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

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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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64 views

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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1answer
50 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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83 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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35 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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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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74 views

C*-Algebra: Positive Operator

Let $\mathcal{A}$ be a C*-algebra. If $A\in\mathcal{A}$ is a selfadjoint element then $A^*A=A^2$ has positive spectrum since: ...
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92 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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112 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
61 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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15 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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23 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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80 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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2answers
52 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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20 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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1answer
51 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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55 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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24 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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57 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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61 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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33 views

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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12 views

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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62 views

Basic Criterion on Selfadjointness

How to prove the following in a neat subjective way: $$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$
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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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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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91 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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95 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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74 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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32 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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82 views

Symmetric Operator $\iff$ Real Spectrum

One the one hand not every symmetric operator has real spectrum: $$A\text{ symmetric}\nRightarrow\sigma(A)\text{ real}$$ (In fact this is true only for the self adjoint ones.) Does the converse fail ...
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Normal Operators: Spectrum vs. Numerical Range

Disclaimer: As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.! Prove that for normal operators the ...
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Adjoint of sum of two operators

Let $A$ be self-adjoint and $B$ symmetric (which means densely defined for me as well) with $A$-bound less than $1$. Does this imply that $(A+iB)^*=A-iB$ ?
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62 views

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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1answer
62 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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Book on periodic Schrödinger operators

I am looking for good books about the spectral theory of periodic (1-dimensional) Schrödinger operators on a compact interval. A good reference I found was Reed/Simon Analysis of Operators (and a ...
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72 views

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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97 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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32 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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187 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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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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72 views

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 ...
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3answers
80 views

Finding spectrum of the operator A

$A:\mathcal{l}_2\rightarrow \mathcal{l}_2:(x_n)_{n=1}^\infty \rightarrow (x_{n+1})_{n=1}^\infty$ (left shift) Find the spectrum and all its parts for the operator A. What should I do?
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1answer
37 views

How to show whether this operator is normal? self-adjoint? unitary?

Consider the Hilbert space $H=l_2$ over the complex field and $A:H\rightarrow H.$ How to show whether this operator is normal? self-adjoint? unitary? ...
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How to show that the spectrum is equal to the range of $y$

How to show that the spectrum of $T_y$ is equal to the range of $y$ Given $y\in C[0,1]$ and $T_y: C[0,1] \rightarrow C[0,1]: x\mapsto x\cdot y$ Any help is appreciated, thanks.
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When can we get discrete spectrum?

Suppose that $T$ is a densely defined closed operator on a separable Hilbert space $H$. Form $N = T^*T$. Assume further that $T$ has a finite dimensional kernel and satisfies the commutation relation ...
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When $A_y$ is invertible?

Given $y\in C[0,1]$ Let $A_y:C[0,1]\rightarrow C[0,1]: x\mapsto xy$ When $A_y$ is invertible? Could you please help.
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32 views

How to find the point spectrum $\sigma_p(A)$ of $A$

How to find the point spectrum $\sigma_p(A)$ of $A$ Consider the Hilbert space $H=l_2$ over the complex field and $A:H\rightarrow H.$ ...
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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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1answer
89 views

Help please eigenvalue of Laplacian

Let $\Omega$ be a bounded subset of $\mathbb R^d$ and let $g\in L^2\left(\Omega \right)$. Let $\left(\lambda_n\right)_{n\in \mathbb{N}}$ be the eigenvalues of the Laplacian operator $\Delta$, and ...
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Doubt about integration over a curve

I'm reading a paper and I have a doubt: Suppose we have a compact Riemannian manifold and an eigenfunction of the Laplace operator $\Delta u=-\lambda u$. Suppose in addition that we have the following ...