49 questions
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### Why is every selfadjoint operator closed?

I've read this theorem multiple times, but never seen a proof: Every selfadjoint operator is closed. But it's always been stated without a proof. Is it somehow obvious? I can't see it immediately ...
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$\newcommand{\adj}{\operatorname{adj}}$Let $A\in \mathbb{M}_n$ ($n \geq\ 2$) be a regular matrix and $\adj(A)$ its adjoint. Prove that if A is regular then $\adj(\adj(A)) = (\det A)^{n-2} A$ (where $... 1answer 521 views ### Prove that every self-adjoint unitary linear operator can be expressed in the form$U\alpha = \beta - \gamma$This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me? Suppose$U$is a self-adjoint unitary linear ... 2answers 4k views ### image of adjoint equals orthogonal complement of kernel [duplicate] Let$T:V\to W$be a linear map of finite-dimensional spaces. Then $${\rm im}(T^{\textstyle*})=({\rm ker}\,T)^\perp\ .\tag{*}$$ I can prove this as follows: $${\rm ker}(T^{\textstyle*})=({\rm im}\,T)... 1answer 114 views ### Compute ad_X, ad_Y, and ad_Z relative to a basis For a lie algebra \mathbb{g} we can define the adjoint representation as: ad: \mathbb{g} \rightarrow End(\mathbb{g}) as the map such that ad_x(y)=[x, y] for all \in \mathbb{g} I am ... 3answers 120 views ### \operatorname{Adj} (\mathbf I_n x-\mathbf A) when \operatorname{rank}(\mathbf A)\le n-2 Let \mathbf B denote an n \times n matrix with r\equiv\operatorname{rank}(\mathbf B). I need to prove the following conjecture: If r \le n - 2, then there exists a polynomial matrix \... 2answers 199 views ### T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc) T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc) I need to know whether it is self adjoint and unitary operator given that x_i\in\mathbb C I am not able to do it please tell me how ... 2answers 295 views ### Is intersection of a dense subspace and a closed subspace of a Hilbert space also Dense? I have a Hilbert space H and a closed operator T defined on its domain D(T) which is dense in H. Also M = \text{range} \ T^n, for some n, is given to be closed. Consider the restriction of ... 1answer 313 views ### Adjoint of a matrix and inverse of a matrix As everyone know that we can use a matrix A to represent an operator T. The adjoint of a matrix A is denoted as A^*, which takes complex conjugate of A and then transpose. My problem ... 3answers 147 views ### Selfadjoint Operator: Basic Criterion For symmetric operators one has:$$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$How to prove this in an unveiling way? 2answers 272 views ### Isometric <=> Left Inverse Adjoint Is it true that:$$T\text{ isometric}\iff T^*\text{ left inverse}$$Obviously:$$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle... 2answers 2k views ### Motivation for adjoint operators in finite dimensional inner-product-spaces Given a finite dimensional inner-product-space$(V,\langle\;,\rangle)$and an endomorphism$A\in\mathrm{End}(V)$we can define its adjoint$A^*$as the only endomorphism such that$\langle Ax, y\...
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Recently I found this statement -- the gradient operator is the adjoint of the minus divergence operator -- in one of my lecture notes. Knowing only a little about functional analysis, I'm looking for ...
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### $T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)

I am trying to prove the following statements: Let $X$ and $Y$ be normed spaces (not necessarily complete) Let $T\in L(X,Y)$ (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote ...
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### Connection between categorical notion of adjunction and dual space/adjoint in vector spaces

I'm an economist, not a mathematician. I've been trying to make sense of some concepts in functional analysis: dual, bidual, adjoint, natural mapping. The definitions of these notions come out of ...
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### Find adjoint operator of an operator T

I would like to find the adjoint operator of $$T\colon L^2([0,1])\to H^1([0,1]),\quad x\mapsto\int\limits_0^t x(s)\, ds.$$ Here $H^1([0,1])$ is the Sobolev space $W^{1,2}([0,1])$. I tried to find ...
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### For positive self adjoint $T$, show $|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$

As in title, $T$ is a positive self adjoint, bounded linear operator on a Hilbert Space $X$ and I'd like to show $$|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$$ Self adjoint ...
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### General Cauchy-Schwarz for adjoint positive operators

I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality: $$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ ...
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### Gelfand Triples / Rigged Hilbert Spaces - Reflexivity necessary?

There have been several questions asked on various aspects of Gelfand triples. However, I have not yet found an answer to the following question: Let $V$ be a Banach space, $H$ be a Hilbert space ...
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### Understanding $\overline{Im\: }A=\bigcap_{A'f=0}\ker f$ proof

Theorem: Let $X,Y$ be Banach spaces and $A$ an linear bounded operator. The closure of the image is $\overline{Im\: }A=\{y\in Y:f(y)=0,\forall f\in Y'$ such that $A'f=0$}. $(A'f)(x)=f(A(x))$ is the ...
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### Invertibility in a finite-dimensional inner product space

Let $T$ be an invertible linear operator on a finite-dimensional inner product space. I just want a hint as to how I should prove that $T^{*}$ is also invertible and $( T^{-1} )^{*} = ( T^{*} )^{-1}$. ...
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### $T=AU \iff T$ is a normal operator on Hilbert space

This is Exercise 16.(c) from Conway's Functional Analysis book. Suppose $H$ is a Hilbert space and $T$ is a compact operator on $H$. Assuming the result that $\exists A$ positive operator and $U$ a ...
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### Adjoint of an integral operator

I'm reading through a text about integral operators and I've come across the following theorem: Let $k:\mathbb{R}^2\rightarrow\mathbb{C}$ be a kernel, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
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### Prove that $\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$. [closed]

Let $T$ be a self adjoint bounded linear operator in a Hilbert space $H$. Prove that $$\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$$
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### Exponential Law for based spaces

I realize most people work in "convenient categories" where this is not an issue. In most topology books there is a proof of the fact that there is a natural homeomorphism of function spaces (with ...
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Obviously something must be wrong in the following reasoning proving that any linear operator $T:X\to Y$ between Banach lattices has a lattice homomorphic adjoint: $\forall a,b\in E':$ $$T'(a\wedge b)=... 1answer 2k views ### Kernel of adjoint and orthogonal complement images Alright, suppose we are given V, a finite dimensional inner product space, and a linear map, T:V \rightarrow V, with its corresponding adjoint, T^\star :V \rightarrow V. I want to show: [im(T)]... 1answer 460 views ### Adjoint differential equations Consider the vector differential equations $$\mathbf{x}^{\prime}=\mathbf{A}(t)\cdot\mathbf{x}\tag{1}$$ and \mathbf{y}^{\prime}=-\mathbf{A}^{\ast}(t)\cdot\... 1answer 169 views ### Finding dimension of range of T*T, where T* is the adjoint Consider T ∈ L(U, V ). U, V are finite dimensional. Show that dim range (T*T) = dim range T. Then, show that if dim range T = dim U then T*T is invertible. T* is the adjoint of T. So it is obvious ... 0answers 31 views ### Does an adjoint of an internal Hom functor of a prounital closed category define a tensor product? A closed category is a category equipped with internal Hom functors along with a unit object. Now this answer shows that if C is a closed category whose internal Hom functor has a left adjoint, ... 1answer 59 views ### Can we find a concrete representation of \iota\iota^\ast y, if \iota is a Hilbert-Schmidt embedding between Hilbert spaces? Let U and H be real Hilbert spaces \iota:U\to H be a Hilbert-Schmidt embedding Q:=\iota\iota^\ast Can we find a concrete representation of Qy for some y\in H? By Riesz' representation ... 1answer 130 views ### Prob. 6, Sec. 3.10 in Kreyszig's functional analysis book: Powers of self-adjoint operators Let H be a Hilbert space. If T \colon H \to H is a bounded self-adjoint linear operator and T \neq 0, then T^n \neq 0 for all n \in \mathbb{N}. How to show this? I've managed to show ... 4answers 257 views ### Let A be a 3\times3 matrix. Given \mathrm{adj}(A), find \det(A). Let A be a 3\times3 matrix such that$$\mathrm{adj}(A) = \begin{pmatrix}3 & -12 & -1 \\ 0 & 3 & 0 \\ -3 & -12 & 2\end{pmatrix}.Find the value of $\det(A)$. I know that ...
Let $K_n$ be Hilbert spaces and define \begin{equation*} K := \bigoplus_{\ell_2} K_n = \left\{ (x_1,x_2,\ldots) \in \bigoplus_{n=1}^\infty K_n : \sum_{n=1}^\infty \|x_n\|^2 < \infty \right\} \end{...