# Questions tagged [self-adjoint-operators]

366 questions

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### What is wrong with this proof that symmetric matrices commute?

Symmetric matrices represent real self-adjoint maps, i.e. linear maps that have the following property: $$\langle\vec{v},f(\vec{w})\rangle=\langle f(\vec{v}),\vec{w}\rangle$$ where $\langle,\rangle$ donates the scalar (dot) product.
Using this…

Pancake_Senpai

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### Calculate the operator norm of $A: L^2[0,1] \to L^2[0,1]$ defined by $(Af)(x):=i\int_0^x f(t)\,dt-\frac{i}{2} \int_0^1 f(t) \,dt$

I want to calculate the operator norm of the operator $A: L^2[0,1] \to L^2[0,1]$ which is defined by $$(Af)(x):=i\int\limits_0^x f(t)\,dt-\frac{i}{2} \int\limits_0^1 f(t)\, dt$$
I've already shown that this operator is compact and selfadjoint. I…

thehardyreader

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### Is every self-adjoint operator bounded?

This is a problem our teacher gave us and I have a feeling he forgot to mention some additional data.
Let $H$ be a Hilbert space and $T : H \to H$ a everywhere-defined linear operator such that $T$ is self-adjoint, i.e. $\forall x,y \in H : \langle…

the_firehawk

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### Orthogonal eigendecomposition of self-adjoint operator with indefinite scalar product

Let $V$ be a real vector space, of finite dimension $d$, equipped with a nondegenerate symmetric bilinear form $q$, and let $A$ be a linear map $V\to V$ that is self-adjoint with respect to $q$, i.e., such that $q(A(x),y)=q(x,A(y))$ for all $x,y \in…

MK7

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### Proving an inequality for operators.

Let $\mathbb P_n$ be the space of all $n \times n$ self-adjoint positive definite matrices. Consider the function $\varphi: \mathbb P_n \longrightarrow \mathbb R$ defined by $$\varphi (A) = -\text {tr}\ (A \log A).$$ Show that for all $t \in (0,1)$…

RKC

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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)\|}^2\le {\|x\|}^2-{\|A(x)\|}^2, \forall x\in…

John

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### Why are self-adjoint operators important?

I am learning about self-adjoint and normal operators.
So far, they have come up in the Spectral theorem, which says self-adjoint operators have an eigenvalue basis and a corresponding diagonal matrix.
Do self-adjoint (or normal) operators have any…

Peter_Pan

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### Is the self-adjoint condition required in the definition of a positive operator?

I'm reading Linear Algebra Done Right and it defines a positive operator $T$ as one which is self adjoint and has the property
$$\langle Tv,v \rangle \geq 0$$
for all $v\in V$.
I am confused as to why the self adjoint condition must be included.…

Fgilan

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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 solve, I will detail in a summarized way:
Considering…

Litafie

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### Self-adjoint bounded operator with finite spectrum implies diagonalisable?

Let $T$ be a self-adjoint bounded operator on a not-necessarily finite dimensional Hilbert space.
Suppose $T$ has finite spectrum. Does it follow that the elements of the spectrum are eigenvalues, and the operator diagonlisable?

JP McCarthy

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### Integral with respect to spectral measure

Let $A:D(A)\subset H \to H$ be a self-adjoint unbounded operator on complex Hilbertspace $H$ with corresponding spectral measure $E:\mathcal{B}(\mathbb{R})\to\mathcal{L}(H)$.
I want to show that an element $u\in H$ with $\Vert u \Vert =1$ and…

user3342072

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### Does symmetry of the Green's function of some operator imply that the operator is hermitian?

I know that if an operator $L$ is hermitian(self-adjoint), then its Green's function is symmetric, but is it true the other way?
In other words, is having a symmetric Green's function a necessary and sufficient condition for $L$ to be hermitian?

Garmekain

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### Convergence of (unbounded) self-adjoint operators

I'm learning about the dynamical convergence (i.e, convergence of the unitary group associated with each operator) and resolvent convergence of (unbounded) self-adjoint densely defined operators. I can understand the proof of the initial results,…

Reinaldo R.

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### What assumptions are needed for compactness and self-adjointness?

I am studying functional analysis and got stuck with a textbook exercise. I greatly appreciate some hints/help or a push in the right direction!
Define for $g\in C^0[-1,1]$ the integral operator $T_g:L^2\to L^2$,…

user974406

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votes

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### 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 Borel-$\sigma$-algebra of subsets of $\left…

Fanatics

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