of joint projection valued measures

Given two positive $\sigma$-finite measures $\mu_{1/2}$ on the spaces $X_{1/2}$ one can define the product measure $\mu_1\otimes\mu_2$ on the product space $X_1\times X_2$. It can be proved that the ...
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96 views

Application of eigenvalueproblems for the wave equation

I'm currently searching for a nice little application of an eigenvalueproblem and found the following for acoustics - but one part doesn't make sense for me. Consider the wave equation to find some ...
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1answer
51 views

Module algebras

Spectrum: For Banach algebra $A$ spectrum is denoted by $\sigma(A)$ and defined as the set of all non-zero bounded linear multiplicative function from $A$ to $\Bbb C$.(Function $\psi:A\to\Bbb C$ is ...
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The spectrum of an unbounded operator

It's well known that the spectrum of a bounded operator on a Banach space is a closed bounded set (and non-empty)on the complex plane. And it's also not hard to find unbounded operators which their ...
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What's the spectrum of this operator in $\ell^2$?

Suppose that $\ell^2 = \biggl\{(x_n)_n \in \mathbb{K}^{\mathbb{N}_0} \biggm| \sum_{n=1}^{\infty}|{x_n}^2| < +\infty \biggr\}$ is a Hilbert-space with the inproduct $\langle\cdot,\cdot\rangle_2: ...
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45 views

Riemann Sphere: Holomorphic Functional Calculus

Why do we consider the holomorphic functional calculus on the Riemann sphere rather than the complex plane only? Is there a serious problem? Moreover isn't any curve encircling the spectrum ones ...
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1answer
96 views

Book: Functional Calculus

Is there a good book that investigates in detail the various kinds of functional calculus? I'm having now some knowledge about unbounded operators and integration but I would like to understand ...
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1answer
87 views

“right shift” il $L^1$

Let $X=L^1(\mathbb{R})$ be the space of Lebesgue integrable functions $f:\mathbb{R}\rightarrow \mathbb{C}$ with the usual norm. Let $T\in B(X)$ be defined by $$(Tf)(t)= f(t+1)$$ I need to find the ...
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33 views

Why is this subspectrum closed

Let $u: X \to X $ be a compact operator on a Banach space $X$ and let $\lambda \in \mathbb C$ be non zero. We know that $u-\lambda$ is Fredholm and that $X=\mathrm{ker}(u-\lambda)^n \oplus ...
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1answer
75 views

Cayley Transform: well defined?

Why is the Cayley backtransformation well-defined: $$A_U:=\imath(1+U)(1-U)^{-1}$$ In general $1-U$ is not invertible for example for $U=1$.
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Operator: not closable!

Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ...
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210 views

Resolvent Set: Definition

Given Banach spaces: $X,Y$ Consider a linear operator: $T:\mathcal{D}(T)\to Y$ (not necessarily bounded nor closed nor closable nor densely defined) Define for the shorthand the shifted operator: ...
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Question about a counterexample concerning compact operators

Does anybody know if the following is true, Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is ...
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103 views

What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
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34 views

Inequality norms

Let $A$ be a bounded linear operator on a Banach space $X$. Can we show that for an arbitrary $n \in \mathbb{N}$ and $x \in X$ such that $\|x\|_X \geq 1$ we have that $$\|A^n x \| \leq \|Ax\|^n.$$ ...
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How to show: $A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$

Let $y\in C[0,1]$ and $A_y : C[0,1]\rightarrow C[0,1]: x\mapsto xy$ How to show: $A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$. Could you please help.
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Spectral theory - how to prove this lemma?

in Anver Friedman, Foundations of Modern Analysis I found a lemma (6.7.3): If A is a self-adjoint operator and $\{E_\lambda\}$ is a spectral family such that $A=\int_m^{M+\varepsilon} \lambda ...
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Operators $A$ such that $e^A$ is norm preserving

Let $X$ be a Banach space. $A$ a bounded operator. We can define the exponential of $A$ by $$e^{A}=\sum_{n=0}^{+\infty}\frac{A^n}{n!},$$ which is also a bounded operator. Is there any sufficient ...