0
votes
1answer
32 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 ...
0
votes
1answer
44 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 ...
1
vote
0answers
23 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 ...
0
votes
0answers
10 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 ...
5
votes
2answers
86 views

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: ...
0
votes
2answers
65 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 ...
1
vote
1answer
53 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 ...
1
vote
0answers
32 views

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 ...
0
votes
1answer
47 views

Positive Operator: Norm Estimate

In class we encountered the statement: $$H\geq C\mathrm{Id},C>0\Rightarrow\|\mathrm{e}^{-\beta H}\|<1,\beta>0$$ How does one prove this? Moreover what about the weakened version: $$H\geq ...
1
vote
1answer
18 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 ...
4
votes
1answer
83 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
99 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 ...
3
votes
0answers
35 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
65 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
60 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
65 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
29 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
15 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
25 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
83 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
230 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
76 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
41 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
100 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
2answers
89 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
128 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
83 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
225 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
40 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
130 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
100 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
434 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 ...
4
votes
1answer
126 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
118 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
288 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
183 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
190 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 = ...
3
votes
2answers
306 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 $ ...
2
votes
2answers
99 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
376 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 ...
1
vote
0answers
235 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
325 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
378 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
105 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
228 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
114 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
109 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
524 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
77 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
104 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 ...