Questions tagged [self-adjoint-operators]

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Summation of two Volterra operators

Let $V \in K (L^2 [0,1])$ be the Volters operator $Vf(s) = \int_0^s f(t) dt$, and let $A = e^{i \theta} V + e^{-i \theta} V^*$ with $0 < \theta < \pi$. Find $\underset{{{\|f\|}_{2}}=1}{\max} <...
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Question on norm of a positive operator.

Let $A$ be a positive operator on a real Hilbert space $\mathcal H.$ Is it always true that $\|A\| =\sup\limits_{\|x\| = 1} \left \langle Ax, x \right \rangle$? I know that for self-adjoint operators ...
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74 views

Prove that $\sigma_{\text {ess}} (A)$ is a closed subset of $\mathbb R.$

Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal H.$ Let $E_A$ be the unique spectral measure associated to $A$ obtained from spectral theory for self-adjoint operators defined on the ...
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matrix of operators properties

Let $T$ a self-adjoint operator bounded from below on $D(T)=H^2(\mathbb R)$ in the ambiant Hilbert space $\mathcal H=L^2(\mathbb R)$, with essential spectrum $[0,+\infty)$ and for discret spectrum ...
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1answer
38 views

Can the positive operator stay monotonicity?

If $A,B$ are two self-adjoint operators on Hilbert space $\mathscr{H}$, and assume $A\ge B \ge0$, which means that $$ ((A-B)f,f)\ge 0, \forall f \in D(A) \cap D(B), $$ and $$ \quad(Af_1,f_1) \ge0, (...
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1answer
73 views

Prove that $A$ is not invertible.

Let $A$ be a positive self-adjoint operator on a Hilbert space $\mathcal H.$ If $\inf\limits_{\|x\|=1} \left \langle Ax, x \right \rangle = 0$ then $A$ cannot be invertible. How do I prove it? Any ...
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33 views

Proof that self-adjoint matrices are Hermitian

There is some confusion about the distinction between self-adjoint and Hermitian matrices. Every answer I've come across posits that "a matrix is Hermitian if and only if it is self-adjoint w.r.t....
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1answer
23 views

Every compact self-adjoint (linear) operator is injective (FALSE)

My purpose is to provide a proof, without using the spectral theorem. Let $T: \scr H \to H$ be a linear compact self-adjoint operator. $T$ self-adjoint implies: $ \langle \phi, T \psi \rangle = \...
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1answer
23 views

Convergence of a monotone sequence of bounded self-adjoint operators on a Hilbert space

Let $H$ be a complex Hilbert space and $(T_n)$ a monotonically nondecreasing sequence of bounded self-adjoint operators on $H$. It is well known that if there exists a bounded self-adjoint operator $K$...
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Are prolate spheroidal wave functions in the Lauguerre-Polya class?

For a $c\in\mathbb{R}^{>0}$, prolate spheroidal wave functions of order zero are defined as the eigenfunctions of the self-adjoint operators $F_c:L^2[-1,1]\rightarrow[-1,1]$ and $Q_c:L^2[-1,1]\...
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3answers
53 views

Positive but not self-adjoint operator

Given ${\frak L}({\scr H}) \equiv \{ A : {\scr H} \to {\scr H} \ \vert \ A$ linear$ \}$, the operator $ T \in {\frak L}({\scr H})$ is said to be positive if: $\langle \psi, T \psi \rangle \geq 0 \...
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1answer
19 views

Property of the selfadjoint operators with eigenvaluesin Hilbert spaces

I have to prove this exercise: Let $H$ be an Hilbert space, $T\in\mathcal L(H)$ a self-adjoint operator and $\lambda$ an eigenvalue of $T$. The operator $T-\lambda I$ is self-adjoint, with $I$ being ...
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1answer
39 views

The Fredholm Alternative for Self-Adjoint Operators

Let $H=(H,(\cdot,\cdot)_H)$ be Hilbert space and $A:D(A) \subset H \longrightarrow H$ be a self-adjoint operator, not necessarily bounded. Question. Is true that $\text{Ker}(A)^{\perp}=\text{Range}(A)$...
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12 views

Exponential of self-adjoint unbounded operator as tensor product of exponentials

Given a self-adjoint operator $h$ on $\mathfrak h$ and $H_N := \sum_{j=1}^Nh_j$ defined on $D(H_N) = D(h^{\otimes N}) \subset \mathfrak h^{\otimes N}$, show that $$\forall t \in \mathbb R: e^{-i\...
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43 views

Consequences of Deficiency indices theorem (Von Neumann theory)

Let $T: \operatorname{dom}(T) \rightarrow \scr H$ be a symmetric operator. $T$ admits self-adjoint extensions $\iff$ $d_+ = d_-$, where $d_\pm = \dim \ker(T^\dagger \pm i \mathbb{I})$ If $d_+ = d_-$,...
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2answers
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If $R$ is an orthogonal reflection, then $\frac{1}{2}(R + I)$ is an orthogonal projection

I consider a Hilbert space $H$ over $\mathbb{C}$, and a reflection $R$ in the set of bounded linear operators on $H$ denoted $B(H)$. I want to prove that if $R$ is an orthogonal reflection, then $\...
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1answer
43 views

Let $T$ be a compact and self adjoint operator on a hilbert space $H$ such that $T$ is not invertible. Prove that $\text{Ker}(T)\ne 0$

Let us assume on contrary that $\text{Ker}(T)=0$. Now we will use a standard result which says $\text{Ker}(T)=\overline{\text{Ran}(T^*)}$ As $T$ is self-adjoint, $T=T^*$. So $\text{Ker}(T)=\overline{\...
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47 views

Findind the Pseudoinverse of a $2\times 2$-matrix (real-valued)

As is said in the title I would like to find the pseudoinverse of the real-valued matrix $$A = \left(\begin{array}{cc}1 & 1\\0 & 0\end{array}\right)$$ I've never done this before but I looked ...
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1answer
74 views

Show that the Cayley transform of a bounded self-adjoint operator on a Hilbert space has the number 1 in its resolvent.

This question is from section 10.6 in Kreyszig's Introductory Functional Analysis text. Let $H$ be a Hilbert space and let $D$ be a dense subspace of $H$. Let $T:D\rightarrow H$ be a self-adjoint ...
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Purely discrete spectrum of a self-adjoint operator

Let $(h,D(h))$ be a self-adjoint operator on the infinite dimensional Hilbert space $\mathfrak{h}$ that is bounded below and also assume that tr$_{\mathfrak{h}}(e^{-h})<\infty$. How can I see that $...
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28 views

Proof Explanation: Proving that $f$ has at least one real eigenvalue

I'm having some trouble understanding one part of the following proof. We want to prove the following statement: Let $V$ be an inner-product, finite-dimensional real vector space. If $f:V \to V$ is a ...
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24 views

How to prove eigenvalues are real when given self-adjoint and positive operators?

Let's say I have some T, self-adjoint operator, as well as S, which is a positive operator on a complex finite-dimensional inner product space. How could I go about proving that all eigenvalues of ST ...
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Spectral decomposition of a self-adjoint operator. Working only with a part of it.

A symmetric matrix can be decomposed as $\boldsymbol{\Sigma}=\boldsymbol{U} \boldsymbol{\Lambda} \boldsymbol{U}^{T}$ where $\boldsymbol{U}$ contain the eigenvectors and $\boldsymbol{\Lambda}$ the ...
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19 views

self-adjointness of positive operator for all $x,y \in \mathcal{H}$

Given a bounded positive operator A defined on the entire complex Hilbert space $\mathcal{H}$ such that $\langle Ax, x \rangle \geq 0$, I want to show that this is self-adjoint, i.e $\langle Ax,y \...
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1answer
46 views

Cayley Transform of self-adjoint operator is unitary

If $A$ is a self-adjoint operator, I want to show that the Cayley transform of $A$ defined as the operator $U=(A-iI)(A+iI)^{-1}$ is unitary. Here's my trial: For all $x,y \in \mathcal{H}$, keeping in ...
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46 views

$ (Ker A)^{\perp} $ is an $ A $-invariant subspace

I am stuck on a lemma: Prove that if $ A $ is self-adjoint, then $ (Ker A)^{\perp} $ is an $ A $-invariant subspace, that is $ A: (Ker A)^{\perp} \cap D(A) \to (Ker A)^{\perp} $. First of all, one ...
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1answer
30 views

$Ran(A-\lambda)^{\perp} = Ker(A-\lambda)$ for self-adjoint operator $A$?

I am reading through Theorem 5.5 in Hislop and Sigal, and I have the following confusion about the proof for item 2. My confusion is, why can you immediately conclude that \begin{equation}\label{key} ...
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32 views

Prove that $P = P_U$ if and only if $P$ is self-adjoint

I was reading Axler's Linear Algebra Done Right and the following question appears as exercise $11$ in chapter $7$, section A: Suppose $P \in \mathcal{L}(V)$ is such that $P^2 = P$. Prove that there ...
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1answer
49 views

Does the spectrum of a bounded from below self-adjoint operator have a lower bound?

Let $A$ be a self-adjoint operator in a complex separable Hilbert space that is unbounded, but bounded from below, i.e. $$\exists m>-\infty, ~\forall f\in D(A) \subsetneq \mathcal H , \langle Af, f ...
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1answer
54 views

proving $\varphi$ is an isometry

Let $\varphi$ be a normal map. If all the eigenvalues of $\varphi$ are real, then $\varphi$ is self-adjoint. Suppose $\varphi = UDU^*$ (with $U$ unitary and $D$ diagonal) and $D=D^*$ only if the ...
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1answer
33 views

Why can a self-adjoint compact operator on a Hilbert space be approximated by the linear combination of rank 1 operators?

Hilbert-Schmidt Theorem says that suppose $A$ is a self-adjoint compact operator on the Hilbert space $X$, then $X$ has an orthonormal basis $\{e_i \,|\, i \in I\}$ ($I$ is the index set) which is ...
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Fix $u\in V$ and $x\in W$. Define $T\in L(V,W)$ by $Tv=\langle v,u\rangle x$ fpr every $v\in V$. Find a formula for T*.

$T^*$: $W$ $\to$ $V$. $\langle$v$,$T*w$\rangle$ $=$ $\langle$$Tv,w$$\rangle$ $=$ $\langle$$\langle$$v$,$u$$\rangle$$x$,$w$$\rangle$ $=$ $\langle$$v$,$u$$\rangle$$\langle$$x$,$w$$\rangle$ $=$ $\langle$$...
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Prove A=A* where $Ax = \sum_{n}a_{n}\langle x, u_n \rangle u_n =0 $ iff $a_{n} \in \mathbb{R}$ [closed]

How to prove A=A* where $Ax = \sum_{n}a_{n}\langle x, u_n \rangle u_n=0 $ iff $a_{n} \in \mathbb{R}$ and $[u_n]$ is an orthonormal sequence? Edit: does it have something to do with the equality: $\...
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1answer
26 views

True or false Self adjoint operator and trace

I have a problem trying to prove the next statement or giving a counterexample If $T:\mathbb{R}^3 \to \mathbb{R}^3$ is self-adjoint such that trace($T^2$)=$0$, then $T=0$. I tried with the canonical ...
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Proof of the self-adjointness property of conditional expectation

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$, $H$ be a $\mathbb R$-Hilbert space and $$\pi X:=\operatorname E\...
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28 views

Does $ A^{1/2}$ is a self-adjoint operator?

I am studying the trace operator, and in the proof of the following theorem, there is a step that I do not understand: Theorem: Let $ H $ be a separable Hilbert space, $\{\phi_n \} _ n$ an ...
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1answer
59 views

Is a spectral projection strongly/norm continuous?

Given a bounded self-adjoint operator $H$ and a number $\mu$, consider the spectral projection $P_\mu$ onto the set $\{x|x\leq\mu\}$. I want to check if the map $\mu\mapsto P_\mu$ is strongly/norm ...
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51 views

Most direct way to prove the domain of $A^2$ is dense.

Let $A\colon \operatorname D(A)\to \mathcal H$ be a (generally unbounded but densely defined) self-adjoint operator in a Hilbert space. $$\operatorname D(A^2):=\{\psi \in \operatorname D(A) \text{ s.t....
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How to find the self-adjoint extension of an unbounded symmetric operator?

I have an unbounded symmetric operator, and I would like to find its self-adjoint extension if possible. First off, what properties does such an operator need in order to have a self-adjoint extension,...
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1answer
88 views

Are the following two inner products on differential forms equal?

There are two inner product on differential forms: $\langle \alpha,\beta\rangle$ induced from Riemannian metric $g$ by defining on 1-forms as dual of vector fields then extending to all differential ...
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Comparing Hamiltonians - Quantum harmonic oscillator

For standard 1D quantum harmonic oscillator we have $H\psi = E_n\psi$ with $E=(n+\frac{1}{2})\hbar\omega$ and $H = \frac{P^2}{2m} + \frac{1}{2}m\omega^2 X^2$ where $X$ is position operator and $P$ is ...
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Is the log of a diagonalizable operator also diagonalizable

Let $\mathscr{H}$ be a separable complex Hilbert space and $H$ be an (unbounded) self-adjoint operator on $\mathscr{H}$ bounded from below, e.g., $H\ge 0$. Suppose that $e^{-H}$ is of trace-class so ...
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1answer
31 views

Can we say that $T^n$ is a positive operator?

Let $\mathcal H$ be a complex Hilbert space and $T \in \mathcal L (\mathcal H)$ be a positive operator. Can we say that $T^n$ is also a positive operator? I am able to prove that if $n$ is even then $...
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1answer
35 views

Hermitian Matrix multiplied by complex numbers

I have a question regarding self-adjoint(hermitian) matrices, and their properties when multiplied by an imaginary number. The matrix in question is: $$\mathbf{A} = c \begin{bmatrix} 0&0&0&...
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40 views

The operator $U$ is unitary

If $A$ is a selft-adjoint operator, then the operator $U$ defined by \begin{eqnarray*} U=(A-iI)(A+iI)^{-1} \end{eqnarray*} Is unitary. I know if $U$ is unitary then $U^{*}U=UU^{*}=I$ But when i tried ...
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Understanding the spectral measure of unbounded self-adjoint operator

Let $\mathcal{H}$ be a Hilbert space and $$\Delta: D(\Delta) \rightarrow \mathcal{H}$$ a (unbounded) self-adjoint operator, densely defined on $D(\Delta) \subset \mathcal{H}$. Denote by $\mu^{\Delta}$ ...
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2answers
25 views

Is every bounded operator normal?

Let $A: {\scr H} \to {\scr H}$ be a linear bounded endomorphism on a Hilbert space $\scr H$, in brief: $A \in {\frak B} ({\scr H})$. If it's true that $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A \...
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1answer
38 views

Spectrum and self-adjointness

For bounded operator on a Hilbert space, if its spectrum is a subset of $\mathbb{R}$, then is this operator self adjoint? If not, what is a counterexample?
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64 views

Essentially self adjoint operators: a verification procedure

$\langle \ .,. \rangle : \scr H \! \times \! \scr H \rightarrow \mathbb C$ is the inner product. Let $T : {\cal D}(T) \rightarrow \scr H $ be a linear symmetric operator, so: ${\cal D} (T)$ is dense ...
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1answer
50 views

Spectrum of a self-adjoint integral operator

Trying to figure out the ingredients for the spectral representation of the following self-adjoint integral operator: $(Tf)(s)=\int_0^1 (s-t)^2 f(t) dt$ A hint in the exercise was to consider $im T$ ...

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