Questions tagged [self-adjoint-operators]

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self-adjoint operator and symmetric operator

we recently learned about self-adjoined operator with the formal definition $ ⟨Tv, w⟩ = ⟨v, Tw⟩$ for every $v, w$ in $V.$ In the other side we talked that self-adjoined can be represented as a ...
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1answer
13 views

Concept of 2-variable function for operators on an $n$-dimensional inner product space

I'm reading the book "Finite-Dimensional Vector Spaces (2nd Ed)" by PR Halmos. The concept of a 2-variable function (or polynomial) for operators is introduced in Theorem 1 of Section 84 on page 171 ...
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functional analysis( self-adjoint operators on a separable Hilbert space )

let $A_n$ and $A$ be self-adjoint operators on a separable Hilbert space, and let $A_nx$ converge to $Ax$ for any $x\in X$. Prove that for any continuous bounded $f$,$ f (A_n) x$ holds and converges ...
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14 views

Boundary Value ODEs Reference Book

Can someone suggest me a book on Boundary Value Problems in ODEs, which start from the general theory, and then go on to specialize for self-adjoint values? All the books I have found discuss the self-...
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1answer
65 views

Self adjoint extensions

I'm taking a course in Functional Analysis using some topics from Kreyszig and Reed & Simon books, I have been asked to solve the following exercise: Let $A$ be a symmetric operator such that $\...
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1answer
23 views

For which $C$ is the linear map $\varphi_C : V \to V, A \mapsto CA$ selfadjoint with scalar product $\langle A, B \rangle = \operatorname{tr}(A^t B)$?

Let $V=M_{n\times n}(\mathbb R)$. By the definition of selfadjoint, I have to show for which $C$, $\langle \varphi_C(A), B\rangle = \langle A,\varphi_C (B)\rangle$ is true $\forall A,B \in V$. In ...
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1answer
38 views

Unitary operator as an exponent of self-adjoint

I'm trying to proof that set of unitary operators on infinite Hilbert space is path-connected. That's why i need to show that for each unitary $U$ there is a self-adjoint $B$ such that $U=e^{iB}$. Any ...
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1answer
35 views

A problem about operators on an Euclidean space

Let $A, B$ be self-adjoint positive operators, $C, D$ be orthogonal operators on an Euclidean space, and $AC = DB$. Prove that $C=D$. I know many properties of self-adjoint and orthogonal ...
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1answer
41 views

Operators on an Euclidean space

Let $A, B$ be self-adjoint positive operators, $C, D$ be orthogonal operators on an Euclidean space, and $AC = DB$. Prove that $A=B \Leftrightarrow AC$ is normal. I know many properties of ...
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1answer
30 views

If $S$ and $T$ are distinct self-adjoint operators on $H$ such that $S\leq T$, can we conclude that $\langle Sx,x\rangle<\langle Tx,x\rangle$?

If $S$ and $T$ are distinct self-adjoint operators a (separable) Hilbert space $H$ such that $S\leq T$, can we conclude that $\langle Sx,x\rangle<\langle Tx,x\rangle$ for all $x\in H$? Here $S\leq ...
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28 views

Is self-adjoint operator necessarily linear?

Let $(H, \langle\cdot, \cdot\rangle)$ be a Hilbert space and $P: H \to H$. In this answer, @gerw said that if $$\forall (x,y) \in H^2:\langle Px, y \rangle = \langle x, Py \rangle,$$ then $P$ is ...
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56 views

Self-adjoint operator as composition of bounded and self-adjoint

My teacher gave me a problem: I have two self-adjoint operators $A$ and $B$. Both of them are greater or equal than $I$, i.e. $A \geq I$ and $B \geq I$. I need to to prove that there is a positive ...
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1answer
13 views

Tail of increasing convergent net of self-adjoint operators is bounded

Let $H$ be a Hilbert space and $(T_\alpha)$ an increasing net of self-adjoint operators that converges (in some topology) to an operator $T$. Then $(T_\alpha)$ is not necessarily norm-bounded I think (...
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42 views

prove $Ax_n - \lambda x_n \to 0$

Given a Hilbert space $H$ over $\mathbb{C}$ and a linear self-adjoint operator $A: H \to H$. Let $(x_n) \subset H$ with $\|x_n \|= 1$ such that $\langle Ax_n,x_n \rangle \to \lambda$ where $|\lambda| ...
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1answer
19 views

Semi-positivity of an Hermitian matrix

I´m triying to show that an $n\times n$ matrix $M$ such that $M_{ij}=e^{\gamma_{ij}}$ where $\gamma_{ji}=\gamma_{ij}^{\ast}$ are complex numbers is a positive semi-definite matrix. I made a proof of ...
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1answer
50 views

Hilbert space self adjoint and idempotent

A linear operator $P: H \to H$ on a Hilbert space $H$ is self-adjoint if for every $f, g \in H$, $\langle Pf, g\rangle=\langle f,Pg\rangle$ and is idempotent if for every $f \in H$, $P(P(f))=P(f)$ (i....
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inequality for positive contraction operator

Let $H$ be a Hilbert space, let $A\in B(H)$ satify $\|A\|\le 1$. If $A$ is positive, i.e. $A$ is a self-adjoint operator and for all $x\in H$, $\langle A(x),x\rangle\ge 0$, proof that $${\|x-A(x)\|}^...
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1answer
44 views

If $P$ is a projection and $T$ is positive operator such that $P\leq T$, then why $P(H)\subset\sqrt{T}(H)$?

Suppose that $H$ is a Hilbert space. If $P\colon H\to H$ is a projection (i.e. $P^{2}=P^{*}=P$) and $T\colon H\to H$ is positive operator (i.e. $T\geq0$) such that $P\leq T$, then why $P(H)\subset\...
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1answer
32 views

If $P,Q$ orthogonal projections on $H$ and $T\colon H\to H$ extension of Hilbert isomorphism $S\colon P(H)\to Q(H)$, then $Q=TT^{*}$.

Let $P,Q\colon H\to H$ orthogonal projections (i.e. $P^{2}=P^{*}=P$ and $Q^{2}=Q^{*}=Q$) on a Hilbert space $H$ and let $S\colon P(H)\to Q(H)$ be an isomorphism of Hilbert spaces. We can extend $S$ to ...
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28 views

If $p^{2}=p^{*}=p$ and $pa^{*}a=paa^{*}=0$, then $a^{*}p+pa=0$.

Suppose that $p$ is a projection (i.e. $p^{2}=p^{*}=p$) in a C*-algebra $A$. Let $a\in A$ be an element such that $pa^{*}a=paa^{*}=0$. I want to prove that $$a^{*}p+pa=0.$$ I tried to express $a$ in ...
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29 views

Stone formula and joint spectral measures for commuting operators

Let $\mathcal{H}$ a Hilbert space and $A$, $B$ two self-adjoint, (possibly) unbounded operators on it with domains $D(A)$, $D(B)$. By the spectral theorem, both operators are uniquely associated with ...
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23 views

About invariant subspaces of resolvent operators

I have a trouble with understanding a proof about one-dimensional Schrodinger operator. $\mathfrak{h}_{0}$ is the Hilbert space of $\mathbb{C}^2$-valued functions $\Phi(\lambda) = (\varphi_{1}(\...
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1answer
50 views

If a operator $A$ in Hilbert space is positive then $A$ is self-adjoint?

Let $H=(H,(\cdot,\cdot))$ be a Hilbert space and $A:D(A)\subset H \longrightarrow H$ a linear operator (not necessarily bounded) such that $\overline{D(A)}=H$ and $A \geq 0$, that is, $$(A(x),x)\geq 0,...
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65 views

If a operator $L$ in a Hilbert space is self-adjoint, then $L$ is coercive?

Let $H=(H,(\cdot, \cdot))$ be a Hilbert space and $L:D(L) \subset H \longrightarrow H$ a linear operator densely defined. If $L$ is self-adjoint operator, then $L$ is coercive, that is, there exists $...
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1answer
43 views

How to show the Laplacian is a self-adjoint linear operator

I'm trying to prove the following: $$\iint_R(\phi_1\nabla^2\phi_2-\phi_2\nabla^2\phi_1)\ dR = 0$$ For $ 0 < x < L, 0 < y < H$. Given $\phi_i(x,y) $ satisfying the boundary conditions $$ \...
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71 views

$\frac{(Tx,x)_H}{(x,x)_H}$ attains maximum if $T$ is compact and self-adjoint

I'm struggling with the following proof and hope some of you can help me: By $H$ we denote a real Hilbert space and let be $T: H \rightarrow H$ be a compact, self-adjoint and linear operator. (i) ...
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1answer
41 views

Eigenvectors of Hermitian matrices over arbitrary fields

Fix a field $k$, and suppose $\gamma$ is an involutory automorphism of $\gamma$ (that is, $\gamma \ne 1$, but $\gamma^2 = 1$). Call a matrix $A$ $\gamma$-Hermitian if ${(A^\gamma)}^T = A$ (where the ...
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1answer
47 views

Spectrum of product of self-adjoint operators contained in $\mathbb{R}$

Let $S,T$ be self-adjoint bounded operators on a complex Hilbert space. In this post, it is shown that $\sigma(ST)\subset\mathbb{R}$. The answerer uses that $\sigma(ST)\cup\{0\}=\sigma(TS)\cup\{0\}$ ...
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74 views

Show that $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$ is bounded.

Let $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$. I want to show that $T$ is bounded by finding an upper bound on its norm $\| T \|$. What I've tried Using Cauchy-Schwarz we have \...
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2answers
70 views

Show that norm of normal operator equals spectral radius via $\| T T^* \| = \| T \|^2 = \| T \|^2$

Let $H$ be a Hilbert spaces and $T \in L(H)$ normal, i.e. $T T^* = T^* T$. Show that $r(T) = \| T \|$, (where $r(T) := \sup_{\lambda \in \sigma(T)} | \lambda |$ is the spectral radius of $T$) by first ...
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49 views

When is the weak limit of self-adjoint invertible operators invertible?

I know Let $(T_n)$ be a sequence of invertible bounded linear operators on a Hilbert space $H$ converging strongly to a bounded linear operator $T.$ If $\sup_n \|T_n^{-1}\| <\infty$ and $T$ has ...
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Can changing the defined inner product change wheather a transformation is Normal or Self Adjoint?

Excercise 1 reffered to Only need answer to d Excercise 1 (in text): Given the (complex) linear space spanned by {$\sin x$, $\cos x$, $e^{2x}$} . On this space we define the linear mapping T as the ...
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Boundary fluxes in quantum mechanics are zero locally?

Stone's theorem is famous for giving a one-to-one correspondence between self-adjoint operators $\{H\}$ and unitary (strongly-continuous) groups $\{e^{-itH}\}$. The unitarity of such a group implies ...
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22 views

How to verify the continuous and discrete spectrum of the linear operator $Ly=-y_{zz}+\left(\frac{1}{4}c^2-f'(U)\right)y$?

Let $u_t-u_{xx}-f(u)=0, x\in (-\infty,\infty)$ with $f(0)=f(1)=0, f'(0)<0, f'(1)<0$. Let $u=U(x-ct), U(-\infty)=0, U(\infty)$ denote a travelling front solution. In this context, the following ...
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49 views

Image self adjoint operator

I am asked to prove the following: For a compact self-adjoint operator T on a Hilbert Space H, show that: $$ \operatorname{Im}(T) \mbox{ is closed} \iff \dim(\operatorname{Im}(T)) < \infty$$ I ...
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Counter-example of a non-selfadjoint operator for which $ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$ does not hold.

Let $\mathcal{H}$ be a Hilbert space and $T \in \mathcal{L(H )}$, if $T$ is selfadjoint, then we have : $$ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$$ I want a ...
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101 views

How to prove an operator is invertible

If A is a bounded, self adjoint operator on a Hilbert space, how can you see that $A^2+I$ is invertible? Am I not sure how to even start with this, since I don't know anything about the inverse of A.....
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“Square-normal” matrices are normal

This is from a practice exam for my quals. Let $A$ be an $n \times n$ complex matrix. Suppose $A$ satisfies the following property: $(AA^\dagger)^2 = (A^\dagger A)^2$ Prove that $A$ is normal, ...
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75 views

Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

Let $(H_k)_{k\in\mathbb{N}}$ be a sequence of Hilbert spaces. Define $$ \oplus_{k\in\mathbb{N}}H_k:=\{(x_k)_k \in\Pi_{k\in\mathbb{N}}H_k \mid \sum_{k\in\mathbb{N}}||x_k||^2<\infty\} $$ with $$ \...
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1answer
119 views

Proof that compact self-adjoint operators have at least one non-zero eigenvector (using something analogous to min-max theorem)

I checked several texts on the spectral theorem for compact self-adjoint operators (like this (PDF) and this (PDF)) on Hilbert spaces. They all mention that the cluster points of the real eigenvalues ...
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1answer
67 views

Show $A = \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact set

The problem Let $\left( E, \langle \cdot, \cdot \rangle \right)$ be a euclidean space of dimension $n$. Let $K$ be a compact subset of $E$ containing a basis $e = (e_1, ..., e_n)$ of $E$. We denote ...
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23 views

What will be the matrix of an anti-selfadjoint transformation in an orthonormal basis?

I am a bit confused by this question. The statement of the question tells me that $(x|Ty) = -(Tx|y)$ and that $B$ is an orthonormal basis. Also, that the field is $C$. On my own, I have found that ...
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1answer
102 views

Show $\sup_{0≤f∈L^2}\frac{\|Af\|_{L^2}}{\|f\|_{L^2}}=\sup_{0≤f∈L^2}\frac{⟨Af,f⟩_{L^2}}{\|f\|_{L^2}^2}$ for self-adjoint nonnegativity-preserving $A$

Let $(E,\mathcal E,\mu)$ be a measure space and $A$ be a self-adjoint bounded linear operator on $L^2(\mu)$. Assume $Af\ge0$ for all $f\in\mathcal L^2(\mu)$ with $f\ge0$. Let $$c_1:=\sup_{\substack{f\...
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32 views

checking if an endomorphism is self-adjoint.

Let $V$ be an inner product space and let $\alpha$ be an endomorphism of $V$. Is the endomorphism $\alpha^*\alpha - \sigma_1$ of $V$ self-adjoint? Could anyone give me a hint on how to check this?
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38 views

Bounded self-adjoint linear operator is injective…

Let $T : H → H$ be a bounded self–adjoint linear operator on a Hilbert space $H$. Suppose the range $R(T)$ is dense in $H$. Prove that $T$ is injective.
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1answer
63 views

Sum of Shift Operators

I have the operator $T: l_2[0,\infty) \to l_2[0, \infty)$ defined by $T=S_l+S_r$. So $T(x_0,x_1,x_2,x_3,...)=(x_1, x_2+x_0,x_3+x_1,x_4+x_2,...)$. I've shown that it's spectrum is a subset of $[-2,2]$....
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1answer
45 views

Does there exist a basis of $\mathbb R^3$ consisting of eigenvectors of T?

True or false (and give a proof): There exists $T ∈ L(\mathbb R^3)$ such that $T$ is not self-adjoint (with respect to the usual inner product) and such that there exists a basis of $\mathbb R^3$ ...
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28 views

What is the self-adjoint operator of the sobolev semi-inner product $\langle u', v' \rangle$ in the general case?

I have a simple question that is bothering me. It is so common, that the answer should be easy to find, but I didn't. I am probably making it more difficult than it is. Consider a domain $\Omega=(0,1)...
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20 views

Optimize the contractivity of a self-adjoint operator on a closed subspace

Let $H$ be a $\mathbb R$-Hilbert space and $A\in\mathfrak L(H)$ be self-adjoint with $$\left\|Ax\right\|_H\le\left\|x\right\|_H\;\;\;\text{for all }x\in H.\tag1$$ Let $K\subseteq H$ be closed and ...
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1answer
106 views

Proving a Self-Adjoint Operator is an Orthogonal Projection

I need to prove that a self-adjoint operator $T \in \mathcal{B}(\mathcal{H})$ is an orthogonal projection if $\sigma(T) = \{0,1\}$. I know this means I have to prove $T$ is idempotent, meaning $T = T^...