Questions tagged [self-adjoint-operators]

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Proving that eigenvalues of real self adjoint endomorphism are real

If $\mathbf{A}\colon V \to V$ is selfadjoint for some finite dimensional vector space $V$ over $\mathbb{C}$ then this follows immediately, since for any nonzero eigenvalue $\lambda $ and a ...
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2 answers
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About kernel of compact self-adjoint operators and separable Hilbert spaces

I relate to this one question about corollary 4.10.2 pag. 198 of "Introduction to Hilbert Spaces - Debnath, Mikusinki" third edition, that states Let $A$ be a compact self-adjoint operator ...
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Question related to a bounded operator in a Hilbert space.

The statement is "Let $A_n$ be a sequence in $B(H)$ and $A\in B(H)$ such that $\|A_n -A\|$ $\rightarrow 0$ and $n \rightarrow 0$ if $A_n$ is self adjoint then $A$ is also self-adjoint." I ...
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1 vote
3 answers
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Proof if $T \in L \left( V,V \right)$ is self-adjoint and b,c $\in \mathbb{R}$ such that $b^2 < 4c$ , then $T^2 + bT + cI$ is a positive operator.

I also have a hint: Use Cauchy–Schwarz inequality. I'm really stuck with this problem. I think that $T^2 + bT + cI \geq 0$ because of $b^2 - 4c < 0$ may help but I don't know how to relate it with ...
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3 votes
2 answers
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Is self-adjoint operator in a Hilbert space is always positive operator?

Let me first define the self-adjoint operator. Let $A$ be a bounded operator in a Hilbert space $H$, then $A$ is said to be a self-adjoint operator if $A^*=A$. And $A$ is known as a positive operator ...
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1 answer
40 views

Relation between resolvent set, self adjoint and open neighborhood

My professor gave me a list of exercises to complete in preparation for the exam ... but i can't solve this question. I have the feel that the solution is simple but in this moment I can't figure it ...
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5 votes
2 answers
162 views

Laplace-Beltrami operator is essentially self-adjoint on a bounded domain

I came across this lecture note The Poincaré inequality on domains on a webpage. In the first section, it claims that $L$ is essentially self-adjoint on $\mathcal{D}^\infty$. It is known that the ...
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How to show that the operator $(-\Delta)^{-1} : C^\beta(\overline{\Omega}) \to C^{\beta}(\overline{\Omega})$ is self-adjoint?

Well, I think that this operator can be seen as a constraint of the operator $(-\Delta)^{-1}:L^2(\Omega) \to L^2(\Omega)$ because $C^\beta(\overline{\Omega}) \subset L^2(\Omega)$ and by the regulary ...
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A basic question about the spectral theorem for self adjoint compact operator..

In the statement of this theorem which i state from the book of conway that if T is a compact self adjoint operator on a hilbert space then T have countable no of non zero eigen value ln and there ...
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1 answer
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Prove that a self-adjoint operator is cyclic iff it is unitary equivalent to a multiplication operator on $L^2$

Let $H$ be a seperable Hilbert space, and $T$ is self-adjoint. I am asked to prove that $T$ is cyclic iff it is unitary equivalent to a multiplication operator on $L^2(X,\mu)$, where $X\subset \mathbb{...
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Calculate the adjoint of the operator $D = \frac{1}{i} \frac{d}{dx}$

Let $p \in \mathbb{N}$, $a, b \in \mathbb{R}$. We define on $L_p (a,b)$ the operator $$D = \frac{1}{i} \frac{d}{dx}$$ with the domain $$\operatorname{dom}D = \left\{ u \in W_p^1 (a,b) \ | \ u (a) = u(...
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1 answer
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Every contractive self-adjoint operator is the weak limit of projectors

I try to show that every self-adjoint operator $A$ on Hilbert space $H$ with $\|A\| \leq 1$ is the weak limit of the projectors. My teacher told me that the most important part is to show that for one-...
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1 answer
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A question on two self-adjoint operators with same spectrum

What could be an example of two self-adjoint operators $T_1,T_2\in \mathcal B(E)$,where $E$ is a hilbert space,such that both of them have cyclic vectors and have same spectrum but they are not ...
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Estimate of the length between two consecutive eigenvalues of a self-ajoint operator

Given a self-adjoint operator with a complete discrete spectrum, I wonder if there are any results or a way to estimate the length between two consecutive eigenvalues of such an operator. (Typically, ...
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Show $P_n A P_n \longrightarrow A$ in the strong resolvent sense.

Suppose we have a smooth domain $\Omega \in \mathbb{R}^n$ equipped with the Euclidean norm and a self-adjoint operator $A: \mathcal H^1_0(\Omega) \to \mathcal H^1_0(\Omega)$ defined as $A = -\Delta + ...
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1 vote
0 answers
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Two equivalent statements for self adjoint endomorphisms

Let $V$ be a complex unitary vector space and $f$ an endomorphism on $V$. Show that following statements are equivalent. $f$ is self-adjoint and all eigenvalues of $f$ are are non-negative real ...
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Is a linear operator $T$ that is normal, normal regardless of the ordered basis while the matrix representation of that operator must be on an ortho..

Is a linear operator $T$ that is normal, normal regardless of the ordered basis while the matrix representation of that operator must be on an orthonormal basis? I was attempting an exercise that ...
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Deficiency indices of the Laplace operator on the unit disk

Let $\mathbb{D}=\{(x,y)\in \mathbb R^2: x^2+y^2\le 1\}$ be the unit disk. Let consider the Laplace operator $\Delta_0$ defined on $C_0^\infty(\mathbb D)$, the space of smooth functions $\mathbb D\to \...
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1 vote
1 answer
108 views

Is the Sturm-Liouville operator self-adjoint?

Let $L$ be the Sturm-Liouville operator with domain $$ D(L)=\Bigg\{\psi\in C^\infty([a,b])\cap L^2([a,b]) : \begin{cases} \alpha\,\psi(a)+\beta\, \psi'(a)&=0\\ \gamma\,\psi(b)+\delta \,\psi'(b)&...
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1 vote
1 answer
74 views

Eigenvalues of compact self adjoint operator that is cyclic

Let $T: H \to H$ be an arbitrary compact self adjoint operator, and $H$ a Hilbert space. I am asked to prove that $T$ being cyclic is equivalent to all of the eigenvalues of $T$ having multiplicity ...
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1 answer
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Compact, Self-Adjoint, Nonnegative operators have at least one Eigenvector

This is a statement presented in my class and I am having trouble to understand the proof given by the Professor: Let $T \in K(H)$, where $K(H)$ represents the space of compact operators on a Hilbert ...
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0 answers
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Show the operator is self-adjoint on a period function

Let $V$ be a vector space of all $2\pi$-periodic function from $\Bbb R$ to $\Bbb C$ with the inner product: $$\langle f,g \rangle =\int_0^{2\pi} \overline{f(x)}g(x)dx$$ $$A:V\to V, f\to \frac{df}{dx}$$...
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operators that are compact and positive, their norm equal to the greatest eigenvalue.

I read this from a post Norm of integral operator in $L_2$ Can someone point me to formal proof of this statement? Thanks
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3 votes
2 answers
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Multiplicative property of trace

Let $T_1 \in \mathcal{L}(V)$ and $T_2 \in \mathcal{L}(V)$ be positive operators. Prove that the trace of their product is non-negative i.e., tr($T_1 T_2) \geq 0$ Attempt 1: Obviously, a positive ...
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0 votes
1 answer
22 views

projection of band limit functions

Define $$ P_{\Omega}f(x) = \frac{1}{2\pi}\int_{\mathbb{R}}e^{i\zeta x}\chi_{\{|\zeta|\leq \Omega\}}\hat{f}(\zeta)d\zeta $$ where $\hat{f}$ is the Fourier transform of $f$. Show that $P_{\Omega}^2 = P_{...
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A Geometrical Derivation of Quantum Mechanics Spin Operators

I'm trying to see if there is a way to geometrically derive a general form for the quantum mechanics spin operators. I'm trying to deduce their commutation relations without using any knowledge of ...
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0 votes
2 answers
37 views

Spectrum of general projection and orthogonal projection

I am trying to think about this, but I seem to be stuck. Suppose $P$ is a projection on a Hilbert space $\mathcal{H}$. If I am just talking about a general projection, where I only know that $P^2=P$, ...
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If $T^2$ is positive(that is $T^2$ is self-adjoint and $\langle T^2v,v\rangle\geq 0$) , when $T$ can be self-adjoint?

Suppose $V$ is a vector space with dimension $n$, I'm doubting that if operator $T^2$ is positive, then $T$ under some situations can be self-adjoint. However, I'm not certain about my reasoning. My ...
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1 vote
2 answers
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If the operator $P$ is positive and $T$ is self-adjoint, then is there exists a positive number $n$ such that $nP+T$ is positive?

$V$ is a finite-dimensional complex inner product space and $P, T\in L(V)$. If the operator $P$ is positive and $T$ is self-adjoint, then is there exists a positive number $n$ such that $nP+T$ is ...
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  • 754
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2 answers
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Prove that T is self-adjoint

Define the linear operator $T:\ell^2 \to\ell^2 $ by $$T(x_1,x_2,x_3,...)= \left(x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,...\right)$$ I need to show that T is self-adjoint. I know that T is self-adjoint iff $...
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1 vote
1 answer
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If $T^*T=T^2$, then prove the operator $T$ is self-adjoint [duplicate]

Let V be an n-dimensional inner product space over C and T ∈ L(V ). Suppose that $T^2 = T^∗T$. Prove that T is self-adjoint. My idea is since $T^*T$ is positive, then all the eigenvalues of it should ...
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2 votes
1 answer
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Unique square root of a positive operator on a Hilbert space

Let $T$ be a positive operator on a Hilbert space $\mathcal{H}$. That is $\langle Tx,x\rangle\geq0$ for all $x$. Then there is a unique positive operator $S$ such that $S^2=T$ and this $S$ is called ...
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2 votes
1 answer
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Is it true that core of $C^0$ one-parameter unitary group generator on Hilbert space is invariant under action of this unitary group?

Let $\mathcal{H}$ be complex Hilbert space and $A:\mathcal{H}\supseteq \mathcal{D}(A)\rightarrow\mathcal{H}$ be self-adjoint operator. As Stone theorem says, operator $A$ generates one parameter $C^0$ ...
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What is a bilinear concomitant (or conjunt) with regards to linear differential operators

Context I am studying self-adjoint eigenfunction problems using [1]. I am working through Example 1 on page 54 in [1]. Example 1 (page 54 in [1]) Suppose we have the linear differential operator $$ L[...
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1 vote
1 answer
23 views

Relative bound of $A^r$ with respect to $A$

Consider a nonnegative unbounded self-adjoint operator $A$ on a Hilbert space $\mathcal{H}$, with domain $\mathcal{D}(A)$; by the spectral theorem, the operator $A^r$ is well-defined for all $r\geq0$. ...
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  • 267
2 votes
1 answer
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Solving Simple PDE by Green's Function, Very Confused By Some Mistake

Suppose I want to solve $u_{xy} = xy$ via Green's Function. This will correspond to the associated PDE $G_{xy} = \delta(x - x_G,\ y - y_G)$, and I want my boundary conditions for this Green problem ...
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In numerical solutions for boundary value problems, what does $\mathfrak{A}^\ast$ mean?

Good day! I am currently reading Hiroshi Fujita's paper on "The Contribution to the Theory of Upper and Lower Bounds in Boundary Value Problems", and I came across a space denoted by $\...
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Finding Boundary Conditions That Make This Green's Function Nice

This question used to be included here as a second point, but I figured the two were better asked as two separate questions. When people use the phrase "the ____ operator is self-adjoint" (i....
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1 answer
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Discrete spectrum for product of operators

Let $A$ and $B$ be two positive-definite self-ajoint operators on a Hilbert space $H$. If the spectrum of both $A$ and $B$ is discrete, can we affirm that the spectrum of $AB$ will be also discrete ? ...
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  • 509
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1 answer
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$TT^* = T^*T$ iff $T = A + iB$ with $A, B$ self-adjoint

I'm working through the ideas in this document from Berkley. They claim (I'm paraphrasing), "$TT^* = T^*T$ if and only if $T$ is of the form $T = R + iM$ where $R$ and $M$ are self-adjoint." ...
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Find a self-adjoint operator with $[0, 1] \subseteq \sigma_{sing} (A)$.

I am asked to find a self-adjoint operator $A$ on a Hilbert space $H$ such that $[0, 1] \subseteq \sigma_{sing} (A)$. Here, $\sigma_{sing}(A) = \sigma (A|_{H_{sing}})$. (EDIT: The definition(s) are ...
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1 answer
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If $\{\varphi_n\}$ is an eigenbasis of $K^\ast K$'s support and $h\perp K^m\varphi_n$ for all $n,m\ge1$, then how do you show $Kh=0$?

$\newcommand{\span}{\operatorname{span}}\newcommand{\im}{\operatorname{Im}}$EDIT: According to the below answer, Royden's construction is wrong. However, the key detail which was omitted by the ...
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1 answer
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How is a self-adjoint operator (hyper-)maximal?

I may be missing something obvious, but is self-adjoint operator in a complex separable Hilbert space (hyper-)maximal? Lemma: Let $A:D(A) \subsetneq\mathcal H\rightarrow \mathcal H$ self-adjoint, i.e. ...
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Nth root of positive semidefinite operator over a Hilbert space.

Prove that every positive semidefinite operator $T\in B(\mathcal{H})$, where $\mathcal{H}$ is an infinite dimensional Hilbert space, has a unique positive-semidefinite $n$th root for every $n\in\Bbb{N}...
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Self-adjoint operators on a Hilbert space

Let $A, B ∈ \mathcal{B}(H)$ be selfadjoint operators on a Hilbert space and assume that $0 ≤ A ≤ B$. Prove that $\|A\| ≤ \|B\|$. I tried to show the claim by using the fact that if $A$ is a bounded ...
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Question about Anderson Localization and a specific theorem (RAGE)

I have a question with a theorem which appears in a text called “Invitation to Random Schrödinger Operators”, in unit 7. Theorem 7.7. Let $H$ be a selfadjoint operator on Hilbert Space, take $\psi \...
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5 votes
1 answer
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Problem on a rank 1 perturbation of an self-adjoint operator

With my teacher we have been trying to solve for a while a problem that consists of two parts, and each one in three sections. The first part, which has to do with the problems that I was able to ...
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Why is the maximal domain $D_\text{max}$ of a Sturm-Liouville operator defined the way it is?

Given a Sturm-Liouville type operator which acts on functions on the interval $(a,b)$ $$T:= \frac{1}{w(x)}\left(-\frac{d}{dx}\left[p(x) \frac{d}{dx}\right]+q(x)\right)$$ where $w$, $p^{-1}$ and $q \in ...
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1 vote
1 answer
137 views

Using the Fredholm alternative to find a solvabilty condition for a PDE [closed]

I'm trying to understand how to use the Fredholm alternative to solve a system of PDEs. The specific problem is as follows
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Proof is self-adjoint

Consider a set of polynomials of degree 2 or lower with complex coefficients, $\mathbb{P}_2(\mathbb{C})$. On this space we define the inner product $$\langle p,q\rangle = p(0)\overline{q(0)} + \int_{0}...
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