Questions tagged [adjoint-operators]

For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).

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8
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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 ...
7
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116 views

Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \frac{d^2 f}{dt^2} + f$.

Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \dfrac{d^2 f}{dt^2} + f$ with $f(0) = 0$ and $f'(1) = 0$. Note: The weight ...
6
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0answers
92 views

Why is it called *adjunction* formula?

Let $X$ be a complex manifold, $Y$ a sub-manifold, and $i \colon Y \to X$ the corresponding embedding. Then one can prove that the corresponding canonical bundles satisfy $$ \omega_X \big|_Y = \...
6
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0answers
1k views

Inverse vs. adjoint operators

I hope this is enough of a question (and less of a start of a discussion) to be allowed here. I am trying to get my head around the ideas of inverse and adjoint operators. To keep it simple, let's ...
5
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0answers
41 views

$T$-invariance of $U$ is equivalent to $T^{*}$-invariance of $U^{\perp}$

Is the Following argument correct? Suppose $T\in\mathcal{L}(V)$ and $U$ is a subspace of $V$. Prove that $U$ is invariant under $T$ if and only if $U^{\perp}$ is invariant under $T^*$. Proof. Given ...
5
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1answer
115 views

Show properties of the linear operator $L((a_n)_n)=\left(\frac{1}{n}*a_n\right)_n$

Let $L\colon \ell^p \rightarrow \ell^p$ such that $L((a_n)_n)= \left(\frac{1}{n}*a_n\right)_n$. 1) Determine $L'$(adjoint operator), $\ker(L)$, $\ker(L')$, $\operatorname{rg}(L)$, $\operatorname{rg}(...
5
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0answers
508 views

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 ...
5
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0answers
875 views

Sum of operator and adjoint is self-adjoint

In abstract Hodge theory there is the following lemma: Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. ...
4
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4answers
51 views

Showing that is a normal operator

Let $H$ is a Hilbert space $I$ is unit operator, $T \in B(H)$ and $\lambda \in \mathbb C$ $T$ is normal operator $\Rightarrow$ $T-\lambda I$ is a normal operator too. I could only write : I must ...
4
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1answer
61 views

Using inner product property to determine if operator is an isomorphism.

Let $\varphi$ be an operator on a $k$-vector space $V$ with an inner product $\langle\cdot,\cdot\rangle$. Suppose that $\langle v,\varphi v\rangle = 0$ for every $v\in V$. If we take $k=\mathbb R$, is ...
3
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0answers
63 views

Show that $T$ has an adjoint, and describe $T^*$ explicitly.

Let $V$ be an inner product space and $ \beta, \gamma$ fixed vectors in $V$. Show that $T \alpha = (\alpha\mid\beta) \gamma$ defines a linear operator on $V$. Show that $T$ has an adjoint, and ...
3
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0answers
37 views

Adjoint of a polynomial in a closed linear operator.

Let $ H $ be a Hilbert space and let $ T $ be a closed densely defined linear operator in $ H $ with domain $ D(T) $ and with nonempty resolvent set. We define the following polynomial in T: $ P(T) :...
3
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70 views

Show that the integral-operator compact?

We have the operator $T$ on $L_2[0,1]$ which is defined as $Tf = y$ where $y$ is the solution to the ODE $y^{\prime \prime} + y^\prime = f$ with boundary conditions $y(0)=0, y(1) = 1$ Show that $T$ ...
3
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1answer
30 views

Calculating $T^*T$ when $T$ have direct sum codomain

Let \begin{equation*} \bigoplus_{ \ell_2} K_n := \{ (x_1,x_2,\cdots) \in \bigoplus_{n=1}^\infty : x_n \in K_n, \sum_{n=1}^{\infty} || x_n ||^2 <\infty \} \end{equation*} where $K_n$ are Hilbert ...
3
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0answers
140 views

Explicit form of generators of a Lie algebra in the adjoint representation

My question can be summarized as: Generators of a Lie group in the adjoint representation can be written as, $$ (T^a_\text{Ad})_{bc} = \text{i}f^{abc}, \tag{1}\label{adj} $$ where $f^{abc}$ ...
3
votes
1answer
167 views

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,...
3
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1answer
99 views

Adjoint representation and representations

I'm studying representation theory in order to have a basis to study quantum field theory. I think the text (my professor's) i'm studying on is pretty confusing. I don't really get the difference ...
3
votes
1answer
135 views

Operator $f \mapsto u(f)$ solution of non-homogeneous Laplace equation is compact and self-adjoint

Let $u : L^2_0(D) \to L^2_0(D): = \lbrace f \in L^2 : \int_D f = 0 \rbrace $ be the linear operator which associates $f$ to $u(f)$ the solution of $$ \begin{cases} \Delta u = f & \text{in } D \\ \...
3
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0answers
23 views

Finding two adjoints, and showing boundedness of operators

Let $H = l_2$ and consider the following operators: $T,S:H \to H$ $Tx = (0,x_1,x_2,\ldots)$ and $Sx = (x_2,x_3,x_4,\ldots)$ Show they are bounded, and find the adjoint of both: For $T$, I have $\|Tx\|...
3
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1answer
254 views

Properties of adjoint matrix in a finite dimensional inner product space

let $V$ be a finite dimensional inner product space. Let $T$ be a linear operator on $V$. Prove that there exists an invertible linear operator $U$ such that $U^{-1}TT^*U = T^*T $ where $T^*$ is ...
3
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0answers
72 views

Do I have the correct mental map for adjoint operators for inner product spaces?

Let $X$, $Y$ be finite dimensional inner product spaces, let $A: X \to Y$ be a linear operator, let $A^*: Y \to X$ be the adjoint operator to the linear operator, defined using $<y, Ax>_Y = <...
3
votes
1answer
101 views

How to prove this is a self-adjoint operator?

I have this operator from $H^1_0$ to $H^1_0$ defined by: $$Au(t)=\int_0^1 G(t,s) f(s,u(s))\mathsf ds$$ where $$G(t,s)=\begin{cases} t(1-s), &t\leq s\\s(1-t), &s\leq t.\end{cases}$$ I want to ...
3
votes
1answer
455 views

$T \in B(X,Y)$ is an isometry if and only if $T^*$ is an isometry

I would like to prove that $T \in \mathscr{B}(X,Y)$ is an isometry of $X$ onto $Y$ if and only if $T^*$ is an isometry of $Y^*$ onto $X^*$. I am not really sure what to do. I started the argument as ...
3
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0answers
216 views

Proving that a certain differential operator is self-adjoint

Consider the differential operator $T:u\mapsto -iu'$ for any $u\in D(T):=\{f\in AC[-\pi,\pi]~|~f'\in L^2(-\pi,\pi),f(-\pi)=f(\pi)\}$; we consider $T$ as a densely-defined operator on $L^2(-\pi,\pi)$. ...
2
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0answers
29 views

Question about Friedrichs Extension when reading Applied Functional Analysis

In the book I read recently(Eberhand Zeidler, Applied Functional Analysis: Applications to Mathematical Physics (Applied Mathematical Sciences 108)), when consider the extension of symmetric operator ...
2
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0answers
40 views

adjoint operators in the vector space of real polynomials

This problem is about the space $V$ of real polynomials in the variables $x$ and $y$. If $f$ is a polynomial, $d_f$ will denote the operator $f(d/dx,d/dy)$ , and $d_f(g)$ will denote the result of ...
2
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0answers
36 views

Find $T^*(x,y)$ with the given inner product

Let $T:\mathbb{C}^2 \to \mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1\overline{y_1}+9x_2\overline{y_2}$. Find $T^*(x,y)$ ...
2
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0answers
30 views

Find $ T^*\left( \begin{smallmatrix} a&b\\ c&d\end{smallmatrix} \right)$

Let $T:\mathbb{C^3} \to \mathcal{M}_{2x2}$ and $T(x,y,z)=\begin{pmatrix} 3x+iy & y-iz\\ iy+z & 2x \end{pmatrix}$ with the inner product $<A, B>=tr(\bar{B}^t, A)$. Find $T^*\begin{...
2
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0answers
38 views

Exercises on derivative operator on subspaces of $L^2$

I got stuck with my exercise. Let $H = L^2(0, 2 \pi; \mathbb{C})$, define the operators $A_1 , A_2$ as follows. $$A_j : H \supset D(A_j) \to H , u \mapsto Au := iu' \quad (j=1,2)$$ Where $D(A_1)...
2
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1answer
87 views

Adjoint conditions in PDEs

Suppose $$L=p(x) \frac{d^2}{dx^2} + r(x) \frac{d}{dx} + q(x).$$ Consider $$\int_a^b vL(u) {\rm d}x \quad \quad \ldots (1)$$ From repeated integration by parts, I got the adjoint operator $L^*$ such ...
2
votes
1answer
90 views

Domains of higher powers of two unbounded self-adjoint operators

Let $A : D(A) \rightarrow H$ and $B : D(B) \rightarrow H$ be two unbounded self-adjoint operators that are densely defined on a Hilbert space $H$. Suppose we know that $D(A) = D(B)$. Are there ...
2
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0answers
101 views

Eigenvectors of self-adjoint operators on a Hilbert space.

I am trying to prove that the eigenvectors of a self-adjoint operator $T$ defined on a Hilbert space $\mathcal{H}$ are orthogonal. I did the following: For any eigenvectors $\mathbf{x}$ and $\...
2
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0answers
22 views

How to determine a function that make these operators, orthogonal projectors

Verify if we can find a function $\beta(x) \in C^\infty$ such as the operators $T_{\pm}$ defined as \begin{equation} (T_{+}f)(x) = \beta(x)[ \beta(x)f(x)+\beta(-x)f(-x)] \end{equation} \begin{...
2
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0answers
106 views

Domain of the adjoint of an elliptic operator

I'm just trying to understand the concept of the adjoint of unbounded operators. Let's look at the operator $Au:=-\epsilon\Delta u +(b\cdot\nabla)u+cu$ as an unbounded Operator on $L^{2}(\Omega)$ ...
2
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0answers
45 views

Adjoint of the following operator: $B: L_2[0,1]\to L_2[0,1]; B(x(t))=e^tx(\sqrt t)$

Adjoint of the following operator:$$B: L_2[0,1]\to L_2[0,1]; \\ B(x(t))=e^tx(\sqrt t)$$ This is what I have tried: Using the knowledge that $$<Bx,y^*>=<x,B^*y^*>$$ I have that : $$&...
2
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0answers
257 views

The image of the adjoint operator in a Hilbert space

Let $H$ be a Hilbert space and $T \in L(H)$ a bounded linear operator.If $\exists c>0$ such that $Re(\langle Tx,x \rangle) \geqslant c||x||^2, \forall x \in H$, then prove that $\text{Ran}(T^*)=H$ ...
2
votes
1answer
395 views

Derivation of the adjoint poisson equation

I'm trying to follow the derivation of the adjoint Poisson equation provided in this video, but I'm getting tripped up on some of the steps that are skipped. Here is my derivation in full. Questions ...
2
votes
1answer
471 views

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\...
2
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0answers
79 views

Prove $T^2 = (T^*)T$ if $T$ is positive-definite

If $V$ is a finite-dimensional inner product space and $T: V\to V$ is an invertible, positive-definite (i.e., $\forall v \in V\backslash \{0\}, \langle Tv, v\rangle > 0$) linear map, prove that $T^...
2
votes
1answer
120 views

Matrix of self-adjoint operator such that every element of the diagonal is $0$.

Let $V$ be a finite dimensional $\mathbb R$-vector space and let $T:V\rightarrow V$ be an self-adjoint operator such that $\text{trace}(T)=0$. Show that there exists an orthonormal basis $B$ such that ...
2
votes
1answer
252 views

Adjoint matrix as pseudo-inverse

I'm very new to signal processing (seismic 1,2 and 3D-signal) and have read many papers recently. One thing I encounter quite often is the use of adjoint matrix. If $d = Am$ where $d$ is the data, $...
2
votes
0answers
86 views

Seemingly strange exercise on unitary Hermitian operator

Let $V$ be a finite dimensional inner product space. Let $T$ be a Hermitian unitary operator. Prove there's a subspace $W$ such that for each $v\in V$ we have $Tv=w-w^\prime$ where $w\in W,w^\...
2
votes
0answers
89 views

explicit self adjoint operator which has no diagonalization

Let a linear operator $T : H \to H$ be diagonalizable if $H$ has an orthonormal basis composed of eigenvectors of $H$ Give an example of an explicit self adjoint operator which has no diagonalization ...
2
votes
1answer
80 views

Prob. 5, Sec. 4.5 in Kreyszig's functional book: The adjoint of the composite of two bounded linear operators

Let $X$, $Y$, and $Z$ be normed spaces, either all real or all complex. Let $T \colon X \to Y$ and $S \colon Y \to Z$ be bounded linear operators. Let $X^\prime$, $Y^\prime$, and $Z^\prime$ denote the ...
2
votes
0answers
32 views

Singular values of the differentation operator

I was trying to solve this exercise but I can't get to answer given in book: Find the singular values of the operator D over $R_2[x]$ (That is polynomials with degree equal or less than 2) define as ...
2
votes
0answers
279 views

adjoint method for computing derivatives

I am curious if anyone has heard of this problem before: Suppose that $u(x,p)$ is a function of $x$ and $p$. These arguments need not be scalars. Let $u(x,p)$ satisfy some differential equation, say:...
2
votes
1answer
30 views

Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$

Let $V$ be a complex inner product space. Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$? If not, ...
2
votes
0answers
397 views

Linear map is diagonalizable iff its adjoint is diagonalizable

Problem Let $V$ be a finite inner product space and let $T:V \to V$ be a linear transformation. Prove that $T$ is diagonalizable if and only if the adjoint transformation $T^{*}$ is diagonalizable. ...
2
votes
1answer
41 views

What is the dual of $A\cap B$

I encountered with some elliptic problem which admits a variational formulation in terms of space $X$ and I need to understand its dual. Suppose that $2<p<\infty$, $\Omega\subset {\mathbb R}^d$ ...
2
votes
0answers
145 views

Relationship between eigenvalues of differential operator and eigenvalues of its adjoint operator.

I am considering $L\phi = -\triangle \phi + u \cdot \nabla \phi$ and its "adjoint" operator $L^* \phi = -\triangle \phi - \nabla \cdot (\phi u)$ on a bounded domain $\Omega \subseteq \mathbb{R}^n$. ...