Questions tagged [adjoint-operators]
For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).
101
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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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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)...
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10
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3
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If a linear operator has an adjoint operator, it is bounded
This is a question I'm struggling with for a while:
Let $H$ be a Hilber space. Let $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle Tx,y\...
user188400
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2
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$A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$
So, $A$ is a $n \times n$ matrix with integer entries. The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .
I know that $A^{-1}= {\rm adj}(A)/{\rm ...
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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 ...
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3
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Prove that if $A$ is regular then $\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A$
$\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 $...
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1
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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 $...
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Proof that every bounded linear operator between hilbert spaces has an adjoint.
As a practice exercises(not an assignment question) for one of the papers I am doing currently at university we are asked to show the following;
I $T:H \rightarrow K$ is a bounded linear operator ...
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6
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1
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Adjoint of a nonlinear operator?
For any linear operator $A$, the adjoint $A^*$ is defined as a linear operator that satisfies
$$\langle v, Au\rangle = \langle A^*v, u\rangle$$
Moreover, one has that $(A^{*})^{*} = A$ (proof here). ...
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3
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Showing derivative operator is self-adjoint
Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain
$$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$
I’m ...
1
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2
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Infinite-dimensional inner product space: if $A \geq 0$ and if $\langle Ax, x\rangle = 0$ for some $x$, then $Ax = 0$.
Exercise 8, Section 82 from PR Halmos's Finite-Dimensional Vector Spaces, 2nd Edition
If $A$ is a positive semidefinite operator, and if $\langle Ax, x\rangle = 0$ for some vector $x$, show that $Ax = ...
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Theorem 1.1 of The mathematics of computerized tomography
In the book "The mathematics of computerized tomography" by Natterer comes the theorem:
Theorem 1.1 $f=A^+g$ is the unique solution of $A^*Af=A^*g$ in $range(A^*)$.
where $A:H\rightarrow K$ ...
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Intuition behind $\ker(T)=\ker(T^*)$ for $T$ a normal operator
Let $T : V \to V$ be a normal operator and $V$ a finite-dimensional vector space. Show that $\ker(T)= \ker(T^*)$ and $\text{im}(T) = \text{im}(T^*)$.
I know how to rigorously show this, but I'm ...
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3
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$\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 $\...
- 1,885
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3
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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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1
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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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Connection between adjoint of a matrix and adjoint of an operator
Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with
$$T(x,y) = \left[ \begin{array}{ccc}
1x+2y \\
3x+4y \end{array} \right] $$
The matrix representation of $T$ is
$$ A= \left[ \begin{array}{ccc}
1 &...
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3
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2
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Why is $T$ normal and idempotent? [duplicate]
Let $V$ be an inner-product space, finite-dimensional over $\mathbb{C}$. Let the operator $T:V\to V$ satisfy
$$T^2 = \frac{1}{2}(T+T^*)\,.$$
I'd like to prove that $T$ is normal, i.e., $T^*T = TT^*$,
...
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1
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Is the operator $T((x_1,x_2,...,x_n))=(x_1,\frac{x_2}2,\frac{x_3}3,...)$ on $\ell^2$ self-adjoint and unitary
Consider the Hilbert space
$$\ell^2=\{(x_1,x_2,...,x_n),x_i\in\mathbb C\text{ for all }i\text{ and }\sum_{i=1}^\infty |x_i|^2<\infty\}$$ with the inner product
$$\langle(x_1,x_2,\dots,x_n)(y_1,y_2,...
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Compute $ad_X$, $ad_Y$, and $ad_Z$ relative to a basis
For a lie algebra $\mathbb{g} $ we can define the adjoint representation as:
$ ad: \mathbb{g} \rightarrow End(\mathbb{g}) $ as the map such that $ad_x(y)=[x, y] $ for all $\in \mathbb{g} $
I am ...
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1
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0
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Computing the Adjoint of an Operator
Can a computer take in a linear operator and set of boundary conditions, and compute the adjoint operator/conditions?
I can find calculators/Mathematica code/etc. to compute the adjoint of a matrix, ...
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2
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$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$
$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$ I need to know whether it is self adjoint and unitary operator given that $x_i\in\mathbb C$
I am not able to do it please tell me how ...
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If $\lambda$ is an eigenvalue of a self-adjoint operator, is $\lambda$ in the resolvent set of $\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$?
Let $A$ be a symmetric linear operator on a $\mathbb R$-Hilbert space $H$ and $\lambda\in\mathbb R$. It's easy to see that $$A\left(\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp\right)\subseteq{\...
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1
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1
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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 ...
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if range $A^* \cap$ range $B^* = 0$ and $B^*A = 0$ then $\text{rank }(A + B) = \text{rank }A + \text{rank }B$
Problem.
Let $A, B \in \mathcal{M}_{m,n}(\mathbb{C})$ such that $\text{range }(A^*) \cap \text{range }B^* = \{0\}$ and $B^*A = 0$. Prove that $\text{rank }(A + B) = \text{rank }A + \text{rank }B$.
My ...
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1
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How can I prove these are the adjoint operators?
Let be $V$ a vector space over the complex numbers with inner product. Let be $T$ and $S$ linear operators in $V$ with adjoint operators $T^*$, and $S^*$ respectively.
I know that it satisfies this:
$\...
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3
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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?
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Showing that $Tp(x)=(1-x^2)p''(x)-2xp'(x) $ is self-adjoint
Let $P_2(\Bbb{R})$ have the inner product $\langle p,q\rangle =\int\limits_{-1}^{1} p(x)q(x)dx$ and consider the operator $Tp(x)=(1-x^2)p''(x)-2xp'(x) = [(1-x^2)p'(x)]'$.
Show that T is self-adjoint.
...
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Isometric <=> Left Inverse Adjoint
Is it true that:
$$T\text{ isometric}\iff T^*\text{ left inverse}$$
Obviously:
$$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$
$$...
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Motivation for adjoint operators in finite dimensional inner-product-spaces
Given a finite dimensional inner-product-space $(V,\langle\;,\rangle)$ and an endomorphism $A\in\mathrm{End}(V)$ we can define its adjoint $A^*$ as the only endomorphism such that $\langle Ax, y\...
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Intuition of Adjoint Operator
A linear operator $T$ on an inner product space $V$ is said to have an adjoint operator $T^*$ on $V$ if $⟨T(u),v⟩=⟨u,T^*(v)⟩$ for every $u,v\in V$
I know how to proof "why this operator exist$(\text{...
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Self-adjoint operator as difference of two positive operators
The problem comes from my functional analysis homework.
Let $H$ be a complex Hilbert space and $A:H \to H$ be a bounded, self-adjoint linear operator. Prove that there exist positive operators $P$ ...
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Gradient operator the adjoint of (minus) divergence operator?
Recently I found this statement -- the gradient operator is the adjoint of the minus divergence operator -- in one of my lecture notes. Knowing only a little about functional analysis, I'm looking for ...
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7
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2
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General Cauchy-Schwarz for adjoint positive operators
I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality:
$$|\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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1
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Adjoint of multiplication by $z$ in a Hilbert Space (Bergman space)
I am learning Hilbert space theory from Halmos' "Introduction to Hilbert space and the theory of spectral multiplicity".
While talking about understanding adjoints (p. 39), he calls special ...
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2
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When is the restriction of a normal operator not normal?
I was proving the spectral theorem for normal operators on finite-dimensional complex vector spaces today during a test, when I arrived at the point in which
If $T\in\operatorname{End}(V)$ is ...
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2
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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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3
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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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1
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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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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$$
Self adjoint ...
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2
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662
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Find adjoint operator of an operator T
I would like to find the adjoint operator of
$$
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 ...
user34632
5
votes
3
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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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1
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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)]...
- 1,104
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3
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Why do we need an orthonormal basis to represent the adjoint of the operator?
For any linear operator on a finite dimensional Inner Product Space, we can get orthonormal basis via Gram Schmidt Process.
But what is the necessity of defining the adjoint of the operator using ...
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If $\|Tv\|=\|T^*v\|$ for all $v\in V$, then $T$ is a normal operator
I have solved a question but I am not sure the last step of the question. If someone can verify it that would be great.
Let $V$ be a finite dimensional vector space with complex inner product. Let $...
4
votes
1
answer
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Adjoint differential equations
Consider the vector differential equations
\begin{equation}
\mathbf{x}^{\prime}=\mathbf{A}(t)\cdot\mathbf{x}\tag{1}
\end{equation}
and
\begin{equation}
\mathbf{y}^{\prime}=-\mathbf{A}^{\ast}(t)\cdot\...
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1
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self-adjoint operator without eigenvalues?
I have a self-adjoint operator $d$ which acts on vector fields defined on $\mathbb{R}^n$. I am interested on its eigenvalues. That is, I study the equation $d(X)-\lambda X=0$.
I have found that if $\...
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2
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How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$?
I need to show that for a bounded linear operator, $T$, on a Hilbert space:
\begin{align*}
\|T\|^2=\|T^*T\|
\end{align*}
All I have so far:
\begin{align*}
\|T^*T\|&=\sup\{|\langle T^*Tf,g \rangle ...
- 389
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votes
1
answer
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Adjoint of a linear mapping mapping into a product of Banach Spaces
I am self studying adjoint operators on Banach spaces. The adjoint of a linear mapping $L:X \rightarrow Y$, where $X,Y$ are Banach spaces, is a unique mapping
\begin{equation}
L^*:Y^{*} \rightarrow X^...
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3
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1
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424
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Unitary operators, inner products and unit operators with Dirac notation
From my notes, I have that for a unitary matrix:
$$\underline{\underline{U}}^\dagger=\underline{\underline{U}}^{-1},\qquad \underline{\underline{U}}^\dagger=\big(\underline{\underline{U}}^{T}\big)^*$$
...
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