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Questions tagged [self-adjoint-operators]

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Is it true that $\sigma_{ac}(\bigoplus_{j = 1}^\infty T_j) = \overline {\bigcup_{j = 1}^\infty \sigma_{ac}(T_j)}$?

Let $(H_j)_{j \in \mathbb N}$ be Hilbert spaces and define self-adjoint operators $T_j : \mathcal D(H_j) \to H_j$. Define the operator $T = \bigoplus_{j = 1}^\infty T_j$ on $\mathcal D(T) = \{(\phi_j)...
George C's user avatar
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Commutators of unbounded operators on Hilbert spaces

Commutation seems to be a tricky business when it comes to unbounded operators, because of the domain questions. I have some trouble understanding the usual material about commutators of unbounded ...
Hugo's user avatar
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Basis of eigenfunction of nonself-adjoint operator

If $H$ is a separable Hilbert space, $A$ is a bounded nonself-adjoint operator, $\{\lambda_n\}_{n\in\mathbb{Z}}$ are the eigenvalues of $A$, and the corresponding eigenfunctions are $\{\psi_n\}_{n\in\...
zeng's user avatar
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Infinite dimensional version of Cauchy interlacing theorem

One formulation/application of the Cauchy interlacing theorem in finite dimensions could be written as follows. Let $X$ be a finite dimensional, real Hilbert space. Let $Q \colon X \to X$ be a ...
Drew Brady's user avatar
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Decomposition of Spectrum of $U^{-1}TU$, where $U$ is unitary and $T$ selfadjoint

Suppose we have two Hilbert spaces $H_1,H_2$ and a densely defined self-adjoint operator $T:H_2\supseteq D(T)\to H_2$ as well as an unitary operator $U:H_1\to H_2$. Define $S:=U^{-1}TU:U^{-1}D(T)\to ...
Föölücks's user avatar
1 vote
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On self-adjoint extensions and multiplicity of eigenvalues

I hope you can help me with the following question. Let $B$ be a densely defined closed symmetric operator on a infinite-dimensional separable complex Hilbert space $\mathcal{H}$ with deficiency ...
Mario Ruiz's user avatar
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1 answer
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A self-adjoint, compact, non-negative, operator is non injective if and only if it is finite rank

Is the following statement true? A self-adjoint non-negative compact operator on a Hilbert space $(\mathcal{H}, \langle \cdot, \cdot \rangle)$ is injective if and only if the closure of its range is ...
DimSum's user avatar
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Linear operator as linear combination of positive operators [closed]

Lemma: Let $V$ be vector space with inner product. If $S\in L(V)$ is self adjoint, there exist positive linear operators $T, M\in L(V)$ such that $TM=MT=0$ and $S=T-M$. If $V$ is a complex vector ...
user926356's user avatar
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How to interpret self-adjoint operators in matrix algebras?

Let $A\in \mathrm{GL}(V)$ be a symmetrical matrix in the $n$-dimensional vector space $V$ equipped with the usual dot product, and $\{\vec{v},\vec{w}\}\subseteq V$. If i assume that $M^*=M^\intercal$ ...
Simón Flavio Ibañez's user avatar
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Problem with an example in Werner Amrein book

There is something I don't understand in example 4.13 of Werner Amrein's book "Hilbert Space Methods in Quantum Mechanics". Just before equation (4.55) he says that: if $\lambda\in\text{...
Thomas Belichick's user avatar
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Is this differential operator self-adjoint?

If $V$ is the space of functions in $C^{\infty}(\mathbb{R})$ such that $f(x)=f(x+2)$ for all $x\in \mathbb{R}.$ Equipping $V$ with the inner product of the space $C[-1,1]$, is the operator $D:f\mapsto ...
user926356's user avatar
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Do Self Adjoint Matrices Always have an Orthonormal Basis?

I have a self-adjoint matrix $$ \begin{bmatrix} 1 & 1+ i \\ 1- i & 2 \end{bmatrix} $$ It is self-adjoint because its transpose conjugate is identical to the original matrix. I get the ...
questionz's user avatar
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Asking for reference about self adjointness

Is there someone that knows and can share any reference about the compactness and the self-adjointness of the operator \begin{equation} \pmb{\sigma}\cdot L \end{equation} on $H^1(\mathbb{R}^{3},\...
Davide's user avatar
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Show that operator is self-adjoint

Show that the operator B in a space of $2 \times 2$ real matrices is self-adjoint $$ BX = X \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ \end{pmatrix} $$ I attempted to apply the theorem that an ...
Gleb Cloudy's user avatar
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Orthogonal Projection in an Enlarged Hilbert Space

Let $(H, \langle \cdot, \cdot \rangle_H)$ and $(U,\langle \cdot, \cdot \rangle_U )$ be Hilbert Spaces such that $H$ embeds into $U$. Let $M$ be a closed subspace of $U$, and define $\mathcal{P}$ to be ...
RiaDoog's user avatar
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Spectrum of self-adjoint operator is the support of the corresponding spectral measure

I have the following question: Let $T$ be a self-adjoint operator on a Hilbert space $H$, and let $P_T$ be the corresponding projection-valued measure on $\mathbb R$. Show that $\sigma_T = \text{supp}...
CauchyChaos's user avatar
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1 answer
42 views

On functions of a self-adjoint operator of the form $U^{-1} A U$

I found in Reed Simon that, since $\mathcal{F}$ (Fourier Transform) is unitary in $L^2(\mathbb{R}^n)$ and the self-adjoint operator (in a suitable domain) $-\Delta = H_{0}$ can be expressed as $H_{0} =...
Alessandro Tassoni's user avatar
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Inequality for inverse of an unbounded self-adjoint operator

Given an unbounded self-adjoint operator $A$ on some Hilbert space $\mathcal{H}$, and $\mu$ a non zero real number: I want to show that \begin{equation} \lVert (\mathbf{A} + i\mu \mathbf{I})^{-1} \...
Alessandro Tassoni's user avatar
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Product of a Hilbert-Schmidt operator and a Kato perturbation of a self-adjoint operator

Let $\mathcal{H}$ be a Hilbert space, $A:\mathcal{H}\rightarrow\mathcal{H}$ a Hilbert-Schmidt operator on it, and $H_0:\mathcal{D}(H_0)\rightarrow\mathcal{H}$ an unbounded self-adjoint operator on it. ...
Davide's user avatar
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1 answer
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exponential of operators - Trotter product formula

Let $H$ be a self-adjoint operator, bounded from below, with spectrum consisting of isolated eigenvalues $E_n (E_0 < E_1 < ...)$ with finite multiplicities $d(n)$, and $Tr[exp(- \beta H)] < \...
b.omega's user avatar
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Weinstein's bound on eigenvalues of self-adjoint operators

Consider a complex, separable Hilbert space $H$ and a (densely defined) self-adjoint operator $A: \mathcal D(A)\to H$. Assume that $A$ admits an orthonormal basis of eigenvectors $(\varphi_n)_{n\in \...
Jakob's user avatar
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1 answer
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Inequality for spectral families of self-adjoint operators

Let $A,B$ be self-adjoint operators on a Hilbert space $\mathcal{H}$ with $D(A) \subset D(B)$ and $A \leq B$. By the spectral theorem, we write : $A = \int_{\mathbb{R}} \lambda \,E^A(d\lambda) \hspace{...
Hugo's user avatar
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2 answers
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Operator norm of powers of bounded normal operators and self adjoint operators on Hilbert space

I saw this problem in Sheldon Axler's Measure,Integration and Real Analysis Ex. $10B$, problem $17$ and $18$. Let's just restrict to the self adjoint case. What I am trying to show is that $||T^{n}||=|...
Dovahkiin's user avatar
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When is the operator associated to a sesquilinear form normal?

I am self-studying from Linear Algebra by Hoffman and Kunze. In chapter 9, it is stated that every sesqui-linear form $f$ on a finite-dimensional inner product space $(V,(\cdot|\cdot))$ can be ...
francescoriccardocrescenzi's user avatar
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1 answer
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$0 ≤ A ≤ I. $ $\iff$ ⟨ψ|A|ψ⟩ ∈ [0, 1] for every unit vector $|ψ⟩ ∈ H$

Given two operators $A $and$ B$, where $A ≤ B$ means the operator $B − A$ is positive semidefnite. (i) $0 ≤ A ≤ I.$ (ii) ⟨ψ|A|ψ⟩ ∈ [0, 1] for every unit vector $|ψ⟩ ∈ H$ are equivalent I am having ...
darkside's user avatar
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Sum of the spectra of two s-a operators

I am asking something a little more general than my previous question. It seems "trivial" but I cannot find this statement anywhere and I have self-doubts. Let $L_1$ and $L_2$ be two self-...
Gateau au fromage's user avatar
1 vote
1 answer
36 views

Find the norm of the following operator

Let $H = L^2([-1,1],\lambda)$ with $\lambda$ the Lebesgue measure. Consider $$ T: H \to H: f \mapsto Tf $$ such that $(Tf)(x) = x^2f(x)$. Compute the norm of $T$. Seems easy enough, but I am only able ...
soggycornflakes's user avatar
2 votes
1 answer
175 views

Question about inclusion of domains of unbounded operators on a Hilbert space

Assume that T be a self-adjoint operator on the Hilbert space $L^2(0,1)=\{f:\int_0^1 |f(x)|^2 dx<\infty\}$ with the domain $D(T)$ satisfying $C^\infty_c(0,1)\subset D(T)$ (here $C^\infty_c(0,1)$ ...
Tony419's user avatar
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1 answer
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Properties of a Product of Commuting Projections

Let $\Pi_1$ and $\Pi_2$ be two positive-semidefinite projections which commute, and let $\Pi=\Pi_1\Pi_2=\Pi_2\Pi_1$. Are the following statements correct? $\quad\Pi\:\preceq\:\Pi_1\,$ and $\,\Pi\:\...
Nick Cooper's user avatar
3 votes
2 answers
64 views

If $P$ is a selfadjoint operator on a Hilbert space satisfying $P^4 = P$, is $P$ an orthogonal projection?

I’ve been trying to solve this problem to no avail. I have proved already that the spectrum is a subset of $\{0,1\}$ just like for orthogonal projections. I’ve also proved $\lVert P \rVert = 1$. ...
Neckverse Herdman's user avatar
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Sum of the eigenspaces of a non compact self-adjoint bounded operator

Let $\mathcal H$ be a separable Hilbert space, and $\mathbf A: \mathcal H\rightarrow\mathcal H$ a bounded self-adjoint yet non-compact operator. Suppose that I can construct a sequence $(\mathbf A_n)_{...
Cyril Soler's user avatar
1 vote
1 answer
94 views

Is the composition of positive, self-adjoint operators on a Hilbert space positive?

Let $A,B$ be self-adjoint, bounded, linear operators on a Hilbert space $H$ over $\mathbb{R}$ such that $\langle Ax,x\rangle\geq 0$ and $\langle Bx,x\rangle\geq 0$ for all $x\in H$. Does it also hold ...
Suim's user avatar
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If $P$ is a projection and $D$ is of trace class, is it true that $\mathrm{tr}(PDP)=\mathrm{tr}(DP)$?

According to Varadarajan$^{1}$, in a separable Hilbert space, If $A$ is of trace class and $B$ is any bounded operator, $AB$ and $BA$ are of trace class; and moreover, $tr(AB) = tr(BA)$. Orthogonal ...
mma's user avatar
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Adjoint, self-adjoint, Which is the proper noun?

I'm currently learning linear algebra, and I was confused by these two terminologies. It seems that adjoint is the tranpose of cofactor matrix, and a self-adjoint operator has a matrix representation ...
AL-CEL's user avatar
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Showing this identity involving the spectrum of a self-adjoint operator

I have the following question/task: Let $X$ be a $C^{*}$-algebra, and let $x \in X$ be self-adjoint. Show that the following holds: $$\lVert (x - \lambda)^{-1}\rVert = \frac{1}{\text{dist}(\lambda, \...
MathGeek's user avatar
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A question in functional analysis about selfadjoint operator.

In Hilbert space $u$, Let $T_1$,$T_2$ is selfadjoint operator, if exit $c>0$ such that $cI\le T_1\le T_2$, prove $T_1$,$T_2$ have a bounded inverse operator and $c^{-1}I\ge T_1^{-1}\ge T_2^{-1}$. I ...
luyao's user avatar
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2 votes
1 answer
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Is this testing positive semi-definiteness of the matrix or of its conjugate?

This may sound pedantic but maybe it helps me clarify some misunderstanding. The setup is as follows: Let $A$ be a complex $n\times n$ matrix which is self-adjoint, i.e. $A^*=A$, wherein $*$ denotes ...
P.Jo's user avatar
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Decomposition of normal operators into selfadjoint ones: $T=A+iB$

Assume we are given a densely defined and closed operator $T$ on a Hilbert space such that $D(T)\cap D(T^*)$ is dense. I am looking for sufficient conditions for $T+T^*$ to be selfadjoint as well as ...
Benjamin's user avatar
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If $A\in B(H)$ commutes with all self-adjoint operators, then $A=\lambda I$ for some $\lambda\in\mathbb{R}$

Let $H$ be a complex Hilbert space and $A\in B(H)$. Could anyone explain why the following assertion is true: if $A$ commutes with all self-adjoint operators, then $A=\lambda I$ for some $\lambda\in\...
OSCAR's user avatar
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0 answers
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Brezis' exercise 8.24.4: there exists a sequence $\left(\lambda_n\right)$ in $\mathbb{R}$ with $\left|\lambda_n\right| \rightarrow \infty$

Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e., Exercise 8.24 Prove that for every $\varepsilon>0$ there exists a constant $C_{\...
Akira's user avatar
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Restriction of a self adjoint operator to a closed subspace.

Let A be a self adjoint operator on a Hilbert space $H$ (possibly unbounded, densely defined with domain $D(A)$ ) and let $F$ be a closed subspace of $H$. My question is: If $D(A)\cap F$ is a dense ...
takuya's user avatar
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1 vote
1 answer
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Norm of Identity Minus Self-Adjoint Operator

I am new to functional analysis and therefore will appreciate help on the below, perhaps very simple question. Let $A$ be a bounded linear self-adjoint operator on a Hilbert space $X$. Suppose that $0 ...
GA-Student's user avatar
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0 answers
31 views

Semigroup solving Schrödinger equation weakly is a unitary group

Thank you for reading! Setup: Let $(\mathcal{H},\langle \cdot |\cdot\rangle)$ be a separable complex Hilbert space, and $A:D(A)\to \mathcal{H}$ be a densely defined symmetric unbounded operator in $\...
crimsonmist's user avatar
1 vote
0 answers
40 views

Spectral theory simply implies that two commuting self-adjoint operators are functional calculi of a third self-adjoint operator?

In some articles that I was reading the fact that if two self-adjoint operators $A,B$ commute then this imples the existence of a third self-adjoint operator $C$ such that $f(C) = \Phi_C(f) = A$ and $...
Felipe Dilho's user avatar
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0 answers
27 views

Property of being Fredholm after applying the Helmholtz operator

Consider a bounded self-adjoint operator $L$ on $L^2(\mathbb{R})$. I want to study its essential spectrum, thus I need to determine when is $L-\lambda I$ Fredholm. Am I right with the statement $$ L-\...
Gateau au fromage's user avatar
2 votes
0 answers
99 views

Laplace-Beltrami operator self-adjoint in context of Markov processes and their infinitesimal generators

Let $d\vec{X} = \vec{\mu}(\vec{X})dt +\sigma(\vec{X})d\vec{B}_t$ be a multidimensional SDE. It has infinitesimal generator $$\mathscr{L}f = \vec{\mu}^T \nabla f + \frac12 \operatorname{Tr}(\Sigma \...
Nap D. Lover's user avatar
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3 votes
1 answer
182 views

Reed and Simon, Fourier Analysis and Self-Adjointness, second corollary to Theorem X.$25$: how to show that $D(A^2)$ is dense in $D(A)$ for its norm?

This question arose while trying to figure out the proof of the second corollary to Theorem X.$25$ in Reed and Simon's Fourier Analysis, Self-Adjointness, stated as follows: Theorem X.$25$: Let $A : ...
Bruno B's user avatar
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1 vote
1 answer
60 views

Powers of operators and spectral gap

Let $A:\mathcal{D}(A)\to\mathcal{H}$ be a self-adjoint operator in a Hilbert space $\mathcal{H}$. Furthermore, assume that its spectrum satisfies $\sigma(A)\in [a,\infty)$ for some $a>0$. Then, ...
B.Hueber's user avatar
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is it an equivalent definition of the Schwartz Space? all real smooth function that $ f^{(k)}(\infty)=0=f^{(k)}(-\infty)$

Background: I have only taken half semester Analysis course It's in a german script from a linear algebra course on scalar products and self-adjoint endomorphisms, and so on. It mentions such a ...
IAN16's user avatar
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3 votes
1 answer
109 views

Spectral theorem and invertible operators

Let $\mathcal{H}$ be a (infinite-dimensional) Hilbert space and $A:\mathcal{D}(A)\to\mathcal{H}$ be a densely-defined self-adjoint linear operator. Furthermore, assume that $A$ is injective and that ...
B.Hueber's user avatar
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