4
votes
1answer
55 views

Spectral decomposition of normal operator

Define $T$ from $L_{2}(R)$ into itself by $T(f)(t)=f(t+1)$. Show that $T$ is normal and finds its spectral decomposition. I've shown that $f$ is normal (in fact it's unitary) but how do I find its ...
4
votes
0answers
90 views

Spectral decomposition of $TT^*$

On $l_{2}$ let $T$ be given by $Te_{n}=\frac{e_{n+1}}{n+1}$ where $(e_{n})_{n\ge1}$ is the canonical orthonormal basis. Find the spectral decomposition of $TT^*$. I find that ...
2
votes
0answers
26 views

In which cases the spectrum of an operator contains only eigenvalues?

Let $X\neq \{0\}$ be a complex normed spaces (not necessarily finite-dimensional) and $T:D(T)\subset X\to X$ a linear operator (not necessarily bounded). I would like to know under what conditions can ...
0
votes
1answer
61 views

Spectral theory and sequences: is this fact a general truth or does it depend on the operator?

Let $\lambda\in\mathbb{R}\setminus\{0\}$, $\textbf{i}$ the imaginary unit, $H$ a Hilbert space, $T:D(T)\subset H\to H$ a invertible densely defined linear operator such that $T^{-1}$ is bounded, ...
4
votes
2answers
46 views

Examples of spectral decompositions

I would like examples of spectral decompositions and how they are obtained for normal compact operators and normal non-compact operators on an infinite dimensional hilbert space. I have googled it, ...
0
votes
1answer
54 views

The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
0
votes
0answers
27 views

A couple of proofs on a spectrum

Let $T$ be a normal bounded operator. Let ${\lambda}$ be in $({\sigma}(T))$. Without invoking general algebra theories, show that: a) $p({\lambda},{\lambda}^*)$ is in $({\sigma}(T))$ for all ...
0
votes
0answers
10 views

characterisation up to unitary equivalence

My book says that the spectral theorem for compact normal operators characterises compact normal operators up to unitary equivalence. It doesn't expand on this so I was wondering what does this mean ...
0
votes
0answers
23 views

The eigenspace of 0 problem

In the spectral theorem for a compact normal operator, do we exclude the eigenspace corresponding to 0 (assuming its an eigenvalue) from the space decomposition. My reason for asking is this: Every ...
0
votes
2answers
70 views

Spectral Theorem for normal operators

I want to prove this in the infinite dimensional Hilbert space case. What is the easiest way to go about this (What do I need to know, what theorems do I need,etc). My aim is to show every normal ...
1
vote
1answer
160 views

Spectrum of operator

Like my previous question, I'm considering the same space and operator: Hilbertspace adjoint But this time I am trying to determine the spectrum of $T$. I feel like I'm messing up my definitions a ...
1
vote
1answer
52 views

Multiplication operator on Hilbert space

i looked to the question Spectrum and point spectrum of this operator. I will go further with asking. We know that $T$ is well-defined iff $(\lambda_n)\in\ell^{\infty}$. But if ...
2
votes
1answer
35 views

Approximate point spectrum and left topological zero divisors

Recall that a left topological zero divisor in a Banach algebra $A$ is an element $a\in A$ such that there exists a sequence of unit vectors $(a_{n})$ in $A$ with $\lim_{n\rightarrow\infty}aa_{n}=0$. ...
2
votes
1answer
85 views

Eigenvalues and adjoint of operator $T(x_k)_{k=1}^{\infty} = (x_{2k})_{k=1}^{\infty}$

Let $T$: $l^2 \rightarrow l^2$ denote the operator \begin{align} T(x_1,x_2,\dots, x_n,\dots) = (x_2,x_4,\dots,x_{2n},\dots). \end{align} There are several questions regarding this operator that I need ...
0
votes
1answer
72 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 ...
1
vote
1answer
92 views

Adjoint of resolvent of self-adjoint, densely-defined operator on a Hilbert space

Let $H$ be a Hilbert space, $T=T^*$ a densely-defined linear operator on $H$. Denote the resolvent set of $T$ as $\rho(T)=\{\lambda\in\mathbb{C}~|~T-\lambda$ has bounded, everywhere-defined inverse}, ...
4
votes
0answers
72 views

For $A$ self-adjoint, $\sup_{|x|=1}\langle Ax,x\rangle = \max \sigma(A)$

For a self-adjoint operator $A$ on a Hilbert space $H$, one has $\sup_{|x|=1}\langle Ax,x \rangle = \max\sigma(A)$. I want to prove this using the spectral theorem. My idea is: Let $a = ...
4
votes
1answer
216 views

An alternate proof of Fuglede's theorem

To prove Fuglede's Theorem for normal operators on a separable Hilbert space, why does it suffice to show that $E(S_1)T E(S_2)=0$ for all disjoint Borel sets $S_1$ and $S_2$, where $E$ is the spectral ...
1
vote
0answers
34 views

How many projectors do two commuting self-adjoints have in their common spectral decomposition?

If $A$ and $B$ are two commuting observables on a Hilbert space of dimension $n$ say. So, $$A = \sum_{j \leq a} \lambda_j P_j $$ $$B = \sum_{i \leq b} \mu_i Q_i $$ $$I_n = \sum_{i \leq b} Q_j = ...
0
votes
0answers
106 views

Symmetry adapted basis function to make the Hamiltonian matrix Block Diagonal.

Can anybody give me a tip to solve this problem? I have large quantum mechanical Hamiltonian, to solve it numerically I have to decompose it into the block diagonal form. To convert the hamiltonian ...
2
votes
1answer
83 views

Spectrum of linear operators

I can't solve the following: i) Let $T:l^2 \rightarrow l^2$ , $Tx=\{ (Tx)_n\}_{n=1}^{\infty}$ given by $$(Tx)_n = \dfrac{1}{2}x_{n-1} + \dfrac{1}{2}x_n.$$ Find $\sigma(T)$. ii) Let $S : l^2 ...
3
votes
1answer
338 views

Spectral Theorem for bounded compact, self-adjoint operators as corollary of Hilbert-Schmidt theorem

I'm following Debnath and Mikusinksi's "Introduction to Hilbert Spaces with Applications" and am trying to understand how the spectral theorem for compact self-adjoint operators is a corollary of the ...
3
votes
1answer
109 views

Spectrum proofs

Let $T$ be a densely defined closed unbounded operator on a Hilbert space $H$. Show that if $\lambda$ is a point in the residual spectrum of $T$, then $\bar{\lambda}$ is in the point spectrum of the ...
2
votes
1answer
112 views

Show for compact operator $K$, if $||Kf|| < ||f|| \forall f$, then $||K|| < 1$.

I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact. Again, the statement is, if $K$ is a compact operator on a Hilbert space ...
9
votes
1answer
259 views

An approximate eigenvalue for $ T \in B(X) $.

This is a problem from Conway’s Functional Analysis: Definition An approximate eigenvalue for $ T \in B(X) $ is a scalar $ \lambda $ such that there is a sequence of unit vectors $ x_{n} \in X $ ...
3
votes
1answer
169 views

Proof that the spectrum of the Dirichlet Laplacian is discrete

Let $\Omega\subset\mathbb{R}^n$ a open bounded set. The Dirichlet laplacian can be defined via it's closed semi-bounded form on $H^1_0(\Omega)$. The fact that it's spectrum is discrete is as far as I ...
3
votes
1answer
175 views

Spectrum in an separable Hilbert space

Let $H$ be a separable Hilbert space with orthonormal basis $\{e_i\}$. Let $(c_n)$ be a bounded sequence of complex numbers and consider the bounded linear operator $T$ on $H$ defined by $$Tx = ...
2
votes
2answers
97 views

If $Lat(\mathcal{A})$ is trivial then $\mathcal{A}'$ consists of scalars.

This is related to Exercise 3 of Section 2.5 of Arveson's book on spectral theory. For those who are interested, we were asked to show the following $\mathcal{A}$ is a Banach *-algebra. ...
4
votes
1answer
343 views

Spectral theorem for unitary operators

I saw in several texts, as a part of the spectral theorem for unitary operators, that given a unitary operator $U$ on a Hilbert space $H$ (say it is separable), $H$ can be decomposed as an orthogonal ...
0
votes
0answers
211 views

Diagonal Dominance and Spectral Radius

For positive semi-definite matrices, $A$ and $B$ with real entries, Let: $X=I-(2Diag(A)-B)^{-1}(A-B)$ The spectral radius $\rho(X) \leq ||X||$. As, $(2 Diag(A)-B)$ becomes a better approximation ...
5
votes
0answers
285 views

Why is the numerical range of an operator convex?

Let $T$ be a Hilbert space operator. Its numerical range is \begin{equation} W(T)=\{\langle Tx,x\rangle:\|x\|=1\}.\end{equation} It is a well-known fact that $W(T)$ is a convex subset of the complex ...
11
votes
1answer
361 views

Quantization of angular momentum: is Dirac's proof wrong?

I'm trying to understand the physicist's proof of the theorem on the spectral structure of angular momentum operators (I'm being told that this proof is due to Dirac). I will refer to Ballentine's ...
2
votes
1answer
101 views

Family of Self-Adjoint Operators that are Multiplications on a Common $L^2(\mu)$?

Suppose that $H$ is some (complex) Hilbert space and that $\{T_\alpha: \alpha \in I\}$ is some collection of bounded self-adjoint operators on $H$. A version of the spectral theorem states that for ...
5
votes
2answers
207 views

Why is the numerical range of a self-adjoint operator an interval?

I was reviewing for a test for functional analysis when I came across the following statement: Let $T$ be a bounded self-adjoint operator on a Hilbert space $H$. Then the numerical range of it is ...
2
votes
1answer
96 views

Characterizations of the form domain for unbounded selfadjoint operators

This question follows from this one and especially from Willie Wong's answer: link. In Reed & Simon's book Methods of modern mathematical physics, vol. I, pag.277, the form domain of a ...
4
votes
0answers
102 views

Relations between spectrum and quadratic forms in the unbounded case

Let $H$ be a complex Hilbert space. If $B$ is a bounded self-adjoint operator on $H$ then its spectrum is a closed and bounded subset of the real line and we can find its extremes in terms of the ...
8
votes
1answer
495 views

What is the use of Spectral Theorem?

Obviously the version for compact and self-adjoint linear operators on Hilbert Spaces is very useful since it decomposes the operators into orthogonal projections. However, the following more general ...
2
votes
0answers
74 views

Prove $\forall$ compact $M:\ M \subset C\quad \exists A:l_2\rightarrow l_2, \sigma(A)=M$ [duplicate]

Possible Duplicate: Operator whose spectrum is given compact set Can spectrum “specify” an operator? Prove that for each nonempty $M$ - compact subset of $\mathbf{C}$ exists ...
3
votes
4answers
102 views

reference for strongly continuous semi-groups

At the moment I am trying to understand the proof of the Fredholm property in Salamon's notes on Floer homology. There I came across the notion of an unbounded operator on a (real) Hilbert space which ...
4
votes
1answer
246 views

How to characterize self-adjoint operators in terms of orthogonal diagonalizability

Have a look at the following excerpt of Tosio Kato (taken from Zeidler Applied functional analysis vol. I): The fundamental quality required of operators representing physical quantities in ...
4
votes
0answers
143 views

When functions, analytically continued, carry over certain properties

Let $ \Omega $ be a sufficiently smooth planar region in $ \mathbb{R}^2 $ with spectrum $ \Gamma $ (the set of eigenvalues of the Laplace operator on functions which vanish on the boundary $ \partial ...
1
vote
2answers
480 views

How do the solutions to the wave and heat equations converge in general?

I would like to check my understanding with someone if possible. When we cover the heat and wave equations, for instance, in "methods" courses at university, they normally restrict the initial ...